NEET Physics Class 12

Electrostatics

1. Electrostatics Introduction

The branch of physics which deals with the electric effect of static charge is called electrostatics.

2. Electric Charge

The charge of a material body or particle is the property (acquired or natural) due to which it produces and experiences electrical and magnetic effects. Some naturally charged particles are electrons, protons, α- particles, etc.

Charge is a derived physical quantity. Charge is measured in coulomb in S.Ι. unit. In practice, we use mC (10^-3C), μC (10^-6C), nC(10^-9C), etc.

C.G.S. unit of charge = electrostatic unit = esu.

1 coulomb = 3 × 10⁹ esu of charge

Dimensional formula of charge = [MºLºT¹Ι¹]

2.1 Electric Charges

It was observed that if two glass rods rubbed with wool or silk cloth are brought closer they repel each other.

The two strands of wool or two pieces of silk cloth, with which the rods were rubbed, also repel each other. However, the glass rod and wool attracted each other. Similarly, two plastic rods rubbed with a cat’s fur repelled each other but attracted the future. Careful studies by different scientists concluded, that there were only two kinds of an entity called the electric charge.

We say that the bodies like glass or plastic rods, silk, fur, and pith balls are electrified. They acquire an electric charge on rubbing.

The experiments on pith balls suggested that there are two kinds of electrification and we find that (1) like charges repel and (2) unlike charges attract each other

The charges were named as positive and negative by the American scientist Benjamin Franklin. By convention,

The charge on a glass rod or cat’s fur is called positive and that on a plastic rod or silk is termed negative. If an object possesses an electric charge, it is said to be electrified or charged. When it has no charge it is said to be neutral

2.2 Properties of Charge

Charge is a scalar quantity: It adds algebraically and represents excess, or deficiency of electrons.

A charge is of two types : (i) Positive charge and (ii) Negative charge Charging a body implies the transfer of charge (electrons) from one body to another. A positively charged body means loss of electrons, i.e., deficiency of electrons. A negatively charged body means an excess of electrons. This also shows that the mass of a negatively charged body > the mass of a positively charged identical body.

A charge is conserved: In an isolated system, the total charge (sum of positive and negative) remains constant whatever change takes place in that system.

A charge is quantized: A charge on anybody always exists in integral multiples of a fundamental unit of electric charge. This unit is equal to the magnitude of the charge on an electron (1e = 1.6 × 10^-19 coulomb). So charge on anybody Q = ± ne, where n is an integer and e is the charge of the electron. Millikan’s oil drop experiment proved the quantization of charge or atomicity of charge

Note : Recently, the existence of particles of charge ±e\(\frac{1}{3} \text { and } \frac{2}{3}\) ±e has been postulated. These particles are called quarks but st, this is not considered as the quantum of charge because these are unstable (They have ve very short span of life).

Like point charges repel each other while unlike point charges attract each other. (vi) Charge is always associated with mass, i.e., charge can not exist without mass though mass can exist without charge. The Particle as photons neneutrinoshich have no (rest) mass can never have a charge charge is relativistically invariant: This means that charge is independent the of frame of reference, i.e., the charge on a body does not change whatever be its speed. This property is worth mentioning as in contrast to charging, the mass of a body depends on its speed and increases with an increase in speed.

A charge at rest produces only an electric field around itself; a charge having uniform motion produces an electric as well as maa magnetic field around itself while a charge having accelerated motion emits electromagnetic radiation.

2.3 Charging of a body

A body can be charged by using) friction, (b) conduction, (c) induction, (d) thermionic ionization or thermionic emission (e) photoelectric effect, and (f) field emission.

Charging by Friction: When a neutral body is rubbed against another neutral body then some electrons are transferred from one body to another body which can hold electrons tightly, draws some electrons, and the body which can not hold electrons tightly, loses electrons. The body that draws electrons becomes negatively charged and the body that loses electrons becomes positively charged.

For example: Suppose a glass rod is rubbed with a silk cloth. As the silk can hold electrons more tightly and a glass rod can hold electrons less tightly (due to their chemical properties), some electrons will leave the glass rod and get transferred to the silk. So in the glass rod t, heir will be a ficiency of electrons, therefore it will become positively charged. In AnInhe silk there will be some extra electrons, so it will become negatively charged

Charging by conduction (flow): There are three types of material in nature

Conductor: Conductors are the material in which the outermost electrons are very loosely bound, so they are free to move (flow). So in a conductor, there are a large number of free electrons.

Example: Metals like Cu, Ag, Fe, Al………….

Insulator or Dielectric or Nonconductor: Non-conductors are the materials in which outermost electrons are very tightly bound so they cannot move (flow). Hence in a non-conductor, there are no free electrons.

Examples are rubber, wood, etc, etc.

Semiconductor: Semiconductors are theories with free electrons but very less in number.

Now release how the charging is done by conduction. In this method, w, we take a charged conductor ‘A’ and an uncharged conductor ‘B’. When both are connected some charge will flow from the body to the uncharged body. If both the conductors are identical and kept at a large distance if connected then the charge will be divided equally in both the conductors otherwise they will flow till their electric potential becomes sathe. A detailed study will be done in the first section of this chapter.

Charging by Induction: To understand this, let’s have an introduction to induction.

We have studied that there are a lot of free electrons in the conductors. When a cha charge article +Q is brought near a neutral conductor. Due to the reaction of the +Q charge, many electrons (–ve charges) come closer and accumulate on the closer surface. On the other hand a,, positive charge (deficiency of electrons) appears on the other surface. The flow of charge continues till there resultant force on the free electrons of the conductor becomes zero. This phenomenon is called induction, and the charges produced are called induced charges.

A body can be charged by induction in the following two ways :

Method 1 :

Step 1.

• Take an isolated neutral conductor.

Step 2.

• Bring a charged rod near to it. Due to the charged rod, charges will be induced on the conductor.

Step 3.

• Connect another neutral conductor with it.
• Due to the attraction of the rod, some free electrons will move from the right conductor to the left conductor, and due to the deficiency of electrons positive charges
• Will appear on the right conductor, and on the left conductor, there will be an e excess of electrons due to transfer from the right conductor.

Step 4.

• Now disconnect the connecting wire and remove the rod.

The first conductor will be negatively charged and the second conductor will be positively charged.

Method 2:

Step 1.

• Take an isolated neutral conductor.

Step 2.

• Bring a charged rod near to it. Due to the charged rod, charges will be induced on the conductor.

Step 3.

• Connect the conductor to the earth (this process is called grounding or earthing).
• Due to the attraction of the rod, some free electrons will move from the earth to the conductor, so in the conductor.
• There will be an excess of electrons due to transfer from the earth, so the net charge on the conductor will be negative.

Step 4.

• Now disconnect the connecting wire. The conductor becomes negatively charged.

Thermionic emission: When the metal is heated at a high temperature then some electrons of metals are ejected and the metal becomes positively charged.

Photoelectric effect: When light of sufficiently high frequency is incident on the metal surface then some electrons gain energy from light and come out of the metal surface and the remaining metal becomes positively charged.

Field emission: When the electric field of large magnitude is applied near the metal surface then some electrons come out from the metal surface and hence the metal gets positively charged.

2.4 Gold Leaf Electroscope (GLE)

A simple apparatus to detect charge on a body is the gold-leaf electroscope

It consists of a vertical metal rod housed in a box, with two thin gold leaves attached to its bottom end. When a charged object touches the metal knob at the top of the rod, charge flows onto the leaves and they diverge.

Solved Examples

Example 1. If a charged body is placed near a neutral conductor, will it attract the conductor or repel it?
Solution :

If a charged body (+ve) is placed left side near a neutral conductor, (–ve) charge will induce at the left surface, and (+ve) charge will induce at the right surface.

• Due to the positively charged body –ve induced charge will feel attraction and the +ve induced charge will feel repulsion.
• But as the – ve induced charge is nearer, so the attractive force will be greater than the repulsive force.
• So the net force on the conductor due to the positively charged body will be attractive. Similarly, we can prove this for the negatively charged bodies also.
• From the above example, we can conclude that. “A charged body can attract a neutral body.” If there is attraction between two bodies then one of them may be neutral.
• But if there is repulsion between two bodies, both must be charged (similarly charged). So “repulsion is the sure test of electrification”.

Example 2. A positively charged body ‘A’ attracts a body ‘B’ then the charge on body ‘B’ may be:

1. Positive
2. Negative
3. Zero
4. Can’t say

Solution: (2, 3)

Example 3. Five styrofoam balls A, B, C, D, and E are used in an experiment. Several experiments are performed on the balls and the following observations are made :

1. Ball A repels C and attracts B.
2. Ball D attracts B and does not affect E.
3. A negatively charged rod attracts both A and E.

For your information, an electrically neutral Styrofoam ball is very sensitive to charge induction and gets attracted considerably, if placed near a charged body. What are the charges, if any, on each ball?

A   B   C   D  E

(1) +   –   +  0   +

(2) +   –  +  +    0

(3) +   –   +  0   0

(4) –   +   –   0  0

Answer: 3

Solution:

From (1), As A repels C, both A and C must be charged similarly. Either both are +ve or both are –ve. As A also attracts B, so charge on B should be opposite of A or B may be an uncharged conductor.

From (2) As D does not affect E, both D and E should be uncharged, and as B attracts uncharged D, B must be charged and D must be on the uncharged conductor.

From (3) a –ve charged rod attracts the charged ball A, so A must be +ve, and from exp. (i) C must also be +ve and B must be –ve.

Example 4. Charge conservation is always valid. Is it also true for mass?
Solution:

No, mass conservation is not always. In some nuclear reactions, some mass is lost and it is converted into energy.

Example 5. What are the differences between charging by induction and charging by conduction?
Solution:

The major differences between the two methods of charging are as follows :

In induction, two bodies are close to each other but do not touch each other while in conduction they touch each other. (or they are connected by a metallic wire)

In induction, the total charge of a body remains unchanged while in conduction it changes.

In induction, the induced charge is always opposite to that of the source charge while in conduction charge on two bodies finally is of the same nature.

Example 6. If a glass rod is rubbed with silk it acquires a positive charge because :

1. Protons are added to it
2. Protons are removed from it
3. Electrons are added to it
4. Electrons are removed from it.

Answer: 4. Electrons are removed from it.

Example 7. How can you charge a metal sphere positively without touching it?
Solution :

Figure (a) shows an uncharged metallic sphere on an insulating stand. Bring a negatively charged rod close to the metallic sphere, as shown in Fig. (b). As the rod is brought close to the sphere, the free electrons in the sphere move away due to repulsion and start piling up at the farther end. The near end becomes positively charged due to a deficit of electrons. This process of charge distribution stops when the net force on the free electrons inside the metal is zero.

Connect the sphere to the ground by a conducting wire. The electrons will flow to the ground while the positive charges at the near end will remain held there due to the attractive force of the negative charges on the rod, as shown in Fig. (c).

Disconnect the sphere from the ground. The positive charge continues to be held at the near end of Fig.(d) Remove the electrified rod. The positive charge will spread uniformly over the sphere as shown in Fig. (e).

Example 8. If 10⁹ electrons move out of one body to another body every second, how much time is required to get a total charge of 1 C on the other body?
Solution:

In one second 10⁹ electrons move out of the body. Therefore the charge given out in one second is 1.6 × 10^-19 × 10⁹ C = 1.6 × 10^-10 C.

The time required to accumulate a charge of 1 C can then be estimated to be

⇒ \(\frac{1 \mathrm{C}}{1.6 \times 10^{-10} \mathrm{C} / \mathrm{s}}=6.25 \times 10^9 \mathrm{~s}=\frac{6.25 \times 10^9}{365 \times 24 \times 3600} \text { years }=198 \text { years. }\)

Thus to collect a charge of one coulomb, from a body from which 109 electrons move out every second, we will need approximately 200 years. One coulomb is, therefore, a very large unit for many practical purposes.

Example 9. How much positive and negative charge is there in a cup of water?
Solution :

Let us assume that the mass of one cup of water is 250 g.
The molecular mass of water is 18g.

One mole(= 6.02 × 10²³ molecules) of water is 18 g. Therefore the number of molecules in one 9

cup of water is \(\frac{250 \times 10^9}{18} \times 6.02 \times 10^{23}\)

Each molecule of water contains two hydrogen atoms and one oxygen atom, i.e., 10 electrons and 10 protons. Hence the total positive and total negative charge has the same magnitude. It is equal to \(\frac{250 \times 10^9}{18} \times 6.02 \times 10^{23} \times 10 \times 1.6 \times 10^{-19} \mathrm{C}=1.34 \times 10^7 \mathrm{C} \text {. }\)

Example 10. Which is bigger, a coulomb or charge on an electron? How many electronic charges form one coulomb of charge?
Solutions :

A coulomb of charge is bigger than the charge on an electron.

The magnitude of the  charge on one electron, e = 1.6 ×10^-19 coulomb

Number of electronic charge in one coulomb, \(n=\frac{q}{e}=\frac{1}{1.6 \times 10^{-19}}=0.625 \times 10^{19}\)

Example 11. Assume that each atom in a copper wire contributes one free electron. Estimate the number of free electrons in a copper wire having a mass of 6.4 g (take the atomic weight of copper to be 64g/mol).
Solutions :

Number of atoms in 64 g of copper = 6.023 × 10²³

Number of atoms in 6.4 g of copper = \(\frac{6.023 \times 10^{23}}{64} \times 6.4=6.023 \times 10^{22}\)

As each atom contributes one free electron, therefore, number of free electrons in copper wire = 6.023 × 1022.

3. Coulomb’s Law (Inverse Square Law)

Based on experiments Coulomb established the following law known as Coulomb’s law. The magnitude of the electrostatic force between two point charges is directly proportional to the product of charges and inversely proportional to the square of the distance between them.

i.e. F ∝ q1q2 and F ∝\(\frac{1}{r^2} \quad \Rightarrow \quad F \propto \frac{q_1 q_2}{r^2} \quad \Rightarrow \quad F=\frac{K q_1 q_2}{r^2}\)

Important points regarding Coulomb’s law :

1. It is applicable only for point charges.
2. The constant of proportionality K in SI units in a vacuum is expressed as \(\frac{1}{4 \pi \varepsilon_0}\)and in any other medium expressed as \(\frac{1}{4 \pi \varepsilon}\) If charges are dipped in a medium then electrostatic force on one 230
3. Charge is \(\frac{1}{4 \pi \varepsilon_0 \varepsilon_r} \frac{q_1 q_2}{r^2} .\) ε₀ and ε are called the permittivity of the vacuum and absolute permittivity of the medium respectively. The ratio ε/ε₀ = εr is called the relative permittivity of the medium, which is a dimensionless quantity.
4. The value of relative permittivity εris constant for medium and can have values between 1 to ∞. For vacuum, by definition, it is equal to 1. For air, it is nearly equal to 1 and may be taken to be equal to 1 for calculations. For metals, the value of εris ∞ and for water is 81. The material in which more charge can induce εr will be higher.
5. The value of \(\frac{1}{4 \pi \varepsilon_0}\) = 9 × 10⁹ Nm²C–2 ⇒ ε₀= 8.855 × 10^-12C2/Nm². 0
6. Dimensional formula of ε is M^-1 L^-3 T⁴ A²
7. The force acting on one point charge due to the other point charge is always along the line joining these two charges. It is equal in magnitude and opposite in direction on two charges, irrespective of the medium, in which they lie.
8. The force is conservative i.e., work done by electrostatic force in moving a point charge along a close loop of any shape is zero.
9. Since the force is a central force, in the absence of any other external force, the angular momentum of one particle w.r.t. the other particle (in a two-particle system) is conserved,
10. Coulomb Low In Vector Form

Let us consider q₁ and q₂are placed at positions r₁ = x₁iˆ + y₁jˆ + z₁kˆand r = x₂ i + y₂j + z₂ k respectively. If we want to calculate coulomb force on q₂ then q₁ will be considered as a source charge and q₂ will be considered as a test charge.

⇒ \(\vec{F}=\frac{\mathrm{kq}_1 q_2}{|\overrightarrow{\mathrm{r}}|^3} \overrightarrow{\mathrm{r}} \text { where } \overrightarrow{\mathrm{r}}\)= position vector of test charge – position vector source charge.

(r= position vector of test charge w.r.t. source charge)

1. When we use this formula in vector form then we have to put the value of charges with their sign.
2. If the force F₁₂ on charge q₁ due to charge q₂, and F₂₁ is force on charge q2due to charge q₁ then \(\overrightarrow{\mathrm{F}}_{12}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}_{21}^2} \hat{r}_{12}=-\overrightarrow{\mathrm{F}}_{21}\)

Example 12. Find out the electrostatic force between two point charges placed in the air (each of +1 C) if they are separated by 1m.
Solution :

⇒ \(F_e=\frac{k q_1 q_2}{r^2}=\frac{9 \times 10^9 \times 1 \times 1}{1^2}=9 \times 10^9 \mathrm{~N}\)

From the above result, we can say that the 1 C charge is too large to realize. In nature, the charge is usually of the order of μC

Example 13. Two particles having charges q¹ and q² when kept at a certain distance, exert a force F on each other. If the distance between the two particles is reduced to half and the charge on each particle is doubled then what will be the force between the particles :
Solution :

⇒ \(F=\frac{k q_1 q_2}{r^2}\)

If q’₁= 2q₁, q’₂= 2q₂ r’ = \(\frac{r}{2}\)

then F’ = \(F^{\prime}=\frac{k q_1^{\prime} q_2^{\prime}}{r^{\prime 2}}=\frac{k\left(2 q_1\right)\left(2 q_2\right)}{\left(\frac{r}{2}\right)^2} \)

F’ = \(\quad F^{\prime}=\frac{16 k q_1 q_2}{r^2}\)

Example 14. A particle of mass m carrying charge q¹ revolves around a fixed charge –q² in a circular path of radius r. Calculate the period of revolution and its speed.
Solution :

⇒ \(\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2}=m r \omega^2=\frac{4 \pi^2 m r}{T^2}, \quad T^2=\frac{\left(4 \pi \varepsilon_0\right) r^2\left(4 \pi^2 m r\right)}{q_1 q_2} \quad \text { or } \quad T=4 \pi r \sqrt{\frac{\varepsilon_0 m r}{q_1 q_2}}\)

and also we can say that \(\frac{q_1 q_2}{4 \pi \varepsilon_0 r^2}=\frac{m v^2}{r} \quad \Rightarrow \quad v=\sqrt{\frac{q_1 q_2}{4 \pi \varepsilon_0 m r}}\)

Example 15. A point charge q_A= + 100 µc is placed at point A (1, 0, 2) m, and another point charge q_B= +200µc is placed at point B (4, 4, 2) m. Find :

1. The magnitude of the Electrostatic interaction force acting between them
2. F_A(force on A due to B) and F_B(force on B due to A) in vector form

Solution :

Value of F : \(|F|=\frac{k q_A q_B}{r^2}=\frac{\left(9 \times 10^9\right)\left(100 \times 10^{-6}\right)\left(200 \times 10^{-6}\right)}{\sqrt{(4-1)^2+(4-0)^2+(2-2)^2}}=7.2 \mathrm{~N}\)

Force on B \(\vec{F}_B=\frac{k q_A q_B}{|\vec{r}|^3} \vec{r}=\frac{\left(9 \times 10^9\right)\left(100 \times 10^{-6}\right)\left(200 \times 10^{-6}\right)}{\sqrt{(4-1)^2+(4-0)^2+(2-2)^2}}[(4-1) \hat{\mathrm{i}}+(4-0) \hat{\mathrm{j}}+(2-2) \hat{k}]\)

⇒ \(7.2\left(\frac{3}{5} \hat{i}+\frac{4}{5} \hat{j}\right) N\)

Similarly \(\vec{F}_A=7.2\left(-\frac{3}{5} \hat{i}-\frac{4}{5} \hat{j}\right) \mathrm{N}\)

Action(F_A) and Reaction (F_B) are equal but in opposite directions.

Example 16. A charged metallic sphere A is suspended by a nylon thread. Another charged metallic sphere B held by an insulating handle is brought close to A such that the distance between their centers is 10 cm, as shown in Fig. (a). The resulting repulsion of A is noted (for example, by shining a beam of light and measuring the deflection of its shadow on a screen). Spheres A and B are touched by uncharged spheres C and D respectively, as shown in Fig. (b). C and D are then removed and B is brought closer to A distance of 5.0 cm between their centers, as shown in Fig. (c). What is the expected repulsion of A based on Coulomb’s law? Spheres A and C and spheres B and D have identical sizes. Ignore the sizes of A and B in comparison to the separation between their centers.

Solution :

Let the original charge on sphere A be q and that on B be q’. At a distance r between their centers, the magnitude of the electrostatic force on each is given by

⇒ \(F=\frac{1}{4 \pi \varepsilon_0} \frac{q^{\prime}}{\mathrm{r}^2}\)

neglecting the sizes of spheres A and B in comparison to r. When an identical but uncharged sphere C touches A, the charges redistribute on A and C, and, by symmetry, each sphere carries a charge q/2. Similarly, after D touches B, the redistributed charge on each is q’/2. Now, if the separation between A and B is halved, the magnitude of the electrostatic force on each is

⇒ \(\mathrm{F}^{\prime}=\frac{1}{4 \pi \varepsilon_0} \frac{(\mathrm{q} / 2)\left(\mathrm{q}^{\prime} / 2\right)}{(\mathrm{r} / 2)^2}=\frac{1}{4 \pi \varepsilon_0} \frac{\left(\mathrm{qq}^{\prime}\right)}{\mathrm{r}^2}=F\)

Thus the electrostatic force on A, due to B, remains unaltered.

4. Principle Of Superposition

The electrostatic force is a two-body interaction i.e. electrical force

between two point charges are independent of the presence or absence of other charges and so the principle of superposition is valid i.e. force on the charged particle due to several point charges is the resultant of forces due to individual point charges.

Consider that n point charges q1, q2, q3, …. are discretely distributed in space. The charges are interacting with each other. Let us find the total force on the charge, say due to all other remaining charges. If the charge q2, q3, …. qnexert forces\(\overrightarrow{\mathrm{F}}_{12}, \overrightarrow{\mathrm{F}}_{13}, \ldots . \overrightarrow{\mathrm{F}}_{1 \mathrm{n}}\)on the charge q1, then according to principle of superposition, the total force on charge q1is given by \(\overrightarrow{\mathrm{F}}_1=\overrightarrow{\mathrm{F}}_{12}+\overrightarrow{\mathrm{F}}_{13}+\ldots \ldots \overrightarrow{\mathrm{F}}_{1 \mathrm{n}}\)

Solved Examples

Example 17. Two point charges of charge value Q and q are placed at a distance of x and x/2 respectively from a third charge of charge value 4q, all charges being in the same straight line. Calculate the magnitude and nature of charge Q, such that the net force experienced by the charge q is zero.
Solution :

Suppose that the charge 4q is located at point A. The charges Q and q are placed at points B and C, such that AB = x and AC = x/2. Also, all the charges lie in the same straight line. We assume that the charges of 4q and q are of the same nature, a say positive.

Then, force on the charge q due to 4q,

⇒ \(\mathrm{F}_{\mathrm{A}}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{4 \mathrm{q} \cdot \mathrm{q}}{(\mathrm{x} / 2)^2}\)

The net force experienced by charge q will be zero only if the charge Q exerts force on the charge q equal and opposite to that exerted by the charge 4q. Thus, the charge Q should exert force FBon charge q equal to FA(in magnitude) and along CA. For this, charge Q has to be positive (i.e. of the nature same as that of 4q or q).

Now, force on the charge q due to charge Q

FB= \(\mathrm{F}_{\mathrm{B}}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{Q} . \mathrm{q}}{(\mathrm{BC})^2}\)

FB=\(\mathrm{F}_{\mathrm{B}}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{Q} . \mathrm{q}}{(\mathrm{x} / 2)^2}\) (along CA)

For net force on the charge q to be zero, FB= FA

⇒ \(\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q . q}{(x / 2)^2}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{4 q \cdot q}{(x / 2)^2}=Q=4 q\)

Example 18. Consider three point charges each having charge q at the vertices of an equilateral triangle of side l. What is the force on a charge Q (with the same sign as q) placed at the centroid of the triangle, as shown in the figure?

Solution :

In the given equilateral triangle ABC of sides of length l, if we draw a perpendicular AD to the side BC, AD = AC cos 30º = ( 3 / 2) l and the distance AO of the centroid O from A is (2/3) AD = Thus, (1/ 3)l. By symmetry AO = BO = CO.

Force \(\overrightarrow{\mathrm{F}}_1\) on Q due to charge q at A = \(\frac{3}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{\ell^2}\) along AO

Force \(\overrightarrow{\mathrm{F}}_2\)on Q due to charge q at B = \(\frac{3}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{\ell^2}\) along BO

Force \(\overrightarrow{\mathrm{F}}_3\)on Q due to charge q at C = \(\frac{3}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{\ell^2}\)along CO

The resultant of forces \(\overrightarrow{\mathrm{F}}_2 and \overrightarrow{\mathrm{F}}_3\) is \(\frac{3}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{\ell^2}\) along OA, by the parallelogram law. Therefore, the total force on Q = \(\frac{3}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{\ell^2}\) = 0, where rˆ is the unit vector along OA.

It is clear also by symmetry that the three forces will sum to zero. Suppose that the resultant force was non-zero but in some direction. Consider what would happen if the system was rotated through 60º about O.

Example 19. Consider the charges q, q, and –q placed at the vertices of an equilateral triangle of side l. Calculate the force on each charge.?
Solution :

The forces acting on charge q at A due to charges q at B and –q at C are \(\overrightarrow{\mathrm{F}}_{\mathrm{AB}}\) along BA and \(\overrightarrow{\mathrm{F}}_{\mathrm{Ac}}\) along AC respectively as shown in Fig.

The force of attraction or repulsion for each pair of charges has the same magnitude 2

⇒ \(\mathrm{F}=\frac{\mathrm{q}^2}{4 \pi \varepsilon_0 \ell^2}\)

⇒ \(\left|F_A\right|=\left|F_B\right|=F\)

⇒ \(\left|F_C\right|=\sqrt{3} F\)

⇒ \(\overrightarrow{\mathrm{F}}_{\mathrm{A}}+\overrightarrow{\mathrm{F}}_{\mathrm{B}}+\overrightarrow{\mathrm{F}}_{\mathrm{C}}=0 .\)

It is interesting to see that the sum of the forces on three charges is zero.

Example 20. Two pith-balls each of mass weighing 10^-4kg are suspended from the same point using silk threads 0.5 m long. On charging the pith balls equally, they are found to repel each other to a distance of 0.6 m. Calculate the charge on each ball. (g = 10m/s²)
Solution :

Consider two pith balls A and B each having charge q and mass 10^-13 kg. When the pith balls are suspended from point S by two threads each 0.5 m long, they repel each other to the distance AB = 0.2 m as shown in Fig.

Each of the two pith-balls is in equilibrium under the action of the following three forces :

1. The electrostatic repulsive force F.
2. The weight acts vertically downwards.
3. The tension T in the string is directed towards point S. The three forces mg, F, and T can be represented by the therefore, according to the  triangle law of forces,

⇒ \(\frac{F}{O A}=\frac{m g}{S O}=\frac{T}{A S}\)…….(1)

Here, \(\mathrm{F}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}^2}{(\mathrm{AB})^2}=9 \times 10^9 \times \frac{\mathrm{q}^2}{(0.6)^2} \mathrm{~N}\)

mg = 10^-4 × 10 = 10^-3 N

From the equation (i), we have

F = mg × \(\frac{O A}{S O} \quad \text { or } \quad 9 \times 10^9 \times \frac{q^2}{(0.6)^2}=10^{-3} \times \frac{0.3}{\sqrt{(0.5)^2-(0.3)^2}} \text { or } \quad q=\sqrt{3} \times 10^{-6} \mathrm{C}\)

Example 21 Three equal point charges of charge +qise moving along a circle of radius R and a point charge –2q is also placed at the center of the circle as (shown in the figure), if charges are revolving with constant and same speed then calculate speed.

Solution :

F₂– 2F₁cos 30 = \(\frac{m v^2}{R} \Rightarrow \frac{K(q)(2 q)}{R^2}-\frac{2\left(K q^2\right)}{(\sqrt{3} R)^2} \cos 30=\frac{m v^2}{R} \Rightarrow v=\sqrt{\frac{k q^2}{R m}}\left[2-\frac{1}{\sqrt{3}}\right]\)

Example 22 Two equally charged identical small metallic spheres A and B repel each other with a force 2 × 10^-5N when placed in air (neglect gravitation attraction). Another identical uncharged sphere C is touched to B and then placed at the midpoint of the line joining A and B. What is the net electrostatic force on C?
Solution :

Let initially the charge on each sphere be q and separation between their centers be r; then according to a given problem.

⇒ \(\mathrm{F}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q} \times \mathrm{q}}{\mathrm{r}^2}=2 \times 10^{-5} \mathrm{~N}\)

When sphere C touches B, the charge of B, q will distribute equally on B and C as spheres are identical conductors, i.e., now charges on spheres;

q_B= q_C= (q/2)

So sphere C will experience a force

⇒ \(\mathrm{F}_{\mathrm{CA}}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}(\mathrm{q} / 2)}{(\mathrm{r} / 2)^2}=2 \mathrm{~F} \text { along } \overrightarrow{\mathrm{AB}}\) due to charge on A

and,\(\mathrm{F}_{\mathrm{CB}}=\frac{1}{4 \pi \varepsilon_0} \frac{(\mathrm{q} / 2)(\mathrm{q} / 2)}{(\mathrm{r} / 2)^2}=\mathrm{F} \text { along } \overrightarrow{\mathrm{BA}}\) due to charge on B

So the net force FCon C due to charges on A and B,

F_C= F_CA – F_CB = 2F – F = 2 × 10–5 N along AB.

Example 23 Five point charges, each of value q are placed on five vertices of a regular hexagon of side L. What is the magnitude of the force on a point charge of value – q coulomb placed at the center of the hexagon?
Solution :

Method: 1

If there had been a sixth charge +q at the remaining vertex of hexagon force due to all the six charges on –q at O would be zero (as the forces due to individual charges will balance each other), i.e.,

⇒ \(\overrightarrow{F_R}=0\)

Now if \(\vec{f}\) is the force due to the sixth charge and \(\vec{f}\) due to the remaining five charges.

⇒ \(\vec{F}+\vec{f}=0 \quad \text { i.e. } \quad \vec{F}=-\vec{f} \text { or, } \quad|F|=|f|=\frac{1}{4 \pi \varepsilon_0} \frac{q \times q}{L^2}=\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{L^2}\)

⇒ \(\vec{F}_{\text {Net }}=\vec{F}_{\mathrm{CO}}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}^2}{\mathrm{~L}^2} \text { along } \mathrm{CO}\)

Method: 2

In the diagram we can see that force due to charges A and D are opposite to each other

Similarly

⇒ \(\vec{F}_{D O}+\vec{F}_{A O}=0\)

⇒ \(\vec{F}_{B O}+\vec{F}_{E O}=0\)

Using (1) and (2)\(\vec{F}_{\text {Net }}=\overrightarrow{\mathrm{F}}_{\mathrm{CO}}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}^2}{\mathrm{~L}^2} \text { along } \mathrm{CO} \text {. }\) along CO. 2

Note: The total charge of A rod cannot be considered to be placed at the center of the rod as we do in mechanics for mass in many problems.

Note: If a >> l then F =\(\frac{K Q q}{a^2}\) behavior of the rod is just like a point charge.

5. Electrostatic Equilibrium

The point where the resultant force on a charged particle becomes zero is called the equilibrium position.

5.1 Stable Equilibrium: A charge is initially in an equilibrium position and is displaced by a small distance. If the charge tries to return to the same equilibrium position then this equilibrium is called the position of stable equilibrium.

5.2 Unstable Equilibrium: If the charge is displaced by a small distance from its equilibrium position and the charge does not tend to return to the same equilibrium position. Instead, it goes away from the equilibrium position.

5.3 Neutral Equilibrium: If the charge is displaced by a small distance and it is still in equilibrium condition then it is called neutral equilibrium.

Solved Examples

Example 24 Two equal positive point charges ‘Q’ are fixed at points B(a, 0) and A(–a, 0). Another test charge q0 is also placed at O(0, 0). Show that the equilibrium at ‘O’ is

1. Stable for displacement along the X-axis.
2. Unstable for displacement along the Y-axis.

Solution :

Initially\(\overrightarrow{\mathrm{F}}_{\mathrm{AO}}+\overrightarrow{\mathrm{F}}_{\mathrm{BO}}=0 \Rightarrow\left|\overrightarrow{\mathrm{F}}_{\mathrm{AO}}\right|=\left|\overrightarrow{\mathrm{F}}_{\mathrm{BO}}\right|=\frac{K Q q_0}{\mathrm{a}^2}\)

When the charge is slightly shifted towards the + x-axis by a small distance Δx, then.

⇒ \(\left|\vec{F}_{\mathrm{AO}}\right|<\left|\dot{\vec{F}}_{\mathrm{BO}}\right|\)

Therefore the particle will move toward the origin (its original position) hence the equilibrium is stable.

When the charge is shifted along the axis

After resolving components net force will be along the y-axis so the particle will not return to its original position so it is an unstable equilibrium. Finally, the charge will move to infinity.

Example 25. Two point charges of charge q₁ and q₂(both of the same sign) and each of mass m are placed such that gravitation attraction between them balances the electrostatic repulsion. Are they in stable equilibrium? If not then what is the nature of equilibrium?
Solution :

In given example :

⇒ \(\frac{\mathrm{Kq}_1 \mathrm{q}_2}{\mathrm{r}^2}=\frac{\mathrm{Gm}^2}{\mathrm{r}^2}\)

We can see that irrespective of the distance between them charges will remain in equilibrium. If now distance is increased or decreased then there is no effect on their equilibrium. Therefore it is a neutral equilibrium.

Example 26. A particle of mass m and charge q is located midway between two fixed charged particles each having a charge q and a distance 2l apart. Prove that the motion of the particle will be SHM if it is displaced slightly along the line connecting them and released. Also, find its time.
Solution :

Let the charge q at the mid-point the displaced slightly to the left.

The force on the displaced charge q due to charge q at A,

F1=\(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}^2}{(\ell+\mathrm{x})^2}\)

The force on the displaced charge q due to charge at B,

F₂= \(\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{(\ell-x)^2}\)

Net restoring force on the displaced charge q.

F = F₂– F₁or F = F₂= \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}^2}{(\ell-\mathrm{x})^2}-\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}^2}{(\ell+\mathrm{x})^2}\)

or F = \(\mathrm{F}=\frac{\mathrm{q}^2}{4 \pi \varepsilon_0}\left[\frac{1}{(\ell-\mathrm{x})^2}-\frac{1}{(\ell+\mathrm{x})^2}\right]=\frac{\mathrm{q}^2}{4 \pi \varepsilon_0} \frac{4 \ell \mathrm{x}}{\left(\ell^2-\mathrm{x}^2\right)^2}\)

Since l >> x, ∴ F = \(\frac{q^2 \ell x}{\pi \varepsilon_0 \ell^4} \text { or } F=\frac{q^2 x}{\pi \varepsilon_0 \ell^3}\)

We see that F ∝ x and it is opposite to the direction of displacement. Therefore, the motion is

SHM. T =\(\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}} \text {, here } \mathrm{k}=\frac{\mathrm{q}^2}{\pi \epsilon_0 \ell^3}=2 \pi \sqrt{\frac{\mathrm{m} \pi \epsilon_0 \ell^3}{\mathrm{q}^2}}\)

Example 27. Two identical charged spheres are suspended by strings of equal length. Each string makes an angle θ with the vertical. When suspended in a liquid of density σ = 0.8 gm/cc, the angle remains the same. What is the dielectric constant of the liquid? (The density of the material of the sphere is ρ = 1.6 gm/cc.)
Solution :

Initially, as the forces acting on each ball are tension T, weight mg, and electric force F, for its equilibrium along vertically,

T cos θ = mg …(1)

and along horizontal

T sin θ = F …(2)

Dividing Eqn. (2) by (1), we have

tan θ = Fmg… (3)

When the balls are suspended in a liquid of density σ and dielectric constant K, the electric force will become (1/K) times, i.e., F’ = (F/K) while weight

mg’ = mg – FB= mg – Vσg [as FB= Vσg, where σ is the density of the material of the sphere]

i.e., mg’ = mg \(\) So for equilibrium of ball,

tan θ’ = \(\frac{F^{\prime}}{m g^{\prime}}=\frac{F}{K m g[1-(\sigma / \rho)]}\)… (4)

According to the given information θ’ = θ; so from equations (4) and (3), we have

K = \(\frac{\rho}{(\rho-\sigma)}=\frac{1.6}{(1.6-0.8)}=2\)

6. Electric Field

The electric field is the region around a charged particle or charged body in which if another charge is placed, it experiences electrostatic force.

6.1 Electric field intensity

E: Electric field intensity at a point is equal to the electrostatic force experienced by a unit positive point charge both in magnitude and direction.

If a test charge q0is placed at a point in an electric field and experiences a force \(\overrightarrow{\mathrm{F}}\) due to some charges (called source charges), the electric field intensity at that point due to source charges is given

⇒ \(\overrightarrow{\mathrm{E}}=\frac{\overrightarrow{\mathrm{F}}}{\mathrm{q}_0} \text {; }\)

If the \(\overrightarrow{\mathrm{E}}\)is to be determined practically then the test charge q0 should be small otherwise it will affect the charge distribution on the source which is producing the electric field and hence modify the quantity which is measured.

Example 28. A positively charged ball hangs from a long silk thread. We wish to measure E at point P in the same horizontal plane as that of the hanging charge. To do so, we put a positive test charge q0 at the point and measure F/q₀. Will F/q₀ be less than, equal to, or greater than E at the point in question?
Solution :

When we try to measure the electric field at point P then after placing the test charge at P it repels the source charge (suspended charge) and the measured value of the electric field

Emeasured =\(\frac{F}{q_0}\) will be less than the actual value Eact that we wanted to measure.

6.2 Properties of electric field intensity E:

It is a vector quantity. Its direction is the same as the force experienced by a positive charge.

The direction of the electric field due to the positive charge is always away from it while due to the negative charge is always towards it.

Its S.Ι. unit is Newton/Coulomb.

Its dimensional formula is [MLT^-3A^-1]

The electric force on a charge q placed in a region of the electric field at a point where the electric field intensity is Eis given by \(\overrightarrow{\mathrm{F}}=\mathrm{q} \overrightarrow{\mathrm{E}} \text {. }\).

The electric force on the point charge is in the same direction as the electric field on a positive charge and in the opposite direction on a negative charge.

It obeys the superposition principle, that is, the field intensity at a point due to a system of charges is a vector sum of the field intensities due to individual point charges.

E = E₁ + E₂ + E₃ + …….

It is produced by source charges. The electric field will be a fixed value at a point unless we change the distribution of source charges.

6.3 Electric field due to a point charge :

Consider that a point charge +q is placed at the origin O of the coordinate frame. Let P be the point,

where the electric field due to the point charge +q is to be determined. Let \(\overline{\mathrm{OP}}=\overrightarrow{\mathrm{r}}\) the position vector of the point P.

To find the electric field at point P, place a vanishingly small positive test charge q0at point P. According to Coulomb’s law, force on the test charge due to charge q is given by :

⇒ \(\overrightarrow{\mathrm{F}}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{qq}}{\mathrm{r}_0} \hat{\mathrm{r}}\)

where rˆis unit vector along OP. If \(\) is the electric field at point P, then

⇒ \(\vec{E}=\frac{\vec{F}}{q_0}=\left(\frac{1}{q_0} \cdot \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q q_0}{r^2} \hat{r}\right) \quad \text { or } \quad \vec{E}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r^2} \hat{r}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r^3} \vec{r}\)

The magnitude of the electric field at point P is given by

⇒ \(E=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r^2}\)

Solved Examples

Example 29. The electrostatic force experienced by –3μC charge placed at point ‘P’ due to a system ‘S’ of fixed point charges as shown in the figure is \(\overrightarrow{\mathrm{F}}=(2 \hat{1} \hat{i}+9 \hat{\mathbf{j}}) \mu \mathrm{N}.\)

1. Find out electric field intensity at point P due to S.
2. If now 2μC charge is placed and –3 μC is removed at point P then the force experienced by it will be.

Solution :

⇒ \(\vec{F}=q \vec{E} \quad \Rightarrow \quad(21 \hat{i}+9 \hat{j}) \mu N=-3 \mu C(\vec{E})\)

⇒ \(\vec{E}=-7 \hat{i}-3 \hat{j} \frac{\mu \mathrm{N}}{C}\)

Since the source charges are not disturbed the electric field intensity at ‘P’ will remain the same.

⇒ \(\vec{F}_{2 \mu C}=+2(\vec{E})=2(-7 \hat{i}-3 \hat{j})=-14 \hat{i}-6 \hat{j} \mu N\)

Example 30. Calculate the electric field intensity which would be just sufficient to balance the weight of a particle of charge –10 μc and mass 10 mg. (take g = 10 ms²)
Solution :

As the force on a charge q in an electric field \( \overrightarrow{\mathrm{E}}\) is

⇒ \(\overrightarrow{\mathrm{F}}_{\mathrm{q}}=\mathrm{q} \overrightarrow{\mathrm{E}}\)

So according to the given problem Fe

⇒ \(\left|\vec{F}_{\mathrm{q}}\right|=|\vec{W}| \quad \text { i.e., } \quad|\mathrm{q}| E=m g\)

⇒ \(E=\frac{\mathrm{mg}}{|\mathrm{q}|}=10 \mathrm{~N} / \mathrm{C} .\) in the downward direction.

List of formulas for Electric Field Intensity due to various types of charge distribution :

Electric Field Due To Point Charge

⇒ \(E=\frac{k q}{r^2} \Rightarrow \quad \vec{E}=\frac{k q}{r^2} \hat{r} \quad \Rightarrow \quad \vec{E}=\frac{k q}{r^3} \vec{r}\)

⇒ \(\overrightarrow{\mathrm{r}}\)= position vector of test point with respect to source charge

⇒ \(\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{r}}_{\text {testpoint }}-\overrightarrow{\mathrm{r}}_{\text {sourcecharge }}\)

Solved Examples

Example 31. Find out the electric field intensity at point A (0, 1m, 2m) due to a point charge –20μC situated at point B( 2m, 0, 1m).
Solution :

⇒ E = \(E=\frac{K Q}{|\vec{r}|^3} \vec{r}=\frac{K Q}{|\vec{r}|^2} \hat{r}\) ⇒r= P.V. of A – P.V. of B (P.V. = Position vector)

⇒ \(=(-\sqrt{2} \hat{i}+\hat{j}+\hat{k}) \quad|\vec{r}|==2\) = 2

⇒ E = \(E=\frac{9 \times 10^9 \times\left(-20 \times 10^{-6}\right)}{8}(-\sqrt{2} \hat{i}+\hat{j}+\hat{k})=-22.5 \times 10^3(-\sqrt{2} \hat{i}+\hat{j}+\hat{k}) N / C .\)

Example 32. Two point charges 2μc and – 2μc are placed at points A and B as shown in the figure. Find out electric field intensity at points C and D. [All the distances are measured in meters].

Solution :

The electric field at point C (E_A, are magnitudes only, and arrows represent directions)

The electric field due to positive charge is away from it while due to negative charge, it is towards the charge. It is clear that E_B> E_A.

∴ E_net = (E_B– E_A) towards negative X-axis

⇒ \(\frac{\mathrm{K}(2 \mu \mathrm{c})}{(\sqrt{2})^2}-\frac{\mathrm{K}(2 \mu \mathrm{c})}{(3 \sqrt{2})^2}\) towards negative X-axis = 8000 (–ˆi) N/C

The electric field at point D :

Since the magnitude of charges are same and also AD = BD

So E_A= E_B

Vertical components of\(\overrightarrow{\mathrm{E}}_{\mathrm{A}} \text { and } \overrightarrow{\mathrm{E}}_{\mathrm{B}}\) cancel each other while horizontal components are in the same direction.

So, E_net = 2E_Acosθ = \(\frac{2 . K(2 \mu c)}{2^2} \cos 45^{\circ}\)

⇒ \(=\frac{\mathrm{K} \times 10^{-6}}{\sqrt{2}}=\frac{9000}{\sqrt{2}} \hat{\mathrm{i}} \mathrm{N} / \mathrm{C} .\)

Example 33. Six equal point charges are placed at the corners of a regular hexagon of side ‘a’. Calculate the electric field intensity at the center of the hexagon.

Solution: Zero

Similarly electric field due to a uniformly charged ring at the center of the ring :

Note :

• The net charge on a conductor remains only on the outer surface of a conductor. This property will be discussed in the article on the conductor. (article no.17)
• On the surface of the isolated spherical conductor, the charge is uniformly distributed.

6.4 Electric Field Due To A Uniformly Charged Ring And Arc.

Example 34. Find out the electric field intensity at the center of a uniformly charged semicircular ring of radius R and linear charge density λ.
Solution :

λ = linear charge density.

The arc is the collection of large no. of point charges. Consider a part of the ring as an element of length Rdθ which subtends an angle dθ at the center of the ring and it lies between θ and θ + dθ

⇒ \(\overrightarrow{d E}=d E_x \hat{i}+d E_y \hat{j} \quad E_x=\int d E_x=0\)

⇒ \(E_y=\int d E_y=\int_0^\pi d E \sin \theta=\frac{K \lambda}{R} \int_0^\pi \sin \theta \cdot d \theta=\frac{2 K \lambda}{R}\)

Example 35. Find out the electric field intensity at the center of a uniformly charged quarter ring of radius R and linear charge density λ.
Solution :

Refer to the previous question \(\overrightarrow{d E}=d E_x \hat{i}+d E_y \hat{j}\) on solving E_net= \(E_{\text {net }}=\frac{\mathrm{K} \lambda}{\mathrm{R}}=(\hat{\mathrm{i}}+\hat{\mathrm{j}}) \text {, }\)

Electric field due to ring on its axis :

⇒ \(E_{\text {net }}=\frac{K Q x}{\left[R^2+x^2\right]^{3 / 2}}\)

E will be max when \(\frac{\mathrm{dE}}{\mathrm{dx}}\) = 0, that is at x= \(\frac{R}{\sqrt{2}} \text { and } E_{\max }=\frac{2 K Q}{3 \sqrt{3} \mathrm{R}^2}\)

Case (1) : if x>>R, E = \(\frac{K Q}{x^2}\)

Case (2) : if x<<R, E = \(\frac{K Q x}{R^3}\) Hence the ring will act like a point charge

Solved Examples

Example 36. Positive charge Q is distributed uniformly over a circular ring of radius R. A point particle having a mass m and a negative charge –q is placed on its axis at a distance x from the center. Find the force on the particle. Assuming x << R, find the period of oscillation of the particle if it is released from there. (Neglect gravity)
Solution :

When the negative charge is shifted at a distance x from the center of the ring along its axis then force acting on the point charge due to the ring:

FE= qE (towards centre) = q \(\left[\frac{K Q x}{\left(R^2+x^2\right)^{3 / 2}}\right]\)

if R >>x then

R² + x² ~ R²

⇒ \(F_E=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Qqx}}{\mathrm{R}^3}\)(Towards centre)

Since restoring force FE ∝ x, therefore the motion of charge the particle will be S.H.M. period of SHM.

T = 2π \(\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}} \quad=2 \pi \sqrt{\frac{\mathrm{m}}{\left(\frac{\mathrm{Qq}}{4 \pi \varepsilon_0 \mathrm{R}^3}\right)}}=\left[\frac{16 \pi^3 \varepsilon_0 \mathrm{mR}^3}{\mathrm{Qq}}\right]^{1 / 2}\)

6.5 Electric Field Due To Uniformly Charged Wire

1. Line charge of finite length: Derivation of expression for intensity of the electric field at a point due to line charge of finite size of uniform linear charge density λ. The perpendicular distance of the point from the line charge is r and lines joining ends of line charge distribution make angle θ₁and θ₂ with the perpendicular line.

Ex = \(\frac{K \lambda}{\mathbf{K}}\)[sinθ₁+ sinθ₂] ………..(1) Kλ

EY= \(\frac{K \lambda}{\mathbf{K}}\)[cosθ₂– cosθ₁]

Net electric field at the point

E_net = \(\sqrt{E_x^2+E_y^2}\)

2. We can derive a result for infinitely long line charge: In the above eq. (1) and (2) if we put θ₁= θ₂ = 90º we can get the required result. 2K

E_net = Ex=\(\frac{2 \mathrm{~K} \lambda}{\mathrm{r}}\)

3. For Semi-infinite wire

θ₁= 90º and θ₂ = 0º so Ex=\(E_x=\frac{K \lambda}{r}, E_y=\frac{K \lambda}{r}\)

Solved Examples

Example 37. A point charge q is placed at a distance r from a very long charge thread of uniform linear charge density λ. Find out the total electric force experienced by the line charge due to the point charge. (Neglect gravity).
Solution :

Force on charge q due to the thread,

F = \(\left(\frac{2 K \lambda}{r}\right) \cdot q\)

By Newton’s 3 law, every action has equal and

opposite reaction so force on the thread = \(\frac{2 K \lambda}{r} \cdot q\) (away from point charge)

6.6 Electric Field Due To Uniformly Charged Infinite Sheet

E_net = \(\frac{\sigma}{2 \varepsilon_0}\)toward normal direction.

Note:

• The direction of the electric field is always perpendicular to the sheet.
• The magnitude of the electric field is independent of the distance from the sheet.

Solved Examples

Example 38. An infinitely large plate of surface charge density +σ is lying in a horizontal xy plane. A particle having charge –qo and mass m is projected from the plate with velocity u making an angle θ with a sheet. Find :

1. The time is taken by the particle to return to the plate.
2. Maximum height achieved by the particle.
3. At what distance will it strike the plate (Neglecting gravitational force on the particle)

Solution :

Electric force acting on the particle Fe= qoE : Fe= (qo)\(\) downward

So acceleration of the particle : a = \(\left(\frac{\sigma}{2 \varepsilon_0}\right)\) = uniform

this acceleration will act like ‘g’ (acceleration due to gravity)

So the particle will perform projectile motion.

T = \(\frac{2 u \sin \theta}{g}=\frac{2 u \sin \theta}{\left(\frac{q_0 \sigma}{2 \varepsilon_0 m}\right)}\)

H = \(H=\frac{u^2 \sin ^2 \theta}{2 g}=\frac{2 u^2 \sin ^2 \theta}{2\left(\frac{q_0 \sigma}{2 \varepsilon_0 m}\right)}\)

R = \(R=\frac{u^2 \sin 2 \theta}{g}=\frac{u^2 \sin 2 \theta}{\left(\frac{q_0 \sigma}{2 \varepsilon_0 m}\right)}\)

Example 39. A block having mass m and charge –q is resting on a frictionless plane at a distance L from a fixed large non-conducting infinite sheet of uniform charge density σ as shown in Figure. Discuss the motion of the block assuming that the collision of the block with the sheet is perfectly elastic. Is it SHM?
Solution :

The situation is shown in Figure. The electric force produced by the sheet will accelerate the block towards the sheet producing an acceleration. Acceleration will be uniform because electric field E due to the sheet is uniform.

⇒ \(a=\frac{F}{m}=\frac{q E}{m} \text {, where } E=\sigma / 2 \varepsilon_0\)

As initially, the block is at rest and acceleration is constant, from the second equation of motion, the time taken by the block to reach the wall

⇒ \(\mathrm{L}=\frac{1}{2} \mathrm{at}^2 \quad \text { i.e., } \quad \mathrm{t}=\sqrt{\frac{2 \mathrm{~L}}{\mathrm{a}}}=\sqrt{\frac{2 \mathrm{~mL}}{\mathrm{aE}}}=\sqrt{\frac{4 \mathrm{~mL} \varepsilon_0}{\mathrm{a} \sigma}}\)

As collision with the wall is perfectly elastic, the block will rebound with the same speed and as now its motion is opposite to the acceleration, it will come to rest after traveling the same distance L in the same time t. After stopping it will be again accelerated towards the wall and so the block will execute oscillatory motion with ‘span’ L and period.

⇒ \(\mathrm{T}=2 \mathrm{t}=2 \sqrt{\frac{2 \mathrm{~mL}}{\mathrm{aE}}}=2 \sqrt{\frac{4 \mathrm{~mL} \varepsilon_0}{\mathrm{a \sigma}}}\)

However, as the restoring force F = qE is constant and not proportional to displacement x, the motion is not simply harmonic.

Example 40. If an isolated infinite sheet contains charge Q₁ on its one surface and charge Q2on its other surface then prove that electric field intensity at a point in front of the sheet will be \(\frac{Q}{2 \mathrm{~A} \varepsilon_{\mathrm{O}}},\), where Q = Q₁+ Q₂

Solution: Electric field at point P :

⇒ \(\overrightarrow{\mathrm{E}}=\overrightarrow{\mathrm{E}}_{\mathrm{Q}_1}+\overrightarrow{\mathrm{E}}_{\mathrm{Q}_2}\)

⇒ \(=\frac{Q_1}{2 A \varepsilon_0} \hat{n}+\frac{Q_2}{2 A \varepsilon_0} \hat{n}=\frac{Q_1+Q_2}{2 A \varepsilon_0} \hat{n}=\frac{Q}{2 A \varepsilon_0} \hat{n}\)

[This shows that the resultant field due to a sheet depends only on the total charge of the sheet and not on the distribution of charge on individual surfaces].

Example 41. Three large conducting parallel sheets are placed at a finite distance from each other as shown in the figure. Find out the electric field intensity at points A, B, C, and D.

Solution: For point A

⇒ \(\vec{E}_{\text {net }}=\vec{E}_Q+\vec{E}_{3 Q}+\vec{E}_{-2 Q}=-\frac{Q}{2 A \varepsilon_0} \hat{i}-\frac{3 Q}{2 A \varepsilon_0} \hat{i}+\frac{2 Q}{2 A \varepsilon_0} \hat{i}=-\frac{Q}{A \varepsilon_0} \hat{i}\)

for point B

⇒ \(\vec{E}_{n e t}=\vec{E}_{3 Q}+\vec{E}_{-2 Q}+\vec{E}_Q=-\frac{3 Q}{2 A \varepsilon_0} \hat{i}+\frac{2 Q}{2 A \varepsilon_0} \hat{i}+\frac{Q}{2 A \varepsilon_0} \hat{i}=0\)

for point C

⇒ \(\vec{E}_{\text {net }}=\vec{E}_Q+\vec{E}_{3 Q}+\vec{E}_{-2 Q}=+\frac{Q}{2 A \varepsilon_0} \hat{i}-\frac{3 Q}{2 A \varepsilon_0} \hat{i}-\frac{2 Q}{2 A \varepsilon_0} \hat{i}=-\frac{2 Q}{A \varepsilon_0} \hat{i}\)

for point D

⇒ \(\overrightarrow{\mathrm{E}}_{\text {net }}=\overrightarrow{\mathrm{E}}_{\mathrm{Q}}+\overrightarrow{\mathrm{E}}_{3 Q}+\overrightarrow{\mathrm{E}}_{-2 Q}=+\frac{\mathrm{Q}}{2 \mathrm{~A} \varepsilon_0} \hat{\mathrm{i}}+\frac{3 Q}{2 \mathrm{~A} \varepsilon_0} \hat{\mathrm{i}}-\frac{2 Q}{2 \mathrm{~A} \varepsilon_0} \hat{\mathrm{i}}=\frac{\mathrm{Q}}{\mathrm{A} \varepsilon_0} \hat{\mathrm{i}}\)

6.7 Electric field due to uniformly charged spherical shell

⇒ \(E=\frac{K Q}{r^2} \quad r \geq R\) ⇒ For the outside points and points on the surface the uniformly

charged spherical shell behaves as a point charge placed at the center

E = 0 r < R

The electric field due to the spherical shell outside is always along the radial direction.

Solved Examples

Example 42. The figure shows a uniformly charged sphere of radius R and total charge Q. A point charge q is situated outside the sphere at a distance r from the center of the sphere. Find out the following :

1. Force acting on the point charge q due to the sphere.

2. Force acting on the sphere due to the point charge.

Solution :

The electric field at the position of a point charge

⇒ \(\vec{E}=\frac{K Q}{r^2} \hat{r} \quad \text { so, } \quad \vec{F}=\frac{K q Q}{r^2} \hat{r} \quad|\vec{F}|=\frac{K q Q}{r^2}\)

Since we know that every action has equal and opposite reactions so

⇒ \(\overrightarrow{\mathrm{F}}_{\text {sphere }}=-\frac{\mathrm{KqQ}}{\mathrm{r}^2} \hat{\mathrm{r}}\)

⇒ \(\left|\overrightarrow{\mathrm{F}}_{\text {sphere }}\right|=\frac{\mathrm{KqQ}}{\mathrm{r}^2}\)

Example 43. The figure shows a uniformly charged thin sphere of total charge Q and radius R. A point charge q is also situated at the center of the sphere. Find out the following :

1. Force on charge q
2. Electric field intensity at A.
3. Electric field intensity at B.

Solution :

The electric field at the center of the uniformly charged hollow sphere = 0 So force on charge q = 0

The electric field at A

⇒ \(\overrightarrow{\mathrm{E}}_{\mathrm{A}}=\overrightarrow{\mathrm{E}}_{\text {Sphere }}+\overrightarrow{\mathrm{E}}_{\mathrm{q}}=0+\frac{\mathrm{Kq}}{\mathrm{r}^2} ; \mathrm{r}=\mathrm{CA}\)

E due to sphere = 0, because the point lies inside the charged hollow sphere.

Electric field \(\)

⇒ \(=\frac{K Q}{r^2} \hat{r}+\frac{K q}{r^2} \hat{r}=\frac{K(Q+q)}{r^2} \hat{r} ; r=C B\)

Note :

Here we can also assume that the total charge of the sphere is concentrated at the center, for the calculation of the electric field at B.

Example 44. Two concentric uniformly charged spherical shells of radius R₁ and R₂(R₂> R₁) have total charges Q₁ and respectively. Derive an expression of the electric field as a function of r for the following positions.

r < R₁

R₁≤ r < R₂

r ≥ R₂

Solution :

For r < R₁,

Therefore point lies inside both spheres

E_net = E_Inner + E_Outer = 0 + 0

For R₁≤ r < R₂

Therefore point lies outside the inner sphere but inside the outer sphere:

E_net= E_Inner + E_Outer

⇒ \(=\frac{K Q_1}{r^2}+0 \quad=\frac{K Q_1}{r^2} \hat{r}\)

For r ≥ R₂

The point lies outside the inner as well as outer sphere, therefore.

E_net = Einner+ Eouter

⇒ \(\frac{K Q_1}{r^2} \hat{r}+\frac{K Q_2}{r^2} \hat{r}=\frac{K\left(Q_1+Q_2\right)}{r^2} \hat{r}\)

Example 45. A spherical shell having charge +Q (uniformly distributed) and a point charge + q₀ is placed as shown. Find the force between the shell and the point charge(r>>R).

Force on the point charge + q₀ due to the shell

⇒ \(=q_0 \vec{E}_{\text {shell }}=\left(q_0\right)\left(\frac{K Q}{r^2}\right) \hat{r}=\frac{K Q q_0}{r^2} \hat{r}\) where rˆis unit vector along OP.

From the action-reaction principle, force on the shell due to the point charge will also be Fshell = \(F_{\text {shell }}=\frac{K Q q_0}{r^2}(-\hat{r})\)

Conclusion – To find the force on a hollow sphere due to outside charges, we can replace the sphere with a point charge kept at the center.

Example 46. Find the force acting between two shells of radius R₁ and R₂ which have uniformly distributed charges Q₁ and Q₂ respectively and the distance between their centre is r.

Solution :

The shells can be replaced by point charges kept at the center so force between them \(F=\frac{K Q_1 Q_2}{r^2}\)

6.8 Electric Field Due To Uniformly Charged Solid Sphere

Derive an expression for electric field due to solid sphere of radius R and total charge Q which is uniformly distributed in the volume, at a point which is at a distance r from center for the given two cases.

r ≥ R

r ≤ R

Assume an elementary concentric shell of charge dq. Due to this shell, the electric field at the point (r > R) will be

⇒ \(\mathrm{dE}=\frac{\mathrm{Kdq}}{\mathrm{r}^2}\)[from above result of hollow sphere]

E_net = \(E_{\text {net }}=\int d E=\frac{K Q}{r^2}\)

For r < R, there will be no electric field due to the shell of radius greater than r, so an electric field at the point will be present only due to shells having a radius less than r.

E´_net = \(E_{\text {net }}^{\prime}=\frac{K Q^{\prime}}{r^2}\)

here Q’ = \(\frac{Q}{\frac{4}{3} \pi R^3} \times \frac{4}{3} \pi r^3=\frac{Q r^3}{R^3}\)

⇒ \(E_{\text {net }}^{\prime}=\frac{K Q^{\prime}}{r^2}=\frac{K Q r}{R^3}\)

Note: The electric field inside and outside the sphere is always in a radial direction.

7. Electric Potential

In an electrostatic field, the electric potential (due to some source charges) at point P is defined as the work done by an external agent in taking a point unit positive charge from a reference point (generally taken at infinity) to that point P without changing its kinetic energy.

7.1 Mathematical Representation :

If (W ∞ → P)ext is the work required in moving a point charge q from infinity to a point P, the electric potential of the point P is

Note :

(W∝ → P)ext can also be called the work done by an external agent against the electric force on a unit positive charge due to the source charge.

Write both W and q with the proper sign.

7.2 Properties :

Potential is a scalar quantity, its value may be positive, negative, or zero.

S.Ι. Unit of potential is volt = \(\frac{\text { joule }}{\text { coulmb }}\) and its dimensional formula is [M¹L2T–3Ι^-1].

The electric potential at a point is also equal to the negative of the work done by the electric field in taking the point charge from the reference point (i.e. infinity) to that point.

Electric potential due to a positive charge is always positive and due to a negative charge, it is always negative except at infinite. (taking V∞= 0).

Potential decreases in the direction of the electric field.

V = V₁+ V₂+ V₃+ …….

7.3 Use of potential :

If we know the potential at some point (in terms of numerical value or terms of formula) then we can find out the work done by electric force when charge moves from point ‘P’ to ∞ by the formula Wel )p ∞ = qVp

Solved Examples

Example 47. A charge 2μC is taken from infinity to a point in an electric field, without changing its velocity. If work done against electrostatic forces is –40μJ then find the potential at that point.
Solution :

⇒ \(V=\frac{W_{\text {ext }}}{q}=\frac{-40 \mu \mathrm{J}}{2 \mu \mathrm{C}}=-20 \mathrm{~V}\)

Example 48. When charge 10 μC is shifted from infinity to a point in an electric field, it is found that work done by electrostatic forces is –10 μJ. If the charge is doubled and taken again from infinity to the same point without accelerating it, then find the amount of work done by an electric field and against an electric field.
Solution :

W_est)∞ p = –well)∞ p= well)p ∞= 10 μJ

because ΔKE = 0

Vp= \(\frac{\left.W_{\text {ext }}\right)_{\infty x p}}{q}=\frac{10 \mu \mathrm{J}}{10 \mu \mathrm{C}}=1 \mathrm{~V}\) = 1V 10 C

So if now the charge is doubled and taken from infinity then

1 = \(\left.\left.\frac{\left.w_{e x t}\right)_{\infty p}}{20 \mu \mathrm{C}} \quad \Rightarrow \quad \mathrm{W}_{\text {exx }}\right)_{\infty \mathrm{P}}=20 \mu \mathrm{J} \quad \Rightarrow \quad \mathrm{W}_{\mathrm{el}}\right)_{\infty \mathrm{P}}=-20 \mu \mathrm{J}\)

Example 49. A charge 3μC is released at rest from a point P where the electric potential is 20 V then its kinetic energy when it reaches infinite is :
Solution :

Wel = ΔK = Kf– 0

Wel)P → ∞= qVP= 60 μJ

so, Kf= 60 μJ

Electric Potential due to various charge distributions are given in the table.

7.4 Potential Due To A Point Charge :

The electrostatic potential at a point in an electric field due to the point charge may be defined as the amount of work done per unit positive test charge in moving it from infinity to that point (without acceleration) against the electrostatic force due to the electric field of a point charge. It is a scalar quantity.

Consider a point charge +q placed at point O. Suppose that V_Ais electric potential at point A, whose distance from the source charge +q is rA

If W∞A is work done in moving a vanishingly small positive test charge from infinity to point A, then

⇒ \(V_A=\frac{W_{\infty A}}{q_0}\)

Derivation :

Consider a positive point charge Q at the origin. We wish to determine the potential at any point A with position vector r from the origin.

Work done in bringing a unit positive test charge from infinity to point A. For Q > 0. The work done against the repulsive force on the test charge is positive.

Since work done is independent of the path, we choose a convenient path – along the radial direction from infinity to point A.

At some intermediate point A’ on the path, the electrostatic force on a unit positive charge is \(\) where
rˆis the unit vector along OP’. Work done by electric field on test charge for small \(\overrightarrow{\mathrm{F}}_{\mathrm{E}}=\frac{\mathrm{Q}}{4 \pi \varepsilon_0 \mathrm{r}^2} \hat{\mathrm{r}}\)displacement as shown in figure.

⇒ \(\overrightarrow{\mathrm{F}}_{\mathrm{ext}}=-\overrightarrow{\mathrm{F}}_{\mathrm{E}}\)

⇒ \(\mathrm{dw}=\vec{F}_{\text {ext }} \cdot d \overrightarrow{\mathrm{r}} \quad \Rightarrow \quad\left(-\vec{F}_E\right) \cdot(\mathrm{d} \overrightarrow{\mathrm{r}})\)

⇒ \(d w=F_E(-d r)\) =( Here r is decreasing so we will take dr negative)

⇒ \(w_{\infty A}=-\int_{\infty}^{r_A} d w=-\int_{\infty}^{r_A} \frac{Q}{4 \pi \varepsilon_0 r^2} d r\)

⇒ \(V_A-V_{\infty}=\frac{Q}{4 \pi \varepsilon_0 r_A}\left(V_{\infty}=0\right)\) (reference point is taken at infinity)

⇒ \(V_{\mathrm{A}}=\frac{\mathrm{Q}}{4 \pi \varepsilon_0 \mathrm{r}_{\mathrm{A}}}\)

In this case, the distance of point A from the charge +Q is denoted by

⇒ \(V=\frac{Q}{4 \pi \varepsilon_0 r}\)

Electric Potential Due To A System Of Charges

Let us now find the electrostatic potential at point P due to a group of point charges q1, q2, q3… flying at distances r1, r2, r3…. from point P (fig.). The electrostatic potential at point P due to these charges is found by calculating the electrostatic potential P due to each charge, considering the other charges to be absent, and then adding up these electrostatic potentials algebraically.

The electrostatic potential at point P due to charge q1, when other charges are considered absent,

⇒ \(V_1=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1}{r_1}\)

Similarly, electrostatic potentials at point P due to the individual charges q2, q3,…..qn(when other charges are absent) are given by

⇒ \(V_2=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_2}{r_2} ; V_3=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_3}{r_3} ; \ldots \ldots . . ; V_n=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_n}{r_n}\)

Hence, the electrostatic potential at point P due to the group of n point charges, V = V1+ V2+ V3+ ….. + Vn

⇒ \(=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1}{r_1}+\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_2}{r_2}+\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_3}{r_3}+\ldots . .+\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_n}{r_n}\)

⇒ \(=\frac{1}{4 \pi \varepsilon_0}\left(\frac{q_1}{r_1}+\frac{q_2}{r_2}+\frac{q_3}{r_3}+\ldots . .+\frac{q_n}{r_n}\right) \Rightarrow \quad V=\frac{1}{4 \pi \varepsilon_0} \sum_{i=1}^n \frac{q_i}{r_i}\)

Solved Examples

Example 50. Four-point charges are placed at the corners of a square of side l to calculate the potential at the center of the square.

Solution: V = 0 at ‘C’.

Example 51. Two point charges 2μC and – 4μC are situated at points (–2m, 0m) and (2 m, 0 m) respectively. Find out potential at point C. (4 m, 0 m) and. D (0 v5m)

Solution: Potential at point C

VC= \(\mathrm{V}_{\mathrm{q}_1}+\mathrm{V}_{\mathrm{q}_2}=\frac{\mathrm{K}(2 \mu \mathrm{C})}{6}+\frac{\mathrm{K}(-4 \mu \mathrm{C})}{2}=\frac{9 \times 10^9 \times 2 \times 10^{-6}}{6}-\frac{9 \times 10^9 \times 4 \times 10^{-6}}{2}=-15000 \mathrm{~V}\)

Similarly, VD= \(V_D=V_{Q_1}+V_{Q_2}=\frac{K(2 \mu C)}{\sqrt{(\sqrt{5})^2+2^2}}+\frac{K(-4 \mu C)}{\sqrt{(\sqrt{5})^2+2^2}}=\frac{K(2 \mu C)}{3}+\frac{K(-4 \mu C)}{3}=-6000 \mathrm{~V} .\)

7.5 Potential due to a ring :

Potential at the center of uniformly charged ring: Potential due to the small element dq

⇒ \(\mathrm{dV}=\frac{\mathrm{Kdq}}{\mathrm{R}}\)

Net Potential \(V=\int \frac{K d q}{R} \quad \Rightarrow \quad V=\frac{K}{R} \int d q=\frac{K q}{R}\)

For non-uniformly charged ring potential at the center is

⇒ \(\mathrm{V}=\frac{\mathrm{Kq} \text { total }}{\mathrm{R}}\)

Potential due to half ring at the center is: Kq V

⇒ \(V=\frac{K q}{R}\)

Potential at the axis of a ring:

V = \(\frac{K Q}{\sqrt{R^2+x^2}}\)

Solved Examples

Example 52. The figure shows two rings having charges Q and – 5Q. Find the Potential difference between A and B (V_A– V_B).

Solution : \(V_A=\frac{K Q}{R}+\frac{K(-\sqrt{5} Q)}{\sqrt{(2 R)^2+(R)^2}} \quad V_B=\frac{K(-\sqrt{5} Q)}{2 R}+\frac{K(Q)}{\sqrt{(R)^2+(R)^2}}\)

From above we can easily find V_A– V_B

Example 53. A point charge q0 is placed at the center of a uniformly charged ring of total charge Q and radius R. If the point charge is slightly displaced with negligible force along the axis of the ring then find out its speed when it reaches a large distance.
Solution :

Only electric force is acting on q0

∴ Wel = ΔK = \(=\frac{1}{2} m v^2\) – 0 ⇒ Now Wel)c→∞= q0 Vc= q0.\(\frac{\mathrm{KQ}}{\mathrm{R}}\)

∴ \(\frac{K q_0 \mathrm{Q}}{\mathrm{R}}=\frac{1}{2} m v^2\)mv2 ⇒ v = \(\sqrt{\frac{2 \mathrm{Kq}_0 \mathrm{Q}}{\mathrm{mR}}}\)

7.6 Potential Due To Uniformly Charged Disc :

∴ \(\mathrm{V}=\frac{\sigma}{2 \varepsilon_0}\left(\sqrt{\mathrm{R}^2+\mathrm{x}^2}-\mathrm{x}\right)\), where σ is the charged density and x is the distance of the point on the axis from the center of the disc, R is the radius of disc.

7.7 Potential Due To Uniformly Charged Spherical Shell :

Derivation of expression for potential due to the uniformly charged hollow sphere of radius R and total charge Q, at a point which is at a distance r from the center for the following situation

r > R

r < R

As the formula of E is easy , we use V = \(V=-\int_{r \rightarrow \infty}^{r=r} \vec{E} \cdot d \vec{r}\)

At outside point (r > R):

⇒ \(V_{\text {out }}=-\int_{r \rightarrow \infty}^{r=r}\left(\frac{K Q}{r^2}\right) d r \Rightarrow V_{\text {out }}=\frac{K Q}{r}=\frac{K Q}{\text { (Distance from centre) }}\)

For the outside point, the hollow sphere acts like a point charge.

Potential at the center of the sphere (r=0) :

As all the charges Are at a distance R from the center,

So V_Center = \(\frac{\mathrm{KQ}}{\mathrm{R}}=\frac{\mathrm{KQ}}{\text { (Radius of the sphere) }}\)

Potential at inside point ( r<R ) :

Suppose we want to find potential at point P, inside the sphere. +Q,

The potential difference between Point P and O :

⇒ \(-\int_0^P \vec{E}_{\text {in }} \cdot d \vec{r} \text { Where } E_{\text {in }}=0\)

Where Ein = 0

So V_p – V_o = 0

⇒ V_p = V_o = \(\frac{\mathrm{KQ}}{\mathrm{R}}\)

⇒ V_IN = \(\frac{\mathrm{KQ}}{\mathrm{R}}=\frac{\mathrm{KQ}}{\text { (Radius of the sphere) }}\)

7.8 Potential Due To Uniformly Charged Solid Sphere :

For R ≥ R (Outside)

⇒ \(V=\frac{K Q}{r}\)

For r ≤ R (inside)

⇒ \(V=\frac{K Q}{2 R^3}\left(3 R^2-r^2\right) \quad \Rightarrow \text { Here } \rho=\frac{Q}{\frac{4}{3} \pi R^3}\)

Example 54. Two concentric spherical shells of radius R₁ and R₂(R₂> R₁) have uniformly distributed charges Q₁ and Q₂ respectively. Find out potential

1. At point A
2. At the surface of the smaller shell (i.e. at point B)
3. At the surface of the larger shell (i.e. at point C)
4. At r ≤ R₁
5. At R₁≤ r ≤ R₂
6. At r ≥ R₂image

Solution :

Using the results of the hollow sphere as given in table 7.4.

⇒ \(V_A=\frac{K Q_1}{R_1}+\frac{K Q_2}{R_2}\)

⇒ \(V_B=\frac{K Q_1}{R_1}+\frac{K Q_2}{R_2}\)

⇒ \(V_c=\frac{K Q_1}{R_2}+\frac{K Q_2}{R_2}\)

For \(r \leq R_1 \quad V=\frac{K Q_1}{R_1}+\frac{K {R_2}\)

For \(R_1 \leq r \leq R_2\)

⇒ \(V=\frac{K Q_1}{r}+\frac{K Q_2}{R_2}\)

For \(r \geq R_2\)

⇒ \(V=\frac{K Q_1}{r}+\frac{K Q_2}{r}\)

Example 55. Two hollow concentric nonconducting spheres of radius a and b (a > b) contain charges Qa and Qbrespectively. Prove that the potential difference between two spheres is independent of the charge on the outer sphere. If the outer sphere is given an extra charge, is there any change in potential difference?
Solution :

Vinner sphere = \(\frac{K Q_b}{b}+\frac{K Q_a}{a}\)

Vouter sphere = \(\frac{K Q_b}{a}+\frac{K Q_a}{a}\)

Vinner sphere – Vouter sphere = \(\frac{K Q_b}{b}-\frac{K Q_b}{a}\)

⇒ \(\Delta V=K Q_b\left[\frac{1}{b}-\frac{1}{a}\right]\)

Which is independent of the charge on the outer sphere.

If the outer sphere is given any extra charge then there will be no change in potential difference.

8. Potential Difference

The potential difference between two points A and B is work done by an external agent against the electric field in taking a unit positive charge from A to B without acceleration (or keeping Kinetic Energy constant or Ki= Kf))

Mathematical Representation :

If (WA → B)ext = work done by an external agent against the electric field in taking the unit charge from A to B

V_B– V_A= \(\left.V_B-V_A=\frac{\left(W_{A \rightarrow B}\right)_{\text {ext }}}{q}\right)_{\Delta K=0}=\frac{-\left(W_{A \rightarrow B}\right)_{\text {electric }}}{q}=\frac{U_B-U_A}{q}=\frac{-\int_A^B \vec{F}_e \cdot \overrightarrow{d r}}{q}=-\int_A^B \vec{E} \cdot d r\)

Note: Take W and q both with a sign

Properties :

The difference in potential between two points is called the potential difference. It is also called voltage.

The potential difference is a scalar quantity. Its S.I. unit is also volt.

If V_A and V_B are the potential of two points A and B, then work done by an external agent in taking the charge q from A to B is (W_est)AB= q (V_B– V_A) or (W_el) AB = q (V_A– V_B).

The potential difference between the two points is independent of the reference point.

8.1 Potential difference in a uniform electric field :

V_B– V_A= – E_AB ⋅

⇒ V_B– V_A= – |E| |AB| cos θ

= – |E| d

= – Ed

d = effective distance between A and B along the electric field.

or we can also say that E = \(\frac{\Delta \mathrm{V}}{\Delta \mathrm{d}}\)

Special Cases :

Case 1. Line AB is parallel to the electric field.

∴ V_A– V_B= Ed

Case 2. Line AB is perpendicular to the electric field.

∴ V_A– V_B= 0 ⇒ V_A= V_D

Note: In the direction of the electric field potential always decreases.

Example 56. 1μC charge is shifted from A to B and it is found that work done by an external force is 40μJ in doing so against electrostatic forces then, find the potential difference V_A– V_B
Solution :

(WAB)ext = q(V_B– V_A) ⇒ 40 μJ = 1μC (V_B– V_B) ⇒ V_A– V_B= – 40

Example 57. A uniform electric field is present in the positive x-direction. If the intensity of the field is 5N/C then find the potential difference (V_B–V_A) between two points A (0m, 2 m) and B (5 m,3 m)
Solution :

V_B– V_A= – E.AB = –25V. Δ

The electric field intensity in uniform electric field, E = \(\frac{\Delta \mathrm{V}}{\Delta \mathrm{d}}\)

Where ΔV = potential difference between two points.

Δd = effective distance between the two points.

(projection of the displacement along the direction of the electric field.)

Example 58. Find out following

1. V_A– V_B
2. V_B– V_C
3. V_C – V_A
4. V_D – V_C
5. V_A– V_D

Arrange the order of potential for points A, B, C, and D.

Solution :

⇒ \(\left|\Delta V_{A B}\right|=E d=20 \times 2 \times 10^{-2}=0.4\)

So, V_A– V_B= 0.4 V because In the direction of the electric field potential always decreases.

⇒ \(\left|\Delta V_{B C}\right|=E d=20 \times 2 \times 10^{-2}=0.4 \quad \text { so, } \quad V_B-V_C=0.4 \mathrm{~V}\)

⇒ \(\left|\Delta \mathrm{V}_{\mathrm{CA}}\right|=\mathrm{Ed}=20 \times 4 \times 10^{-2}=0.8 \quad \text { so, } \quad \mathrm{V}_{\mathrm{C}}-\mathrm{V}_{\mathrm{A}}=-0.8 \mathrm{~V}\)

because In the direction of the electric field potential always decreases.

⇒ \(\left|\Delta V_{D C}\right|=E d=20 \times 0=0so, \quad V_D-V_C=0\)

because the effective distance between D and C is zero.

⇒ \(\left|\Delta V_{A D}\right|=E d=20 \times 4 \times 10^{-2}=0.8 \quad \text { so, } \quad V_A-V_D=0.8 \mathrm{~V}\)

because In the direction of the electric field potential always decreases.

The order of potential

V_A> V_B> V_C= V_D.

8.2 Potential difference due to infinitely long wire :

Derivation of expression for the potential difference between two points, which have perpendicular distance rAand from infinitely long line charge of uniform linear charge density λ.

From the definition of potential difference

⇒ \(V_{A B}=V_B-V_A=-\int_{r_A}^{r_B} \vec{E} \cdot \overrightarrow{d r}=-\int_{r_A}^{r_E} \frac{2 K \lambda}{r} \hat{r} \cdot \overrightarrow{d r}\)

⇒ \(V_{\mathrm{AB}}=-2 \mathrm{~K} \lambda \ell \mathrm{n}\left(\frac{\mathrm{r}_{\mathrm{B}}}{\mathrm{r}_{\mathrm{A}}}\right)\)

8.3 Potential Difference Due To Infinitely Long Thin Sheet:

Derivation of expression for the potential difference between two points, having separation d in the direction perpendicularly to a very large uniformly charged thin sheet of uniform surface charge density σ.

Let the points A and B have perpendicular distances A and B respectively then from the definition of potential difference.

⇒ \(V_{A B}=V_B-V_A=-\int_{r_A}^{T_B} \vec{E} \cdot \overrightarrow{d r}=-\int_{r_A}^{r_B} \frac{\sigma}{2 \varepsilon_0} \hat{r} \cdot \overrightarrow{d r} \Rightarrow V_{A B}=-\frac{\sigma}{2 \varepsilon_0}\left(r_B-r_A\right) \quad=-\frac{\sigma d}{2 \varepsilon_0}\)

9. Equipotential Surface :

9.1 Equipotential Surface Definition: If the potential of a surface (imaginary or physically existing) is the same throughout then such surface is known as an equipotential surface.

9.2 Properties of Equipotential Surfaces :

The following properties are associated with the equipotential surfaces :

No work was done in moving a test charge over an equipotential surface. Let A and B be two points on an equipotential surface (Fig. ). If a positive test charge q0is moved from point A to B, then work done in moving the test charge is related to the electrostatic potential difference between the two points as

⇒ \(V_B-V_A=\frac{W_{A B}}{q_0}\)

Since the two points A and B are on the same equipotential surface, V_B– V_A= 0

∴ \(\frac{W_{A B}}{q_0}=0\)

Hence, no work is done in moving a test charge between two points on an equipotential surface.

The electric field is always at right angles to the equipotential surface Since the work done in moving a test charge between two points on an equipotential surface is zero, the displacement of the test charge and the force applied to it must be perpendicular to each other. Since displacement is along the equipotential surface, and force on test charge is \(\mathrm{q}_0 \overrightarrow{\mathrm{E}}\), then the electric field \((\vec{E})\) must be at right angles to the equipotential surface.

The equipotential surfaces help to distinguish regions of strong fields from those of weak fields. We know that

E = \(-\frac{\mathrm{dV}}{\mathrm{dr}}\) or dr = \(-\frac{\mathrm{dV}}{\mathrm{E}}\)

For the same change in the value of dV i.e. dV = constant, we have

⇒ \(\mathrm{dr} \propto \frac{1}{\mathrm{E}}\)

i.e. the spacing between the equipotential surfaces will be denser in the regions, where the electric field is stronger and vice-versa. Therefore, the equipotential surfaces are closer together, where the electric field is stronger, and farther apart, where the field is weaker.

The equipotential surfaces tell the direction of the electric field.

Again E = \(-\frac{\mathrm{dV}}{\mathrm{dr}}\)

The negative sign tells that the electric field is directed in the direction of electric potential with distance. Therefore, the direction of the electric field is from the equipotential surfaces that are close to each other to those that are more and farther away from each other, provided such surfaces have been drawn for the same change in the value of DV.

No two equipotential surfaces can intersect each other.

In case, two equipotential surfaces intersect each other, then at their point of intersection, there will be two values of electric potential. As it is not possible, the two equipotential surfaces can not intersect each other.

9.3 Examples of equipotential surfaces :

For a uniform electric field: In a uniform electric field, the strength and direction of the field are the same at every point inside it.

In a uniform electric field, equipotential surfaces differing by the same amount of potential difference will be equidistant from each other.

For an isolated point charge: The electric field due to an isolated point charge is radial and varies inversely as the square of the distance from the charge. The potential at all the points equidistant from the charge is the same. All such points lie on the surface of a spherical shell, such that the charge lies at its center. Therefore, for a point charge, equipotential surfaces will be a series of concentric spherical shells.

For a system of two-point charges: the dotted lines represent electric lines of force. The thick circles around the charges represent equipotential surfaces due to individual charges.

Line charge: Equipotential surfaces have curved surfaces as that of coaxial cylinders of different radii.

Solved Examples

Example 59. Some equipotential surfaces are shown in the figure. What can you say about the magnitude and the direction of the electric field?

Solution: Here we can say that the electric will be perpendicular to equipotential surfaces.

Also \(|\vec{E}|=\frac{\Delta V}{\Delta d}\)

where ΔV = potential difference between two equipotential surfaces. Δd = perpendicular distance between two equipotential surfaces.

So \(|\vec{E}|=\frac{10}{\left(10 \sin 30^{\circ}\right) \times 10^{-2}}=200 \mathrm{~V} / \mathrm{m}\)

Now there are two perpendicular directions either direction 1 or direction 2 as shown in the figure, but since we know that in the direction of the electric field electric potential decreases so the correct direction is direction 2.

Hence E = 200 V/m, making an angle 120° with the x-axis

Example 60. The figure shows some equipotential surfaces produced by some charges. At which point the value of the electric field is greatest?

Solution :

E is larger where equipotential surfaces are closer. ELOF are ⊥ to equipotential surfaces. In the figure we can see that for point B they are closer so E at point B is the maximum

11. Electrostatic Potential Energy

11.1 Electrostatic potential energy of a point charge due to many charges: The electrostatic potential energy of a point charge at a point in an electric field is the work done in taking the charge from a reference point (generally at infinity) to that point without acceleration (or keeping KE const. or Ki= Kf)).

Its Mathematical formula is

U = W∝P)_ext acc = 0 = qV = – WP∝)el

Here q is the charge whose potential energy is being calculated and V is the potential at its position due to the source charges.

Note: Always put q and V with a sign.

11.2 Properties :

Electric potential energy is a scalar quantity but may be positive, negative, or zero.

Its unit is the same as the unit of work or energy that is the joule (in the S.Ι. system).

Some times energy is also given in electron-volts 1eV = 1.6 × 10^-19 J

Electric potential energy depends on the reference point. (Generally, Potential Energy at r= ∞ is taken zero)

Solved Examples

Example 61 The four identical charges q each are placed at the corners of a square of side a. Find the potential energy of one of the charges due to the remaining charges.

Solution :

The electric potential of point A due to the charges placed at B, C, and D is

⇒ \(\mathrm{V}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{\mathrm{a}}+\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{\sqrt{2 \mathrm{a}}}+\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{\mathrm{a}}=\frac{1}{4 \pi \varepsilon_0}\left(2+\frac{1}{\sqrt{2}}\right) \frac{\mathrm{q}}{\mathrm{a}}\)

∴ Potential energy of the charge at A is = qV =\(\frac{1}{4 \pi \varepsilon_0}\left(2+\frac{1}{\sqrt{2}}\right) \frac{q^2}{\mathrm{a}}\)

Example 62. A particle of mass 40 mg and carrying a charge 5 × 10^-9 C is moving directly towards a fixed positive point charge of magnitude 10^-8 C. When it is at a distance of 10 cm from the fixed point charge it has a speed of 50 cm/s. At what distance from the fixed point charge will the particle come momentarily to rest? Is the acceleration constant during the motion?
Solution :

If the particle comes to rest momentarily at a distance r from the fixed charge, then from the conservation of energy we have

⇒ \(\frac{1}{2} m u^2+\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{\mathrm{a}}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{\mathrm{r}}\)

Substituting the given data, we get

⇒ \(\frac{1}{2} \times 40 \times 10^{-6} \times \frac{1}{2} \times \frac{1}{2}=9 \times 10^9 \times 5 \times 10^{-8} \times 10^{-9}\left[\frac{1}{r}-10\right]\)

or, \(\frac{1}{r}-10=\frac{5 \times 10^{-6}}{9 \times 5 \times 10^{-8}}=\frac{100}{9} \quad \Rightarrow \frac{1}{r}=\frac{190}{9} \quad \Rightarrow r=\frac{9}{190} \mathrm{~m}\)

or, i.e., r = 4.7 × 10^-2 m

As here, F = \(\mathrm{F}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{\mathrm{r}^2} \quad \text { so } \quad \text { acc. }=\frac{\mathrm{F}}{\mathrm{m}} \propto \frac{1}{\mathrm{r}^2}\)

i.e., acceleration is not constant during the motion.

Example 63. A proton moves from a large distance with a speed u m/s directly towards a free proton originally at rest. Find the distance of the closest approach for the two protons in terms of the mass of proton m and its charge e.
Solution :

As here the particle at rest is free to move, when one particle approaches the other, due to electrostatic repulsion other will also start moving so the velocity of the first particle will decrease while of other will increase and at the closest approach both will move with same velocity.

So if v is the common velocity of each particle at closest approach, then by ‘conservation of momentum’ of the two protons system.

mu = mv + mv i.e., v = \(\frac{1}{2} u\)

And by conservation of energy

⇒ \(\frac{1}{2} m u^2=\frac{1}{2} m v^2+\frac{1}{2} m v^2+\frac{1}{4 \pi \varepsilon_0} \frac{e^2}{r}\)

r = \(\frac{\mathrm{e}^2}{\pi \mathrm{m} \varepsilon_0 \mathrm{u}^2}\)

12. Electrostatic Potential Energy Of A System Of Charges

(This concept is useful when more than one charges move.) It is the work done by an external agent against the internal electric field required to make a system of charges in a particular configuration from infinite separation without accelerating it.

12.1 Types of system of charge

• Point charge system
• Continuous charge system.

12.2 Derivation For A System Of Point Charges:

Keep all the charges at infinity. Now bring the charges one by one to their corresponding position and find the work required. PE of the system is an algebric sum of all the works.

Let W1= work done in bringing the first charge

W2= work done in bringing the second charge against force due to 1st charge.

W3 work done in bringing the third charge against force due to the 1st and 2nd charges. n(n 1)

PE = W₁+ W₂ + W₃ + …… . (This will contain\(\frac{n(n-1)}{2}={ }^n C_2\)terms)

Method of calculation (to be used in problems)

U = sum of the interaction energies of the charges.

= (U₁₂ + U₁₃ + …….. + U_1n) + (U₂₃ + U₂₄ + …….. + U_2n) + (U₂₄ + U₃₅ + …….. + U_3n)

The method of calculation is useful for symmetrical point charge systems. Find the PE of each charge due to the rest of the charges.

If U₁= PE of the first charge due to all other charges.

= (U₁₂ + U₁₃ + …….. + U_1n)

U₂= PE of the second charge due to all other charges.

= (U₂₁ + U₂₃ + …….. + U_2n) then U = PE of the system

⇒ \(=\frac{U_1+U_2+\ldots . . U_n}{2}\)

Solved Examples

Example 64 Find out the potential energy of the two-point charge system having q1 and q2 charges separated by distance r.
Solution :

Let both the charges be placed at a very large separation initially.

Let W₁= work done in bringing charge q₁ in absence of q₂= q(V_f– V_i) = 0

W₂= work done in bringing charge q₂ in presence of q₁= q(V_f– V_i) = q1(Kq2/r – 0)

PE = W₁+ W₂ = 0 + Kq₁q₂/ r = Kq₁q₂/ r

Example 65 Figure shows an arrangement of three-point charges. The total potential energy of this arrangement is zero. Calculate the ratio \(\frac{\mathrm{q}}{\mathrm{Q}} .\)

Solution : Usys= \(\frac{1}{4 \pi \varepsilon_0}\left[\frac{-q Q}{r}+\frac{(+q)(+q)}{2 r}+\frac{Q(-q)}{r}\right]=0\)

Example 66. Two point charges each of mass m and charge q are released when they are at a distance r from each other. What is the speed of each charged particle when they are at a distance of 2r?
Solution :

According to momentum conservation both the charge particles will move with the same speed now applying energy conservation.

⇒ \(0+0+\frac{K q^2}{r}=2 \frac{1}{2} m v^2+\frac{K q^2}{2 r} \Rightarrow \quad v=\sqrt{\frac{K q^2}{2 r m}}\)

Example 67. Two charged particles each having equal charges 2 × 10^-5 C are brought from infinity to within a separation of 10 cm. Calculate the increase in potential energy during the process and the work required for this purpose.
Solution :

ΔU = U_f– U_I = U_f– 0 = U_f

We have to simply calculate the electrostatic potential energy of the given system of charges

ΔU = Uf= \(\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r}=\frac{9 \times 10^9 \times 2 \times 10^{-5} \times 2 \times 10^{-5} \times 100}{10} \mathrm{~J}=36 \mathrm{~J}\)

work required = 36 J.

Example 68. Three equal charges q are placed at the corners of an equilateral triangle of side a.

1. Find out the potential energy of the charge system.
2. Calculate the work required to decrease the side of the triangle to a/2.
3. If the charges are released from the shown position and each of them has the same mass m then find the speed of each particle when they lie on the triangle of side 2a.

Solution :

Method I (Derivation)

Assume all the charges are at infinity initially.

work done in putting charge q at corner A

W₁= q (v_f– v_i) = q (0 – 0)

Since potential at A is zero in the absence of charges, work done in putting q at corner B in the presence of charge at A :

⇒ \(\mathrm{W}_2=\left(\frac{\mathrm{Kq}}{\mathrm{a}}-0\right)=\frac{\mathrm{Kq}^2}{\mathrm{a}}\)

Similarly, work done in putting charge q at corner C in the presence of charge at A and B.

⇒ \(\mathrm{W}_3=\mathrm{q}\left(\mathrm{v}_{\mathrm{f}}-v_{\mathrm{i}}\right) \quad=\mathrm{q}\left[\left(\frac{\mathrm{Kq}}{\mathrm{a}}+\frac{\mathrm{Kq}}{\mathrm{a}}\right)-0\right]\)

So-net potential energy PE = \(=W_1+W_2+W_3=0+\frac{K q^2}{a}+\frac{2 K q^2}{a}=\frac{3 K q^2}{a}\)

Method 2 (using direct formula)

U = U₁₂ + U₁₃ + U₂₃ = \(\frac{K^2}{a}+\frac{K q^2}{a}+\frac{K q^2}{a}=\frac{3 K q^2}{a}\)

Work required to decrease the sides

W = U_f– U_i = \(\frac{3 K q^2}{a / 2}-\frac{3 K q^2}{a}=\frac{3 K q^2}{a}\)

Work done by electrostatic forces = change is the kinetic energy of particles.

U_i– U_f= K_f– K_i ⇒ \(\frac{3 K q^2}{a}-\frac{3 K q^2}{2 a}=3\left(\frac{1}{2} m v^2\right)-0 \Rightarrow \quad v=\sqrt{\frac{K q^2}{a m}}\)

Example 69 Four identical point charges q are placed at four corners of a square of side a. Find out the potential energy of the charge system

Solution :

Method 1 (using direct formula) :

U = U₁₂ + U₁₃ + U₁₄ + U₂₃ + U₂₄ + U₃₄

⇒ \(=\frac{K^2}{a}+\frac{K^2}{a \sqrt{2}}+\frac{K^2}{a}+\frac{K^2}{a}+\frac{K^2}{a \sqrt{2}}+\frac{K^2}{a}=\left[\frac{4 K^2}{a}+\frac{2 K q^2}{a \sqrt{2}}\right]=\frac{2 K q^2}{a}\left[2+\frac{1}{\sqrt{2}}\right]\)

Method 2 [using U = 2(U₁+ U₂+ ……)] :

U₁= total P.E. of charge at corner 1 due to all other charges

U₂= total P.E. of charge at corner 2 due to all other charges

U₃= total P.E. of charge at corner 3 due to all other charges

U₄= total P.E. of charge at corner 4 due to all other charges

Due to symmetry U₁= U₂= U₃= U₄

U_net = \(\frac{\mathrm{U}_1+\mathrm{U}_2+\mathrm{U}_3+\mathrm{U}_4}{2}=2 \mathrm{U}_1=2\left[\frac{\mathrm{Kq}^2}{\mathrm{a}}+\frac{\mathrm{Kq}^2}{\mathrm{a}}+\frac{\mathrm{Kq}^2}{\sqrt{2} \mathrm{a}}\right]=\frac{2 \mathrm{Kq}^2}{\mathrm{a}}\left[2+\frac{1}{\sqrt{2}}\right]\)

Example 70. Six equal point charges q are placed at six corners of a hexagon of side a. Find out the potential energy of the charge system

Solution :

Unit = \(\frac{\mathrm{U}_1+\mathrm{U}_2+\mathrm{U}_3+\mathrm{U}_4+\mathrm{U}_5+\mathrm{U}_6}{2}\)

Due to symmetry U₁= U₂= U₃= U₄= U₅= U₆ so U_net = 3U₁= \(\frac{3 K q^2}{a}\left[2+\frac{2}{\sqrt{3}}+\frac{1}{2}\right]\)

12.3 Electric Potential Energy For Continues Charge System :

This energy is also known as self-energy.

P.E. (Self Energy) of a uniformly charged spherical shell:-

Uself = \(\frac{K Q^2}{2 R}\)

Self-energy of the uniformly charged solid sphere :

For a solid sphere P.E. is Uself = \(\frac{3}{5} \frac{K Q^2}{R}\)

Example 71 A spherical shell of radius R with uniform charge q is expanded to a radius 2R. Find the work performed by the electric forces and external agents against electric forces in this process.
Solution :

W_ext = U_f– U_i= \(\frac{q^2}{16 \pi \varepsilon_0 R}-\frac{q^2}{8 \pi \varepsilon_0 R}=-\frac{q^2}{16 \pi \varepsilon_0 R}\)

W_ext = U_f– U_i= \(\frac{q^2}{8 \pi \varepsilon_0 R}-\frac{q^2}{16 \pi \varepsilon_0 R}=\frac{q^2}{16 \pi \varepsilon_0 R}\)

Example 72 Two nonconducting hollow uniformly charged spheres of radii and R2 with charge and Q2 respectively are placed at a distance r. Find out the total energy of the system.

Solution : Utotal = Uself + UInteraction = \(\frac{\mathrm{Q}_1^2}{8 \pi \varepsilon_0 \mathrm{R}_1}+\frac{\mathrm{Q}_2^2}{8 \pi \varepsilon_0 \mathrm{R}_2}+\frac{\mathrm{Q}_1 \mathrm{Q}_2}{4 \pi \varepsilon_0 \mathrm{r}}\)

Example 73. Two concentric spherical shells of radius R1 and R2(R2> R1) have uniformly distributed charges Q1 and Q2respectively. Find out the total energy of the system.

Solution : Utotal = Uself 1 + Uself 2 + UInteraction = \(\frac{\mathrm{Q}_1^2}{8 \pi \varepsilon_0 \mathrm{R}_1}+\frac{\mathrm{Q}_2^2}{8 \pi \varepsilon_0 \mathrm{R}_2}+\frac{\mathrm{Q}_1 \mathrm{Q}_2}{4 \pi \varepsilon_0 \mathrm{R}_2}\)

12.4 Energy Density :

Energy Density Definition: Energy density is defined as energy stored in unit volume in any electric field. Its mathematical formula is given as following Energy density =\(\frac{1}{2}\) where E = electric field intensity at that point ε = ε₀εrelectric permittivity of medium

Example 74 Find out energy stored in an imaginary cubical volume of side a in front of an infinitely large nonconducting sheet of uniform charge density σ.
Solution :

Energy stored

⇒ \(\mathrm{U}=\int \frac{1}{2} \varepsilon_0 \mathrm{E}^2 \mathrm{dV}\)where dV is small volume = \(\frac{1}{2} \varepsilon_0 \mathrm{E}^2 \int \mathrm{dV}\)

⇒ \(\frac{1}{2} \varepsilon_0 \frac{\sigma^2}{4 \varepsilon_0^2} \cdot a^3 \cdot=\frac{\sigma^2 a^3}{8 \varepsilon_0}\)

13. Relation Between Electric Field Intensity And Electric Potential

13.1 For uniform electric field :

The potential difference between two points A and B

VB – VA = – E . AB

13.2 Nonuniform electric field

⇒ \(E_x=-\frac{\partial V}{\partial x}, E_y=-\frac{\partial V}{\partial y}, E_z=-\frac{\partial V}{\partial z} \quad \Rightarrow \quad \vec{E}=E_x \hat{i}+E_y \hat{j}+E_z \hat{k}\)

⇒ \(-\left[\hat{\mathrm{i}} \frac{\partial}{\partial \mathbf{x}} \mathrm{V}+\hat{\mathrm{j}} \frac{\partial}{\partial \mathbf{y}} \mathrm{V}+\hat{\mathrm{k}} \frac{\partial}{\partial \mathbf{z}} \mathrm{V}\right]=-\left[\hat{\mathrm{i}} \frac{\partial}{\partial \mathbf{x}}+\hat{\mathrm{j}} \frac{\partial}{\partial \mathbf{y}}+\hat{\mathrm{k}} \frac{\partial}{\partial \mathbf{z}}\right] \mathrm{V}=-\nabla \mathrm{V}=- \text { grad } V\)

Where\(\frac{\partial V}{\partial X}\)= derivative of V with respect to x (keeping y and z constant)

⇒ \(\frac{\partial V}{\partial y}\)= derivative of V with respect to y (keeping z and x constant)

⇒ \(\frac{\partial \mathrm{V}}{\partial \mathrm{z}}\)= derivative of V with respect to z (keeping x and y constant)

13.3 If Electric Potential And Electric Field Depends Only On One Coordinate, Say R :

⇒ \(\overrightarrow{\mathrm{E}}=-\frac{\partial \mathrm{V}}{\partial \mathrm{r}} \hat{\mathrm{r}}\)

where rˆis a unit vector along increasing r.

⇒ \(\int d V=-\int \vec{E} \cdot \overrightarrow{d r} \quad \Rightarrow \quad V_B-V_A=-\int_{r_A}^{r_B} \vec{E} \cdot \overrightarrow{d r}\)

⇒ \(\overrightarrow{\mathrm{dr}}\)is along the increasing direction of r.

The potential of a point V = \(V=-\int_{\infty}^r \vec{E} \cdot \overrightarrow{d r}\)

Example 75 A uniform electric field is along x – the x-axis. The potential difference VA– VB = 10 V between two points A (2m, 3m) and B (4m, 8m). Find the electric field intensity.

Solution : E = \(\frac{\Delta \mathrm{V}}{\Delta \mathrm{d}}=\frac{10}{2}\) = 5 V / m. It is along the + ve x-axis.

Example 76 V = x2 + y , Find \(\vec{E}\)
Solution :

⇒ \(\frac{\partial V}{\partial x}=2 x, \frac{\partial V}{\partial y}=1 \quad \text { and } \frac{\partial V}{\partial z}=0\)

⇒ \(\vec{E}=-\left(\hat{i} \frac{\partial V}{\partial x}+\hat{j} \frac{\partial V}{\partial y}+\hat{k} \frac{\partial V}{\partial z}\right)=-(2 x \hat{i}+\hat{j})\) Electric field is nonuniform.

Example 77 For given \(\vec{E}=2 x \hat{i}+3 y \hat{j}\) find the potential at (x, y) if V at origin is 5 volts.
Solution :

⇒ \(\int_5^V d V=-\int \vec{E} \cdot \overline{d r}=-\int_0^x E_x d x-\int_0^y E_y d y\)

⇒ \(V-5=-\frac{2 x^2}{2}-\frac{3 y^2}{2} \Rightarrow V=-\frac{2 x^2}{2}-\frac{3 y^2}{2}+5\)

14. Electric Dipole

14.1 Electric Dipole

If two point charges are equal in magnitude q and opposite in sign separated by a distance a such that the distance of field point r>>a, the system is called a dipole. The electric dipole moment is defined as a vector quantity having magnitude p = (q × a) and direction from negative charge to positive charge.

Note: [In chemistry, the direction of dipole moment is assumed to be from positive to negative charge.] The C.G.S unit of electric dipole moment is debye which is defined as the dipole moment of two equal and opposite point charges each having charge 10–10 frankline and separation of 1 Å, i.e.,

1 debye (D) = 10–10 × 10–8 = 10–18 Fr × cm

1D = 10–18 ×\(\frac{C}{3 \times 10^9}\) × 10–2 m = 3.3 × 10–30 C × m.

S.I. Unit is coulomb × metre = C . m

Solved Examples

Example 78 A system has two charges qA= 2.5 × 10–7 C and qB= – 2.5 × 10–7 C located at points A : (0, 0, – 0.15 m) and B ; (0, 0, + 0.15 m) respectively. What is the net charge and electric dipole moment of the system?
Solution :

Net charge = 2.5 × 10–7 – 2.5 × 10–7 = 0

Electric dipole moment,

P = (Magnitude of charge) × (Separation between charges)

= 2.5 × 10–7[0.15 + 0.15] C m = 7.5 × 10–8 C m

The direction of the dipole moment is from B to A.

14.2 Electric Field Intensity Due to Dipole :

At the axial point:-

⇒ \(\vec{E}=\frac{K q}{\left(r-\frac{a}{2}\right)}-\frac{K q}{\left(r+\frac{a}{2}\right)^2} \text { along the } \hat{P}=\frac{K q(2 r a)}{\left(r^2-\frac{a^2}{4}\right)^2} \hat{P}\)

If r >> a then

⇒ \(\vec{E}=\frac{K q 2 r a}{r^4} \hat{P}=\frac{2 K \vec{P}}{r^3},\)

As the direction of the electric field at the axial position is along the dipole moment \((\vec{P})\)

⇒ \(\vec{E}_{\text {axial }}=\frac{2 K \vec{P}}{r^3}\)

The electric field at the perpendicular Bisector (Equitorial Position)

E_net= \(2 \mathrm{E} \cos \theta \text { (along }-\hat{P} \text { ) }\)

⇒ \(\overrightarrow{\mathrm{E}}_{\text {net }}=2\left(\frac{\mathrm{Kq}}{\left(\sqrt{r^2+\left(\frac{a}{2}\right)^2}\right)^2}\right) \frac{\frac{a}{2}}{\sqrt{r^2+\left(\frac{a}{2}\right)^2}}(-\hat{P})=2 \frac{K q a}{\left(r^2+\left(\frac{a}{2}\right)^2\right)^{3 / 2}}(-\hat{P})\)

If r >> a then

⇒ \(\vec{E}_{\text {net }}=\frac{K P}{r^3}(-\hat{P})\)

As the direction of \(\vec{E}\) at equitorial position is opposite of \(\vec{P}\) so we can write in vector form:

⇒ \(\vec{E}_{\text {eqt }}=-\frac{K \vec{P}}{r^3}\)

Electric field at general point (r, θ) :

⇒ \(\mathrm{E}_{\text {net }}=\frac{\mathrm{KP}}{\mathrm{r}^3} \sqrt{1+3 \cos ^2 \theta} ; \quad \tan \phi=\frac{\tan \theta}{2}\)

Solved Examples

Example 79 The electric field due to a short dipole at a distance r, on the axial line, from its midpoint is the same as that of the electric field at a distance r’, on the equatorial line, from its mid-point. Determine the ratio \(\)
Solution :

⇒ \(\frac{1}{4 \pi \varepsilon_0} \frac{2 \mathrm{p}}{\mathrm{r}^3}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{p}}{\mathrm{r}^{\prime 3}} \text { or } \quad \frac{2}{\mathrm{r}^3}=\frac{1}{\mathrm{r}^3} \quad \text { or } \quad \frac{\mathrm{r}^3}{\mathrm{r}^3}=2 \quad \text { or } \quad \frac{\mathrm{r}}{\mathrm{r}^{\prime}}=2^{1 / 3}\)

Example 80 Two charges, each of 5 μC but opposite in sign, are placed 4 cm apart. Calculate the electric field intensity of a point that is at a distance of 4 cm from the midpoint on the axial line of the dipole.
Solution :

We can not use the formula of short dipole here because the distance of the point is comparable to the distance between the two point charges.

q = 5 × 10–6 C, a = 4 ×10–2 m, r = 4 × 10–2 m

image

E_ess= E++ E– = \(\frac{\mathrm{K}(5 \mu \mathrm{C})}{(2 \mathrm{~cm})^2}-\frac{\mathrm{K}(5 \mu \mathrm{C})}{(6 \mathrm{~cm})^2}=\frac{144}{144 \times 10^{-8}} N \mathrm{NC}^{-1}=10^8 \mathrm{~N} \mathrm{C}^{-1}\)

Example 81 Two charges ± 10 μC are placed 5 × 10–3 m apart. Determine the electric field at a point Q which is 0.15 m away from O, on the equatorial line.
Solution: In the given problem, r >> an

∴ E = \(\frac{1}{4 \pi \varepsilon_0} \frac{p}{r^3}=\frac{1}{4 \pi \varepsilon_0} \frac{q(a)}{r^3}\)

or E = 9 × 109 \(\frac{10 \times 10^{-6} \times 5 \times 10^{-3}}{0.15 \times 0.15 \times 0.15} \mathrm{NC}^{-1}\)

= 1.33 ×105 NC–1

14.3 Electric Potential due to a small dipole :

Potential at axial position :

V = \(\frac{K q}{\left(r-\frac{a}{2}\right)}+\frac{K(-q)}{\left(r+\frac{a}{2}\right)}\)

V = \(\frac{K q a}{\left(r^2-\left(\frac{a}{2}\right)^2\right)}\)

If r >> a than

V = \(\frac{\text { Kqa }}{r^2}\)

where qa = p

V_axial= \(\frac{K P}{r^2}\)

Potential at equatorial position :

V = \(\frac{K q}{\sqrt{r^2+\left(\frac{a}{2}\right)^2}}+\frac{K(-q)}{\sqrt{r^2+\left(\frac{a}{2}\right)^2}}=0\)

Veqt = 0

Potential at general point (r,θ) :

V = \(\frac{K(\vec{P} \cdot \vec{r})}{r^3}\)

Example 82

1. Find the potential at points A and B due to the small charge – system fixed near the origin. (the distance between the charges is negligible).
2. Find work done to bring a test charge from point A to point B, slowly. All parameters are in S.I. units.

Solution :

The dipole moment of the system is

⇒ \(\vec{P}=(q a) \hat{i}+(q a) \hat{j}\)

Potential at point A due to the dipole

VA = \(K \frac{(\vec{P} \cdot \vec{r})}{r^B}=\frac{K[(q a) \hat{i}+(q a) \hat{j}] \cdot(4 \hat{i}+3 \hat{j})}{5^3}=\frac{k(q a)}{125}(7)\)

⇒ V_B= \(\frac{K[(q a) \hat{i}+(q a) \hat{j}] \cdot(3 \hat{i}-4 \hat{j})}{(5)^3}=\frac{K(q a)}{125}\)

WA → B = UB– UA= q0(V_B– V_A) =\(\left[-\frac{\mathrm{K}(\mathrm{qa})}{125}-\left(\frac{\mathrm{K}(\mathrm{qa})(7)}{125}\right)\right] \Rightarrow \mathrm{W}_{\mathrm{A} \rightarrow \mathrm{B}}=\frac{\mathrm{Kqq}_0 \mathrm{a}}{125} \text { (8) }\)

14.4 Dipole in a uniform electric field

Dipole is placed along the electric field :

In this case, F_net = 0, τ_net = 0 so it is an equilibrium state. And it is a stable equilibrium position.

If the dipole is placed at θ angle from \(\vec{E}\):

In this case F_net = 0 but

Net torque τ = (qEsinθ) (a)

Here qa = P ⇒ τ = PE sinθ in vector form τ= \(\vec{P} \cdot \vec{E}\)

Example 83 A dipole is formed by two point charges –q and +q, each of mass m, and both the point charges are connected by a rod of length l and mass m1. This dipole is placed in a uniform electric field E. If the dipole is disturbed by a small angle θ from a stable equilibrium position, prove that its motion will be almost SHM. Also, find its period.
Solution :

If the dipole is disturbed by θ angle,

τ_net = –PE sinθ (here – ve sign indicates that the direction of the torque is opposite of θ). If θ is very small, sinθ = θ

τ_net = –(PE)θ

τ_net∝ (–θ) so motion will be almost SHM.

T = 2π \(\sqrt{\frac{\mathrm{I}}{\mathrm{K}}}\)

The potential energy of a dipole placed in a uniform electric field :

UB– UA= \(-\int_A^B \vec{F} \cdot \overrightarrow{d r} \quad \text { Here } \quad U_B-U_A=-\int_A^B \vec{\tau} \cdot \overrightarrow{d \theta}\)

In the case of dipole, at θ = 90°, P.E. is assumed to be zero.

Uθ– U90° =\(-\int_{\theta=90^{\circ}}^{\theta=\theta}(-P E \sin \theta)(d \theta)\)

(As the direction of torque is opposite of θ)

Uθ– 0 = – PE cos θ

θ = 90° is chosen as a reference so that the lower limit comes out to be zero.

Uθ= \(-\vec{P} \cdot \vec{E}\)

From the potential energy curve, we can conclude :

at θ = 0, there is a minimum of P.E. so it is a stable equilibrium position.

at θ = 180°, there is a maxima of P.E. so it is a position of unstable equilibrium.

Example 84 Two point masses of mass m and equal and opposite charge of magnitude q are attached on the corners of a non-conducting uniform rod of mass m and the system is released from rest in uniform electric field E as shown in figure from θ = 53°

1. Find the angular acceleration of the rod just after releasing
2. What will be the angular velocity of the rod when it passes through stable equilibrium?
3. Find the work required to rotate the system by 180°.

Solution :

t_net= PE sin53° = I α

⇒ \(\alpha=\frac{(\mathrm{q} \ell) \mathrm{E}\left(\frac{4}{5}\right)}{\frac{\mathrm{m} \ell^2}{12}+\mathrm{m}\left(\frac{\ell}{2}\right)^2+\mathrm{m}\left(\frac{\ell}{2}\right)^2}=\frac{48 \mathrm{qE}}{35 \mathrm{~m} \ell}\)

From energy conservation :

Ki+ Ui= Kf+ Uf

0 + (– PE cos 53°) = \(\frac{1}{2}\)Ιω2 + (–PE cos 0°)

where I = \(\mathrm{I}=\frac{\mathrm{m} \ell^2}{12}+\mathrm{m}\left(\frac{\ell}{2}\right)^2+\mathrm{m}\left(\frac{\ell}{2}\right)^2 \quad \Rightarrow \quad \omega=\sqrt{\frac{48 \mathrm{qE}}{35 \mathrm{~m} \ell}}\)

W_ext= Uf– Ui

W_ext = (–PE cos(180° + 53°)) – (–PEcos 53°)

W_ext = (ql)E\(\left(\frac{4}{5}\right)+(q \ell) E\left(\frac{4}{5}\right) \quad \Rightarrow \quad W_{\text {ext }}=\left(\frac{8}{5}\right) q \ell E\)

15. Electric Lines Of Force (ELOF)

The line of force in an electric field is an imaginary line, the tangent to which at any point on it represents the direction of the electric field at the given point.

15.1 Properties :

A line of force originates from a positive charge and terminates on a negative charge. If there is only one positive charge then lines start from a positive charge and terminate at ∞. If there is only one negative charge then lines start from ∞ and terminate at a negative charge.

Two lines of force never intersect each other because there cannot be two directions of \(\vec{E}\)

Electric lines of force produced by static charges do not form a closed loop.

If lines of force make a closed loop, then work done to move a +q charge along the loop will be non-zero. So it will not be a conservative field. So these types of lines of force are not possible in electrostatics.

The Number of lines per unit area (line density) represents the magnitude of the electric field.

If lines are dense, ⇒ E will be more

If Lines are rare, ⇒ E will be less, and if E = O, no line of force will be found here

The number of lines originating (terminating) is proportional to the charge.

Example 85 If several electric lines of force from charge q is 10 then find out several electric lines of force from 2q charge.
Solution :

No. of ELOF ∝ charge

10 ∝ q ⇒ 20 ∝ 2q

• So the number of ELOF will be 20.
• Electric lines of force end or start perpendicularly on the surface of a conductor.
• Electric lines of force never enter into conductors.

Example 86 Some electric lines of force are shown in the figure, for points A and B

1. E_A> E_B
2. EB> E_A
3. V_A> V_B
4. V_B> B_A

Solution: lines are denser at B so E_A> EBIn the direction of the Electric field, potential decreases so V_A> V_B

Example 87 If a charge is released in an electric field, will it follow lines of force?
Solution :

Case I :

If lines of force are parallel (in a uniform electric field):-

In this type of field, if a charge is released, the force on it will be QoE and its direction will be along \(\vec{E}\). So the charge will move in a straight line, along the lines of force.

Case 2: –

If lines of force are curved (in a non-uniform electric field):-

The charge will not follow lines of force

Example 88 A charge + Q is fixed at a distance of d in front of an infinite metal plate. Draw the lines of force indicating the directions.
Solution :

There will be induced charge on two surfaces of the conducting plate, so ELOF will start from the +Q charge and terminate at the conductor and then will again start from the other surface of the conductor.

15.2 Solid Angle :

A solid angle is a measure of a cone. Consider the intersection of the given cone with a sphere of radius R. The solid angle ΔΩ of the cone is defined to be equal to ΔS/R2, where ΔS is the area on the sphere cut out by the cone.

16. Electric Flux

Consider some surface in an electric field The electric flux of the field over the \(\overrightarrow{\mathrm{E}}\). Let us select a small area element \(\overrightarrow{\mathrm{dS}}\) on this surface.

area element is given by dφE= \(\overrightarrow{\mathrm{E}} \overrightarrow{\mathrm{dS}}\)

The direction of \(\overrightarrow{\mathrm{dS}}\) is normal to the surface. It is along ˆn

or dφE= EdS cos θ or dφE= (E cos θ) dS or dφE= EndS

where En is the component of the electric field in the direction of \(\overrightarrow{\mathrm{dS}}\).

The electric flux over the whole area is given by φE= \(\)

If the electric field is uniform over that area then φE=

Special Cases :

Case I: If the electric field is normal to the surface, then the angle of the electric field \(\vec{E}\) with normal will be zero

So φ = ES cos 0

φ = ES

Case II: If the electric field is parallel to the surface (glazing), then the angle made by \(\vec{E}\) with normal = 90º

So φ = ES cos 90º = 0

16.1 Physical Meaning :

The electric flux through a surface inside an electric field represents the total number of electric lines of force crossing the surface. It is a property of the electric field

16.2 Unit

• The SI unit of electric flux is Nm2 C–1(gauss) or J m C–1.
• Electric flux is a scalar quantity. (It can be positive, negative, or zero)

Example 89. If the electric field is given by \((6 \hat{i}+3 \hat{j}+4 \hat{k}) N / C\) calculate the electric flux through a surface of area 20 m2 lying in YZ plane.
Solution :

Here,\(\overrightarrow{\mathrm{E}}=6 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}\)

They are vectors representing the surface of area 20 units in the YZ-plane given by

⇒ \(\overrightarrow{\mathrm{S}}=20 \hat{\mathrm{i}}\)

Therefore, electric flux through the surface,

⇒ \(\phi=\vec{F} \cdot \vec{S}=(6 \hat{i}+3 \hat{j}+4 \hat{k}) \cdot 20 \hat{i}=120 \mathrm{~N}-\mathrm{m}^2 / \mathrm{C}\)

Example 90. A rectangular surface of sides 10 cm and 15 cm is placed inside a uniform electric field of 25 Vm–1, such that normal to the surface makes an angle of 60º with the direction of the electric field. Find the flux of the electric field through the rectangular surface.
Solution :

The flux through the rectangular surface given by

φ =\(\overrightarrow{\mathrm{E}} . \Delta \overrightarrow{\mathrm{S}}=\) EΔ S cos

Here, E = 25 V m–1;

ΔS = 10 × 15 = 150 cm2 = 150 × 10–4 m2

and θ = 60º

∴ φ = 25 × 150 × 10–4 cos 60º = \(\frac{3 \sqrt{3}}{16} \mathrm{Nm}^2 \mathrm{C}^{-1}\)

Example 91 The electric field in a region is given by \(\vec{E}=\frac{3}{5} E_0 \vec{i}+\frac{4}{5} E_0 \vec{j}\) with E₀= 2.0 × 103 N/C. Find the flux of this field through a rectangular surface of area 0.2m2 parallel to the Y–Z plane.
Solution :

⇒ \(\phi_E=\vec{E} \cdot \vec{S}=\left(\frac{3}{5} E_0 \vec{i}+\frac{4}{5} E_0 \vec{j}\right) \cdot(0.2 \hat{i})=240 \frac{N-m^2}{C}\)

Example 92 A point charge Q is placed at the corner of a square of side a, then find the flux through the square.

Solution :

The electric field due to Q at any point of the square will be along the plane of the square and the electric field line is perpendicular to the square; so φ = 0.

In other words, we can say that no line is crossing the square so flux = 0.

Case-III: Curved surface in uniform electric field Suppose a circular surface of radius R is placed in a uniform electric field as shown.

Flux passing through the surface φ = E (πR²)

Now suppose, a hemispherical surface is placed in the electric field flux through the hemispherical surface

φ = ∫ Eds cos θ φ = E ∫ ds cos θ

where ∫ ds cos θ is a projection of the spherical surface Area on the base.

∫ds cosθ = πR²

so φ = E(πR²) = same Ans. as in previous case

so we can conclude that

If the number of electric field lines passing through two surfaces are same, then flux passing through these surfaces will also be the same, irrespective of the shape of the surface

φ1= φ2= φ3= E(πR²)

Case 4:

Flux through a closed surface :

Suppose there is a spherical surface and a charge ‘placed at the center. flux through the spherical surface

φ = \(\int \vec{E} \cdot \overrightarrow{d s}=\int E d s \quad \text { as } \vec{E} \text { is along } \overrightarrow{d s} \text { (normal) }\)

φ= \(\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\mathrm{Q}}{\mathrm{R}^2} \int \quad \text { ds } \quad \text { where } \int \quad \mathrm{ds}=4 \pi \mathrm{R}^2\)

φ = \(\left(\frac{1}{4 \pi \mathrm{R}^2} \frac{\mathrm{Q}}{\mathrm{R}^2}\right)\left(4 \pi \mathrm{R}^2\right) \Rightarrow \phi=\frac{\mathrm{Q}}{\varepsilon_{\mathrm{o}}}\)

Now if the charge Q is enclosed by any other closed surface, still same lines of force will still pass through the surface.

So here also flux will be φ = \(\frac{\mathrm{Q}}{\varepsilon_{\mathrm{o}}}\), that’s what Gauss Theorem is. ο

17. Gauss’s Law In Electrostatics Or Gauss’s Theorem

This law was stated by a mathematician Karl F Gauss. This law gives the relation between the electric field at a point on a closed surface and the net charge enclosed by that surface. This surface is called the Gaussian surface. It is a closed hypothetical surface. Its validity is shown by experiments. It is used to determine the electric field due to some symmetric charge distributions.

17.1 Statement and Details :

Gauss’s law is stated as given below.

The surface integral of the electric field intensity over any closed hypothetical surface (called Gaussian surface) in free space is equal to \(\frac{1}{\varepsilon_0}\) times the total charge enclosed within the surface. Here, ε₀is the permittivity of free space.

If S is the Gaussian surface and \(\sum_{i=1}^n q_i\)is the total charge enclosed by the Gaussian surface, then according to Gauss’s law,

⇒ \(\phi_E=\int \vec{E} \cdot \overrightarrow{d S}=\frac{1}{\varepsilon_0} \sum_{i=1}^n q_i\)

The circle on the sign of integration indicates that the integration is to be carried out over the closed surface.

Note :

Flux through the Gaussian surface is independent of its shape.

Flux through the Gaussian surface depends only on the total charge present inside the Gaussian surface.

Flux through the Gaussian surface is independent of the position of charges in the ide Gaussian surface.

Electric field intensity at the Gaussian surface is due to all the charges present inside as well as outside the Gaussian surface.

In a close surface, incoming flux is taken negatively while outgoing flux is taken positively because ˆis taken positively in an outward direction.

In a Gaussian surface φ = 0 does not imply E = 0 at every point of the surface but E = 0 at every point implies φ = 0.

Example 93 Find out flux through the given Gaussian surface.

Solution : φ = \(\frac{\mathrm{Q}_{\mathrm{in}}}{\varepsilon_0}=\frac{2 \mu \mathrm{C}-3 \mu \mathrm{C}+4 \mu \mathrm{C}}{\varepsilon_0}=\frac{3 \times 10^{-6}}{\varepsilon_0} \mathrm{Nm}^2 / \mathrm{C}\)

Example 94 If a point charge q is placed at the centre of a cube then find out flux through any one surface of the cube.
Solution :

Flux through 6 surfaces = \(\frac{\mathrm{q}}{\varepsilon_0}\). Since all the surfaces are symmetrical 0

so, flux through one surfaces = \(\frac{1}{6} \frac{\mathrm{q}}{\varepsilon_0}\)

17.2 Flux through open surfaces using Gauss’s Theorem :

Example 95 A point charge +q is placed at the centre of curvature of a hemisphere. Find flux through the hemispherical surface.

Solution :

Let’s put an upper-half hemisphere. Now flux passing through the entire sphere = \(=\frac{\mathrm{q}}{\varepsilon_0}\)

As the charge q is symmetrical to the upper half and lower half hemispheres, so half-half flux will emit from both surfaces.

Example 96 A charge Q is placed at a distance a/2 above the center of a horizontal, square surface of the edge as shown in the figure. Find the flux of the electric field through the square surface.

Solution: We can consider imaginary faces of the cube such that the charge lies at the center of the cube. Due to symmetry, we can say that flux through the given area (which is one face of the cube)

φ = \(\phi=\frac{\mathrm{Q}}{6 \varepsilon_0}\)

Example 97 Find flux through the hemispherical surface

Solution :

Flux through the hemispherical surface due to +q = \(\frac{\mathrm{q}}{2 \varepsilon_0}\) (we have seen in previous examples)

Flux through the hemispherical surface due to +q0charge = 0, because due to +q0charge field lines entering the surface = field lines coming out of the surface.

17.3 Finding qin from flux :

Solved Examples

Example 98 Flux (in S.I. units) coming out and entering a closed surface is shown in the figure. Find the charge enclosed by the closed surface.

Solution :

Net flux through the closed surface = + 20 + 30 + 10 -15 = 45 N.m2/c from Gauss`s theorem

φ_net = \(\frac{q_{\text {in }}}{\varepsilon_0} \quad \Rightarrow \quad 45=\frac{q_{\text {in }}}{\varepsilon_0} \quad \Rightarrow \quad q_{\text {in }}=(45) \varepsilon_0\)

17.4 Finding electric field from Gauss`s Theorem :

From Gauss`s theorem, we can say

⇒ \(\int \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{ds}}=\phi_{\text {net }}=\frac{\mathrm{q}_{\mathrm{in}}}{\varepsilon_0}\)

17.4.1 Finding E due to a spherical shell:-

The electric field outside the Sphere :

Since the electric field due to a shell will be radially outwards. So let’s choose a spherical Gaussian surface. Gauss`s theorem for this spherical Gauss`s surface,

⇒ \(\int \overrightarrow{\mathrm{E}} \overrightarrow{\mathrm{ds}}=\phi_{\text {net }}=\frac{\mathrm{q}_{\text {in }}}{\varepsilon_{\mathrm{o}}}=\frac{\mathrm{q}}{\varepsilon_{\mathrm{o}}}\)

⇒ \(\int|\vec{E}||\overrightarrow{d s}| \cos 0\) (because the E→is normal to the surface)

⇒ \(E \int \mathrm{ds}\) (because the value of E is constant at the surface) E

E (4πr²) (ds ∫total area of the spherical surface = 4 πr²)

⇒ E (4πr²) = \(\frac{q_{\text {in }}}{\varepsilon_0} \quad \Rightarrow \quad E_{\text {out }}=\frac{\mathrm{q}}{4 \pi \varepsilon_0 r^2}\)

The electric field inside a spherical shell :

Let’s choose a spherical Gaussian surface inside the shell. Applying Gauss`s theorem for this surface q, R

⇒ \(\int \vec{E} \overrightarrow{\mathrm{ds}}=\phi_{\text {net }}=\frac{\mathrm{q}_{\text {in }}}{\varepsilon_0}=0\)

⇒ \(\int|\vec{E} \| \overrightarrow{d s}| \cos 0\)

⇒ \(E \int d s\)

↓
E (4πr²) ⇒ E (4πr²) = 0 ⇒ Ein = 0

17.4.2 Electric field due to solid sphere (having uniformly distributed charge Q and radius R) :

The electric field outside the sphere :

The direction of the electric field is radially outwards, so we will choose a spherical Gaussian surface Applying Gauss`s theorem

⇒ \(\int \overrightarrow{\mathrm{E}} \overrightarrow{\mathrm{ds}}=\phi_{\text {net }}=\frac{\mathrm{q}_{\text {in }}}{\varepsilon_{\mathrm{o}}}=\frac{\mathrm{Q}}{\varepsilon_{\mathrm{o}}}\)

⇒ \(\int|\vec{E} \| \overrightarrow{d s}| \cos 0\)

⇒ \(\mathrm{E} \int \mathrm{ds}\)

⇒ \(E\left(4 \pi r^2\right)\)

⇒ E (4πr²) = \(\frac{\mathrm{Q}}{\varepsilon_{\mathrm{o}}} \Rightarrow \mathrm{E}_{\text {out }}=\frac{\mathrm{Q}}{4 \pi \varepsilon_{\mathrm{o}} \mathrm{r}^2}\)

The electric field inside a solid sphere :

For this choose a spherical Gaussian surface inside the solid sphere Applying Gauss`s theorem to this surface

⇒ \(\int^{\downarrow}Ed s\)

⇒ \(E\left(4 \pi r^2\right) \quad \Rightarrow \quad E\left(4 \pi r^2\right)=\frac{q r^3}{\varepsilon_0 R^3}\)

⇒ \(E=\frac{q r}{4 \pi \varepsilon_0 R^3} \Rightarrow E_{\text {in }}=\frac{k Q}{R^3} r\)

17.4.3 Electric field due to infinite line charge (having uniformly distributed charged of charge density λ ) :

The electric field due to infinite wire is radial so we will choose a cylindrical Gaussian surface as shown in the figure.

⇒ \(\frac{\mathrm{q}_{\mathrm{in}}}{\varepsilon_{\mathrm{o}}}=\frac{\lambda \ell}{\varepsilon_{\mathrm{o}}}\)

φ₃ = \(\int \vec{E} \cdot \overrightarrow{d s} \quad=\int E d s=E \int d s=E(2 \pi r \ell)\)

E (2πrl) = \(\frac{\lambda \ell}{\varepsilon_o}\)

E = \(\frac{\lambda}{2 \pi \varepsilon_0 r}=\frac{2 \mathrm{k} \lambda}{\mathrm{r}}\)

17.4.4 Electric field due to infinity long charged tube (having uniform surface charge density σ and radius R)):

E outside the tube:- let’s choose a cylindrical Gaussian surface

φ_net = \(\frac{\mathrm{q}_{\text {in }}}{\varepsilon_{\mathrm{o}}}=\frac{\sigma 2 \pi \mathrm{R} \ell}{\varepsilon_{\mathrm{o}}}\)

About × 2πrl = \(\frac{\sigma 2 \pi \mathrm{R} \ell}{\varepsilon_{\mathrm{o}}}\)

E = \(\frac{\sigma \mathrm{R}}{\mathrm{r} \varepsilon_0}\)

E inside the tube :

let’s choose a cylindrical Gaussian surface inside the tube.

φ_net = \(\phi_{\text {net }}=\frac{\mathrm{q}_{\text {in }}}{\varepsilon_{\mathrm{o}}}=0 \quad \text { So } \quad \mathrm{E}_{\text {in }}=0\)

17.4.5 E due to infinitely long solid cylinder of radius R (having uniformly distributed charge in volume (charge density ρ)) :

E at outside point:-

Let’s choose a cylindrical Gaussian surface. Applying Gauss`s theorem

E × 2πrl = \(\mathrm{E} \times 2 \pi \mathrm{r} \ell=\frac{\mathrm{q}_{\text {in }}}{\varepsilon_{\mathrm{o}}}=\frac{\rho \times \pi \mathrm{R}^2 \ell}{\varepsilon_{\mathrm{o}}} \quad \Rightarrow \quad \mathrm{E}_{\text {out }}=\frac{\rho \mathrm{R}^2}{2 \mathrm{r} \varepsilon_0}\)

E at inside point :

let’s choose a cylindrical Gaussian surface inside the solid cylinder. Applying Gauss`s theorem

E × 2πrl= \(\frac{q_{\text {in }}}{\varepsilon_{\mathrm{o}}}=\frac{\rho \times \pi \mathrm{r}^2 \ell}{\varepsilon_{\mathrm{o}}} \quad \Rightarrow \quad \mathrm{E}_{\text {in }}=\frac{\rho \mathrm{r}}{2 \varepsilon_0}\)

18. Conductor

18.1 Conductor and its properties [For electrostatic condition]

• Conductors are materials that contain a large number of free electrons that can move freely inside the conductor.
• Ιn electrostatics conductors are always equipotential surfaces.
• Charge always resides on the outer surface of the conductor.

If there is a cavity inside the conductor having no charge then a charge will always reside only on the outer surface of the conductor.

• The electric field is always perpendicular to the conducting surface.
• Electric lines of force never enter into conductors.
• Electric field intensity near the conducting surface is given by the formula

⇒ \(\overrightarrow{\mathrm{E}}=\frac{\sigma}{\varepsilon_0} \hat{n}\)

⇒ \(\overrightarrow{E_A}=\frac{\sigma_{\mathrm{A}}}{\varepsilon_0} \hat{n} ; \overrightarrow{E_{\mathrm{B}}}=\frac{\sigma_{\mathrm{B}}}{\varepsilon_0} \hat{n} \text { and } \overrightarrow{\mathrm{E}_{\mathrm{C}}}=\frac{\sigma_{\mathrm{C}}}{\varepsilon_0} \hat{n}\)

When a conductor is grounded its potential becomes zero.

When an isolated conductor is grounded then its charge becomes zero.

When two conductors are connected there will be charge flow till their potential becomes equal

Electric pressure: Electric pressure at the surface of a conductor is given by formula P = \(\frac{\sigma^2}{2 \varepsilon_0}\)where σ is the local surface charge density.

18.2 Finding field due to a conductor

Suppose we have a conductor and at any ‘A’, local surface charge density = σ. We have to find an electric field just outside the conductor surface.

For this let’s consider a small cylindrical Gaussian surface, which is partly inside and partly outside the conductor surface, as shown in the figure. It has a small cross-section area ds and negligible height.

Applying Gauss’s theorem to this surface

So, \(\mathrm{Eds}=\frac{\sigma \mathrm{ds}}{\varepsilon_0} \quad \mathrm{E}=\frac{\sigma}{\varepsilon_0}\)

Electric field just outside the surface of conductor E = \(\mathrm{E}=\frac{\sigma}{\varepsilon_0}\) direction will be normal to the surface in vector form \(\overrightarrow{\mathrm{E}}=\frac{\sigma}{\varepsilon_0} \hat{\mathbf{n}}\)(here n = unit vector normal to the conductor surface)

18.3 Electrostatic pressure at the surface of the conductor

Electrostatic pressure at the surface of the conductor in \(\mathrm{P}=\frac{\sigma^2}{2 \varepsilon_0}\)

where σ = local surface charge density.

Electrostatic shielding

Consider a conductor with a cavity of any shape and size, with no charges inside the cavity. The electric field inside the cavity is zero, whatever the charge on the conductor and the external

Any cavity in a conductor remains shielded from outside electric influence: the field inside the cavity is always zero (If cava it has no charge). This is known as electrostatic shielding.

This effect can be made use of in protecting sensitive instruments from outside electrical influence.

18.4 Electric field due to a conducting and nonconducting uniformly charge infinite sheets

Example 99 Prove that if an isolated (isolated means no charges are near the sheet) large conducting sheet is given a charge then the charge distributes equally on its two surfaces.
Solution :

Let there be x charge on the left side of the sheet and Q–x charge on the right side of the sheet. Since point P lies inside the conductor so

EP= O

⇒ \(\frac{x}{2 A \varepsilon_O}-\frac{Q-x}{2 A \varepsilon_0}=0 \quad \Rightarrow \frac{2 x}{2 A \varepsilon_0}=\frac{Q}{2 A \varepsilon_0} \quad \Rightarrow x=\frac{Q}{2} \quad Q-x=\frac{Q}{2}\)

So charges are equally distributed on both sides

Example 100 If an isolated infinite sheet contains charge Q₁on its one surface and charge Q₂on its other surface then prove that electric field intensity at a point in front of the sheet will be\(\frac{Q}{2 A \varepsilon_0}\) where Q = Q₁+ Q₂

Solution:

The electric field at point P :

⇒ \(\vec{E}=\vec{E}_{Q_1}+\vec{E}_{Q_2}\)

⇒ \(\frac{Q_1}{2 A \varepsilon_0} \hat{n}+\frac{Q_2}{2 A \varepsilon_0} \hat{n}=\frac{Q_1+Q_2}{2 A \varepsilon_0} \hat{n}=\frac{Q}{2 A \varepsilon_0} \hat{n}\)

[This shows that the resultant field due to a sheet depends only on the total charge of the sheet and not on the distribution of charge on individual surfaces].

Example 101 Two large parallel conducting sheets (placed at a finite distance ) are given charges Q and 2Q respectively. Find out charges appearing on all the surfaces.

Solution :

Let there be x amount of charge on the left side of the first plate, so on its right side charge will be Q–x, similarly for the second plate there is y charge on the left side, and 2Q – y charge is on the right side of the second plate Ep= 0 ( By the property of conductor)

⇒ \(\frac{x}{2 A \varepsilon_{\mathrm{o}}}-\left\{\frac{\mathrm{Q}-\mathrm{x}}{2 \mathrm{~A} \varepsilon_{\mathrm{o}}}+\frac{\mathrm{y}}{2 \mathrm{~A} \varepsilon_{\mathrm{o}}}+\frac{2 \mathrm{Q}-\mathrm{y}}{2 \mathrm{~A} \varepsilon_{\mathrm{o}}}\right\}=0\)

we can also say that the charge on the left side of P = charge on the right side of P

x = Q – x + y + 2Q – y ⇒ x = \(x=\frac{3 Q}{2}, Q-x=\frac{-Q}{2}\)

Similarly for point Q:

x + Q – x + y = 2Q – y ⇒ y = Q/2 , 2Q – y = 3Q/2

So final charge distribution of plates is: –

Example 102 An isolated conducting sheet of area A and carrying a charge Q is placed in a uniform electric field E, such that the electric field is perpendicular to the sheet and covers all the sheets. Find out charges appearing on its two surfaces.

Solution: Let there be x charge on the left side of the plate and Q – x charge on the right side of the plate EP= 0

⇒ \(\frac{X}{2 \mathrm{~A} \varepsilon_0}+E=\frac{Q-X}{2 \mathrm{~A} \varepsilon_0}\)

⇒\(\frac{\mathrm{x}}{\mathrm{A} \varepsilon_0}=\frac{\mathrm{Q}}{2 \mathrm{~A} \varepsilon_0}-\mathrm{E} \quad \Rightarrow \quad \mathrm{x}=\frac{\mathrm{Q}}{2}-E A \varepsilon_0 \text { and } \mathrm{Q}-\mathrm{x}=\frac{\mathrm{Q}}{2}+E A \varepsilon_0\)

So charge on one side is \(\frac{Q}{2}\)– EAε_o and other side \(\frac{\mathrm{Q}}{2}+\mathrm{EA} \varepsilon_0\)

Note: Solve this question for Q = 0 without using the above answer and match that answer with the answers that you will get by putting Q = 0 in the above answer.

18.5 Some other important results for a closed conductor.

If a charge q is kept in the cavity then –q will be induced on the inner surface and +q will be induced on the outer surface of the conductor (it can be proved using Gauss theorem)

If a charge q is kept inside the cavity of a conductor and the conductor is given a charge Q then –q charge will be induced on the inner surface and the total charge on the outer surface will be q + Q. (it can be proved using Gauss theorem)

The resultant field, due to q (which is inside the cavity) and induced charge on S₁, at any point outside S₁(like B, C) is zero. The resultant field due to q + Q on and any other charge outside S₂, at any point inside of surface S₂(like A, B) is zero

The resulting field in a charge-free cavity in a closed conductor is zero. There can be charges outside the conductor and on the surface also. Then also this result is true. No charge will be induced on the innermost surface of the conductor.

Charge distribution for different types of cavities in conductors

Using the result that \(\)charges are not at the geometrical center Eresin the conducting material should be zero and using result (iii) We can show

Note: In all cases charge on inner surface S₁= –q and on outer surface S₂= q. The distribution of charge on ‘S₁’ will not change even if some charges are kept outside the conductor (i.e. outside the surface S₂). However the charge distribution on ‘S₂’ may change if some charges(s) are/are kept outside the conductor.

Sharing of charges: Two conducting hollow spherical shells of radii R₁ and R₂ having charges Q₁ and Q₂respectively and separated by a large distance, are joined by a conducting wire Let the final charges on spheres be q₁ and q₂ respectively.

Potential on both spherical shells becomes equal after joining, therefore

⇒ \(\frac{K q_1}{R_1}=\frac{K q_2}{R_2} \Rightarrow \frac{q_1}{q_2}=\frac{R_1}{R_2}\)……(i) 2

and, q₁+ q₂= Q₁+ Q₂……(ii)

from (i) and (ii) q₁= \(q_1=\frac{\left(Q_1+Q_2\right) R_1}{R_1+R_2} \quad q_2=\frac{\left(Q_1+Q_2\right) R_2}{R_1+R_2}\)

ratio of charges \(\frac{q_1}{q_2}=\frac{R_1}{R_2} \quad \Rightarrow \quad \frac{\sigma_1 4 \pi R_1^2}{\sigma_2 4 \pi R_2^2}=\frac{R_1}{R_2}\)

ratio of surface charge densities \(\frac{\sigma_1}{\sigma_2}=\frac{R_2}{R_1}}\)

Ratio of final charges \(\frac{q_1}{q_2}=\frac{R_1}{R_2}\)

Ratio of final surface charge densities.\(\frac{\sigma_1}{\sigma_2}=\frac{R_2}{R_1}\)

Example 103 The two conducting spherical shells are joined by a conducting wire and cut after some time when the charge stops flowing. Find out the charge on each sphere after that.

Solution :

After cutting the wire, the potential of both the shells is equal

Thus, potential of inner shell Vin = \(\frac{K x}{R}+\frac{K(-2 Q-x)}{2 R}=\frac{k(x-2 Q)}{2 R}\) and potential of outer shell 2R

⇒ \(V_{\text {out }}=\frac{K x}{2 R}+\frac{K(-2 Q-x)}{2 R}=\frac{-K Q}{R}\)

As Vout = Vin ⇒\(\frac{-K R}{R}=\frac{K(x-2 Q)}{2 R}\) ⇒ –2Q = x – 2Q ⇒ x = 0

So charge on inner spherical shell = 0 and outer spherical shell = – 2Q.

Example 104 Two conducting hollow spherical shells of radii R and 2R carry charges – Q and 3Q respectively. How much charge will flow into the earth if the inner shell is grounded?

Solution: When the inner shell is grounded to the Earth then the potential of the inner shell will become zero because the potential of the Earth is taken to be zero.

⇒ \(\frac{K x}{R}+\frac{K 3 Q}{2 R}=0 \quad \Rightarrow \quad x=\frac{-3 Q}{2},\), the charge that has increased

⇒ \(=\frac{-3 Q}{2}-(-Q)=\frac{-Q}{2}\)hence charge flows into the Earth = \(\frac{Q}{2}\)

Example 105 An isolated conducting sphere of charge Q and radius R is connected to a similar uncharged sphere (kept at a large distance) by using a high-resistance wire. After a long time, what is the amount of heat loss?
Solution :

When two conducting spheres of equal radius are connected charge is equally distributed on them (Result VI). So we can say that heat loss in the system

ΔH = Ui– Uf

⇒ \(=\left(\frac{\mathrm{Q}^2}{8 \pi \varepsilon_0 \mathrm{R}}-0\right)-\left(\frac{\mathrm{Q}^2 / 4}{8 \pi \varepsilon_0 \mathrm{R}}+\frac{\mathrm{Q}^2 / 4}{8 \pi \varepsilon_0 \mathrm{R}}\right)=\frac{\mathrm{Q}^2}{16 \pi \varepsilon_0 \mathrm{R}}\)

10. Van De Graff Generator

This is a machine that can build up high voltages of the order of a few million volts. The resulting large electric fields are used to accelerate charged particles (electrons, protons, ions) to high energies needed for experiments to probe the small-scale structure of matter.

Designed by R.J. Van de Graaff in 1931.

It is an electrostatic generator capable of generating a very high potential of the order of 5 × 106 V.

This high potential is used in accelerating the charged particles.

Principle :

• It is based on the following two electrostatic phenomena :
• The electric discharge takes place in air or gases readily at pointed conductors.
• If a hollow conductor is in contact with another conductor, which lies inside the hollow conductor. Then, the charge is supplied to the the inner conductor. The charge immediately shifts to the outer surface of the hollow conductor.
• Consider a large spherical conducting shell A having radius R and charge +Q, potential inside the shell is constant and it is equal to that at its surface.
• Therefore, the  potential inside the charged conducting shell A,
• \(\mathrm{V}_1=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{Q}}{\mathrm{R}}\)
• Suppose that a small conducting sphere B having radius r and charge +q is placed at the center of shell A.
• Then, potential due to the sphere B at the surface of shell A,
• \(\mathrm{V}_2=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}}{\mathrm{R}}\)
• and potential due to the sphere B at its surface, 1 q
• \(V_3=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r}\)

Thus, the total potential at the surface of shell A due to the charges Q and q,

• \(V_A=V_1+V_2=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{R}+\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{R} \quad \text { or } \quad V_A=\frac{1}{4 \pi \varepsilon_0}\left(\frac{Q}{R}+\frac{q}{R}\right)\)
• and the total potential at the surface of sphere B due to the charges Q and q,
• \(V_B=V_1+V_3=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{R}+\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r} \quad \text { or } \quad V_B=\frac{1}{4 \pi \varepsilon_0}\left(\frac{Q}{R}+\frac{q}{r}\right)\)
• It follows that VB> VA. Hence, the potential difference between the sphere and the shell,
• \(V=V_B-V_A=\frac{1}{4 \pi \varepsilon_0}\left(\frac{Q}{R}+\frac{q}{r}\right)-\frac{1}{4 \pi \varepsilon_0}\left(\frac{Q}{R}+\frac{\mathrm{q}}{\mathrm{R}}\right) \Rightarrow \quad V=\frac{1}{4 \pi \varepsilon_0} \cdot \mathrm{q}\left(\frac{1}{r}-\frac{1}{R}\right)\)
• It follows that the potential difference between the sphere and the shell is independent of the charge Q on the shell. Therefore, if the sphere is connected to the shell by a wire, the charge supplied to the sphere will immediately flow to the shell.
• It is because the potential of the sphere is higher than that of the shell and the charge always flows from higher to lower potential.
• It forms the basic principle of the Van de Graaff generator.

Construction :

Working :

• An endless belt of insulating material is made to run on two pulleys P₁ and P₂ with the help of an electric motor.
• The metal comb C₁, called spray comb is held potential with the help of E.H.T. source (≈ 104 V), it produces ions in its vicinity. The positive ions get sprayed on the belt due to the repulsive action of comb C₁.
• These positive ions are carried upward by the moving belt. A comb C₂, called a collecting comb is positioned near the upper end of the belt, such that the pointed ends touch the belt and the other end is in contact with the inner surface of the metallic sphere S. The comb C₂ collects the positive ions and transfers them to the metallic sphere.
• The charge transferred by the comb immediately moves onto the outer surface of the hollow sphere. As the belt goes on moving, the accumulation of positive charge on the sphere also keeps on taking place continuously and its potential rises considerably.
• With the increase of charge on the sphere, its leakage due to ionization of surrounding air also becomes faster.
• The maximum potential to which the sphere can be raised is reached when the rate of loss of charge due to leakage becomes equal to the rate at which the charge is transferred to the sphere.
• To prevent the leakage of charge from the sphere, the generator is completely enclosed inside an earth-connected steel tank, which is filled with air under pressure.
• If the charged particles, such as protons, neutrons, etc. are now generated in the discharge tube D with the lower end earthed and the upper end inside the hollow sphere, they get accelerated in a downward direction along the length of the tube. At the other end, they come to hit the target with large kinetic energy.
• Van de Graaff generator of this type was installed at the Carnegie Institute in Washington in 1937. One such generator was installed at the Indian Institute of Technology, Kanpur in 1970 and it accelerates particles to 2 MeV energy.

Problem 1. Two charges of Q each are placed at two opposite corners of a square. A charge q is placed at each of the other two corners.

1. If the resultant force on Q is zero, how are Q and q related?
2. Could q be chosen to make the resultant force on each charge zero?

Solution : (a) Let at a square ABCD charge be placed as shown

Now forces on charge Q (at point A) due to other charges are \(\) respectively shown in the figure.

Fnet on Q = \(\)(at point A)

But Fnet = 0 So, ΣFx= 0

ΣFx= – FQQ cos45° – FQq

⇒ \( \frac{K Q^2}{(\sqrt{2} a)^2} \cdot \frac{1}{\sqrt{2}}+\frac{K Q q}{a^2}=0 \quad \Rightarrow\)

For the resultant force on each charge to be zero :

From previous data, force on charge Q is zero when q = \(-\frac{Q}{2 \sqrt{2}}\) if for this value of charge q, force on q is zero then and only then the value of q exists for which the resultant force on each charge is zero.

Force on q:-

Forces on charge q (at point D) due to other three charges are \(\overrightarrow{\mathrm{F}}_{\mathrm{qQ}}, \overrightarrow{\mathrm{F}}_{\mathrm{qq}} \text { and } \overrightarrow{\mathrm{F}}_{\mathrm{qQ}}\) respectively shown in figure.

The net force on charge q:-

⇒ \(\vec{F}_{\text {net }}=\vec{F}_{\mathrm{qq}}+\overrightarrow{\mathrm{F}}_{\mathrm{qQ}}+\overrightarrow{\mathrm{F}}_{\mathrm{qQ}} \quad \text { But } \overrightarrow{\mathrm{F}}_{\text {net }}=0\)

But So, ΣFx= 0

ΣFx= \(-\frac{K q^2}{(\sqrt{2} a)^2} \cdot \frac{1}{\sqrt{2}}-\frac{K Q q}{(a)^2} \Rightarrow q=-2 \sqrt{2} Q\)

But from previous condition, q = \(-\frac{Q}{2 \sqrt{2}}\)

So, no value of q makes the resultant force on each charge zero.

Problem 2. The figure shows a uniformly charged thin non-conducting sphere of total charge Q and radius R. If point charge q is situated at point ‘A’ which is at a distance r < R from the center of the sphere then find out the following

1. Force acting on charge q.
2. The electric field at the center of the sphere.
3. The electric field at point B.

Solution :

The electric field inside a hollow sphere = 0

∴ Force on charge q.

F = qE = q × 0 = 0

The net electric field at the center of the sphere

E_net = \(\overrightarrow{\mathrm{E}}_1+\overrightarrow{\mathrm{E}}_2\)

E₁= Field due to sphere = 0

E₂= field due to this charge = \(=\frac{\mathrm{Kq}}{\mathrm{r}^2}\)

E_net = \(\frac{\mathrm{Kq}}{\mathrm{r}^2}\)

Electric field at B due to charge on sphere\(\overrightarrow{\mathrm{E}}_1=\frac{\mathrm{KQ}}{\mathrm{r}_1^2} \hat{\mathrm{r}}_1\)

and due to charge q at A ,\(\vec{E}_2=\frac{K q}{r_2{ }^2} \hat{r}_2 \quad \text { So, } \vec{E}_{\text {net }}=\vec{E}_1+\vec{E}_2 \quad=\frac{K Q}{r_1{ }^2} \hat{r}_1+\frac{K q}{r_2{ }^2} \hat{r}_2\)

where r₁= CB and r₂= AB

Problem 3. The figure shows two concentric spheres of radius and R₂(R₂> R₁) which contain uniformly distributed charges Q and –Q respectively. Find out electric field intensities at the following positions :

1. r < R₁
2. R₁≤ r < R₂
3. r ≥ R₂

Solution :

Net electric field = E₁+ E₂

E₁= field due to the sphere of radius R₁

E₂= field due to the sphere of radius R₂

E₁= 0, E₂= 0

E_net = 0

E₁= \(\frac{\mathrm{KQ}}{\mathrm{r}^2}, \mathrm{E}_2=0 \quad \Rightarrow \quad \overrightarrow{\mathrm{E}}=\frac{\mathrm{Kq}}{\mathrm{r}^2} \hat{r}\)

⇒ \(\vec{E}_1=\frac{K q}{r^2} \hat{r} \vec{E}_2=\frac{K q}{r^2}(-\hat{r}) \quad \Rightarrow \quad \vec{E}_{\text {net }}=\vec{E}_1+\vec{E}_2=0\)

Problem 4. Three identical spheres each having a charge q (uniformly distributed) and radius R, are kept in such a way that each touches the other two. Find the magnitude of the electric force on any sphere due to the other two.
Solution :

Given three identical spheres each having a charge q and radius R are kept as shown:-

For any external point; the sphere behaves like a point charge. So it becomes a triangle having point charges on its corner.

⇒ \(\left|\vec{F}_{\mathrm{qq}}\right|=\frac{\mathrm{kq}^2}{4 \mathrm{R}^2}\)

So net force (F) = 2.\(\frac{k q^2}{4 R^2} \cdot \cos \frac{60}{2}=2 \cdot \frac{k q^2}{4 R^2} \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{4} \frac{k q^2}{R^2}\)

Problem 5. A uniform electric field of 10 N/C exists in the vertically downward direction. Find the increase in the electric potential as one goes up through a height of 50cm.
Solution :

E =\(-\frac{\mathrm{dv}}{\mathrm{dr}} \Rightarrow \mathrm{dv}=-\overrightarrow{\mathrm{E}} \cdot \mathrm{d} \overrightarrow{\mathrm{r}}\)

⇒ \(\overrightarrow{\mathrm{E}}=\text { constant } \quad \Rightarrow \quad \Delta v=-\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\Delta r}\)

⇒ \(\Delta v=-10(-\hat{j}) \cdot\left(50 \times 10^{-2}\right) \hat{j}\) = 5 volts.

Problem 6. An electric field of 20 N/C exists along the x-axis in space. Calculate the potential difference V_B – V_A where the points A and B are given by –

1. A = (0,0) ; B = (4m , 2m)
2. A = (4m,2m) ; B = (6m , 5m)
3. A = (0,0) ; B = (6m , 5m)

Solution: Electric fields-axis axis mean \(\vec{E}=20 \hat{i}\)

1. \(\left|\Delta V_{A B}\right|=\vec{E} \cdot \vec{d}=20 \hat{i} \cdot 4 \hat{i}=80 \mathrm{~V} \Rightarrow \quad V_B-V_A=-80 \mathrm{~V}\)
2. \(\left|\Delta V_{B C}\right|=\vec{E} \cdot \vec{d}=20 \hat{i} \cdot 2 \hat{i}=40 \text { volt } \quad \Rightarrow \quad V_C-V_B=-40 V\)
3. \(\left|\Delta V_{A C}\right|=\vec{E} \cdot \vec{d}=20 \hat{i} \cdot 6 \hat{i}=120 \text { volt } \quad \Rightarrow \quad V_c-V_A=-120 \mathrm{~V}\)

Problem 7. A point charge of charge –q and mass m is released with negligible speed from a distance 3Ron the axis of a fixed uniformly charged ring of charge Q and radius R. Find out its velocity when it reaches the center of the ring.

Solution :

As potential due to uniform charged ring at its axis (at x distance)

⇒ \(V=\frac{k Q}{\sqrt{R^2+x^2}}\)

So potential at point A due to ring

⇒ \(V_1=\frac{k Q}{\sqrt{R^2+3 R^2}}=\frac{k Q}{2 R}\)

So potential energy of charge –q at point A

⇒ \(\text { P.E.E. }=\frac{-k Q q}{2 R}\)

and potential at point B V2= \(V_2=\frac{k Q}{R}\)

So potential energy of charge –q at point B

⇒ \(P. E_{.2}=\frac{-k Q q}{R}\)

Now by energy conservation P.E.1+ K.E.1= P.E2+ K.E2

⇒ \(\frac{-k Q q}{2 R}+0=\frac{-k Q q}{R}+\frac{1}{2} m v^2 \quad \Rightarrow \quad v^2=\frac{k Q q}{m R}\)

So the velocity of charge – q at point B v \(v=\sqrt{\frac{k Q q}{m R}}\)

Problem 8. Four small point charges each of equal magnitude q are placed at four corners of a regular tetrahedron of side a. Find out the potential energy of the charge system

Solution: Potential energy of the system :

U = U₁₂ + U₁₃ + U₁₄ + U₂₃ + U24+ U₃₄

⇒ \(U=\frac{-k q^2}{a}+\frac{k q^2}{a}+\frac{-k q^2}{a}+\frac{-k q^2}{a}+\frac{k q^2}{a}+\frac{-k q^2}{a}\)

Total potential energy of this charge system U = \(\frac{-2 k q^2}{a}\)

Problem 9. If V = x₂y + y₂z then find \(\) at (x, y, z)
Solution :

Given V = x2y + y2z and \(\vec{E}=-\frac{\partial v}{\partial r}\)

⇒ \(\overrightarrow{\mathrm{E}}=-\left[\frac{\partial \mathrm{V}}{\partial \mathbf{x}} \hat{\mathbf{i}}+\frac{\partial \mathrm{V}}{\partial \mathbf{y}} \hat{\mathbf{j}}+\frac{\partial \mathrm{V}}{\partial \mathbf{z}} \hat{\mathbf{k}}\right]\)

⇒ \(\vec{E}=-\left[2 x y \hat{i}+\left(x^2+2 y z\right) \hat{j}+y^2 \hat{k}\right]\)

Problem 10. If E = 2r2then find V(r)
Solution :

Given E = 2r2

we know that \(\int d v=-\int \vec{E} \cdot d r=-\int 2 r^2 d r\)

⇒ \(V(r)=\frac{-2 r^3}{3}+c\)

Problem 11. A charge Q is uniformly distributed over a rod of length l. Consider a hypothetical cube of edge l with the center of the cube at one end of the rod. Find the minimum possible flux of the electric field through the entire surface of the cube.
Solution :

According to Gauss law: flux depends upon charge inside the closed hypothetical surface so for minimum possible flux through the entire surface of the cube = the charge inside should be minimal.

Linear charge density of rod = \(\frac{Q}{\ell}\) and minimum length of rod inside the cube = \(\frac{\ell}{2}\)

So charge inside the cube = \(\frac{\ell}{2} \cdot \frac{Q}{\ell}=\frac{Q}{2}\)

So flux through the entire surface of the cube = \(\frac{\Sigma \mathrm{q}}{\epsilon_{\mathrm{o}}}=\frac{\mathrm{Q}}{2 \epsilon_0}\)

Problem 12. A charge Q is placed at a corner of a cube. Find the flux of the electric field through the six surfaces of the cube.

Solution :

By Gauss law, φ = \(\frac{\mathrm{q}_{\text {in }}}{\varepsilon_0}\)

Here, since Q is kept at the corner so only \(\frac{q}{8}\) charge is inside the cube. (since complete

charge can be enclosed by 8 such cubes) ∴ qin = \(\frac{Q}{8}\)

So, φ = \(\phi=\frac{q_{\text {in }}}{\varepsilon_0}=\frac{Q}{8 \varepsilon_0}\)

Problem 13. An isolated conducting sphere of charge Q and radius R is grounded by using a high-resistance wire. What is the amount of heat loss?

Solution :

When the sphere is ground, its potential becomes zero which means all charge goes to earth (due to the sphere being conducting and isolated) so all energy in the sphere is converted into heat, a total of 2

Heat loss = \(\frac{k Q^2}{2 R}\)

Problem 14. Two uncharged and parallel conducting sheets each of area A are placed in a uniform electric field E at a finite distance from each other. Such that the electric field is perpendicular to the sheets and covers all the sheets. Find out charges appearing on its two surfaces.

Solution :

Plates are conducting so the net electric field inside these plates should be zero. So, the electric field due to induced charges (on the surface of the plate) balances the outside electric field.

Here \(\overrightarrow{\mathrm{E}}_{\mathrm{i}}\) = induced electrid field

For both plates \(\overrightarrow{\mathrm{E}}_{\mathrm{i}}+\overrightarrow{\mathrm{E}}=0 \quad \Rightarrow \quad \overrightarrow{\mathrm{E}}_{\mathrm{i}}=-\overrightarrow{\mathrm{E}}\) …………….. (1)

Let charge induced on surfaces are +Q and – Q then

⇒ \(\left|\vec{E}_i\right|=\frac{Q}{A \varepsilon_0}\)

by equation (1)

⇒ \(\frac{Q}{A \varepsilon_0}=E\) ⇒ Q = AEε₀

Problem 15. A positive charge q is placed in front of a conducting solid cube at a distance d from its center. Find the electric field at the center of the cube due to the charges appearing on its surface.
Solution :

Here Ei= electric field due to induced charges

and Eq= electric field due to charge q

We know that the net electric field in a conducting cavity is equal to zero.

This means E= 0 at the center of the cube

⇒ \(\overrightarrow{\mathrm{E}}_{\mathrm{i}}+\overrightarrow{\mathrm{E}}_{\mathrm{q}}=\)

⇒ \(\vec{E}_i=-\vec{E}_i\)

⇒ \(\vec{E}_i=-\frac{k q}{d^2} \overrightarrow{\mathrm{PO}}\)

Chapter 4 Nuclear Physics Multiple Choice Questions Exercise 1 Section (A): Properties Of The Nucleus

Question 1. The mass number of a nucleus is

1. Always less than its atomic number
2. Always more than its atomic number
3. Equal to its atomic number
4. Sometimes more than and sometimes equal to its atomic number

Answer: 4. Sometimes more than and sometimes equal to its atomic number

Question 2. The stable nucleus that has a radius 1/3 that of os189 is –

1. ₃Li⁷
2. ₂He⁴
3. ₅B¹⁰
4. ₆C¹²

Answer: 1. ₃Li⁷

Question 3. For a uranium nucleus, how does its mass vary with volume?

1. m ∝ v
2. M ∝ 1/v
3. M ∝ \(\sqrt{V}\)
4. M ∝ v²

Answer: 1. m ∝ v

Question 4. The graph of ln (R/R₀) versus ln A (r = radius of a nucleus and a = its mass number) is

1. A straight line
2. A parabola
3. An ellipse
4. None of them

Answer: 1. A straight line

Question 5. 1 Amu is equivalent to:

1. 931 Mev
2. 0.51ev
3. 9.31 mev
4. 1.02 mev

Answer: 1. 931 Mev

Question 6. If the mass number for an element is m and the atomic number is z, then several neutrons will be :

1. M – z
2. Z –m
3. M + z
4. Z

Answer: 1. M – z

Question 7. Which of the following particles has a similar mass to an electron?

1. Proton
2. Neutron
3. Positron
4. Neutrino

Answer: 3. Positron

Question 8. Different atoms of the same element which have different masses but have the same chemical properties are called:

1. Isochoric
2. Isotope
3. Isobar
4. Isobaric

Answer: 2. Isotope

Question 9. The Mass-energy equation e = mc² was given by

1. Newton
2. Kepler
3. Einstein
4. Millikan

Answer: 3. Einstein

Question 10. The mass numbers of nuclei a and b are respectively 135 and 5. The ratio of their radii is:

1. 1: 27
2. 27 :1
3. 1 : 3
4. 3: 1

Answer: 2. 27 :1

Question 11. If the nucleus 125 13 52 ai has a nuclear radius of about 3.6 fm, then it would have its radius approximately as :

1. 6.0 FM
2. 9.6 FM
3. 12.0 FM
4. 4.8 FM

Answer: 1. 6.0 FM

Question 12. Two nuclei have their mass numbers in the ratio of 1 : 3. The ratio of their nuclear densities would be

1. 1 : 3
2. 3: 1
3. 1/3 : 1
4. 1: 1

Answer: 4. 1/3: 1

Question 13. A nucleus disintegrates into two nuclear parts which have their velocities in the ratio 2: 1. The ratio of their nuclear sizes will be:

1. 21/3 : 1
2. 1: 31/2
3. 31/2 : 1
4. 1: 21/3

Answer: 4. 1: 21/3

Question 14. If the radius of the 125 13 52 al the nucleus is estimated to be 3.6 fermis, then the radius of the nucleus is near:

1. 6 Fermi
2. 8 Fermi
3. 4 Fermi
4. 5 Fermi

Answer: 1. 6 Fermi

Question 15. The uncle of which one of the following pairs of nuclei are isotones:-

1. ₃₄Se⁷⁴, ₃₁Ga⁷¹
2. ₃₈Sr⁸⁴, ₃₈Sr⁸⁶
3. ₄₂Mo⁹²,₄₀Zr⁹²
4. ₂₀Ca⁴⁰,₁₆S³²

Answer: 1. ₃₄Se⁷⁴, ₃₁Ga⁷¹

Question 16. The range of a nuclear force is approximate –

1. 2 × 10^–10 M
2. 1.5 × 10^–20 m
3. 7.2 × 10^–4 m
4. 1.4 × 10^–15 m

Answer: 4. 1.4 × 10^–15 m

Question 17. The order of magnitude of the density of the uranium nucleus is, (m
p = 1.67 × 10–27 kg):

1. 1020 Kg m^–3
2. 1017 Kg m^–3
3. 1014 Kg m^–3
4. 1011 Kg m^–3

Answer: 2.1017 Kg m^–3

Question 18. Which has the highest penetrating power?

1. γ-Rays
2. β-Rays
3. α-Rays
4. Cathode rays

Answer: 1. γ-Rays

Question 19. The penetrating power is minimal for

1. α– Rays
2. β – Rays
3. ϒ – Rays
4. X – rays

Answer: 1. α– Rays

Chapter 4 Nuclear Physics Multiple Choice Questions Section (B): Mass Defect And Binding Energy

Question 1. Two protons are kept at a separation of 50Å. F n is the nuclear force and F e is the electrostatic force between them, then

1. F_n>>F_e
2. F_n=F_e
3. F_n<<F_e
4. F_nF_e

Answer: 3. F_n<<F_e

Question 2. Masses of nucleus, neutron, and protons are M, n m, and m p respectively. If the nucleus has been divided into neutrons and protons, then

1. M=(A-Z)m_n+Zm_p
2. M=ZM_n+(A-Z)m_p
3. M<A(A-X)m_n+Xm_p
4. M>(A-z)m_n+Zm_p

Answer: 2. M=ZM_n+(A-Z)m_p

Question 3. As the mass number A increases, the binding energy per nucleon in a nucleus

1. Increases
2. Decreases
3. Remains The Same
4. Varies In A Way That Depends On The Actual Value Of A.

Answer: 4. Varies In A Way That Depends On The Actual Value Of A.

Question 4. Which of the following is a wrong description of the binding energy of a nucleus?

1. It is the energy required to break a nucleus into its constituent nucleons.
2. It is the energy released when free nucleons combine to form a nucleus
3. It is the sum of the rest of the mass energies of its nucleons minus the rest of the mass energy of the nucleus
4. It is the sum of the kinetic energy of all the nucleons in the nucleus

Answer: 4. It is the sum of the kinetic energy of all the nucleons in the nucleus

Question 5. The energy of the reaction Li7 + p → 2 He4 is (the binding energy per nucleon in Li7 and He4 nuclei are 5.60 and 7.06 MeV respectively.)

1. 17.3 MeV
2. 1.73 MeV
3. 1.46 MeV
4. Depends On Binding Energy Of Proton

Answer: 1. 17.3 MeV

Question 6. Let Fpp, F on, and F nn denote the magnitudes of the nuclear force by a proton on a proton, by a proton on a neutron, and by a neutron on a neutron respectively. When the separation is 1 FM,

1. F_pp > F_pn = F _nn
2. F_pp= F_pn = F_nn
3. F_pp> F_pn > F_nn
4. F_pp< F_pn = F_nn

Answer: 2. F_pp= F_pn = F_nn

Question 7. The binding energies of two nuclei Pⁿ and Q²ⁿ and x and y joules. If 2x > y then the energy released in the reaction Pⁿ + Pⁿ → Q2ⁿ, will be

1. 2x + y
2. 2x – y
3. –(2x – y)
4. x + y

Answer: 3. –(2x – y)

Question 8. ₁H¹ + ₁H¹ + ₁H²→ X + ₁e⁰ + energy .The emitted particle is-

1. Neutron
2. Proton
3. α-particle
4. Neutrino

Answer: 3. α-particle

Question 9. In the following equation, particle X will be ₆C¹¹ → 5B¹¹ + β¹ + X

1. Neutron
2. Antineutrino
3. Neutrino
4. Proton

Answer: 3. Neutrino

Question 10. The mass of a proton is 1.0073 u and that of the neutron is 1.0087 u (u = atomic mass unit). The binding 24He energy of is (Given:- helium nucleus mass 4.0015 u)

1. 0.0305 J
2. 0.0305 erg
3. 28.4 MeV
4. 0.061 U

Answer: 3. 28.4 MeV

Question 11. The mass number of a nucleus is

1. Always Less Than Its Atomic Number
2. Always More Than Its Atomic Number
3. Sometimes Equal To Its Atomic Number
4. Sometimes Less Than And Sometimes More Than Its Atomic Number

Answer: 3. Sometimes Equal To Its Atomic Number

Question 12. For the stability of any nucleus

1. Binding Energy Per Nucleon Will Be More
2. Binding Energy Per Nucleon Will Be Less
3. Number Of Electrons Will Be More
4. None Of The Above

Answer: 1. Binding Energy Per Nucleon Will Be More

Question 13. IF Mo is the mass of an oxygen isotope 8O17, Mp and Mn are the masses of a proton and a neutron, respectively the nuclear binding energy of the isotope is

1. (M0 – 8Mp) C²
2. ( Mo – 8MP – 9Mn) C²
3. Moc²
4. (Mo – 17 Mn) C²

Answer: ( Mo – 8MP – 9Mn) C²

Question 14. If in a nuclear fusion process, the masses of the fusing nuclei are m 1 and m2 and the mass of the resultant nucleus is m 3, then

1. m₃ = |m₁ – m₂|
2. m₃ < (m₁ + m₂)
3. m₃ > (m₁ + m₂)
4. m₃ = m₁ + m₂)

Answer: 2. m₃ < (m₁ + m₂)

Question 15. M p denotes the mass of a proton and M n that of a neutron. A given nucleus, of binding energy B, contains Z protons and N neutrons. The mass M (N, Z) of the nucleus is given (c is the velocity of light)

1. M(N, Z) = NM n + ZM p + B/c²
2. M(N, Z) = NM n + ZM p – B/c²
3. M(N, Z) = NM n + ZM p + B/c²
4. M(N, Z) = NM n + ZM p – B/c²

Answer: 2. M(N, Z) = NM n + ZM p – B/c²

Question 16. In the reaction \({ }_1^2 \mathrm{H}+{ }_1^3 \mathrm{H} \rightarrow{ }_2^4 \mathrm{HE}+{ }_0^1 \mathrm{n} .\). If the binding energies of H, H and He are respectively a, b, and c (in MEV), then the energy (in MeV released in this reaction is)

1. a + b + c
2. c + a + b
3. c – (a + b)
4. a + b + c

Answer: 3. c – (a + b)

Question 17. The binding energy of deuteron is 2.2 MeV and that of He is 28 MeV. If two deuterons are fused to 4 form one 2 He then the energy released is:-

1. 25.8 MeV
2. 23.6 MeV
3. 19.2 MeV
4. 30.2 MeV

Answer: 2. 23.6 MeV

Question 18. If the binding energy per nucleon in and nuclei are 5.60 MeV and 7.06 MeV respectively, then in the reaction \(\mathrm{p}+{ }_3^7 \mathrm{Li} \rightarrow 2{ }_2^4 \mathrm{He}\) energy of proton must be :

1. 39.2 MeV
2. 28.24 MeV
3. 17.28 MeV
4. 1.46 MeV

Answer: 3. 17.28 MeV

Question 19. If Mp is the mass of an oxygen isotope 8O17, M p and MN are the masses of a proton and a neutron respectively, the nuclear binding energy of the isotope is:

1. (M_ο – 8MP)C²
2. (M_o – 8MP – 9MN)C²
3. M_oC²
4. (M_o – 17M N)C²

Answer: 2. M_oC²

Question 20. Binding energy per nucleon is of the order of –

1. 7.6 eV
2. 7.6 μeV
3. 7.6 MeV
4. 7.6 KeV

Answer: 3. 7.6 MeV

Question 21. A free neutron decays to a proton but a free proton does not decay to a neutron. This is because

1. A neutron Is A Composite Particle Made Of A Proton And An Electron Whereas Proton Is a Fundamental Particle
2. Neutron Is An Uncharged Particle Whereas Proton Is A Charged Particle
3. Neutron Has Larger Rest Mass Than The Proton
4. Weak Forces Can Operate In A Neutron But Not In A Proton.

Answer: 3. Neutron Has Larger Rest Mass Than The Proton

Question 22. M P and MN are masses of proton and neutron, respectively, at rest. If they combine to form a deuterium nucleus. The mass of the nucleus will be:

1. Less Than M_p
2. Less Than (M_p + M_n)
3. Less Than (M_p + 2m_n)
4. Greater Than (M_p + 2m_n)

Answer: 2. Less Than (M_p + M_n)

Question 23. The figure shows a plot of binding energy per nucleon (B.E/A) vs mass number (A) for nuclei. Four nuclei, P, Q, R, and S are indicated on the curve. The process that would release energy is

1. R → 2S
2. P → Q + S
3. P → 2R
4. Q → R + S

Answer: 3. P → 2R

Question 24. A positron of 1MeV collides with an electron of 1 MeV and gets annihilated and the reaction produces two-ray photons. If the effective mass of each photon is 0.0016 amu, then the energy of each ray photon is about-

1. 1.5 MeV
2. 3 MeV
3. 6 MeV
4. 2 MeV

Answer: 1. 1.5 MeV

Question 25. Masses of two isobars \({ }_{29}^{64} \mathrm{Cu} \text { and }{ }_{30}^{64} \mathrm{Zn}\) 63.9298 u and 63.9292 u respectively. It can be concluded from these data that:

1. Both the isobars are stable
2. 64Zn is radioactive, decaying to 64Cu through -decay
3. 64Cu is radioactive, decaying to 64Zn through -decay
4. 64Cu is radioactive, decaying to 64Zn through -decay

Answer: 4. 64Cu is radioactive, decaying to 64Zn through -decay

Question 26. The binding energy per nucleon vs. mass number curve for nuclei is shown in the figure. W, X, Y, and Z are four nuclei indicated on the curve. The process that would release energy is :

1. Y → 2Z
2. W → X + Z
3. W → 2Y
4. X → Y + Z

Answer: 3. W → 2Y

Chapter 4 Nuclear Physics Multiple Choice Questions Section (C): Radioactive Decay And Displacement Law

Question 1. An α-particle is bombarded on 14N. As a result, a 17O nucleus is formed and a particle is emitted. This article is a

1. Neutron
2. Proton
3. Electron
4. Positron

Answer: 2. Proton

Question 2. A free neutron decays into a proton, an electron, and:

1. A neutrino
2. An antineutrino
3. An α-article
4. A α-particle

Answer: 2. An antineutrino

Question 3. The specific activity (per gm) of radium is near –

1. 1 Bq
2. 1 Ci
3. 3.7 × 1010 Ci
4. 1 mCi

Answer: 2. 3.7 × 1010 Ci

Question 4. When a β¯ -particle is emitted from a nucleus, the neutron-proton ratio :

1. Is Decreased
2. Is Increased
3. Remains The Same
4. First Then (2)

Answer: 1. Is Decreased

Question 5. In one α and 2β-emissions:

1. Mass Number Reduces By 2
2. Mass Number Reduces By 6
3. Atomic Number Reduces By 2
4. Atomic Number Remains Unchanged

Answer: 4. Atomic Number Remains Unchanged

Question 6. Which ray contains (+Ve) charge particle :

1. α-rays
2. β-rays
3. γ-rays
4. X-rays

Answer: 1. α-rays

Question 7. Which of the following cannot be bombarded to disintegrate a nucleus –

1. α-ray
2. γ-ray
3. β-ray
4. Laser

Answer: 4. Laser

Question 8. Which of the following is a correct statement?

1. Beta Rays Are the Same As Cathode Rays.
2. Gamma Rays Are High Energy Neutrons.
3. Alpha Particles Are Singly-Ionized Helium Atoms.
4. Protons And Neutrons Have Exactly The Same Mass.

Answer: 1. Beta Rays Are Same As Cathode Rays.

Question 9. A deutron is bombarded on a ₇O¹⁶ nucleus then α-particle is emitted then the product nucleus is :

1. ₇N¹²
2. ₅B¹⁰
3. ₄Be⁹
4. ₇N¹⁴

Answer: 4. ₇N¹⁴

Question 10. An α– particle is bombarded on N¹⁴ As. a result, an O¹⁷ -nucleus is formed and a particle X is emitted. The particle X is

1. Neutron
2. Proton
3. Electron
4. Positron

Answer: 2. Proton

Question 11. In the reaction \({ }_{92} X^{234} \longrightarrow_{87} Y^{222}\) How many α-particles and β-particles are emitted?

1. 3 and 5
2. 5 and 3
3. 3 and 3
4. 3 and 1

Answer: 4. 3 and 1

Question 12. Which of the following radiations has the least wavelength?

1. γ-rays
2. β-rays
3. α-rays
4. X-rays

Answer: 1. γ-rays

Question 13. When U 238 nucleus originally at rest, decays by emitting an alpha particle having a speed u, the recoil speed of the residual nucleus is

1. \(\frac{4 u}{238}\)
2. \(-\frac{4 u}{234}\)
3. \(\frac{4 u}{234}\)
4. \(-\frac{4 \mathrm{u}}{238}\)

Answer: 3. \(\frac{4 u}{234}\)

Question 14. A nucleus with Z = 92 emits the following in a sequence : α, α, β–, β–, α, α, α, α; β–, β–, α. The Z of the resulting nucleus is:

1. 76
2. 78
3. 82
4. 74

Answer: 2. 78

Question 15. A nuclear reaction given by \(X^A \rightarrow_{z+1} Y^A+{ }_{-1} \mathrm{e}^0+\bar{v}\)

1. β-decay
2. γ-decay
3. fusion
4. fission

Answer: 1. β-decay

Question 16. In the radioactive decay process, the negatively charged emitted  – particles are

1. The Electrons Present Inside The Nucleus
2. The Electrons Produced Inside As A Result Of The Decay Of Neutrons Inside The Nucleus
3. The Electrons Produced As A Result Of Collisions Between Atoms
4. The Electrons Orbiting Around The Nucleus

Answer: 2. The Electrons Produced Inside As A Result Of The Decay Of Neutrons Inside The Nucleus

Question 17. Ub the disintegration series \(\underset{92}{238} \mathrm{U}{\alpha} X \xrightarrow{\beta^{-}} Y \underset{Z}{A}\) the values of Z and A respectively will be

1. 92,236
2. 88,230
3. 90,234
4. 31,234

Answer: 1. 92,236

Question 18. A nucleus represented by the symbol has:-

1. Z protons and A – Z neutrons
2. Z protons and A neutrons
3. A protons and Z – A neutrons
4. Z protons and A – Z PROTONS

Answer: 3. A protons and Z – A neutrons

Question 19. In the radioactive decay process, the negatively charged emitted -particles are:

1. The electrons present inside the nucleus
2. The electrons produced as a result of the decay of neutrons inside the nucleus
3. The electrons produced as a result of collisions between atoms
4. The electrons orbiting around the nucleus

Answer: 3. The electrons produced as a result of collisions between atoms

Question 20. A nuclear transformation is denoted by \(\mathrm{X}(\mathrm{n}, \alpha) \rightarrow{ }_3^7 \mathrm{Li}\). Which of the following is the nucleus of element X?

1. \({ }_6^{12} \mathrm{C}\)
2. \({ }_5^{10} \mathrm{~B}\)
3. \({ }_5^9 \mathrm{~B}\)
4. \({ }_4^{11} \mathrm{Be}\)

Answer: 2. \({ }_5^{10} \mathrm{~B}\)

Question 21. When ₃Li⁷ nuclei are bombarded by protons, and the resultant nuclei are 4Be⁸, the emitted particles will be

1. Neutrons
2. Alpha Particles
3. Beta Particles
4. Gamma Photons

Answer: 4. Gamma Photons

Question 22. The ‘rad’ is the correct unit used to report the measurement of

1. The Rate Of Decay Of Radioactive Source
2. The Ability Of A Beam Of Gamma Ray Photons To Produce Ions In A Target
3. The Energy Delivered By Radiation To A Target.
4. The Biological Effect Of Radiation

Answer: 4. The Biological Effect Of Radiation

Question 23. In gamma-ray emission from a nucleus:

1. Both The Neutron Number And The Proton Number Change
2. There Is No Change In The Proton Number And The Neutron Number
3. Only The Neutron Number Changes
4. Only The Proton Number Changes

Answer: 2. There Is No Change In The Proton Number And The Neutron Number

Question 24. Bombardment of a neutron \({ }_0 \mathrm{n}^1+{ }_5 \mathrm{~B}^{10} \rightarrow{ }_2 \mathrm{He}^4+\mathrm{x}\) on boron, forms a nucleus x with emission of  particle Nuclear x is –

1. ₆C¹²
2. ₃Li⁶
3. ₃Li⁷
4. ₄Be⁹

Answer: 3. ₃Li⁷

Question 25. 22Ne nucleus, after absorbing energy, decays into two -particles and an unknown nucleus. The unknown nucleus is:

1. Nitrogen
2. Carbon&14
3. Boron&12
4. Oxygen

Answer: 2. Carbon&14

Question 26. Consider a sample of a pure beta-active material

1. All the beta particles emitted have the same energy
2. The beta particles originally exist inside the nucleus and are ejected at the time of beta decay
3. The antineutrino emitted in a beta decay has zero rest mass and hence zero momentum.
4. The active nucleus changes to one of its isobars after the beta decay

Answer: 4. The active nucleus changes to one of its isobars after the beta decay

Question 27. \(\mathrm{X}+\mathrm{n} \rightarrow \alpha+{ }_3 \mathrm{Li}^7\) then X will be :-

1. \({ }_5^{10} \mathrm{~B}\)
2. \({ }_5^9 \mathrm{~B}\)
3. \({ }_4^{11} \mathrm{~B}\)
4. \({ }_2^4 \mathrm{He}\)

Answer: 1. \({ }_5^{10} \mathrm{~B}\)

Question 28. M n and M p represent the mass of neutron and proton respectively An element having mass M has N neutron and Z-protons, then the correct relation will be:-

1. M < {N.mn + Z.Mp}
2. M > {N.mn + Z.M p}
3. M = {N.mn + Z.M p}
4. M = N {.mn + Mp}

Answer: 1. M < {N.mn + Z.Mp}

Question 29. In the nucleus of an atom, neutrons are in excess, then emitted particles are:

1. Neutron
2. Electron
3. Proton
4. Positron

Answer: 2. Electron

Question 30. A nuclei X with mass number A and charge number Z disintegrates into one α-particle and one β-particle. The resulting R has atomic mass and atomic number, equal to:

1. (A – Z) and (Z – 1)
2. (A – Z) and (Z – 2)
3. (A – and (A – 2)
4. (A – and (Z – 1)

Answer: 4. (A – and (Z – 1)

Question 31. In gamma-ray emission from a nucleus

1. Both The Neutron Number And The Proton Number Change
2. There Is No Change In The Proton Number And The Neutron Number
3. Only The Neutron Number Changes
4. Only The Proton Number Changes

Answer: 1. Both The Neutron Number And The Proton Number Change

Question 32. Which one of the following is a possible nuclear reaction?

1. \({ }_5^{10} \mathrm{~B}+{ }_2^4 \mathrm{He} \longrightarrow{ }_7^{13} \mathrm{~N}+{ }_1^1 \mathrm{H}\)
2. \({ }_{11}^{23} \mathrm{Na}+{ }_1^1 \mathrm{H} \longrightarrow{ }_{10}^{20} \mathrm{Ne}+{ }_2^4 \mathrm{He}\)
3. \({ }_{93}^{239} \mathrm{~Np} \longrightarrow{ }_{94}^{239} \mathrm{pu}+\mathrm{B}^{-}+\overline{\mathrm{v}}\)
4. \({ }_7^{11} \mathrm{~N}+{ }_1^1 \mathrm{H} \longrightarrow{ }_6^{12} \mathrm{C}+\mathrm{B}^{-}+\overline{\mathrm{v}}\)

Answer: 3. \({ }_{93}^{239} \mathrm{~Np} \longrightarrow{ }_{94}^{239} \mathrm{pu}+\mathrm{B}^{-}+\overline{\mathrm{v}}\)

Question 33. In an α-decay, the Kinetic energy of the Q particle is 48 MeV and the Q-value of the reaction is 50 MeV. The mass number of the mother nucleus is:- (Assume that the daughter nucleus is in the ground state)

1. 96
2. 100
3. 104
4. None of these

Answer: 2. 100

Question 34. Protons and singly ionized atoms of U²³⁵ & U²³⁸ are passed in turn (which means one after the other and not at the same time) through a velocity selector and then enter a uniform magnetic field. The protons describe semicircles of radius 10 mm. The separation between the ions of U235 and U238 after describing the semicircle is given by

1. 60mm
2. 30mm
3. 2350mm
4. 2380mm

Answer: 1. 60mm

Question 35. Which of the following processes represents a gamma decay?

1. \({ }^A X_Z+\gamma \longrightarrow{ }^A X_{Z-1}+a+b\)
2. \({ }^A X_Z+{ }^1 n_0 \longrightarrow A-3 X_{Z-2}+c\)
3. \({ }^A X_Z \longrightarrow{ }^A X_Z+f\)
4. \({ }^A X_Z \longrightarrow{ }^A X_Z+f\)

Answer: 3. \({ }^A X_Z \longrightarrow{ }^A X_Z+f\)

Question 36. A nucleus with mass number 220 initially at rest emits an α-particle. If the Q value of the reaction is 5.5 MeV, calculate the kinetic energy of the α-particle

1. 4.4 MeV
2. 5.4 MeV
3. 5.6 MeV
4. 6.5 MeV

Answer: 2. 5.4 MeV

Chapter 4 Nuclear Physics Multiple Choice Questions Section (D): Statistical Law Of Radioactive Decay

Question 1. In one average-life

1. Half The Active Nuclei Decay
2. Less Than Half The Active Nuclei Decay
3. More Than Half The Active Nuclei Decay
4. All The Nuclei Decay

Answer: 3. More Than Half The Active Nuclei Decay

Question 2. A freshly prepared radioactive source of half-life 2h emits radiation of intensity which is 64 times the permissible safe level. The minimum time after which it would be possible to work safely with this source is –

1. 6 h
2. 12 h
3. 24 h
4. 128 h

Answer: 2. 12 h

Question 3. 10 grams of 57Co kept in an open container decays β–a particle with a half-life of 270 days. The weight of the material inside the container after 540 days will be very nearly –

1. 10 g
2. 7.5 g
3. 5 g
4. 2.5 g

Answer: 2. 7.5 g

Question 4. After a time equal to four half-lives, the amount of radioactive material remaining undecayed is –

1. 6.25 %
2. 12.50 %
3. 25.0 %
4. 50.0 %

Answer: 1. 6.25 %

Question 5. The decay constant of the parent nuclide in the Uranium series is. Then the decay constant of the stable end product of the series will be –

1. λ/238
2. λ/206
3. λ/208
4. zero

Answer: 4. zero

Question 6. The half-life of thorium (Th 23is 1.4 × 1010 years. Then the fraction of thorium atoms decaying per year is very nearly –

1. 1 × 10–11μ
2. 4.95 × 10–11
3. 0.69 × 10–11
4. 7.14 × 10–11

Answer: 1. 1 × 10–11

Question 7. The half-life of 215At is 100 μs. The time taken for the radioactivity of a sample of 215At to decay to 1/16th of its initial value is :

1. 400 μs
2. 6.3 μs
3. 40 μs
4. 300 μs

Answer: 4. 300 μs

Question 8. Two identical samples (same material and same amount) P and Q of a radioactive substance having mean life T are observed to have activities A P & AQ respectively at the time of observation. If P is older than Q, then the difference in their ages is:

1. \(\mathrm{T} \ell \mathrm{n}\left(\frac{\mathrm{A}_{\mathrm{P}}}{\mathrm{A}_{\mathrm{Q}}}\right)\)
2. \(T \ell n\left(\frac{A_Q}{A_P}\right)\)
3. \(\frac{1}{T} \ln \left(\frac{A_P}{A_Q}\right)\)
4. \(T\left(\frac{A_P}{A_Q}\right)\)

Answer: 1. \(\mathrm{T} \ell \mathrm{n}\left(\frac{\mathrm{A}_{\mathrm{P}}}{\mathrm{A}_{\mathrm{Q}}}\right)\)

Question 9. Two radioactive sources A and B initially contain an equal number of radioactive atoms. Source A has a half-life of 1 hour and source B has a half-life of 2 hours. At the end of 2 hours, the ratio of the rate of disintegration of A to that of B is:

1. 1: 2
2. 2: 1
3. 1: 1
4. 1: 4

Answer: 4. 1: 4

Question 10. If 10% of a radioactive material decays in 5 days then the amount of original material left after 15 days is about –

1. 65%
2. 73%
3. 70%
4. 63%

Answer: 2. 73%

Question 11. The half-life of radioactive Polonium (Po) is 138.6 days. For ten lakh Polonium atoms, the number of disintegrations in 24 hours is –

1. 2000
2. 3000
3. 4000
4. 5000

Answer: 4. 5000

Question 12. Half-lives of two radioactive substances A and B are respectively 20 min and 40 min. Initially, the samples of A and B have an equal number of nuclei. After 80 min the ratio of the remaining number of A and B nuclei is:

1. 1: 16
2. 4: 1
3. 1: 4
4. 1: 1

Answer: 1. 1: 16

Question 13. A nucleus n Xm emits one and two particles. The resulting nucleus is :

1. Nx^m–4
2. N –2 Y^{m – 4}
3. N – 4 Z ^{m – 4}
4. None Of These

Answer: 3. N – 4 Z ^{m – 4}

Question 14. The half-life of a radioactive element is 12.5 Hours and its quantity is 256 gm. After how much time its \quantity will remain 1 gm:-

1. 50 Hrs
2. 100 Hrs
3. 150 Hrs
4. 200 Hrs

Answer: 2. 100 Hrs

Question 15. If the half-life of a substance is 38 days and its quantity is 10.38 g. The quantity remaining after 19 days will be:

1. 0.151g
2. 7.0 g
3. 0.51g
4. 0.16 g

Answer: 1. 0.151g

Question 16. Remaining quantity (in %) of the radioactive element after 5 half-lives:

1. 4.125%
2. 3.125%
3. 31.1%
4. 42.125%

Answer: 2. 3.125%

Question 17. When neutrons are bvaomardexd on ⎯⎯→ 5B10 then +on1 X + 2He4, X is :

1. 3Li7
2. 3Li6
3. 5Be8
4. 2Li7

Answer: 1. 3Li7

Question 18. A sample of radioactive element containing 4 × 1016 active nuclei. The half-life of the element is 10 days, then the number of decayed nuclei after 30 days:-

1. 0.5 × 1016
2. 2 × 1016
3. 3.5 × 1016
4. 1 × 1016

Answer: 3. 3.5 × 1016

Question 19. A sample of radioactive element has a mass of 10 gm at an instant t = 0. The approximate mass of this element in the sample after two mean lives is:_

1. 1.35 gm
2. 2.50 gm
3. 3.70 gm
4. 6.30 gm

Answer: 1. 1.35 gm

Question 20. The decay constant of a radioactive substance is. The life and mean life of substance are respectively given by

1. \(\frac{1}{\lambda} \text { and } \frac{\log _e 2}{\lambda}\)
2. \(\frac{\log _e 2}{\lambda} \text { and } \frac{1}{\lambda}\)
3. \(\frac{\lambda}{\log _e 2} \text { and } \frac{1}{\lambda}\)
4. \(\frac{\lambda}{\log _e 2} \text { and } 2 \lambda\)

Answer: 2. \(\frac{\log _e 2}{\lambda} \text { and } \frac{1}{\lambda}\)

Question 21. The half-life of a certain radioactive substance is 12 days. The time taken for 8th of the sample to decay is

1. 36 days
2. 12 days
3. 4 days
4. 24 days

Answer: 1. 36 days

Question 22. If N 0 is the original mass of the substance of half-life period tl/2 = 5 years, then the amount of substance left after 15 years is:

1. N 0 / 8
2. N 0 / 16
3. N 0 / 2
4. N 0 / 4

Answer: 1. N 0 / 8

Question 23. The atomic bomb was first made by

1. Otto Hahn
2. Fermi
3. Oppenheimer
4. Taylor

Answer: 1. Otto has

Question 24. A radioactive element has a half-life of 3.6 days. At what time will it be left 1/32nd undecayed?

1. 4 days
2. 12 days
3. 18 days
4. 24 days

Answer: 3. 18 days

Question 25. If a sample of 16 g radioactive substance disintegrates to 1g in 120 days, then what will be the half-life of the sample?

1. 15 days
2. 7.5 days
3. 30 days
4. 60 days

Answer: 3. 30 days

Question 26. n α-particles per second are being emitted by N atoms of a radioactive element. The half-life of element will be

1. \(\left(\frac{\mathrm{n}}{\mathrm{N}}\right) \mathrm{s}\)
2. \(\left(\frac{N}{n}\right) s\)
3. \(\frac{0.693 \mathrm{~N}}{n} s\)
4. \(\frac{0.693 n}{N} s\)

Answer: 3. \(\frac{0.693 \mathrm{~N}}{n} s\)

Question 27. In a sample of radioactive material, what percentage of the initial number of active nuclei will decay during one mean life?

1. 37%
2. 50%
3. 63%
4. 69.3%

Answer: 3. 63%

Question 28. If the half-life of any sample of a radioactive substance is 4 days, then the fraction of the sample will remain undecayed after 2 days and will be

1. \(\sqrt{2}\)
2. \(\frac{1}{\sqrt{2}}\)
3. \(\frac{\sqrt{2}-1}{\sqrt{2}}\)
4. \(\frac{1}{2}\)

Answer: 2. \(\frac{1}{\sqrt{2}}\)

Question 29. A nucleus with Z = 92 emits the following in a sequence: α β–, β –, α α α β-; β–, β–, α, α+, β+, α: The Z of the resulting nucleus is

1. 76
2. 80
3. 82
4. 74

Answer: 2. 80

Question 30. If a radioactive substance decays \(\frac{1}{16} \text { th }\) of its original amount in 2 h, then the half-life of that substance is

1. 15 min
2. 30 min
3. 45 min
4. None Of These

Answer: 2. 30 min

Question 31. If N 0 is the original mass of the substance of half-life period T1/2 = 5 ye, then the amount of substance left after 15 yr is

1. N 0 /8
2. N 0/16
3. N 0/2
4. N 0/4

Answer: 1. N 0 /8

Question 32. The half-life of radium is about 1600 years. Of 100g of radium existing now, 25g will remain undocked after:-

1. 6400 years
2. 2400 years
3. 3200 years
4. 4800 years

Answer: 3. 3200 years

Question 33. In a radioactive material the activity at time t1 is R1 and at a later time t2, it is R2. If the decay constant of the material is , then

1. R₁ = R₂ e–λ(t₁–t₂)
2. R₁ = R₂ eλ(t₁–t₂)
3. R₁ = R₂ (t₁ / t₂)
4. R₁ = R₂

Answer: 1.R₁ = R₂ e–λ(t₁–t₂)

Question 34. The atomic bomb was first made by

1. Otto Hahn
2. Fermi
3. Oppenheimer
4. Taylor

Answer: 1. Otto hahn

Question 35. Two radioactive materials X₁ and X₂ have decay constants 5λ and λ  respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of X₁ to that of X₂ will be e after a time

1. \(\frac{1}{2} \lambda\)
2. \(\frac{1}{4 \lambda}\)
3. \(\frac{e}{\lambda}\)

Answer: 3. \(\frac{e}{\lambda}\)

Question 36. Starting with a sample of pure 66Cu, 7/8 of it decays into Zn in 15 minutes. The corresponding half-life is:

1. 10 minute
2. 15 minute
3. 5 minute
4. 7 minute

Answer: 3. 5 minute

Question 37. The activity of the Po sample is 5 millicuries. Half-life of Po is 138 days, what amount of Po was initially taken? (Avogadro’s no. = 6.02×1026 Per k mole)

1. 3.18 × 1015 atoms
2. 3.18 × 1013 atoms
3. 3.18 × 1016 atoms
4. 3.18 × 1014 atoms

Answer: 1. 3.18 × 1015 atoms

Question 38. The half-life period of a radioactive element X is the same as the mean-life time of another radioactive element Y. Initially both of them have the same number of atoms. Then

1. X and Y have the same decay rate initially
2. X and Y decay at the same rate always
3. Y will decay at a faster rate than X
4. X will decay at a faster rate than Y

Answer: 3. Y will decay at a faster rate than X

Question 39. A radioactive element ThA (₈₄Po²¹⁶) can undergo α and βis a type of disintegration with half-lives, T1 and T 2 respectively. Then the half-life of ThA is

1. T1 + T2
2. T1 T2
3. T1 – T2
4. \(\frac{\mathrm{T}_1 \mathrm{~T}_2}{\mathrm{~T}_1+\mathrm{T}_2}\)

Answer: 4. \(\frac{\mathrm{T}_1 \mathrm{~T}_2}{\mathrm{~T}_1+\mathrm{T}_2}\)

Question 40. The mean lives of radioactive substances are 1620 years and 405 years for -emission and -emission respectively. Then the time in which three-fourths of a sample will decay is –

1. 224 years
2. 324 years
3. 449 years
4. 810 years

Answer: 3. 449 years

Question 41. Two radioactive materials X 1 and X2 have decay constants 10 and  respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of X 1 to that of X2 will be 1/e after a time.

1. 1/(10)
2. 1/(11)
3. 11/(10)
4. 1/(9)

Answer: 4. 1/(10)

Question 42. A sample of radioactive material has mass m, decay constant, and molecular weight M. Avogadro constant = N The initial activity of the sample is

1. \(\frac{\lambda \mathrm{m}}{\mathrm{M}}\)
2. \(\frac{\lambda \mathrm{mN}_{\mathrm{A}}}{\mathrm{M}}\)
3. \(m N_A e^\lambda\)

Answer: 3. \(m N_A e^\lambda\)

Question 43. The activity of a radioactive element is 103 dis/sec. Its half-life is 1 sec. After 3 sec. its activity will be :

1. 1000 dis/sec
2. 250 dis/sec
3. 125 dis/sec
4. none of these

Answer: 3. 125 dis/sec

Question 44. A 280-day-old sample of a radioactive substance has the activity of 6000 DPS. In the next 140 days, activity falls to 3000 dps. The initial activity of the sample would have been

1. 9000
2. 24000
3. 12,000
4. 18,000

Answer: 2. 24000

Question 45. A radioactive sample at any instant has a disintegration rate of 5000 disintegrations per minute. After
5 min, the rate is 1250 disintegrations per min. then the decay constant (per – minute) is

1. 0.4 In 2
2. 0.2 In2
3. 0.1 In 2
4. 0.8 In2

Answer: 4. 0.8 In2

Question 46. A radioactive sample consists of two distinct species having an equal number of atoms initially. The mean lifetime of one species is and that of the other is 5. The decay products in both cases are stable. A plot is made of the total number of radioactive nuclei as a function of time. Which of the following figures best represents the form of this plot?

Question 47. A 280-day-old sample of a radioactive substance has an activity of 6000 DPS. In the next 140 days, activity falls to 3000 dps. The initial activity of the sample would have been

1. 9000
2. 24000
3. 12,000
4. 18,000

Answer: 1. 9000

Question 48. A radioactive sample at any instant has a disintegration rate of 5000 disintegrations per minute. After 5 min, the rate is 1250 disintegrations per minute. then the decay constant (per – minute) is

1. 0.4 In 2
2. 0.2 In2
3. 0.1 In 2
4. 0.8 In2

Answer: 1. 0.4 In 2

Question 49. Radioactivity is –

1. Irreversible Process
2. Spontaneous Disintegration Process
3. Not Effected By Temperature Or Pressure
4. Process Obeying All Of The Above

Answer: 4. Process Obeying All Of The Above

Chapter 4 Nuclear Physics Multiple Choice Questions Section (E): Nuclear Fission And Fusion

Question 1. If the mass of the fissionable material is less than the critical mass, then

1. Fission And Chain Reactions Both Are Impossible
2. Fission Is Possible But Chain Reaction Is Impossible
3. Fission Is Impossible But Chain Reaction Is Possible
4. Fission And Chain Reaction Are Possible.

Answer: 2. Fission Is Possible But Chain Reaction Is Impossible

Question 2. Which of the following materials is used for controlling the fission

1. Heavy Water
2. Graphite
3. Cadmium
4. Beryllium Oxide

Answer: 3. Cadmium

Question 3. The atomic reactor is based on

1. Controlled Chain Reaction
2. Uncontrolled Chain Reaction
3. Nuclear Fission
4. Nuclear Fusion

Answer: 1. Controlled Chain Reaction

Question 4. Thermal neutron means

1. Neutron Being Heated
2. The Energy Of These Neutrons Is Equal To The Energy Of Neutrons in a heated atom
3. These neutrons have the Energy Of A Neutron In A Nucleus At a normal temperature
4. Such Neutrons Gather Energy Released In The Fission Process

Answer: 3. These neutrons have the Energy Of A Neutron In A Nucleus At a normal temperature

Question 5. \({ }_{92} U^{235}\) nucleus ab sorbs a slow neutron and undergoes fission into 54X139 and 38Sr94 nuclei.The other particles produced in this fission process are

1. 1β and 1α
2. 2β and 1 neutron
3. 2 Neutrons
4. 3 Neutrons

Answer: 4. 3 Neutrons

Question 6. Two lithium 6Li nuclei in a lithium vapor at room temperature do not combine to form a carbon 12C nucleus because

1. A Lithium Nucleus Is More Tightly Bound Than A Carbon Nucleus
2. The Carbon Nucleus Is An Unstable Particle
3. It Is Not Energetically Favourable
4. Coulomb Repulsion Does Not Allow The Nuclei To Come Very Close

Answer: 4. Coulomb Repulsion Does Not Allow The Nuclei To Come Very Close

Question 7. Choose the true statement.

1. The energy released per unit mass is more in fission than in fusion
2. The energy released per atom is more in fusion than in fission.
3. The energy released per unit mass is more in fusion and that per atom is more in fission.
4. Both fission and fusion produce the same amount of energy per atom as well as per unit mass.

Answer: The energy released per unit mass is more in fusion and that per atom is more in fission

Question 8. A fusion reaction is possible at high temperatures because –

1. Atoms Are Ionised At High Temperature
2. Molecules Break-Up At High Temperature
3. Nuclei Break-Up At High Temperature
4. Kinetic Energy Is High Enough To Overcome Repulsion Between Nuclei.

Answer: 4. Kinetic Energy Is High Enough To Overcome Repulsion Between Nuclei.

Question 9. In a uranium reactor whose thermal power is P = 100 MW, if the average number of neutrons liberated in each nuclear splitting is 2.5. Each splitting is assumed to release an energy E = 200 MeV. The number of neutrons generated per unit of time is –

1. 4 × 1018 s–1
2. 8 × 1023 s–1
3. 8 × 1019 s–1
4. 16 × 1018 s–1

Answer: 4. 16 × 1018 s–1

Question 10. Assume that the nuclear binding energy per nucleon (B/A) versus mass number (A) is as shown in the figure. Use this plot to choose the correct choice given below.

1. Fusion Of Two Nuclei With Mass Numbers Lying In The Range Of 1 < A < 50 Will Release Energy
2. Fusion Of Two Nuclei With Mass Numbers Lying In The Range Of 51 < A < 100 Will Release Energy
3. Fission Of A Nucleus Lying In The Mass Range Of 100 < A < 200 Will Release Energy When Broken Into Two Equal Fragments
4. Both and (2)

Answer: 3. Fission Of A Nucleus Lying In The Mass Range Of 100 < A < 200 Will Release Energy When Broken Into Two Equal Fragments

Question 11. A fission reaction is given by \({ }_{92}^{236} \mathrm{U} \rightarrow{ }_{54}^{140} \mathrm{Xe}+{ }_{38}^{94} \mathrm{Sr}+\mathrm{x}+\mathrm{y}\) where x and y are two particle considering \({ }_{92}^{236} U\) to be at rest, the kinetic energies of the products are denoted by KXe, KSr, Kx (2MeV) and ky (2mev), respectively. let the binding energies per nucleon of ,\({ }_{92}^{236} \mathrm{U},{ }_{54}^{140} \mathrm{Xe}\) and \({ }_{38}^{94} \mathrm{Sr}\) MeV, 8.5 MeV and 8.5 MeV, respectively. Considering different conservation laws, the correct option(s)
is(are)

1. \(x=n, y=n, K_{s r}=129 \mathrm{MeV}, K_{\mathrm{xe}}=86 \mathrm{MeV}\)
2. \(x=p, y=e-, K_{s t}=129 \mathrm{MeV}, \mathrm{K}_{\mathrm{xe}}=86 \mathrm{MeV}\)
3. \(x=p, y=n, K_{s r}=129 \mathrm{MeV}, K_{x_e}=86 \mathrm{MeV}\)
4. \(x=n, y=n, K_{s r}=86 \mathrm{MeV}, K_{x e}=129 \mathrm{MeV}\)

Answer: 1. \(x=n, y=n, K_{s r}=129 \mathrm{MeV}, K_{\mathrm{xe}}=86 \mathrm{MeV}\)

Question 12. In a fission reaction \({ }_{92}^{236} \mathrm{U} \longrightarrow{ }^{117} \mathrm{X}+{ }^{117} \mathrm{Y}+n+n\) the average binding energy per nucleon of X and Y is 8.5 MeV whereas that of 236U is 7.6 MeV. The
total energy liberated will be about \({ }_{92}^{236} \mathrm{U} \longrightarrow{ }^{117} \mathrm{X}+{ }^{117} \mathrm{Y}+n+n\)

1. 200ev
2. 2 me v
3. 200 me v
4. 20000 meV

Answer: 3. 200 me v

Question 13. Energy is released in nuclear fission due to

1. Few mass is converted into energy
2. The total binding energy of fragments is more than the B. E. of the parental element
3. The total B.E. of fragments is less than the B.E. of parental element
4. The total B.E. of fragments is equal to the B.E. of the parental elements is

Answer: 2. Total B.E. of fragments is less than the B.E. of parental element

Question 14. Boron rods in nuclear reactors are used as a:

1. Moderator
2. Control Rods
3. Coolant
4. Protective Shield

Answer: 2. Control Rods

Question 15. 200 Me V energy is obtained by fission of 1 nuclei of 92U 235, to obtain 1 kW energy number of fission per second will be:

1. 3.215 × 1013
2. 3.215 × 1014
3. 3.215 × 1015
4. 3.215 × 1016

Answer: 1. 3.215 × 1013

Question 16. The best moderator for neutrons is –

1. Beryllium Oxide
2. Pure Water
3. Heavy Water
4. Graphite

Answer: 3. Heavy Water

Question 17. Which of the following are suitable for the fusion process:-

1. Light nuclei
2. heavy nuclei
3. Element must be lying in the middle of the periodic table
4. Middle elements, which are lying on the binding energy curve

Answer: 1. Light nuclei

Question 18. Solar energy is mainly caused due to:-

1. Burning of hydrogen in the oxygen
2. Fusion of uranium present in the sun
3. Fusion of protons during the synthesis of heavier elements
4. Gravitational contraction

Answer: 3. Fusion of protons during synthesis of heavier elements

Question 19. Which one of the following acts as a neutron absorber in a nuclear reactor?

1. Cd-rod
2. Heavy water (D2O)
3. Graphite
4. Distilled water (H2O)

Answer: 1. Cd-rod

Question 20. The functions of mediators in nuclear reactors are:

1. Decrease The Speed Of Neutrons
2. Increase The Speed Of Neutrons
3. Decrease The Speed Of Electrons
4. Decrease The Speed Of Electrons

Answer: 1. Decrease The Speed Of Neutrons

Question 21. A chain reaction in the fission of uranium is possible, because:

1. Two Intermediate Sized Nuclear Fragments Are Formed
2. Three Neutrons Are Given Out In Each Fission
3. Fragments In Fission Are Radioactive
4. Large Amount Of Energy Is Released

Answer: 2. Three Neutrons Are Given Out In Each Fission

Question 22. Nuclear fusion is common to the pair:

1. Thermonuclear Rector, Uranium-Based Nuclear Reactor
2. Energy Production In Sun, Uranium-Based Nuclear Reactor
3. Energy Production Of Heavy Nuclei Hydrogen Bomb
4. Disintegration Of Heavy Nuclei Hydrogen Bomb

Answer: 3. Energy Production Of Heavy Nuclei Hydrogen Bomb

Question 23. IN any fission process the ratio \(\frac{\text { mass of fission products }}{\text { maas of parent nucleus }}\)

1. Greater than 1
2. Depends on the mass of the parent nucleus
3. Less than 1
4. Less than 1

Answer: 3. Less than 1

Question 24. Fission of nuclei is possible because the binding energy in nucleons in them-

1. Decreases with mass number at low mass numbers
2. Increases with mass number at low mass numbers
3. Decreases with mass number at high mass numbers
4. Increases with mass number at high mass numbers

Answer: 3. Decreases with mass number at high mass numbers

Question 25. In the nuclear fusion reaction, \({ }_1^2 \mathrm{H}+{ }_1^3 \mathrm{H} \longrightarrow{ }_2^4 \mathrm{He}+\mathrm{n}\) given that the repulsive potential energy between the two nuclei is ~ 7.7 × 10 –14 J, the temperature at which the gases must be heated to initiate the reaction is nearly (Boltzmann’s constant k = 1.38 × 10 –23 J/K):

1. 107K
2. 105K
3. 103K
4. 109K

Answer: 4. 109K

Question 26. The operation of a nuclear reactor is said to be critical if the multiplication factor (k) has a value

1. 1
2. 1.5
3. 2.1
4. 2.5

Answer: 1. 1

Question 27. This question contains Statement 1 and Statement 2. Of the four choices given after the statements, choose the one that best describes the two statements.  Statement-1: Energy is released when heavy nuclei undergo fission or light nuclei undergo fusion. and Statement 2: For heavy nuclei, the binding energy per nucleon increases with increasing Z while for light nuclei it decreases with increasing Z.

1. Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1
2. Statment-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1
3. Statement-1 is true, Statement-2 is false
4. Statement-1 is false, Statement-2 is true

Answer: 3. Statement-1 is true, Statement-2 is false

Chapter 4 Nuclear Physics Multiple Choice Questions Exercise 2

Question 1. The dynamic mass of an electron having rest mass m0 and moving with speed 0.8 c is

1. 0.6 m₀
2. 0.8 m₀
3. 3 m₀
4. 1.25 m₀

Answer: 3. 3 m₀

Question 2. An alpha nucleus of energy 2 mv2 bombards a heavy closet approach for the alpha nuclieus will be proportional to

1. v²
2. 1/M
3. 1/ v⁴
4. 1 / ze

Answer: 2. 1/M

Question 3. An α-particle of energy 5 MeV is scattered through 180º by a fixed uranium nucleus. The distance of the closest approach is of the order of:

1. 1 Å
2. 10–10 cm
3. 10-12 cm
4. 10–15 cm

Answer: 3. 10-12 cm

Question 4. Helium nuclei combine to form an oxygen nucleus. The energy released in the reaction is if mO = 15.9994 amu and mHe = 4.0026 amu

1. 10.24 MeV
2. 0 MeV
3. 5.24 MeV
4. 4 MeV

Answer: 1. 10.24 MeV

Question 5. A nucleus z X has mass represented by M(A, Z). If M p and M n denote the mass of proton and neutron respectively and BE the binding energy (in MeV), then:

1. \(B E=\left[M(A, Z)-Z M_p-(A-Z) M_n\right] c^2\)
2. \(B E=\left[Z M_p+(A-Z) M_n-M(A, Z)\right] c^2\)
3. \(B E=\left[Z M_p+A M_n-M(A, Z)\right] c^2\)
4. \(B E=M(A, Z)-Z M_p-(A-Z) M_n\)

Answer: 2. \(B E=\left[Z M_p+(A-Z) M_n-M(A, Z)\right] c^2\)

Question 6. If M (A, Z), M p and M n A denote the masses of the nucleus Z X, proton, and neutron respectively in units of u (1 u = 931.5 MeV/cand BE represents its binding energy in MeV, then

1. M (A, Z) = ZM p + (A – Z) Mn – BE/c²
2. M (A, Z) = ZM p + (A – Z) Mn – BE
3. M (A, Z) = ZM p + (A – Z) Mn – BE
4. M (A, Z) = ZM p + (A – Z) Mn + BE/c2

Answer: 1. M (A, Z) = ZM p + (A – Z) Mn – BE/c²

Question 7. The binding energy per nucleon of the deuteron and helium nuclei is 1.1 MeV and 7 MeV respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is:

1. 13.9 MeV
2. 26.9 MeV
3. 23.6 MeV
4. 19.2 MeV

Answer: 3. 23.6 MeV

Question 8. ₆C¹¹ undergoes a decay by emitted β+ then writes its complete equation. Given the mass value of m(5C1= 11.011434 u, m(5B1= 11.009305 u. m e = 0.000548 u and 1 u = 931.5 MeV/c² Calculate the Q-value of reaction.

1. 0.962 Mev
2. 09.62Mev
3. 096.2MeV
4. 962.0MeV

Answer: 1. 0.962 Mev

Question 9. The atomic weight of boron is 10.81 and it has two isotopes 5B10 and 5B11 in nature would be:

1. 19: 81
2. 10: 11
3. 15: 16
4. 81: 19

Answer: 1. 19: 81

Question 10. A nucleus of mass number 332 after many disintegrations α and β radiations, decays into another nucleus whose mass number is 220 and atomic number is 86. The numbers of α and β radiations will be:

1. 4, 0
2. 3, 6
3. 3, 2
4. 2, 1

Answer: 3. 3, 2

Question 11. A radioactive sample at any instant has a disintegration rate of 5000 disintegrations per minute. After 5 minutes, the rate is 1250 disintegrations per minute. Then, the decay constant (per minute) is :

1. 0.4 ln 2
2. 0.2 ln 2
3. 0.1 ln 2
4. 0.8 ln 2

Answer: 1. 0.4 ln 2

Question 12. At time t = 0, some radioactive gas is injected into a sealed vessel. At time T, some more of the same gas is injected into the same vessel. Which one of the following graphs best represents the variation of the logarithm of activity A of the gas with time t?

Answer: 2.

Question 13. N atoms of a radioactive element emit n alpha particles per second at an instant. Then the half-life of the element is

1. \(\frac{\mathrm{n}}{\mathrm{N}} \mathrm{sec} \text {. }\)
2. \(1.44 \frac{\mathrm{n}}{\mathrm{N}} \mathrm{sec} .\)
3. \(0.69 \frac{\mathrm{n}}{\mathrm{N}} \text { sec. }\)
4. \(0.69 \frac{N}{n} \text { sec. }\)

Answer: 4. \(0.69 \frac{N}{n} \text { sec. }\)

Question 14. Two isotopes P and Q of atomic weight 10 and 20, respectively are mixed in equal amounts by weight. After 20 days their weight ratio is found to be 1: 4. Isotope P has a half-life of 10 days. The half-life of isotope Q is

1. Zero
2. 5 days
3. 20 days
4. Infinite

Answer: 4. Infinite

Question 15. The half-life of a radioactive substance is 4 days. Its 100 g is kept for 16 days. After this period, the amount of substance that remained was:

1. 25 g
2. 15 g
3. 10 g
4. 6.25 g

Answer: 4. 6.25 g

Question 16. Two radioactive substances A and B have decay constants 5 and respectively. At = 0 they have the same number of nuclei. The ratio of number of nuclei of A to those of B will be \(\left(\frac{1}{e}\right)^2\) after a time interval:

1. \(\frac{1}{4 \lambda}\)
2. 4
3. 2
4. \(\frac{1}{2 \lambda}\)

Answer: 4. \(\frac{1}{2 \lambda}\)

Question 17. The half-life period of a radio-active element X is the same as the mean lifetime of another radio-active element Y. Initially they have the same number of atoms. Then:

1. X will decay faster than Y
2. Y will decay faster than X
3. X and Y have the same decay rate initially
4. X and Y decay at the same rate always

Answer: 2. Y will decay faster than X

Question 18. How much uranium is required per day in a nuclear reactor of power capacity of 1 MW

1. 15 mg
2. 1.05 gm
3. 105 gm
4. 10.5 kg

Answer: 2. 1.05 gm

Question 19. Complete the equation for the following fission process \({ }_{92} \mathrm{U}^{235}+{ }_0 \mathrm{n}^1 \longrightarrow{ }_{38} \mathrm{Sr}^{90}+\ldots \ldots.\)

1. \({ }_{54} X^{143}+3{ }_0 n\)
2. \({ }_{54} X e^{145}\)
3. \({ }_{57} \mathrm{Xe}^{142}\)
4. \({ }_{54} \mathrm{Xe}^{142}+{ }_0 \mathrm{n}^1\)

Answer: 1. \({ }_{54} X^{143}+3{ }_0 n\)

Chapter 4 Nuclear Physics Multiple Choice Questions Part – 1: Neet / Aipmt Question (Previous Years)

Question 1. In the nuclear decay given below \({ }_Z^A X \longrightarrow{ }_{Z+1}^A Y \longrightarrow{ }_{Z-1}^{A-4} B^{\circ} \longrightarrow{ }_{Z-1}^{A-4} B \text {, }\)  the particles emitted in the sequence are

1. β,α, ϒ
2. ϒ,β,α
3. β,γ,α
4. α,β,γ

Answer: 1. β,α, ϒ

Question 2. The number of beta particles emitted by a radioactive substance is twice the number of alpha particles emitted by it. The resulting daughter is an

1. Isobar Of Parent
2. Isomar Of Parent
3. Isotone Of Parent
4. Isotope Of Parent

Answer: 4. Isotope Of Parent

Question 3. A radioactive nucleus X converts into to stable nucleus Y. Half-life of X is 50 yr. Calculate the age of the radioactive sample when the radio of X and Y is 1:15.

1. 200 yr
2. 350 yr
3. 150 yr
4. 250 yr

Answer: 1. 200 yr

Question 4. The mass of a 3Li nucleus is 0.042 u less than the sum of the masses of all its nucleons. The binding 73Li energy per nucleon of a nucleus is nearly

1. 46 MeV
2. 5.6 MeV
3. 3.9 MeV
4. 23 MeV

Answer: 2. 5.6 MeV

Question 5. The activity of a radioactive sample is measured as N₀ counts per minute at t = 0 and N₀/e counts per minute at t = 5 minutes. The time (in minutes) at which the activity reduces to half its value is

1. \(\log _e \frac{2}{5}[latex]
2. [latex]\frac{5}{\log _e 2}\)
3. 5 log₁₀2
4. 5 log_e2

Answer: 4. 5 log₁₀2

Question 6. 2 An alpha nucleus of energy 2 bombards a heavy nuclear target of charge Ze. Then the distance of the closest approach for the alpha nucleus will be proportional to

1. \(\frac{1}{\mathrm{Ze}}\)
2. u²
3. \(\frac{1}{\mathrm{~m}}\)
4. \(\frac{1}{u^4}\)

Answer: 3. \(\frac{1}{\mathrm{~m}}\)

Question 7. The decay constant of a radioisotope is λ. If A₁ and A2 are its activities at times t₁ and t₂ respectively, the number of nuclei that have decayed during the time (t₂ – t

1. A₁t₁ – A₂t₂
2. A₁ – A₂
3. (A₁ – A₂)/λ
4. λ(A₁ – A₂)

Answer: 3. (A₁ – A₂)/λ

Question 8. The binding energy per nucleon in deuterium and helium nuclei are 1.1 MeV and 7.0 MeV, respectively. When two deuterium nuclei fuse to form a helium nucleus the energy released in the fusion is

1. 23.6 MeV
2. 2.2 MeV
3. 28.0 MeV
4. 30.2 MeV

Answer: 1. 23.6 MeV

Question 9. Two radioactive nuclei P and Q, in a given sample decay into a stable nucleolus R. At time t = 0, many P species are 4 N₀ and that of Q are N₀. The half-life of P (for conversion to R) is 1 minute whereas that of Q is 2 minutes. Initially, there are no nuclei of R present in the sample. When the number of nuclei of P and Q are equal, the number of nuclei of R present in the sample would be –

1. 9N₀
2. 3N₀
3. 2 5N₀
4. 22N₀

Answer: 2. 3N₀

Question 10. The half life of a radioactive isotope ‘X’ is 50 years. It decays to another element ‘Y’ which is stable. The two elements ‘X’ and ‘Y’ were found to be in the ratio of 1: 15 in a sample of a given rock. The age of the rock was estimated to be:

1. 150 years
2. 200 years
3. 250 years
4. 100 years

Answer: 2. 200 years

Question 11. The power obtained in a reactor using U 235 disintegration is 1000 kW. The mass decay of U 235 per hour is:

1. 10 microgram
2. 20 microgram
3. 40 microgram
4. 1 microgram

Answer: 3. 40 microgram

Question 12. A radioactive nucleus of mass M emits a photon of frequency  and the nucleus recoils. The recoil energy will be:

1. Mc² – hν
2. h²v² / 2Mc²
3. zero
4. hν

Answer: 2. h²v² / 2Mc²

Question 13. A nucleus \({ }_n^m x\) emits one –particle and two – particles. The resulting nucleus is :

1. \({ }_{n-4}^{m-6} Z\)
2. \(\mathrm{m}_{\mathrm{m}}-6\)
3. \({ }_n^{m-4} X\)
4. \({ }_{n-2}^{m-4} Y\)

Answer: 3. \({ }_n^{m-4} X\)

Question 14. Fusion reaction takes place at high temperatures because:

1. Nuclei Break Up At High Temperature
2. Atoms Get Ionised At High Temperature
3. Kinetic Energy Is High Enough To Overcome The Coulomb Repulsion Between Nuclei
4. Molecules Break Up At High Temperature

Answer: 3. Kinetic Energy Is High Enough To Overcome The Coulomb Repulsion Between Nuclei

Question 15. If the nuclear radius of 27 Al is 3.6 Fermi, the approximate nuclear radius of 64 Cu in Fermi is:

1. 2.4
2. 1.2
3. 4.8
4. 3.6

Answer: 3. 4.8

Question 16. A mixture consists of two radioactive materials A₁ and A₂ with half-lives of 20s and 10 s respectively. Initially, the mixture has 40 g of A1 and 160 g of A₂. The amount of the two in the mixture will become equal after:

1. 60 s
2. 80 s
3. 20 s
4. 40 s

Answer: 4. 40 s

Question 17. The half life of a radioactive nucleus is 50 days. The time interval (t₂ – between the time t₂ when 3 of 1 it has decayed and the time t1 when 3 of it had decayed is:

1. 30 days
2. 50 days
3. 60 days
4. 15 days

Answer: 2. 50 days

Question 18. A certain mass of Hydrogen is changed to Helium by the process of fusion. The Mass defect in the fusion reaction is 0.02866 u. The energy liberated per u is : (given 1u = 931 MeV)

1. 26.7 MeV
2. 6.675 MeV
3. 13.35MeV
4. 2.67 MeV

Answer: 2. 6.675 MeV

Question 19. The half life of a radioactive isotope ‘X’ is 20 years. It decays to another element ‘Y’ which is stable. The two elements ‘X’ and ‘Y’ were found to be in the ratio 1: 7 in a sample of a given rock. The age of the rock is estimated to be:

1. 60 years
2. 80 years
3. 100 years
4. 40 years

Answer: 1. 60 years

Question 20. The Binding energy per nucleon of\({ }_3^7 \mathrm{Li} \text { and }{ }_2^4 \mathrm{He}\)  and nucleon are 5.60 MeV and 7.06 MeV, respectively. In 7 1 4 4 the nuclear reaction \({ }_3^7 \mathrm{Li}+{ }_1^1 \mathrm{H} \rightarrow{ }_2^4 \mathrm{He}+{ }_2^4 \mathrm{He}+\mathrm{Q}\), the value of energy Q released is

1. 19.6MeV
2. –2.4 MeV
3. 8.4 MeV
4. 17.3 MeV

Answer: 4. 17.3 MeV

Question 21. A radioisotope ‘X’ with a half-life of 1.4 × 109 years decays to ‘Y’ which is stable. A sample of the rock from a cave was found to contain ‘X’ and ‘Y’ in the ratio 1: 7. The age of the rock is

1. 1.96 × 109 years
2. 3.92 × 109 years
3. 4.20 × 109 years
4. 8.40 × 109 years

Answer: 3. 4.20 × 109 years

Question 22. If radius of the \({ }_{12}^{27} \mathrm{Al}\) Al Te nucleus is taken to be \(\mathrm{R}_{\mathrm{Al}^{\prime}}\) the the radius of \({ }_{53}^{125} \mathrm{Te}\) nucleus is nearly:

1. \(\frac{5}{3} R_{A l}\)
2. \(\frac{3}{5} R_{A I}\)
3. \(\left(\frac{13}{53}\right)^{1 / 3} \mathrm{R}_{\mathrm{Al}}\)
4. \(\left(\frac{53}{13}\right)^{1 / 3} R_{\mathrm{Al}}\)

Answer: 1. \(\frac{5}{3} R_{A l}\)

Question 23. The half-life of a radioactive substance is 30 minutes. The time (in minutes) taken between 40% decay and 85% decay of the same radioactive substance is :

1. 60
2. 15
3. 30
4. 45

Answer: 1. 60

Question 24. Radioactive material ‘A’ has a decay constant of ‘8 ’ and material ‘B’ has a decay constant of ‘λ’. Initially, they have same number of nuclei. After what time, the ratio of many nuclei of material ‘B’ to that ‘A’ will be \(\frac{1}{\mathrm{e}}\)?

1. \(\frac{1}{\lambda}\)
2. \(\frac{1}{7 \lambda}\)
3. \(\frac{1}{8 \lambda}\)
4. \(\frac{1}{9 \lambda}\)

Answer: 2. \(\frac{1}{7 \lambda}\)

Question 25. For a radioactive material, the half-life is 10 minutes. If initially there are 600 number of nuclei, the time taken (in minutes) for the disintegration of 450 nuclei is:

1. 20
2. 15
3. 30
4. 10

Answer: 1. 20

Question 26. The rate of radioactive disintegration at an instant for a radioactive sample of half life 2.2 × 109 s is 1010 s–1. The number of radioactive atoms in that sample at that instant is

1. 3.17×1020
2. 3.17×1017
3. 3.17×1018
4. 3.17×1019

Answer: 4. 3.17×1019

Question 27. The total energy of an electron in an atom in an orbit is –3.4 eV. Its kinetic and potential energies are, respectively:

1. –3.4 eV, –3.4 eV
2. –3.4 eV, –6.8 eV
3. 3.4 eV, –6.8 eV
4. 3.4 eV, 3.4 eV

Answer: 3. 3.4 eV, –6.8 eV

Question 28. The radius of the first permitted Bohr orbit for the electron, in a hydrogen atom equals 0.51 Å and its− ground state energy equals –13.6 eV. If the electron in the hydrogen atom is replaced by muon [charge same as electron and mass 207 me], the first Bohr radius and ground state energy will be:

1. 0.53 × 10–13 m, –3.6 eV
2. 25.6 × 10–13 m, –2.8 eV
3. 2.56 × 10–13 m, –2.8 eV
4. 2.56 × 10–13 m, –13.6 eV

Answer: 3. 2.56 × 10–13 m, –2.8 eV

Question 29. The total energy of an electron in the nth stationary orbit of the hydrogen atom can be obtained by

1. \(E_n=\frac{13.6}{n^2} e V\)
2. \(E_n=-\frac{13.6}{n^2} e V\)
3. \(\mathrm{E}_{\mathrm{n}}=-\frac{1.36}{\mathrm{n}^2} \mathrm{eV}\)
4. En=-13.6xn2ev

Answer: 2. \(E_n=-\frac{13.6}{n^2} e V\)

Question 30. For which one of the following. Bohr model is not valid?

1. Singly Ionized Neon Atom (Ne+)
2. Hydrogen Atom
3. Singly Ionized Helium Atom (He+)
4. Deuteron Atom

Answer: 1. Singly Ionized Neon Atom (Ne+)

Question 31. What happens to the mass number and atomic number of an element when it emits -radiation?

1. The mass number decreases by four and the atomic number decreases by two.
2. Mass number and atomic number remain unchanged.
3. The mass number remains unchanged while the atomic number decreases by one.
4. Mass number increases by four and atomic number increases by two.

Answer: 2. Mass number and atomic number remain unchanged.

Question 32. The half life of a radioactive sample undergoing -decay is 1.4 × 1017s. If the number of nuclei in the sample is 2.0 × 1021, the activity of the sample is nearly:

1. 104 Bq
2. 105 Bq
3. 106 Bq
4. 103 B

Answer: 1. 104 Bq

Question 33. When a uranium isotope \({ }_{92}^{235} U\) is bombarded with a neutron it generate \({ }_{36}^{89} \mathrm{Kr}\) three neutrons and

1. \({ }_{36}^{103} \mathrm{Kr}[latex]
2. [latex]{ }_{56}^{144} B a\)
3. \({ }_{40}^{91} \mathrm{Zr}\)
4. \({ }_{36}^{101} \mathrm{Kr}\)

Answer: 2. \({ }_{56}^{144} B a\)

Question 34. The energy equivalent of 0.5 g of a substance is

1. 0.5x1013J
2. 4.5x1016J
3. 4.5x1013J
4. 1.5x1013J

Answer: 3. 4.5x1013J

Question 35. A nucleus with mass number 240 breaks into two fragments each of mass number 120, the binding energy per nucleon of unfragmented nuclei is 7.6 MeV while that of fragmentsis 8.5 MeV. The total gain in the Binding Energy in the process is

1. 9.4MeV
2. 804Mev
3. 216MeV
4. 0.9MeV

Answer: 3. 216MeV

Question 36. A 9. Radioactive nucleus \({ }_{\mathrm{Z}}^{\mathrm{A}} \mathrm{X}\) undergoes spontaneous decay in the sequence \({ }_{\mathrm{z}}^{\mathrm{A}} \mathrm{X} \rightarrow_{\mathrm{z}-1} \mathrm{~B} \rightarrow_{\mathrm{Z}-3} \mathrm{C} \rightarrow_{\mathrm{z}-2} \mathrm{D}\) where Z is the atomic number of element X. The possible decay particles
in the sequence are:

1. \(\alpha, \beta^{+}, \beta^{-}\)
2. \(\beta^{+}, \alpha, \beta^{-}\)
3. \(\beta^{-}, \alpha, \beta^{+}\)
4. \(\alpha, \beta^{-}, \beta^{+}\)

Answer: 2. \(\beta^{+}, \alpha, \beta^{-}\)

Question 37. The half-life of a radioactive nuclide is 100 hours. The fraction of original activity that will remain after 150 hours would be

1. \(\frac{1}{2 \sqrt{2}}\)
2. \(\frac{2}{3}\)
3. \(\frac{2}{3 \sqrt{2}}\)
4. 1/2

Answer: 1. \(\frac{1}{2 \sqrt{2}}\)

Chapter 4 Nuclear Physics Multiple Choice Questions Part – 2: Jee (Main) / Aieee Problems (Previous Years)

Question 1. The above is a plot of binding energy per nucleon Eb, against the nuclear mass M; A, B, C, D, and E, correspond to different nuclei. Consider four reactions:

1. A + B → C + e
2. C → A + B +e
3. D + E → F + e and
4. F → D + E + e,

Answer: 4. F → D + E + e,

Question 2. The half life of a radioactive substance is 20 minutes. The approximate time interval (t2 – between the time t2 when \(\frac{2}{3}\) of it has decayed and time t1 when \(\frac{1}{3}\) of it had decayed is:

1. 7 min
2. 14 min
3. 20 min
4. 28 min

Answer: 3. 20 min

Question 3. Statement – 1: A nucleus having energy E1 decays by b– emission to a daughter nucleus having energy E 2, but the b– rays are emitted with a continuous energy spectrum having endpoint energy E1 – E2. Statement – 2: To conserve energy and momentum in B-decay at least three particles must take part in the transformation.

1. Statement 1 is correct but statement 2 is not correct.
2. Statement-1 and statement-2 both are correct and statement-2 is the correct explanation of statement-1.
3. Statement-1 is correct, statement-2 is correct and statement-2 is not the correct explanation of statement-1.
4. Statement-1 is incorrect, and statement-2 is correct.

Answer: 2. Statement-1 and statement-2 both are correct and statement-2 is the correct explanation of statement-1.

Question 4. Assume that a neutron breaks into a proton and an electron. The energy released during this process is: (mass of neutron = 1.6725 × 10–27 kg, Mass of proton = 1.6725 × 10–27 kg, mass of electron = 9 × 10–31 kg)

1. 0.73 MeV
2. 7.10 MeV
3. 6.30 MeV
4. 5.4 MeV

Answer: 1. 0.73 MeV

Question 5. In a hydrogen-like atom, electrons make a transition from an energy level with quantum number n to another with a quantum number (n–1). If n>>1, the frequency of radiation emitted is proportional to :

1. \(\frac{1}{n}\)
2. \(\frac{1}{n^2}\)
3. \(\frac{1}{n 3 / 2}\)

Answer: 4. \(\frac{1}{n 3 / 2}\)

Question 6. Half-lives of two radioactive elements A and B are 20 minutes and 40 minutes, respectively, initially the samples have equal numbers of nuclei. After 80 minutes, the ratio of decayed numbers of A and B nuclei will be:

1. 4:1
2. 1:4
3. 5:4
4. 1:16

Answer: 3. 5:4

Question 7. A radioactive nucleus A with a half-life T, decays into a nucleus B. At t = 0, there is no nucleus B. At some time t, the ratio of the number of B to that of A is 0.3. Then, t is given by

1. \(t=\frac{T}{\log (1.3)}\)
2. \(\mathrm{t}=\frac{\mathrm{T}}{2} \frac{\log 2}{\log 1.3}\)
3. \(\mathrm{t}=\mathrm{T} \frac{\log 1.3}{\log 2}\)
4. t=T log (1.3)

Answer: 3. \(\mathrm{t}=\mathrm{T} \frac{\log 1.3}{\log 2}\)

Question 8. A sample of radioactive material A, which has an activity of 10 mCi(1 Ci = 3.7 × 10 10 decays/s), has twice the number of nuclei as another sample of a different radioactive material B which has an activity of 20 mCi. The correct choices for half-lives of A and B would then be respectively:

1. 10 days and 40 days
2. 20 days and 5 days
3. 5 days and 10 days
4. 20 days and 10 days

Answer: 1. 10 days and 40 days

Question 9. At a given instant, say t = 0, two radioactive substances A and B have equal activities. The ratio \(\frac{R_B}{R_A}\) of Their Activities after time itself decays with time \(t \text { as } e^{-3 t}\) If the half-life of A is n2, the half-life of B is :

1. \(\frac{\ell n 2}{4}\)
2. 2ln2
3. 4ln2
4. \(\frac{\ell \mathrm{n} 2}{2}\)

Answer: 2. 2ln2

Question 10. Using a nuclear counter the count rate of emitted particles from a radioactive source is measured. At t = 0, it was 1600 counts per second and at t = 8 seconds it was 100 counts per second. The count rate observed, as counts per second, at t = 6 seconds is close to:

1. 150
2. 400
3. 360
4. 2000

Answer: 2. 400

Question 11. Consider the nuclear fission \(\mathrm{Ne}^{20} \rightarrow 2 \mathrm{He}^4+\mathrm{C}^{12}\)  Given that the binding energy/nucleon of Ne²⁰, He⁴, and C¹² are, respectively, 8.03 MeV, 7.07 MeV, and 7.86 MeV, identify the correct statement:

1. Energy Of 11.9 Mev Has To Be Supplied
2. 8.3 Mev Energy Will Be Released
3. Energy Of 12.4 Mev Will Be Supplied
4. Energy Of 3.6 Mev Will Be Released

Answer: 3. Energy Of 12.4 Mev Will Be Supplied

Question 12. In a radioactive decay chain, the initial nucleus is \({ }_{90}^{232} \mathrm{Th}.\) At the end there are 6 β-particles and 4 βparticles that are emitted. If the end nucleus is

1. A=202;X=8
2. A=208;Z=80
3. A=200;Z=81
4. A=208; Z=82

Answer: 4. A=208; Z=82

Chapter 4 Nuclear Physics Multiple Choice Questions Self Practice Paper

Question 1. Binding Energy per nucleon of a fixed nucleus XA is 6 MeV. It absorbs a neutron moving with KE = 2 MeV, and converts into Y at the ground state, emitting a photon of energy 1 MeV. The Binding Energy per nucleon of Y (in MeV) is

1. \(\frac{(6 A+1)}{(A+1)}\)
2. \(\frac{(6 A-1)}{(A+1)}\)
3. 7
4. \(\frac{7}{6}\)

Answer: 2. \(\frac{(6 A-1)}{(A+1)}\)

Question 2. The half life of a radioactive substance ‘A’ is 4 days. The probability that a nucleus will decay in two-half

1. \(\frac{1}{4}\)
2. \(\frac{3}{4}\)
3. \(\frac{1}{2}\)
4. 1

Answer: 3. \(\frac{1}{4}\)

Question 3. To determine the half life of a radioactive element, a student plots a graph of \(\ln \left|\frac{d N(t)}{d t}\right|\) versus t. Here \(\frac{\mathrm{dN}(\mathrm{t})}{\mathrm{dt}}\) is the rate of radioactive decay at time t. If the number of radioactive nuclei of this element decreases by a factor of p after 4.16 years, the value of p is:

1. 8
2. 6
3. 7
4. 9

Answer: 1. 8

Question 4. The activity of a freshly prepared radioactive sample is 1010 disintegrations per second, whose mean life is 109 s. The mass of an atom of this radioisotope is 10–25 kg. The mass (in mg) of the radioactive sample is

1. 8
2. 3
3. 5
4. 1

Answer: 4. 1

Question 5. An accident in a nuclear laboratory resulted in the deposition of a certain amount of radioactive material with a half-life of 18 days inside the laboratory. Tests revealed that the radiation was 64 times more than the permissible level required for the safe operation of the laboratory. What is the minimum number of days after which the laboratory can be considered safe for use?

1. 64
2. 90
3. 108
4. 12

Answer: 3. 108

Question 6. The energy spectrum of b -particles (number N(E) as a function of -energy E) emitted from a radioactive source is:

Answer: 4.

Question 7. The ‘rad’ is the correct unit used to report the measurement of

1. The Rate Of Decay Of Radioactive Source
2. The Ability Of A Beam Of Gamma Ray Photons To Produce Ions In A Target
3. The Energy Delivered By Radiation To A Target.
4. The Biological Effect Of Radiation

Answer: 4. The Biological Effect Of Radiation

Chapter 7 Geometrical Optics Exercise 1 Multiple Choice Questions And Answers

Chapter 7 Geometrical Optics Plane Mirror

Question 1. A clock hung on a wall has marks instead of numbers on its dial. On the opposite wall there is a mirror, and the image of the clock in the mirror if read, indicates the time as 8: 20. What is the time on the clock

1. 3: 40
2. 4: 40
3. 5: 20
4. 4: 20

Answer: 1. 3: 40

Question 2. A ray of light incident on a plane mirror at an angle of incidence of 30°. The deviation produced by the mirror is

1. 30°
2. 60°
3. 90°
4. 120°

Answer: 4. 120°

Question 3. The image of a real object formed by a plane mirror is

1. Erect, real, and of equal size
2. Erect, virtual, and of equal size
3. Inverted, real, and of equal size
4. Inverted, virtual, and of equal size

Answer: 2. Erect, virtual, and of equal size

Question 4. Two mirrors are inclined at an angle θ as shown in the figure. A light ray is incident parallel to one of the mirrors. Light will start retracing its path after the third reflection if :

1. θ = 45
2. θ = 30
3. θ= 60
4. All three

Answer: 2. θ = 30

Question 5. If an object is placed symmetrically between two plane mirrors, inclined at an angle of 72°, then the total number of images formed is

1. 5
2. 4
3. 2
4. Infinite

Answer: 2. 4

Question 6. A man 180 cm high stands in front of a plane mirror. His eyes are at a height of 170 cm from the floor. Then the minimum length of the plane mirror for him to see his full-length image is

1. 90 cm
2. 180 cm
3. 45 cm
4. 360 cm

Answer: 1. 90 cm

Question 7. A thick plane mirror shows a number of images of the filament of an electric bulb. Of these, the brightest image is the

1. First
2. Second
3. Last
4. Fourth

Answer: 2. Second

Question 8. Two vertical plane mirrors are inclined at an angle of 60° with each other. A ray of light traveling horizontally is reflected first from one mirror and then from the other. The resultant deviation is

1. 60°
2. 100°
3. 180°
4. 240°

Answer: 4. 240°

Question 9. When a plane mirror is placed horizontally on level ground at a distance of 60m from the foot of a tower, the top of the tower and its image in the mirror subtend an angle of 90° at the eye. The height of the tower will be

1. 30 m
2. 60 m
3. 90 m
4. 120 m

Answer: 2. 60 m

Question 10. An object is at a distance of 0.5 m in front of a plane mirror. The distance between the object and the image is

1. 0.5 m
2. 1 m
3. 0.25 m
4. 1.5 m

Answer: 2. 1 m

Question 11. The light reflected by a plane mirror may form a real image

1. If the ray incident on the mirror is diverging
2. If the rays incident on the mirror are converging
3. If the object is placed very close to the mirror
4. Under no circumstances

Answer: 2. If the rays incident on the mirror are converging

Question 12. Two plane mirrors are inclined to each other at an angle of 600. If a ray of light incident on the first mirror is parallel to the second mirror, it is reflected from the second mirror

1. Perpendicular to the first mirror
2. Parallel to the first mirror
3. Parallel to the second mirror
4. Perpendicular to the second mirror

Answer: 2. Parallel to the first mirror

Question 13. It is desired to photograph the image of an object placed at a distance of 3 m from a plane mirror. The camera, which is at a distance of 4.5 m from the mirror should be focused for a distance of

1. 3 m
2. 4.5 m
3. 6 m
4. 7.5 m

Answer: 4. 7.5 m

Question 14. An unnumbered wall clock shows time 04: 25: 37, where 1st term represents hours, 2nd represents minutes and the last term represents seconds. What time will its image in a plane mirror show?

1. 08: 35: 23
2. 07: 35: 23
3. 07: 34: 23
4. None of these

Answer: 3. 07: 34: 23

Question 15. Two plane mirrors are parallel to each other and spaced 20 cm apart. An object is kept in between them at 15 cm from A. Out of the following at which point(s) image(s) is/are not formed in mirror A (distance measured from mirror A):

1. 15 cm
2. 25 cm
3. 45 cm
4. 55 cm

Answer: 3. 45 cm

Question 16. A thick mirror produces a number of images of an object. The brightest images are

1. First
2. Second
3. Third
4. Last one

Answer: 2. Second

Question 17. If two mirrors are kept at 60° from each other, then the number of images formed by them is

1. 5
2. 6
3. 7
4. 8

Answer: 1. 5

Question 18. To get three images of a single object, one should have two plane mirrors at an angle of

1. 60°
2. 90°
3. 120°
4. 30°

Answer: 2. 90°

Question 19. A man is 6 feet tall. In order to see his entire image, he requires a plane mirror of minimum length equal to:

1. 6 ft
2. 12 ft
3. 2 ft
4. 3 ft

Answer: 4. 3 ft

Chapter 7 Geometrical Optics Spherical Mirror

Question 1. A convex mirror has a focal length of f. A real object is placed at a distance f in front of it from the pole, then it produces an image at

1. Infinity
2. f
3. f/2
4. 2f

Answer: 3. f/2

Question 2. The image formed by the convex mirror of a focal length 30 cm is a quarter of the size of the object. Then the distance of the object from the mirror is-

1. 30 cm
2. 90 cm
3. 120 cm
4. 60 cm

Answer: 2. 90 cm

Question 3. The largest distance of the image of a real object from a convex mirror of focal length 10 cm can be

1. 20 cm
2. Infinite
3. 10 cm
4. Depends on the position of the object

Answer: 3. 10 cm

Question 4. In the case of the concave mirror, the minimum distance between a real object and its real image is

1. f
2. 2f
3. 4f
4. Zero

Answer: 4. Zero

Question 5. A luminous point object is moving along the principal axis of a concave mirror of focal length 12 cm towards it. When its distance from the mirror is 20 cm its velocity is 4 cm/s. The velocity of the image in cm/s at that instant is

1. 6, towards the mirror
2. 6, away from the mirror
3. 9, away from the mirror
4. 9, towards the mirror.

Answer: 3. 9, away from the mirror

Question 6. Which of the following can form an erect, virtual, diminished image?

1. Plane mirror
2. Concave mirror
3. Convex mirror
4. None of these

Answer: 3. Convex mirror

Question 7. The focal length of a concave mirror is 20 cm. Determine where an object must be placed to form an image magnified two times when the image is real

1. 30cm from the mirror
2. 10cm from the mirror
3. 20cm from the mirror
4. 15cm from the mirror

Answer: 1. 30cm from the mirror

Question 8. A convex mirror of focal length f forms an image that is 1/n times the object. The distance of the object from the mirror is

1. (n-1) f
2. \(\left(\frac{n-1}{n}\right) f\)
3. \(\left(\frac{n+1}{n}\right) f\)
4. (n+1) f

Answer: 1. (n-1) f

Question 9. Which of the following could not produce a virtual image

1. Plane mirror
2. Convex mirror
3. Concave mirror
4. All the above can produce a virtual image

Answer: 4. All the above can produce a virtual image

Question 10. The field of view is maximum for

1. Plane mirror
2. Concave mirror
3. Convex mirror
4. Cylindrical mirror

Answer: 3. Convex mirror

Question 11. The focal length of a concave mirror is f and the distance from the object to the principal focus is x. The ratio of the size of the image to the size of the object is

1. \(\frac{f+x}{f}\)
2. \(\frac{f}{x}\)
3. \(\sqrt{\frac{f}{x}}\)
4. \(\frac{f^2}{x^2}\)

Answer: 2. \(\frac{f}{x}\)

Question 12. The image formed by a convex mirror is

1. Virtual
2. Real
3. Enlarged
4. Inverted

Answer: 1. Virtual

Question 13. The image formed by a convex mirror of a focal length of 30 cm is a quarter of the size of the object. The distance of the object from the mirror is

1. 30 cm
2. 90 cm
3. 120 cm
4. 60 cm

Answer: 2. 90 cm

Question 14. A person sees his virtual image by holding a mirror very close to the face. When he moves the mirror away from his face, the image becomes inverted. What type of mirror he is using

1. Plane mirror
2. Convex mirror
3. Concave mirror
4. None of these

Answer: 3. Concave mirror

Question 15. A square ABCD of side 1mm is kept at a distance of 15 cm in front of the concave mirror as shown in the figure. The focal length of the mirror is 10 cm. The length of the perimeter of its image will be(nearly):

1. 8 mm
2. 2 mm
3. 12 mm
4. 6 mm

Answer: 3. 12 mm

Question 16. A particle is moving towards a fixed spherical mirror. The image:

1. Must move away from the mirror
2. Must move towards the mirror
3. May move towards the mirror
4. Will move towards the mirror, only if the mirror is convex.

Answer: 3. May move towards the mirror

Question 17. The distance of an object from the focus of a convex mirror of radius of curvature ‘a’ is ‘b’. Then the distance of the image from the focus is:

1. b²/ 4a
2. a / b²
3. a²/ 4b
4. 4b / a²

Answer: 3. a² / 4b

Question 18. An object is placed at a distance u from a concave mirror and its real image is received on a screen placed at a distance of v from the mirror. If f is the focal length of the mirror, then the graph between 1/v versus 1/u is

Answer: 2

Chapter 7 Geometrical Optics Refraction In General, Refraction At Plane Surface And Total Internal Reflection

Question 1. Total internal reflection occurs in waves when a wave enters-

1. Glass from air
2. Air from vacuum
3. Water from air
4. Air from water

Answer: 4. Air from water

Question 2. An object is placed at a 24 cm distance above the surface of a lake. If water has a refractive index of 4/3, then at what distance from the lake surface, a fish will sight the object

1. 32 cm above the surface of the water
2. 18 cm over the surface of the water
3. 6 cm over the surface of the water
4. 6 cm below the surface of the water

Answer: 1. 32 cm above the surface of water

Question 3. The time taken to cross a 4 mm window glass with a refractive index of 1.5 will be

1. 2 x 10^-8 sec
2. 2 x 10⁸ sec
3. 2 x 10^-11 sec
4. 2 x 10¹¹ sec

Answer: 3. 2 x 10^-11 sec

Question 4. The total internal reflection of a beam of light occurs when a beam of light enters- [i_c = critical angle, i = angle of incidence]

1. Rarer medium from a denser one and i < i_c
2. Rarer medium from a denser i > i_c
3. Denser medium from rarer i < i_c
4. Denser medium from a rarer i > i_c

Answer: 2. Rarer medium from a denser i > i_c

Question 5. A bubble in the glass slab [μ = 1.5] when viewed from one side appears at 5 cm and at 2 cm from another side the thickness of the slab is

1. 3.75 cm
2. 23 cm
3. 10.5 cm
4. 1.5 cm

Answer: 3. 10.5 cm

Question 6. A light wave travels from glass to water. The refractive index for glass and water are and — respectively. The value of the critical angle will be:

1. \(\sin ^{-1}\left(\frac{1}{2}\right)\)
2. \(\sin ^{-1}\left(\frac{9}{8}\right)\)
3. \(\sin ^{-1}\left(\frac{8}{9}\right)\)
4. \(\sin ^{-1}\left(\frac{5}{7}\right)\)

Answer: 3. \(\sin ^{-1}\left(\frac{8}{9}\right)\)

Question 7. The wavelength of light in a vacuum is 6000 Å and in a medium, it is 4000 Å. The refractive index of the medium is:

1. 2.4
2. 1.5
3. 1.2
4. 0.67

Answer: 2. 1.5

Question 8. A beam of light is converging towards a point. A plane parallel plate of glass of thickness t, the refractive index μ is introduced in the path of the beam. The convergent point is shifted by (assume near normal incidence):

1. \(t\left(1-\frac{1}{\mu}\right)\) away
2. \(t\left(1+\frac{1}{\mu}\right)\) away
3. \(t\left(1-\frac{1}{\mu}\right)\) nearer
4. \(t\left(1+\frac{1}{\mu}\right)\) nearer

Answer: 1. \(t\left(1-\frac{1}{\mu}\right)\) away

Question 9. When a beam of light goes from a denser medium (μ_d) to a rarer medium (μ_r), then it is generally observed that the magnitude of the angle of incidence is half that of the angle of refraction. Then magnitude of incident angle will be- (here μ = μ_d/μ_c)

1. \(2 \sin ^{-1}\left(\frac{\mu}{2}\right)\)
2. \(2 \cos ^{-1} \mu\)
3. \(\cos ^{-1}\left(\frac{\mu}{2}\right)\)
4. \(2 \cos ^{-1}\left(\frac{\mu}{2}\right)\)

Answer: 2. \(2 \cos ^{-1} \mu\)

Question 10. To an observer on the earth, the stars appear to twinkle. This can be ascribed to

1. The fact that stars do not emit light continuously
2. Frequent absorption of starlight by their own atmosphere
3. Frequent absorption of starlight by the Earth’s atmosphere
4. The refractive index fluctuations in the Earth’s atmosphere

Answer: 4. The refractive index fluctuations in the earth’s atmosphere

Question 11. The refractive index of a certain glass is 1.5 for light whose wavelength in vacuum is 6000 Å. The wavelength of this light, when it passes through glass, is

1. 4000 Å
2. 6000 Å
3. 9000 Å
4. 15000 Å

Answer: 1. 4000 Å

Question 12. When light travels from one medium to the other of which the refractive index is different, then which of the following will change

1. Frequency, wavelength, and velocity
2. Frequency and wavelength
3. Frequency and velocity
4. Wavelength and velocity

Answer: 4. Wavelength and velocity

Question 13. A monochromatic beam of light passes from a denser medium into a rarer medium. As a result

1. Its velocity increases
2. Its velocity decreases
3. Its frequency decreases
4. Its wavelength decreases

Answer: 1. Its velocity increases

Question 14. A rectangular tank of depth 8 meters is full of water (μ = 4/3), the bottom is seen at the depth

1. 6m
2. 8/3 m
3. 8 cm
4. 10 cm

Answer: 1. 6m

Question 15. If _iμ_j represents the refractive index when a light ray goes from medium i to medium j, then the product ₂μ₁ X ₃μ₂ X ₄μ₃ is equal to

1. \({ }_3 \mu_1\)
2. \({ }_3 \mu_2\)
3. \(\frac{1}{{ }_1 \mu_4}\)
4. \({ }_4 \mu_2\)

Answer: 3. \(\frac{1}{{ }_1 \mu_4}\)

Question 16. The wavelength of light diminishes μ times (μ = 1.33 for water) in a medium. A driver from inside the water looks at an object whose natural color is green. He sees the object as

1. Green
2. Blue
3. Yellow
4. Red

Answer: 1. Green

Question 17. A diver in a swimming pool wants to signal his distress to a person lying on the edge of the pool by flashing his waterproof flashlight

1. He must direct the beam vertically upwards
2. He has to direct the beam horizontally
3. He has to direct the beam at an angle to the vertical which is slightly less than the critical angle of incidence for total internal reflection
4. He has to direct the beam at an angle to the vertical which is slightly more than the critical angle of incidence for the total internal reflection

Answer: 3. He has to direct the beam at an angle to the vertical which is slightly less than the critical angle of incidence for total internal reflection

Question 18. The wavelength of light in two liquids ‘x’ and ‘y’ is 3500 Å and 7000 Å, then the critical angle of x relative to y will be

1. 60°
2. 45°
3. 30°
4. 15°

Answer: 3. 30°

Question 19. Total internal reflection of a ray of light is possible when the (i_c = critical angle, i = angle of incidence)

1. Ray goes from denser medium to rarer medium and i < i_c
2. Ray goes from denser medium to rarer medium and i > i_c
3. Ray goes from rarer medium to denser medium and i > i_c
4. Ray goes from rarer medium to denser medium and i < i_c

Answer: 2. Ray goes from denser medium to rarer medium and i > i_c

Question 20. A diver at a depth of 12 m in water (μ = 4/3) sees the sky in a cone of semi-vertical angle

1. sin^-1(4/3)
2. tan^-1(4/3)
3. sin^-1(3/4)
4. 90°

Answer: 3. sin^-1(3/4)

Question 21. The critical angle for a diamond (refractive index = 2) is

1. About 20°
2. 60°
3. 45°
4. 30°

Answer: 4. 30°

Question 22. The reason for the shining of air bubbles in water is

1. Diffraction of light
2. Dispersion of light
3. Scattering of light
4. Total internal reflection of light

Answer: 4. Total internal reflection of light

Question 23. ‘Mirage’ is a phenomenon due to

1. Reflection of light
2. Refraction of light
3. Total internal reflection of light
4. Diffraction of light

Answer: 3. Total internal reflection of light

Question 24. Given that the velocity of light in quartz =1.5 x 10⁸ m/s and the velocity of light in glycerine = (9/4) x 10⁸ m/s. Now a slab made of quartz is placed in glycerine as shown. The shift of the object produced by the slab is

1. 6 cm
2. 3.55 cm
3. 9 cm
4. 2 cm

Answer: 1. 6 cm

Question 25. A ray of light passes from a vacuum into a medium of refractive index n. If the angle of incidence is twice the angle of refraction, then the angle of incidence is:

1. cos^-1 (n/2)
2. sin^-1 (n/2)
3. 2 cos^-1 (n/2)
4. 2 sin^-1 (n/2)

Answer: 3. 2 cos^-1 (n/2)

Question 26. The critical angle of light going from medium A to medium B is θ. The speed of light in medium A is v. The speed of light in medium B is:

1. \(\frac{v}{\sin \theta}\)
2. \(v \sin \theta\)
3. \(v \cot \theta\)
4. \(v \tan \theta[latex]

Answer: 1. [latex]\frac{v}{\sin \theta}\)

Question 27. A ray of light passes through four transparent media with refractive indices μ₁, μ₂, μ_3, and μ₄ as shown in the figure. The surfaces of all media are parallel. If the emergent ray CD is parallel to the incident ray AB, we must have:

1. μ₁ = μ₂
2. μ₂ = μ₃
3. μ₃ = μ₄
4. μ₄ = μ₁

Answer: 4. μ₄ = μ₁

Question 28. If the critical angle for total internal reflection from a medium to vacuum is 30°, then the speed of light in the medium is

1. 6 x 10⁸ m/s
2. 3 x 10⁸ m/s
3. 2 x 10⁸ m/s
4. 1.5 x 10⁸ m/s

Answer: 4. 1.5 x 10⁸ m/s

Question 29. If a glass rod is immersed in a liquid of the same refractive index, then it will appear

1. Bent
2. Longer
3. Shorter
4. Invisible

Answer: 4. Invisible

Question 30. A metal coin is at the bottom of a beaker filled with a liquid of refractive index 4/3 to a height of 6 cm. To an observer looking from above the surface of the liquid, the coin will appear at a depth of :

1. 7.5 cm
2. 6.75 cm
3. 4.5 cm
4. 1.5 cm

Answer: 3. 4.5 cm

Question 31. A transparent cube with a 15 cm edge contains a small air bubble. Its apparent depth when viewed through one face is 6 cm and when viewed through the opposite face is 4 cm. Then the refractive index of the material of the cube is

1. 2.0
2. 2.5
3. 1.6
4. 1.5

Answer: 4. 1.5

Question 32. If the critical angle for total internal reflection from a medium to vacuum is 30°, the velocity of light in a medium is

1. 3 x 10⁸ m/s
2. 1.5 x 10⁸ m/s
3. 6 x 10⁸ m/s
4. 10⁸ m/s

Answer: 2. 1.5 x 10⁸ m/s

Question 33. For the given incident ray as shown in the figure, the condition of the total internal reflection of the ray will be satisfied if the refractive index of the block is:

1. \(n>\frac{\sqrt{3}+1}{2}\)
2. \(n<\frac{\sqrt{3}+1}{2}\)
3. \(n>\sqrt{\frac{3}{2}}\)
4. \(n<\sqrt{\frac{3}{2}}\)

Answer: 3. \(n>\sqrt{\frac{3}{2}}\)

Question 34. A small coin is resting on the bottom of a beaker filled with a liquid. A ray of light from the coin travels up to the surface of the liquid and moves along its surface How fast is the light traveling in the liquid?

1. 1.8 × 10⁸ m/s
2. 2.4 × 10⁸ m/s
3. 3.0 × 10⁸ m/s
4. 1.2 × 10⁸ m/s

Answer: 1. 1.8 × 10⁸ m/s

Question 35. Which of the following is used in optical fibers?

1. Total internal reflection
2. Scattering
3. Diffraction
4. Refraction

Answer: 1. Total internal reflection

Question 36. Transmission of light in optical fiber is due to the:

1. Scattering
2. Diffraction
3. Polarisation
4. Multiple total internal reflections

Answer: 4. Multiple total internal reflections

Question 37. Which of the following will remain constant in the phenomenon of refraction of light?

1. Wavelength
2. Velocity
3. Frequency
4. None

Answer: 3. Frequency

Question 38. The phenomena of total internal reflection are seen when the angle of incidence is:

1. 90°
2. Greater than the critical angle
3. Equal to the critical angle
4. 0°

Answer: 2. Greater than critical angle

Question 39. A ray of light from a denser medium strikes a rarer medium of an angle of incidence i. the reflected and refracted rays make an angle of 900 with each other. The angle of reflection and refraction are r and r’. The critical angle is

1. sin^-1 [tan r]
2. sin^-1 [cot i]
3. sin^-1 [tan r’]
4. sin^-1 [sin r’]

Answer: 1. sin^-1 [cot i]

Question 40. A ray of light travels from an optically denser to a rarer medium. The critical angle for the two media is c. The maximum possible deviation of the ray will be–

1. 2c
2. π/2 – c
3. π – c
4. π – 2c

Answer: 2. π/2 – c

Question 41. A ray of light traveling in water is incident on its surface open to air. The angle of incidence is θ, which is less than the critical angle. Then there will be :

1. Only a reflected ray and no refracted ray
2. Only a refracted ray and no reflected ray
3. A reflected ray and a refracted ray and the angle between them would be less than 180° – 2θ
4. A reflected ray and a refracted ray and the angle between them would be greater than 180° – 2θ.

Answer: 3. A reflected ray and a refracted ray and the angle between them would be less than 180° – 2θ

Question 42. A light beam is traveling from Region 1 to Region 4. The refractive index in Regions 1, 2, 3, and 4 are n₀, and, respectively. The angle of incidence θ for which the beam just misses entering Region 4 is

1. \(\sin ^{-1}\left(\frac{3}{4}\right)\)
2. \(\sin ^{-1}\left(\frac{1}{8}\right)\)
3. \(\sin ^{-1}\left(\frac{1}{4}\right)\)
4. \(\sin ^{-1}\left(\frac{1}{3}\right)\)

Answer: 2. \(\sin ^{-1}\left(\frac{1}{8}\right)\)

Question 43. The wavelength of light in two liquids x and y are 3500 Å and 7000 Å, then the critical angle of x relative to y will be

1. 60°
2. 45°
3. 30°
4. 15°

Answer: 3. 30°

Question 44. The correct thickness of glass having _aμ_g = 1.5, which permits an equal number of wavelengths as that of an 18 cm long column of water is- [_aμ_g = 4/3]

1. 12 cm
2. 16 cm
3. 18 cm
4. 24 cm

Answer: 2. 16 cm

Question 45. Dimension of \(\frac{1}{\mu_0 \varepsilon_0}\) is (where symbols have usual meaning):

1. [LT^-1]
2. [L^-1T]
3. [L^-2T]
4. [L²T^-2]

Answer: 4. [L²T^-2]

Question 46. When a ray of light is reflected from a denser medium interface, then the following changes

1. Wavelength
2. Phase
3. Frequency
4. Speed

Answer: 2. Phase

Question 47. If the refractive index of a medium is 1.5, then the velocity of light in that medium will be:

1. 10 x 10⁸
2. 2 x 10⁸
3. 3 x 10⁸
4. 4 x 10⁸

Answer: 2. 2 x 10⁸

Question 48. Light enters into glass from the air then it:

1. Frequency increases
2. Frequency decreases
3. Wavelength increases
4. Wavelength decreases

Answer: 4. Wavelength decreases

Question 49. The value of the refractive index for any medium is

1. \(1 / \sqrt{\mu_r \varepsilon_r}\)
2. \(\sqrt{\mu_{\mathrm{r}} \varepsilon_{\mathrm{r}}}\)
3. \(\sqrt{\mu_{\mathrm{r}} / \varepsilon_{\mathrm{r}}}\)
4. \(\sqrt{\varepsilon_{\mathrm{r}} / \mu_{\mathrm{r}}}\)

Answer: 2. \(\sqrt{\mu_{\mathrm{r}} \varepsilon_{\mathrm{r}}}\)

Chapter 7 Geometrical Optics Refraction By Prism

Question 1. A ray of light is incident normally on one of the faces of a prism apex angle of 30° and refractive index √2. The angle of deviation of the ray is:

1. 15°
2. 22.5°
3. 0°
4. 12.5°

Answer: 1. 15°

Question 2. The refractive index of the material of the prism of 60º angle is √2. At what angle the ray of light be incident on it so that minimum deviation takes place?

1. 45º
2. 60º
3. 30º
4. 75º

Answer: 1. 45º

Question 3. A ray of light is incident at an angle of 60° on one face of a prism which has an apex angle of 30°. The ray emerging out of the prism makes an angle of 30° with the incident ray. The refractive index of the material of the prism is

1. √2
2. √3
3. 1.5
4. 1.6

Answer: 2. √3

Question 4. If the critical angle for the medium of the prism is C and the angle of the prism is A, then there will be no emergent ray when

1. A < 2C
2. A = 2C
3. A > 2C
4. A ≥ 2C

Answer: 3. A > 2C

Question 5. A ray of monochromatic light is incident on one refracting face of a prism of angle 75°. It passes through the prism and is incident on the other face at the critical angle. If the refractive index of the material of the prism is √2, the angle of incidence on the first face of the prism is

1. 30°
2. 45°
3. 60°
4. 0°

Answer: 2. 45°

Question 6. A ray of light is incident at an angle i on a surface of a prism of small angle A and emerges normally from the opposite surface. If the refractive index of the material of the prism is μ, the angle of incidence i is nearly equal to:

1. A/μ
2. A/(2 μ)
3. μ A
4. μ A/2

Answer: 3. μ A

Question 7. A prism having an apex angle of 4° and a refractive index of 1.50 is located in front of a vertical plane mirror as shown. A horizontal ray of light is incident on the prism. The total angle through which the ray is deviated is:

1. 4° clockwise
2. 178° clockwise
3. 2° clockwise
4. 8° clockwise

Answer: 2. 178° clockwise

Question 8. The critical angle between an equilateral prism and air is 45°. If the incident ray is perpendicular to the refracting surface, then

1. After deviation, it will emerge from the second refracting surface
2. It is totally reflected on the second surface and emerges perpendicularly from the third surface in air
3. It is totally reflected from the second and third refracting surfaces and finally emerges from the first surface
4. It is totally reflected from all three sides of the prism and never emerges out

Answer: 2. It is totally reflected on the second surface and emerges out perpendicularly from the third surface in the air

Question 9. When light rays are incident on a prism at an angle of 45°, the minimum deviation is obtained. If the refractive index of the material of the prism is √2, then the angle of the prism will be

1. 30°
2. 40°
3. 50°
4. 60°

Answer: 4. 60°

Question 10. The refractive index of a prism for a monochromatic wave is √2 and its refracting angle is 60°. For minimum deviation, the angle of incidence will be

1. 30°
2. 45°
3. 60°
4. 75°

Answer: 2. 45°

Question 11. A parallel beam of monochromatic light is incident at one surface of an equilateral prism. The angle of incidence is 55° and the angle of emergence is 46°. The angle of minimum deviation will be

1. Less than 41°
2. Equal to 41°
3. More than 41°
4. None of the above

Answer: 1. Less than 41°

Question 12. The minimum refractive index of a material, of a prism of apex angle 90°, for which light cannot be transmitted for any value of i:

1. √3
2. 1.5
3. √2
4. None of these

Answer: 3. √2

Question 13. A horizontal light ray passes through a prism (μ = 1.5) of angle 4°. Further, it is incident on a plane mirror M that has been placed vertically. By what angle the mirror is rotated so that the ray after reflection becomes horizontal?

1. 1°
2. 2°
3. 4°
4. 8°

Answer: 1. 1°

Question 14. For a prism of refractive index √3, the angle of the prism is equal to the angle of minimum deviation. The value of the angle of the prism is

1. 60°
2. 50°
3. 45°
4. 30°

Answer: 1. 60°

Question 15. An equilateral prism is kept on a horizontal surface. A typical ray of light PQRS is shown in the figure. For minimum deviation

1. The ray PQ must be horizontal
2. The ray RS must be horizontal
3. The ray QR must be horizontal
4. Any one of them can be horizontal

Answer: 3. The ray QR must be horizontal

Question 16. A prism has a refracting angle of 60°. If it produces a minimum deviation of 30° in an incident ray, the angle of incidence is

1. 15°
2. 30°
3. 45°
4. 60°

Answer: 3. 45°

Question 17. A thin prism P of angle 4° made of glass of refractive index 1.54 is combined with a thin prism Q made of glass of refractive index 1.72 to produce dispersion without deviation. The angle of prism Q is:

1. 4°
2. 3°
3. 2.6°
4. 5.3°

Answer: 2. 3°

Question 18. A prism has a refracting angle of 60°. A ray of given monochromatic light suffers a minimum deviation of 38° in passing through the prism refractive index of the material of the prism is:

1. 1.5094
2. 1.3056
3. 0.7849
4. 2.425

Answer: 1. 1.5094

Question 19. An equilateral prism has p = √3. Its angle of minimum deviation will be:

1. 30°
2. 60°
3. 120°
4. 45°

Answer: 2. 60°

Question 20. The deviation is maximum for which color?

1. violet
2. Red
3. Blue
4. Green

Answer: 1. violet

Question 21. The refractive indices of violet and red light are 1.54 and 1.52 respectively. If the angle of the prism is 10°, the angular dispersion is

1. 0.02°
2. 0.2°
3. 3.06°
4. 30.6°

Answer: 2. 0.2°

Question 22. A microscope is focused on a mark on a piece of paper and then a slab of glass of thickness 3 cm and refractive index 1.5 is placed over the mark. How should the microscope be moved to get the mark in focus again?

1. 1 cm upward
2. 4.5 cm downward
3. 1 cm downward
4. 2 cm upward

Answer: 1. 1 cm upward

Question 23. The refractive index of glass is 1.520 for red light and 1.525 for blue light. Let D₁, and D₂ be angles of minimum deviation for red and blue light respectively in a prism of this glass. Then,

1. D₁ can be less than or greater than D₂ depending on the angle of the prism
2. D₁ > D₂
3. D₁ < D₂
4. D₁ = D₂

Answer: 3. D₁ = D₂

Question 24. A light ray is incident perpendicularly to one face to a 90º prism and is totally internally reflected at the glass-air interface. If the angle of reflection is 45º, we conclude that the refractive index n is

1. \(\mathrm{n}<\frac{1}{\sqrt{2}}\)
2. \(n>\sqrt{2}\)
3. \(n>\frac{1}{\sqrt{2}}\)
4. \(n<\sqrt{2}\)

Answer: 2. \(n>\sqrt{2}\)

Chapter 7 Geometrical Optics Refraction By Spherical Surface

Question 1. An object is placed at a distance of 20 cm, in a rarer medium, from the pole of a convex spherical refracting surface of a radius of curvature 10 cm. If the refractive index of the rarer medium is 1 and of the refracting medium is 2, then the position of the image is at

1. (40/3) cm from the pole and inside the denser medium
2. 40 cm from the pole and inside the denser medium.
3. (40/3) cm from the pole and outside the denser medium
4. 40 cm from the pole and outside the denser medium.

Answer: 2. 40 cm from the pole and inside the denser medium.

Question 2. There is a small black dot at the center C of a solid glass sphere of refractive index μ. When seen from outside, the dot will appear to be located:

1. Away from C for all values of μ
2. At C for all values of μ
3. At C for μ = 1.5, but away from C for μ ≠ 1.5
4. At C only for ≤ μ ≤ 1.5.

Answer: 2. At C for all values of μ

Question 3. The image for the converging beam after refraction through the curved surface is formed at:

1. \(x=40 \mathrm{~cm}\)
2. \(x=\mathrm{cm}\)
3. \(x=\frac{40}{3}-\frac{40}{3} \mathrm{~cm}\)
4. \(x=\frac{180}{7} \mathrm{~cm}\)

Answer: 1. \(x=40 \mathrm{~cm}\)

Question 4. In the figure shown a point object O is placed in air. A spherical boundary of radius of curvature 1.0 m separates two media. AB is the principal axis. The refractive index above AB is 1.6 and below AB is 2.0. The separation between the images formed due to refraction at the spherical surface is:

1. 12 m
2. 20 m
3. 14 m
4. 10 m

Answer: 1. 12 m

Chapter 7 Geometrical Optics Lens

Question 1. A convex lens is dipped in a liquid whose refractive index is equal to the refractive index of the lens. Then its focal length will

1. Become zero
2. Become infinite
3. Become small, but non-zero
4. Remain unchanged

Answer: 2. Become infinite

Question 2. A biconvex lens with equal radii of curvature has a refractive index of 1.6 and a focal length of 10 cm. Its radius of curvature will be:

1. 20 cm
2. 16 cm
3. 10 cm
4. 12 cm

Answer: 4. 12 cm

Question 3. A convex lens forms a real image 9 cm long (high) on a screen. Without altering the position of the object and the screen, the lens is displaced and we get again a real image 4 cm long (high) on the screen. Then the length of the object is

1. 9 cm
2. 4 cm
3. 6 cm
4. 36 cm

Answer: 3. 6 cm

Question 4. An object is placed at a distance of 5 cm from a convex lens of focal length 10 cm, then the image is

1. Real, diminished, and at a distance of 10 cm from the lens.
2. Real, enlarged, and at a distance of 10 cm from the lens.
3. Virtual, enlarged, and at a distance of 10 cm from the lens.
4. Virtual, diminished, and at a distance of 10/3 cm from the lens.

Answer: 3. Virtual, enlarged, and at a distance of 10 cm from the lens.

Question 5. Inside water, an air bubble behaves-

1. Always like a convex lens
2. Always like a concave lens
3. Always like a slab of equal thickness
4. Sometimes concave and sometimes like a convex lens

Answer: 2. Always like a concave lens

Question 6. A lens behaves as a converging lens in air and a diverging lens in water. The refractive index of the material is (refractive index of water = 1.33)

1. Equal to unity
2. Equal to 1.33
3. Between unity and 1.33
4. Greater than 1.33

Answer: 3. Between unity and 1.33

Question 7. In the figure given below, there are two convex lenses L₁ and L₂ having focal lengths of f₁ and f₂ respectively. The distance between L₁ and L₂ will be

1. f₁
2. f₂
3. f₁ + f₂
4. f₁-f₂

Answer: 3. f₁ + f₂

Question 8. A virtual erect image by a diverging lens is represented by (u, v, f are coordinates)

Answer: 4

Question 9. What should be the value of distance d so that the final image is formed on the object itself? (focal lengths of the lenses are written on the lenses).

1. 10 cm
2. 20 cm
3. 5 cm
4. None of these

Answer: 1. 10 cm

Question 10. A thin linear object of size 1 mm is kept along the principal axis of a convex lens of focal length 10 cm. The object is 15 cm from the lens. The length of the image is:

1. 1 mm
2. 4 mm
3. 2 mm
4. 8 mm

Answer: 2. 4 mm

Question 11. A glass lens is placed in a medium in which it is found to behave like a glass plate. The refractive index of the medium will be:

1. Greater than the refractive index of glass
2. Smaller than the refractive index of glass
3. Equal to the refractive index of glass
4. No case will be possible from the above

Answer: 3. Equal to the refractive index of glass

Question 12. A divergent lens will produce

• Always a virtual image
• Always real image
• Sometimes real and sometimes virtual
• None of above

Answer: 1. Always a virtual image

Question 13. The minimum distance between an object and its real image formed by a convex lens is

1. 1.5 f
2. 2 f
3. 2.5 f
4. 4 f

Answer: 4. 4 f

Question 14. A biconvex lens forms a real image of an object placed perpendicular to its principal axis. Suppose the radii of curvature of the lens tend to infinity. Then the image would

1. Disappear
2. Remain as real image still
3. Be virtual and of the same size as the object
4. Suffer from aberrations

Answer: 3. Suffer from aberrations

Question 15. The radius of curvature of the convex surface of a thin plano-convex lens is 15 cm and the refractive index of its material is 1.6. The power of the lens will be

1. +1D
2. -2D
3. +3D
4. +4D

Answer: 4. +4D

Question 16. A lens is placed between a source of light and a wall. It forms images of areas A₁ and A₂ on the wall for its two different positions. The area of the source of light is

1. \(\frac{A_1+A_2}{2}\)
2. \(\left[\frac{1}{A_1}+\frac{1}{A_2}\right]^{-1}\)
3. \(\sqrt{A_1 A_2}\)
4. \(\left[\frac{\sqrt{A_1}+\sqrt{A_2}}{2}\right]^2\)

Answer: 3. \(\left[\frac{\sqrt{A_1}+\sqrt{A_2}}{2}\right]^2\)

Question 17. If the central portion of a convex lens is wrapped in black paper as shown in the figure

1. No image will be formed by the remaining portion of the lens
2. The full image will be formed but it will be less bright
3. The central portion of the image will be missing
4. There will be two images each produced by one of the exposed portions of the lens

Answer: 2. The full image will be formed but it will be less bright

Question 18. A convex lens forms a real image of a point object placed on its principal axis. If the upper half of the lens is painted black, the image will

1. Be shifted downwards
2. Be shifted upwards
3. Not be shifted
4. Shift on the principal axis

Answer: 3. Not be shifted

Question 19. In the figure, an air lens of radii of curvature 10 cm (R₁ = R₂ = 10 cm) is cut in a cylinder of glass (p = 1.5). The focal length and the nature of the lens is

1. 15 cm, concave
2. 15 cm, convex
3. ∞, neither concave nor convex
4. 0, concave

Answer: 1. 15 cm, concave

Question 20. A lens made of glass with a refractive index 1.5 has a focal length of 10 cm in air and 50 cm when completely immersed in a liquid. Then the refractive index of the liquid is

1. 1.36
2. 1.33
3. 1.30
4. 1.38

Answer: 1. 1.36

Question 21. The correct conclusion that can be drawn from these figures is

1. μ₁ < μ but μ < μ₂
2. μ₁ > μ but μ < μ₂
3. μ₁ = μ but μ < μ₂
4. μ₁ = μ but μ₂ < μ

Answer: 3. μ₁ = μ but μ < μ₂

Question 22. A magnifying glass is to be used at the fixed object distance of 1 inch. if it is to produce an erect image magnified 5 times, its focal length should be

1. 0.2 inch
2. 0.8 inch
3. 1.25 inch
4. 5 inch

Answer: 3. 1.25 inch

Question 23. A convex lens of focal length f produces a virtual image n times the size of the object. Then the distance of the object from the lens is

1. \(\frac{n-1}{n} f\)
2. \(\frac{n+1}{n} f\)
3. \(\frac{\mathrm{f}}{\mathrm{n}}\)
4. \(\frac{n}{n-1} f\)

Answer: 1. \(\frac{n-1}{n} f\)

Question 24. The focal length of a convex lens made from a material of refractive index 1.52 is 10 cm when placed in air. If it is immersed in carbon disulfide of refractive index 1.68, then its focal length and nature will be

1. + 36.4 cm, convex lens.
2. – 36.4 cm, concave lens.
3. + 54.6 cm, convex lens.
4. – 54.6 cm, concave lens.

Answer: 4. – 54.6 cm, concave lens.

Question 25. Two symmetric double convex lenses A and B have the same focal length, but the radii of curvature differ so that, R_A = 0.9 R_B. If n_A = 1.63, find n_B.

1. 1.7
2. 1.6
3. 1.5
4. 4/3

Answer: 1. 1.7

Question 26. When a lens of power P (in the air) made of material of refractive index μ is immersed in a liquid of refractive index μ₀. Then the power of the lens is:

1. \(\frac{\mu-1}{\mu-\mu_0} \mathrm{P}\)
2. \(\frac{\mu-\mu_0}{\mu-1} \mathrm{P}\)
3. \(\frac{\mu-\mu_0}{\mu-1} \cdot \frac{P}{\mu_0}\)
4. None of these

Answer: 3. \(\frac{\mu-\mu_0}{\mu-1} \cdot \frac{P}{\mu_0}\)

Question 27. A thin symmetrical double convex lens of power P is cut into three parts, as shown in the figure. Power of A is:

1. 2P
2. P/2
3. P/3
4. P

Answer: 4. P

Question 28. A convex lens makes a real image of 4 cm long (height) on a screen. When the lens is shifted to a new position without disturbing the object or the screen, we get a real image on the screen, which is 16 cm long. The length (height) of the object is:

1. 1/4 cm
2. 8 cm
3. 20 cm
4. 2 cm

Answer: 2. 8 cm

Question 29. An object is placed at a distance of 10 cm from a convex lens of power 5D. Find the position of the image:

1. -40 cm
2. 30 cm
3. 20 cm
4. -10 cm

Answer: 3. 20 cm

Question 30. The radius of the convex surface of the plano-convex lens is 20 cm and the refractive index of the material of the lens is 1.5 The focal length is :

1. 30 cm
2. 50 cm
3. 20 cm
4. 40 cm

Answer: 4. 40 cm

Question 31. An asymmetric double convex lens is cut in two equal parts by a plane perpendicular to the principal axis. If the power of the original lens is 4D, the power of a cut lens will be

1. 2D
2. 3D
3. 4D
4. 5D

Answer: 1. 2D

Question 32. A point object is placed at the focus of a double concave lens. The image is formed

1. At infinity
2. Between the focus and the lens
3. At focus
4. Between the focus and infinity

Answer: 2. Between the focus and the lens

Question 33. A thin convex lens with a refractive index of 1.5 is placed in a liquid with a refractive index of 2.0. Then the power of the lens in air is 10 D. Then in the liquid its power will be

1. 20 D
2. 10 D
3. -10 D
4. -5 D

Answer: 4. -5 D

Question 34. A body is located on a wall. Its image of equal size is to be obtained on a parallel wall with the help of a convex lens. The lens is placed at a distance d ahead of the second wall, then the required focal length will be:

1. Only \(\frac{d}{4}\)
2. Only \(\frac{d}{2}\)
3. More than Only \(\frac{d}{4}\) but less than Only \(\frac{d}{2}\)
4. Less than Only \(\frac{d}{4}\)

Answer: 2. Only \(\frac{d}{2}\)

Question 35. An equiconvex lens is cut into two halves along (1) XOX’ and (2) YOY’ as shown in the figure. Let f, f ’,f ’’ be the focal lengths of the complete lens, of each half in case (1), and of each half in case (2), respectively.

Choose the correct statement from the following

1. f ’ = f, f ’’ = f
2. f ’ = 2f, f ’’ = 2f
3. f ’ = f, f ’’ = 2f
4. f ’ = 2f, f ’’ = f

Answer: 3. f ’ = f, f ’’ = 2f

Question 36. A convex lens is dipped in a liquid whose refractive index is equal to the refractive index of the lens. Then its focal length will:

1. Become small, but non-zero
2. Remain unchanged
3. Become zero
4. Become infinite

Answer: 4. Become infinite

Question 37. A student measures the focal length of a convex lens by putting an object pin at a distance |u| from the lens and measuring the distance ‘v’ of the image pin. The graph between ‘u’ and ‘v’ plotted by the student should look like

Answer: 2

Question 38. The image formed on the retina of the eye is proportional to:

1. Size of object
2. Area of object
3. \(\frac{\text { size of object }}{\text { size of image }}\)
4. \(\frac{\text { size of image }}{\text { size of object }}\)

Answer: 1. Size of object

Question 39. In order to obtain an image on the wall of a bulb at a distance d from the wall a convex lens is placed between the bulb and wall. The focal length of the lens will be

1. d/2
2. Between d/2 and d/4
3. More then d/2
4. Less than d/4

Answer: 4. Less than d/4

Question 40. An object is placed 12 cm to the left of a converging lens of a focal length of 8 cm. Another converging lens of 6 cm focal length is placed at a distance of 30 cm from the object to the right of the first lens. The second lens will produce

1. A virtual enlarged image
2. No image
3. A real inverted image
4. A real enlarged image

Answer: 3. A real inverted image

Question 41. A biconvex lens of focal length f forms a circular image of radius r of the sun in the focal plane. Then which option is correct:

1. πr² ∝ f
2. πr² ∝ f²
3. If a lower half part is covered by a black sheet, then the area of the image is equal to πr²/2
4. If f is doubled, the intensity will increase

Answer: 2. πr² ∝ f²

Question 42. A slide projector magnified a film of 100 cm² on a screen. If linear magnification is 4, then the area of the magnified image will be:

1. 1600 cm²
2. 800 cm²
3. 400 cm²
4. 200 cm²

Answer: 1. 1600 cm²

Chapter 7 Geometrical Optics Combination Of Thin Lens Or Lens And Mirrors

Question 1. A convex lens of focal length 25 cm and a concave lens of focal length 20 cm are mounted coaxially separated by a distance of d cm. If the power of the combination is zero, d is equal to

1. 45
2. 30
3. 15
4. 5

Answer: 4. 5

Question 2. A convex lens of power 4D and a concave lens of power 3D are placed in contact, the equivalent power of combination:

1. 1D
2. D
3. 7D
4. D

Answer: 1. 1D

Question 3. Two thin lenses of power +5D and -2D are placed in contact with each other. The focal length of the combination will behave like a

1. Convex lens of focal length 3m
2. A concave lens of focal length 0.33m
3. Convex lens of focal length 0.33m
4. None of the above

Answer: 3. Convex lens of focal length 0.33m

Question 4. The focal length of a plano-concave lens is -10 cm, then its focal length when its plane surface is polished is (n = 3/2):

1. 20 cm
2. -5 cm
3. 5 cm
4. None of these

Answer: 3. 5 cm

Question 5. A combination of two thin lenses with focal lengths f₁ and f₂ respectively forms an image of a distant object at a distance of 60 cm when lenses are in contact. The position of this image shifts by 30 cm towards the combination when two lenses are separated by 10 cm. The corresponding values of f₁ and f₂ are

1. 30 cm, -60 cm
2. 20 cm, -30cm
3. 15cm, -20 cm
4. 12 cm, -15 cm

Answer: 2. 20 cm, -30cm

Question 6. A plano convex lens (f = 20 cm) is silvered at the plane surface. Now f will be

1. 20 cm
2. 40 cm
3. 30 cm
4. 10 cm

Answer: 4. 10 cm

Question 7. A luminous object is placed at a distance of 30 cm from a convex lens of focal length 20 cm. On the other side of the lens, at what distance from the lens must a convex mirror of radius of curvature 10 cm be placed in order to have an erect image of the object coincident with it?

1. 12 cm
2. 30 cm
3. 50 cm
4. 60cm

Answer: 3. 50 cm

Question 8. The plane surface of a plano-convex lens of focal length f is silvered. It will behave as:

1. Plane mirror
2. Convex mirror of focal length 2f
3. A concave mirror of focal length f/2
4. None of the above

Answer: 3. Concave mirror of focal length f/2

Question 9. A concave lens of focal length 20 cm placed in contact with a plane mirror acts as a:

1. Convex mirror of focal length 10 cm
2. A concave mirror of focal length 40 cm
3. A concave mirror of focal length 60 cm
4. A concave mirror of focal length 10 cm

Answer: 1. Convex mirror of focal length 10 cm

Question 10. A plano-convex lens of focal length 10 cm is silvered at its plane face. The distance d at which an object must be placed in order to get its image on itself is:

1. 5 cm
2. 20 cm
3. 10 cm
4. 2.5 cm

Answer: 3. 10 cm

Question 11. In figure (1) a thin lens of focal length 10 cm is shown. The lens is cut into two equal parts, and the parts are arranged as shown in Figure (2). An object AB of height 1 cm is placed at a distance of 7.5 cm from the arrangement. The height of the final image will be:

1. 0.5 cm
2. 2 cm
3. 1 cm
4. 4 cm

Answer: 2. 2 cm

Question 12. A convex lens of focal length 40 cm is kept in contact with a concave lens of focal length 25 cm. The power of the combination is

1. \(-6.5 \mathrm{D}\)
2. \(+6.5 \mathrm{D}\)
3. \(\frac{f_0}{f_c}-1.5 D\)
4. \(\frac{f_c}{f_o}+1.5 \mathrm{D}\)

Answer: 3. \(\frac{f_0}{f_c}-1.5 D\)

Question 13. The plane faces of two identical plano-convex lenses, each having a focal length of 40 cm, are placed against each other to form a common convex lens. The distance from this lens at which an object must be placed to obtain a real inverted image with magnification equal to unity is

1. 80 cm
2. 40 cm
3. 20 cm
4. 160 cm

Answer: 2. 40 cm

Question 14. A convex lens and a concave lens, each having the same focal length of 25 cm, are put in contact to form a combination of lenses. The power in diopters of the combination is:

1. 25
2. 50
3. Infinite
4. Zero

Answer: 4. Zero

Question 15. A plano-convex lens with a refractive index of 1.5 and a radius of curvature of 30 cm is silvered at the curved surface. Now this lens has been used to form the image of an object. At what distance from this lens an object be placed in order to have a real image of the size of the object?

1. 20 cm
2. 30 cm
3. 60 cm
4. 80 cm

Answer: 1. 20 cm

Question 16. A plano-convex lens is made of a material with a refractive index p = 1.5. The radius of curvature of the curved surface of the lens is 20 cm. If its plane surface is silvered, the focal length of the silvered lens will be:

1. 10 cm
2. 20 cm
3. 40 cm
4. 80 cm

Answer: 2. 20 cm

Question 17. power of the combination of two lenses of focal lengths 20 cm and 25 cm respectively:

1. + 1 D
2. + 9 D
3. – 1 D
4. – 9 D

Answer: 2. + 9 D

Chapter 7 Geometrical Optics Dispersion Of Light

Question 1. Dispersive power of a prism depends on-

1. Material
2. Prism angle
3. Shape of prism
4. Angle on incidence

Answer: 1. Material

Question 2. When light is passed through a prism, the color which deviates least is:

1. Red
2. violet
3. Blue
4. Green

Answer: 1. Red

Question 3. If the refractive index of red, violet, and yellow lights are 1.42, 1.62, and 1.50 respectively for a medium, its dispersive power will be

1. 0.4
2. 0.3
3. 0.2
4. 0.1

Answer: 1. 0.4

Question 4. Two thin lenses, one convex of focal length 30 cm and the other concave of focal length 10cm are put into contact. If this combination is equivalent to an achromatic lens then the ratio of dispersive powers (ω₁,/ω₂) of the above two lenses is

1. 1/3
2. – 3
3. 3
4. – 1/3

Answer: 3. 3

Question 5. The colors are characterized by which of the following characteristics of light-

1. Frequency
2. Amplitude
3. Wavelength
4. Velocity

Answer: 1. Frequency

Question 6. The dispersion of light in a medium implies that :

1. Lights of different wavelengths travel at different speeds in the medium
2. Lights of different frequencies travel at different speeds in the medium
3. The refractive index of a medium is different for different wavelengths
4. All of the above.

Answer: 4. All of the above.

Question 7. The critical angle of light passing from glass to air is minimal for

1. Red
2. Green
3. Yellow
4. Violet

Answer: 4. Violet

Question 8. A plane glass slab is placed over various colored letters. The letter which appears to be raised the least is:

1. Violet
2. Yellow
3. Red
4. Green

Answer: 3. Red

Question 9. A medium has n_v = 1.56, n_r = 1.44. Then its dispersive power is:

1. 3/50
2. 6/25
3. 0.03
4. None of these

Answer: 2. 6/25

Question 10. All the listed things below are made of flint glass. Which one of these has the greatest dispersive power (ω)?

1. Prism
2. Glass slab
3. Biconvex lens
4. All have same

Answer: 4. All have the same

Question 11. Light of wavelength 4000 Å is incident at a small angle on a prism of apex angle 4°. The prism has n_v = 1.5 and n_r = 1.48. The angle of dispersion produced by the prism in this light is:

1. 0.2°
2. 0.08°
3. 0.192°
4. None of these

Answer: 4. None of these

Question 12. When white light passes through a glass prism, one gets a spectrum on the other side of the prism. In the emergent beam, the ray which is deviating least is or Deviation by a prism is lowest for

1. Violet ray
2. Green ray
3. Red ray
4. Yellow ray

Answer: 3. Red ray

Question 13. A spectrum is formed by a prism of dispersive power ω. If the angle of deviation is ‘δ’, then the angular dispersion is

1. ω/δ
2. δ/ω
3. 1/ ωδ
4. ωδ

Answer: 4. ωδ

Question 14. When white light passes through the achromatic combination of prisms, then what is observed

1. Only deviation
2. Only dispersion
3. Deviation and dispersion
4. None of the above

Answer: 1. Only deviation

Question 15. An achromatic combination of lenses is formed by joining

1. 2 convex lenses
2. 2 concave lenses
3. 1 convex lens and 1 concave lens
4. Convex lens and plane mirror

Answer: 3. 1 convex lens and 1 concave lens

Question 16. An achromatic convergent doublet of two lenses in contact has a power of + 2D. The convex lens has power + 5 D. What is the ratio of the dispersive powers of the convergent and divergent lenses?

1. 2: 5
2. 3: 5
3. 5: 2
4. 5: 3

Answer: 2. 3: 5

Question 17. The ratio of the angle of minimum deviation of a prism when dipped in water and when in the air will be (_aμ_g = 3/2 and _aμ_w = 4/3) If the prism angle is very small

1. 1/8
2. 1/2
3. 3/4
4. 1/4

Answer: 4. 1/4

Question 18. The respective angles of the flint and crown glass prisms are A’ and A. They are to be used for dispersion without deviation, then the ratio of their angles A’/A will be

1. \(-\frac{\left(\mu_y-1\right)}{\left(\mu_y{ }^{\prime}-1\right)}\)
2. \(-\frac{\left(\mu_y{ }^{\prime}-1\right)}{\left(\mu_y-1\right)}\)
3. \(\left(\mu_y{ }^{\prime}-1\right)\)
4. \(\left(\mu_y-1\right)\)

Answer: 1. \(-\frac{\left(\mu_y-1\right)}{\left(\mu_y{ }^{\prime}-1\right)}\)

Question 19. The focal length of a convex lens will be the maximum for

1. Blue light
2. Yellow light
3. Greenlight
4. Red light

Answer: 4. Red light

Question 20. Rainbows are formed by:

1. Reflection and diffraction
2. Refraction and scattering
3. Dispersion and total internal reflection
4. Interference only

Answer: 3. Dispersion and total internal reflection

Question 21. The ratio of the refractive index of red light to blue light in air is

1. Less than unity
2. Equal to unity
3. Greater than unity
4. Less as well as greater than unity depending upon the experimental arrangement

Answer: 1. Less than unity

Question 22. The refractive index of a piece of transparent quartz is the greatest for

1. Red light
2. Violet light
3. Greenlight
4. Yellow light

Answer: 2. Violet light

Question 23. The refractive index for a material for infrared light is

1. Equal to that of ultraviolet light
2. Less than that for ultraviolet light
3. Equal to that for the red color of light
4. Greater than that for ultraviolet light

Answer: 2. Less than that for ultraviolet light

Question 24. With respect to air critical angle in a medium for the light of red color λ₁ is θ. Other facts remain the same, the critical angle for the light of yellow color [λ₂] will be

1. θ
2. More than θ
3. Less than θ
4. None of these

Answer: 3. Less than θ

Question 25. A beam of light composed of red and green rays is incident obliquely at a point on the face of a rectangular glass slab. When coming out of the opposite parallel face, the red and green rays emerge from:

1. Two points propagating in two different non-parallel directions
2. Two points propagating in two different parallel directions
3. One point propagating in two different directions
4. One point propagating in the same direction

Answer: 2. Two points propagating in two different parallel directions

Chapter 7 Geometrical Optics Defects Of Vision

Question 1. A shortsighted person can read a book clearly at a distance of 10 cm from the eyes. The lenses required to read the book kept at 60cm are :

1. Convex lenses of focal length 30 cm
2. Convex lenses of focal length 30 cm
3. Convex lenses of focal length 12 cm
4. Concave lenses of focal length 12 cm

Answer: 4. Concave lenses of focal length 12 cm

Question 2. A person can’t see the objects clearly placed at a distance of more than 40 cm. He is advised to use a lens of power –

1. + 2.5 D
2. – 2.5 D
3. + 0.4 D
4. – 0.4 D

Answer: 2. – 2.5 D

Question 3. A person can see clearly only up to a distance of 25cm. He wants to read a book placed at a distance of 50cm. What kind of lens does he require for his spectacles and what must be its power?

1. Concave, – 1.0 D
2. Convex, + 1.5 D
3. Concave, – 2.0 D
4. Convex, + 2.0 D

Answer: 4. Convex, + 2.0 D

Question 4. Astigmatism (for the human eye) can be removed by using

1. Concave lens
2. Convex lens
3. Cylindrical lens
4. Prismatic lens

Answer: 3. Cylindrical lens

Question 5. The circular part in the center of the retina is called

1. Blind spot
2. Yellow spot
3. Red spot
4. None of the above

Answer: 2. Yellow spot

Question 6. A person cannot see distinctly at a distance of less than one meter. Calculate the power of the lens that he should use to read a book at a distance of 25 cm

1. +3.0 D
2. +0.125D
3. -3.0 D
4. +4.0 D

Answer: 1. +3.0 D

Question 7. A person who can see things most clearly at a distance of 10 cm, requires spectacles to enable them to see clearly things at a distance of 30 cm. What should be the focal length of the spectacles?

1. 15 cm (concave)
2. 15 cm (convex)
3. 10 cm
4. 0

Answer: 1. 15 cm (concave)

Question 8. The power of a lens used by a short-sighted person is – 2 D. Find the maximum distance of an object, which he can see without spectacles

1. 25 cm
2. 50 cm
3. 100 cm
4. 10 cm

Answer: 2. 50 cm

Chapter 7 Geometrical Optics Optical Instrument

Question 1. A simple microscope has a focal length of 5 cm. The magnification at the least distance of distinct vision is-

1. 1
2. 5
3. 4
4. 6

Answer: 4. 6

Question 2. In a compound microscope, the intermediate image is –

1. Virtual, erect, and magnified
2. Real, erect, and magnified
3. Real, inverted, and magnified
4. Virtual, erect, and reduced

Answer: 3. Real, inverted, and magnified

Question 3. The resolving power of a telescope is more when its objective lens has

1. Greater focal length
2. Smaller focal length
3. Greater diameter
4. Smaller diameter

Answer: 3. Smaller focal length

Question 4. A Galileo telescope has an objective of focal length 100 cm & magnifying power of 50. The distance between the two lenses in normal adjustment will be

1. 150 cm
2. 100 cm
3. 98 cm
4. 200 cm

Answer: 3. 98 cm

Question 5. The convex lens is used in

1. Microscope
2. Telescope
3. Projector
4. All of the above

Answer: 4. All of the above

Question 6. The magnifying power of a simple microscope can be increased if an eyepiece of:

1. A shorter focal length is used
2. A longer focal length is used
3. A shorter diameter is used
4. A longer diameter is used

Answer: 1. Shorter focal length is used

Question 7. The focal length of the objective of a microscope is

1. Arbitrary
2. Less than the focal length of the eyepiece
3. Equal to the focal length of the eyepiece
4. Greater than the focal length of the eyepiece

Answer: 2. Less than the focal length of the eyepiece

Question 8. The resolving power of a microscope depends upon

1. The focal length and aperture of the eye lens
2. The focal lengths of the objective and the eye lens
3. The apertures of the objective and the eye lens
4. The wavelength of light illuminating the object

Answer: 4. The wavelength of light illuminating the object

Question 9. An astronomical telescope has an eyepiece of focal length 5 cm. If the angular magnification in normal adjustment is 10, when the final image is at least a distance of distinct vision (25cm) from the eyepiece, then angular magnification will be:

1. 10
2. 12
3. 50
4. 60

Answer: 2. 12

Question 10. A person with a defective sight is using a lens having a power of +2D. The lens he is using is

1. A concave lens with f = 0.5 m
2. A convex lens with f = 2.0 m
3. A concave lens with f = 0.2 m
4. A convex lens with f = 0.5 m

Answer: 4. Convex lens with f = 0.5 m

Question 11. The focal lengths of the objective and eye lens of a microscope are 1 cm and 5 cm respectively. If the magnifying power for the relaxed eye is 45, then the length of the tube is

1. 30 cm
2. 25 cm
3. 15 cm
4. 12 cm

Answer: 3. 15 cm

Question 12. In a compound microscope magnification will be large, if the focal length of the eyepiece is:

1. Large
2. Smaller
3. Equal to that of objective
4. Less than that of objective

Answer: 2. Smaller

Question 13. The focal length of the objective lens of a compound microscope is :

1. Equal to the focal length of its eyepiece
2. Less than the focal length of the eyepiece
3. Greater than the focal length of the eyepiece
4. Any of the above three

Answer: 2. Less than the focal length of the eyepiece

Question 14. Microscope is an optical instrument which :

1. Enlarges the object
2. Increases the visual angle formed by the object at the eye
3. Decreases the visual angle formed by the object at the eye
4. Brings the object nearer

Answer: 2. Increases the visual angle formed by the object at the eye

Question 15. For which of the following colors, the magnifying power of a microscope will be maximum:

1. White colour
2. Red colour
3. Violet colour
4. Yellow colour

Answer: 3. Violet colour

Question 16. The length of the compound microscope is 14 cm. The magnifying power for the relaxed eye is 25. If the focal length of the eye lens is 5 cm, then the object distance for the objective lens will be:

1. 1.8 cm
2. 1.5 cm
3. 2.1 cm
4. 2.4 cm

Answer: 1. 1.8 cm

Question 17. If the focal length of the objective and eye lens are 1.2 cm and 3 cm respectively and the object is put 1.25cm away from the objective lens the final image is formed at infinity. The magnifying power of the microscope is:

1. 150
2. 200
3. 250
4. 400

Answer: 2. 200

Question 18. When the object is self-luminous, the resolving power of a microscope is given by the expression:

1. \(\frac{2 \mu \sin \theta}{1.22 \lambda}\)
2. \(\frac{\mu \sin \theta}{\lambda}\)
3. \(\frac{2 \mu \cos \theta}{1.22 \lambda}\)
4. \(\frac{2 \mu}{\lambda}\)

Answer: 1. \(\frac{2 \mu \sin \theta}{1.22 \lambda}\)

Question 19. the focal length of the objective and eye lens of an astronomical telescope are respectively 2m and 5 cm. The final image is formed at (1) the least distance of distinct vision and (2) infinity. The magnifying power in both cases will be:

1. –48, –40
2. –40, –48
3. –40, 48
4. –48, 40

Answer: 1. –48, –40

Question 20. For observing a cricket match, a binocular is preferred to a terrestrial telescope because:

1. The binocular gives the proper three-dimensional view
2. The binocular has a shorter length
3. The telescope does not give an erect image
4. Telescopes have chromatic aberrations

Answer: 1. The binocular gives the proper three-dimensional view

Question 21. An astronomical telescope has an angular magnification of magnitude 5 for distant objects. The separation between the objective and the eyepiece is 36 cm and the final image is formed at infinity. The focal length f₀ of the objective and the focal length f_e of the eyepiece are:

1. f₀ = 45 cm and f_e = –9 cm
2. f₀ = 7.2 cm and f_e = 5 cm
3. f₀ = 50 cm and f_e = 10 cm
4. f₀ = 30 cm and f_e = 6 cm

Answer: 4. f₀ = 30 cm and f_e = 6 cm

Question 22. In the Galilean telescope, if the powers of an objective and eye lens are respectively +1.25 D and -20D, then for relaxed vision, the length and magnification will be:

1. 21.25 cm and 16
2. 75 cm and 20
3. 75 cm and 16
4. 8.5 cm and 21.25

Answer: 3. 75 cm and 16

Question 23. An astronomical telescope has two lenses of focal powers 0.5 D and 20D. Then its magnifying power will be:

1. 8
2. 20
3. 30
4. 40

Answer: 4. 40

Question 24. The magnifying power of the objective of a compound microscope is 7. If the magnifying power of a microscope is 35, the magnifying power of the eye lens will be:

1. 5
2. 30
3. 35
4. 28

Answer: 1. 5

Question 25. For a telescope, larger the diameter of the objective lens

1. Greater is the resolving power
2. Smaller is the resolving power
3. Greater is the magnifying power
4. Smaller is the magnifying power

Answer: 1. Greater is the resolving power

Question 26. The focal length of the objective and eye-piece of a telescope are respectively 100 cm and 2 cm. The moon subtends an angle of 0.5° at the eye. If it is looked through the telescope, the angle subtended by the moon’s image will be

1. 100°
2. 50°
3. 25°
4. 10°

Answer: 3. 25°

Question 27. The focal lengths of the objective and eyepiece of a microscope are 4 cm and 8 cm respectively and the distance of the object from the objective lens is 4.5 cm then magnifying power is:

1. 18
2. 32
3. 64
4. 20

Answer: 2. 32

Question 28. When the length of the tube of the microscope is increased then the magnifying power :

1. Decreases
2. Increases
3. Remains unchanged
4. May decrease or increase

Answer: 1. Decreases

Question 29. An electron microscope is better than an optical microscope, because of :

1. More resolving power
2. Comfortable use
3. Low purchasing cost
4. Observation can be taken fastly

Answer: 1. More resolving power

Question 30. For a compound microscope, the focal lengths of the object lens and eye lens are f₀ and f_e respectively, then magnification will be done by a microscope when

1. f₀ = f_e
2. f₀ > f_e
3. f₀ < f_e
4. None of these

Answer: 3. f₀ < f_e

Question 31. An astronomical telescope has a large aperture to

1. Reduce spherical aberration
2. Have high resolution
3. Increase the span of observation
4. Have low dispersion

Answer: 2. Have high resolution

Question 32. The wavelength of light used in an optical instrument is λ₁ = 4000 Å and λ₂ = 5000 Å, then the ratio of their respective resolving powers (corresponding to λ₁ and λ₂) is

1. 16: 25
2. 9: 1
3. 4: 5
4. 5: 4

Answer: 4. 5: 4

Question 33. The image formed by an objective of a compound microscope is

1. Virtual and diminished
2. Real and diminished
3. Real and enlarged
4. Virtual and enlarged

Answer: 3. Real and enlarged

Chapter 7 Geometrical Optics Exercise 2 Multiple Choice Questions And Answers

Question 1. A person’s eye is at a height of 1.5 m. He stands in front of a 0.3m long plane mirror which is 0.8 m above the ground. The length of the image he sees of himself is:

1. 1.5m
2. 1.0m
3. 0.8m
4. 0.6m

Answer: 4. 0.6m

Question 2. A point object is kept in front of a plane mirror. The plane mirror is performing SHM of amplitude 2 cm. The plane mirror moves along the x-axis and the x-axis is normal to the mirror. The amplitude of the mirror is such that the object is always in front of the mirror. The amplitude of the SHM of the image is

1. Zero
2. 2 cm
3. 4 cm
4. 1 cm

Answer: 3. 4 cm

Question 3. In the figure shown, the image of a real object is formed at the point I. AB is the principal axis of the mirror. The mirror must be:

1. Concave and placed towards the right of the I
2. Concave and placed towards the left of the I
3. Convex and placed towards the right of the I
4. Convex and placed towards the left of I

Answer: 2. Concave and placed towards the left of I

Question 4. AB is an incident beam of light and DC is a reflected beam (the number of reflections for this may be 1 or more than 1) of light. AB and DC are separated by some distance (maybe large). It is possible by placing what type of mirror on the right side.

1. Two plane mirror
2. One concave mirror
3. One prism
4. All of the above

Answer: 4. All of the above

Question 5. A point object at 15 cm from a concave mirror of a radius of curvature of 20 cm is made to oscillate along the principal axis with an amplitude of 2 mm. The amplitude of its image will be

1. 2 mm
2. 4 mm
3. 8 mm
4. 16 mm

Answer: 3. 8 mm

Question 6. The distance between an object and its doubly magnified image by a concave mirror is: [Assume f = focal length]

1. 3 f/2
2. 2 f/3
3. 3 f
4. Depends on whether the image is real or virtual.

Answer: 1. 3 f/2

Question 7. A ray of light is incident on a parallel slab of thickness t and refractive index n. If the angle of incidence θ is small, then the displacement in the incident and emergent ray will be:

1. \(\frac{t \theta(n-1)}{n}\)
2. \(\frac{t \theta}{n}\)
3. \(\frac{t \theta n}{n-1}\)
4. None

Answer: 1. \(\frac{t \theta(n-1)}{n}\)

Question 8. A concave spherical surface with a radius of curvature 10 cm separates two mediums X and Y of refractive indices 4/3 and 3/2 respectively. The Centre of curvature of the surface lies in the medium X. An object is placed in medium X.

1. Image is always real
2. The image is real if the object’s distance is greater than 90 cm.
3. Image is always virtual
4. The image is virtual only if the object’s distance is less than 90 cm.

Answer: 3. Image is always virtual

Question 9. The observer ‘O’ sees the distance AB as infinitely large. If the refractive index of the liquid is μ₁ and that of glass is μ₂, then \(\frac{\mu_1}{\mu_2}\) is:

1. 2
2. 1/2
3. 4
4. None of these

Answer: 1. 2

Question 10. In the given figure a plano-concave lens is placed on a paper on which a flower is drawn. How far above its actual position does the flower appear to be?

1. 10 cm
2. 15 cm
3. 50 cm
4. None of these

Answer: 1. 10 cm

Question 11. The focal lengths of the objective & the eyepiece of a compound microscope are 1 cm and 5 cm respectively. An object placed at a distance of 1.1 cm from the objective has its final image formed at 25 cm from the eyepiece. The length of the microscope tube is:

1. 6.1 cm
2. 49/8 cm
3. 6 cm
4. 91/6 cm

Answer: 4. 91/6 cm

Question 12. A concave mirror of focal length 15 cm forms an image having twice the linear dimension (height) of the object. The position of the object when the image is virtual will be:

1. 45 cm
2. 30 cm
3. 7.5 cm
4. 22.5 cm

Answer: 3. 7.5 cm

Question 13. Four lenses are made from the same type of glass. The radius of curvature of each face is given below. Which will have the greatest positive power?

1. 10 cm convex and 15 concave
2. 5 cm convex and 10 cm concave
3. 15 cm convex and plane
4. 20 cm convex and 30 cm concave

Answer: 2. 5 cm convex and 10 cm concave

Question 14. A prism of refractive index has a refracting angle of 60°. Ar what angle a ray must be incident on it so that it suffers a minimum deviation?

1. 45°
2. 60°
3. 90°
4. 180°

Answer: 1. 45°

Question 15. The refractive index of the material of the prism and liquid are 1.56 and 1.32 respectively. What will be the value of θ for the following refraction?

1. \(\sin \theta>\frac{13}{11}\)
2. \(\sin \theta>\frac{11}{13}\)
3. \(\sin \theta>\frac{\sqrt{3}}{2}\)
4. \(\sin \theta>\frac{1}{\sqrt{2}}\)

Answer: 2. \(\sin \theta>\frac{11}{13}\)

Question 16. In a laboratory four convex lenses L₁, L₂, L₃, and L₄ of focal lengths 2, 4, 6, and 8 cm, respectively are available. Two of these lenses form a telescope of length 10 cm and magnifying power 4. The objective and eye lenses are respectively

1. L₂, L₃,
2. L₂, L₄
3. L₁, L₂,
4. L₄, L₁

Answer: 4. L₄, L₁

Question 17. Distance stars are viewed with the help of an astronomical telescope. The angular separation between two stars can be just resolved by the telescope.

1. Is independent of the diameter of the aperture of the telescope
2. Increases with the increases in the diameter of the aperture of the telescope
3. Decreases with the increase in the diameter of the telescope aperture
4. Increases quadratically with the diameter of the telescope aperture

Answer: 3. Decreases with the increase in the diameter of the telescope aperture

Question 18. The refractive index of water is 5/3. A light source is placed in water at a depth of 4 m. Then what must be the minimum radius of the disc placed on the water surface so that the light of the source can be stopped?

1. 3 m
2. 4 m
3. 5 m
4. ∞

Answer: 1. 3 m

Question 19. Statement-1: The formula connecting u, v, and f for a spherical mirror is valid only for mirrors whose sizes are very small compared to their radii of curvature.

Statement-2: Laws of reflection are strictly valid for plane surfaces, but not for large spherical surfaces.

1. Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
2. Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
3. Statement-1 is True, Statement-2 is False
4. Statement-1 is False, and Statement-2 is True.

Answer: 3. Statement-1 is True, Statement-2 is False

Chapter 7 Geometrical Optics Exercise 3 Multiple Choice Questions And Answers

Question 1. A boy is trying to start a fire by focusing sunlight on a piece of paper using a biconvex lens with a focal length of 10 cm. The diameter of the sun is 1.39 x 10⁹ m and its mean distance from the earth is 1.5 x 10¹¹ m. What is the diameter of the sun’s image on the paper?

1. 9.2 x 10^-4 m
2. 6.5 x 10^-4 m
3. 6.5 x 10^-5 m
4. 12.4 x 10^-4 m

Answer: 1. 9.2 x 10^-4 m

Question 2. Two thin lenses of focal lengths f₁ and f₂ are in contact and coaxial. The power of the combination is:

1. \(\sqrt{\frac{f_1}{f_2}}\)
2. \(\sqrt{\frac{f_2}{f_1}}\)
3. \(\frac{f_1-f_2}{f_2 f_1}\)
4. \(\frac{f_1+f_2}{f_1 f_2}\)

Answer: 4. \(\frac{f_1+f_2}{f_1 f_2}\)

Question 3. A ray of light traveling in a transparent medium of refractive index μ falls on a surface separating the medium from the air at an angle of incidence of 45°. For which of the following values of μ the ray can undergo total internal reflection?

1. μ = 1.33
2. μ = 1.40
3. μ = 1.50
4. μ = 1.25

Answer: 3. μ = 1.50

Question 4. A lens having focal length f and aperture of diameter d forms an image of intensity I. The aperture of diameter in the central region of the lens is covered by black paper. The focal length of the lens and intensity of the image now will be respectively

1. f and \(\frac{I}{4}\)
2. \(\frac{3 \mathrm{f}}{4}\) and \(\frac{I}{2}\)
3. f and \(\frac{3 I}{4}\)
4. \(\frac{\mathrm{f}}{2}\) and \(\frac{\mathrm{I}}{2}\)

Answer: 3. f and \(\frac{3 I}{4}\)

Question 5. The speed of light in media M₁ and M₂ are 1.5 × 10⁸ m/s and 2.0 × 10⁸ m/s respectively. A ray of light enters from medium M₂ to M₂ at an incidence angle i. If the ray suffers total internal reflection, the value of i is

1. Equal to \(\sin ^{-1}\left(\frac{2}{3}\right)\)
2. Equal to or less than \(\sin ^{-1}\left(\frac{3}{5}\right)\)
3. Equal to or greater than \(\sin ^{-1}\left(\frac{3}{4}\right)\)
4. Less than \(\sin ^{-1}\left(\frac{2}{3}\right)\)

Answer: 3. Equal to or greater than \(\sin ^{-1}\left(\frac{3}{4}\right)\)

Question 6. A ray of light is incident on a 60° prism at the minimum deviation position. The angle of refraction at the first face (i.e., incident face) of the prism is

1. Zero
2. 30°
3. 45°
4. 60°

Answer: 2. 30°

Question 7. Which of these is not due to total internal reflection?

1. Working with optical fiber
2. Difference between the apparent and real depth of the pond
3. Mirage on hot summer days
4. Brilliance of diamond

Answer: 2. Difference between apparent and real depth of the pond

Question 8. The dimensions of (μ₀∈₀)-½are:

1. [L^1/2 T^-1/2]
2. [L^-1 T]
3. [L T^-1]
4. [L^-1/2 T^1/2]

Answer: 3. [L T^-1]

Question 9. A biconvex lens has a radius of curvature of magnitude 20 cm. Which one of the following options describes best the image formed of an object of height 2 cm placed 30 cm from the lens? (Refractive index of lens material = 1.5)

1. Virtual, upright, height = 1 cm
2. Virtual, upright, height = 0.5 cm
3. Real, inverted, height = 4 cm
4. Real, inverted, height = 1cm

Answer: 3. Real, inverted, height = 4 cm

Question 10. A thin prism of angle 15º made of glass of refractive index µ₁ = 1.5 is combined with another prism of glass of refractive index µ₂ = 1.75. The combination of the prism produces dispersion without deviation. The angle of the second prism should be:

1. 7º
2. 10º
3. 12º
4. 5º

Answer: 2. 10º

Question 11. A conversing beam of rays is incident on a diverging lens. Having passed through the lens the rays intersect at a point 15 cm from the lens on the opposite side. If the lens is removed the point where the rays meet will move 5 cm closer to the lens. The focal length of the lens is:

1. –10 cm
2. 20 cm
3. –30 cm
4. 5 cm

Answer: 3. –30 cm

Question 12. When a biconvex lens of glass having a refractive index of 1.47 is dipped in a liquid, it acts as a plane sheet of glass. This implies that the liquid must have a refractive index.

1. Equal to that of glass
2. Less then one
3. Greater than that of glass
4. Less than that of glass

Answer: 1. Equal to that of glass

Question 13. A ray of light is incident at an angle of incidence, i, on one face of the prism of angle A (assumed to be small), and emerges normally from the opposite face. If the refractive index of the prism is μ, the angle of incidence i, is nearly equal to:

1. μA
2. A/μ
3. A/2μ

Answer: 1. μA

Question 14. A concave mirror of focal length ‘f₁’ is placed at a distance of ‘d’ from a convex lens of focal length ‘f₂’ A beam of light coming from infinity and falling on this convex lens – concave mirror combination returns to infinity. The distance ‘d’ must equal:

1. f₁ + f₂
2. –f₁ + f₂
3. 2f₁ + f₂
4. –2f₁ + f₂

Answer: 3. 2f₁ + f₂

Question 15. The magnifying power of a telescope is 9. When it is adjusted for parallel rays the distance between the objective and eyepiece is 20 cm. The focal lengths of lenses are:

1. 10 cm, 10 cm
2. 15 cm, 5 cm
3. 18 cm, 2 cm
4. 11 cm, 9 cm

Answer: 3. 18 cm, 2 cm

Question 16. The dimensions of (μ₀ε₀)^-1/2 are:

1. [L^1/2T^-1/2]
2. [L^-1T]
3. [LT^-1]
4. [L^1/2T^1/2]

Answer: 3. [LT^-1]

Question 17. A plano convex lens fits exactly into a plano concave lens. Their plane surfaces are parallel to each other. If lenses are made of different materials of refractive indices μ₁ and μ₂ and R is the radius of curvature of the curved surface of the lenses, then the focal length of the combination is:

1. \(\frac{\mathrm{R}}{2\left(\mu_1-\mu_2\right)}\)
2. \(\frac{R}{\left(\mu_1-\mu_2\right)}\)
3. \(\frac{2 R}{\left(\mu_2-\mu_1\right)}\)
4. \(\frac{\text { [NEET }}{2\left(\mu_1+\mu_2\right)}\)

Answer: 2. \(\frac{R}{\left(\mu_1-\mu_2\right)}\)

Question 18. For a normal eye, the cornea of the eye provides a converging power of 40 D, and the least converging power of the eye lens behind the cornea is 20 D. Using this information, the distance between the retina and the cornea-eye lens can be estimated to be:

1. 2.5 cm
2. 1.67 cm
3. 1.5 cm
4. 5 cm

Answer: 2. 1.67 cm

Question 19. If the focal length of the objective lens is increased then the magnifying power of:

1. The microscope will increase but that of the telescope decrease
2. Microscope and telescope both will increase
3. Microscope and telescope both will decrease
4. Microscope will decrease but that of telescope will increase.