NEET Physics Class 11

NEET Physics Solutions For Class 11 Centrifuge Force

A Centrifuge: A centrifuge works on the principle of centrifugal force.

• The centrifuge consists of two steel tubes suspended from the ends of a horizontal bar which can be rotated at high speed in a horizontal plane by an electric motor.
• The tubes are filled with the liquid and the bar is set into rotation.
• Due to rotational motion, the tubes get tied and finally become horizontal.
• Due to heavy mass, the heavier particles experience more centrifugal force than that of the liquid particles. Therefore, is then stopped so that the tubes become vertical.

Question 1. Two balls of equal masses are attached to a string at a distance of 1 m and 2 m from one end as shown in Fig. The string with masses is then moved in a horizontal circle with constant speed. Find the ratio of the tension T₁ and T₂.
Answer:

Let the balls of the two circles be r₁ and r₂. The linear speeds of the two masses are v₁= ωr₁, v₂= ωr₂

where ω is the angular speed of the circular motion. The tension in the strings is such that

⇒ \(T_2=\frac{m v_2^2}{r_2}=m \omega^2 r_2\)

Centrifugal Force in NEET Physics Class 11: Explanation and Solutions

∴ \(T_1=m \omega^2 r_1+T_2=m \omega^2\left(r_1+r_2\right)\)

∴ \(\frac{T_1}{T_2}=\frac{r_1+r_2}{r_2}=\frac{1+2}{2}=\frac{3}{2}\)

Conical Pendulum:

(This is the best Question of uniform circular motion)

A conical pendulum consists of a body attached to a string, such that it can revolve in a horizontal circle with uniform speed. The string traces out a cone in the space.

The force acting on the bob is

1. Tension T
2. Weight mg

The horizontal component T sinθ of the tension T provides the centripetal force and the vertical component T cos θ balances the weight to bob

∴ T sinθ = and T cosθ = mg

From these equations ………(1)

and tan θ = ………(2)

If h = height of conical pendulum tanθ = = ………(3)

From (2) and (3)

The time period of the revolution

Hints To Solve Numerical Problems (UCM)

1. First, show all forces acting on a particle
2. Resolve these forces along radius and tangent.
3. The resultant force along the radial direction provides the necessary centripetal force.
4. Resultant force along tangent = Ma_r= 0 (a_r= tangential acceleration)

Question 1. A vertical rod is rotating about its axis with a uniform angular speed ω. A simple pendulum of length l is attached to its upper end what is its inclination with the rod?
Answer:

Let the radius of the circle in which the bob is rotating be, the tension in the string is T, the weight of the bob mg, and the inclination of the string θ. Then T cos θ balances the weight mg and T sin θ provides the centripetal force necessary for circular motion.

That is –

T cos θ = mg and T sin θ = mω² x

but x = l sin θ

∴ T = mω² l

and \(\cos \theta=\frac{\mathrm{mg}}{\mathrm{T}}=\frac{\mathrm{mg}}{\mathrm{m} \omega^2 \ell}\) or \(\theta=\cos ^{-1}\left(\frac{g}{\omega^2 \ell}\right)\)

Question 2. A circular loop has a small bead that can slide on it without friction. The radius of the loop is r. Keeping the loop vertically it is rotated about a vertical diameter at a constant angular speed ω. What is the value of angle θ, when the bead is in dynamic equilibrium?
Answer:

Centripetal force is provided by the horizontal component of the normal reaction N. The vertical component balances the weight. Thus

N sin θ = mω²x and N cos θ = mg

Also x = r sin θ ⇒ N = mω²r

cos θ = \(\frac{g}{\omega^2 r}\) or \(\theta=\cos ^{-1}\left(\frac{g}{r \omega^2}\right)\)

Question 3. A particle of mass m slides down from the vertex of the semihemisphere, without any initial velocity. At what height from the horizontal will the particle leave the sphere?
Answer:

Let the particle leave the sphere at height h, \(\frac{\mathrm{mv}^2}{\mathrm{R}}=\mathrm{mg} \cos \theta-\mathrm{N}\)

When the particle leaves the sphere N = 0,

⇒ \(\frac{m v^2}{R}=m g \cos \theta \Rightarrow v^2=g R \cos \theta\)

According to law of conservation of energy ( K . E.+ P. E.) at A =( K . E.+ P. E.) at B

⇒ \(o+m g R=\frac{1}{2} m v^2+m g h \Rightarrow v^2=2 g(R-h)\)

From 1 and 2 h = \(\frac{2 R}{3}\) Also cosθ = 2/3

Question 4. A particle describes a horizontal circle of radius r in a funnel-type vessel of the frictionless surface with half one angle θ. If the mass of the particle is m, then in dynamical equilibrium the speed of the particle must be –
Answer:

The normal reaction N and weight mg are the only forces acting on the particle (inertial frame view), the N is making an angle \(\left(\frac{\pi}{2}-\theta\right)\) with the vertical.

The vertical component of N balances the weight mg and the horizontal component provides the centripetal force required for circular motion.

Thus

⇒ \(N \cos \left(\frac{\pi}{2}-\theta\right)=m g\)

⇒ \(N \sin \left(\frac{\pi}{2}-\theta\right)=\frac{m^2}{r}\)

or N sin θ = \(\frac{m v^2}{r} \mathrm{mg}\)

N cos θ = tan θ = \(\frac{\mathrm{rg}}{\mathrm{v}^2}\) on dividing we get ,

so \(v=\sqrt{\frac{\mathrm{rg}}{\tan \theta}}\)

Question 5. Prove that a motor car moving over a

1. Convex bridge is lighter than the same car resting on the same bridge.
2. The concave bridge is heavier than the same car resting on the same bridge.

Answer:

Apparent weight of car = N (normal reaction)

1. Convex bridge

The motion of the motor car over a convex bridge is the motion along the segment of a circle. The centripetal force is provided by the difference in weight mg of the car and the normal reaction N of the bridge.

∴ \(\mathrm{mg}-\mathrm{N}=\frac{\mathrm{mv}^2}{\mathrm{r}}\)

or N = mg – \(\frac{m v^2}{r}\)

Clearly N < mg, i.e., the apparent weight of the moving car is less than the weight of the stationary car.

2. Concave bridge N – mg = \(\frac{m v^2}{r}\)

Apparent weight N = mg + \(\frac{m v^2}{r}\)

Motion In Vertical Circle:

Motion of a body suspended by string: This is the best Question of non-uniform circular motion.

Suppose a particle of mass m is attached to an inexcusable light string of length r. The particle is moving in a vertical circle of radius r, about a fixed point O.

At lost point A velocity of particle = u (in a horizontal direction)

After covering ∠θ velocity of particle = v (at point B)

Resolve weight (mg) into two components

1. mg cos θ (along radial direction)
2. mg sin θ (tangential direction)

Then force T – mg cos θ provides the necessary centripetal force

T – mg cosθ = \(\frac{m v^2}{\mathrm{r}}\) ……(1)

Δ OCB cos θ = \(\frac{r-h}{r}\) ……(2)

or h = r (1 – cosθ)

By conservation of energy at points A and B

⇒ \(\frac{1}{2} m u^2=\frac{1}{2} m v^2+m g h \text { or } u^2=v^2+2 g h\)

or v² = u² + 2gh ……..(3)

Substitute value of cos θ and v² in equn. (1)

⇒ \(T-m g\left[\frac{r-h}{r}\right]=\frac{m}{r}\left(u^2-2 g h\right) \text { or } T=\frac{m}{r}\left[u^2-2 g h+g r-g h\right) \text { or } T=\frac{m}{r}\left[u^2+g r-3 g h\right]\) …………(4)

1. If velocity becomes zero at height h₁

⇒ \(O=u^2-2 g h,\) or \(\mathrm{h}_1\)

=\(\frac{\mathrm{u}^2}{2 \mathrm{~g}}\) ………..

2. If tension becomes zero at height h₂

⇒ \(\mathrm{O}=\frac{\mathrm{m}}{\mathrm{r}}\left[\mathrm{u}^2+\mathrm{gr}-3 \mathrm{g} \mathrm{h}_2\right]\)

or \(\mathrm{u}^2+\mathrm{gr}-3 \mathrm{gh}_2=0\)

or \(h_2=\frac{u^2+g r}{3 g}\) …….. (4)

NEET Physics Centrifugal Force: Key Concepts and Solutions

3. Case of oscillation

It v = 0, T ≠ 0 then h₁< h₂

⇒ \(\frac{u^2}{2 g}<\frac{u^2+g r}{3 g}\)

⇒ \(3 u^2<2 u^2+2 g r\)

⇒ \(\mathrm{u}^2<2 \mathrm{gr}\)

⇒ \(u<\sqrt{2 g r}\)

4. Case Of Leaving The Circle

If v = 0, T = 0 then h₁ < h₂

⇒ \(\frac{u^2}{2 g}>\frac{u^2+g r}{3 g}\)

⇒ \(\sqrt{5 \mathrm{gr}}>\mathrm{u}>\sqrt{2 \mathrm{gr}}\)

Case of leaving the circle

⇒ \(3 \mathrm{u}^2>2 \mathrm{u}^2+2 \mathrm{gr}\)

⇒ \(\mathrm{u}^2>2 \mathrm{gr}\)

⇒ \(u>\sqrt{2 g r}\)

⇒ \(\sqrt{5 \mathrm{gr}}>\mathrm{u}>\sqrt{2 \mathrm{gr}}\)

5. Case Of Complete The Circle

Case of completing the circle or looping the loop

NEET Physics Centrifugal Force: Solved Questions and Explanations

⇒ \(u \geq \sqrt{5 \mathrm{gr}}\)

T > 0

v ≠ 0

Special Note

The same conditions apply if a particle moves inside a smooth spherical shell of radius R. The only difference is that the tension is replaced by the normal reaction N.

This is shown in the figure given below \(v=\sqrt{g R} \quad N=0\)

1. Condition of looping the loop is u ≥ \(\sqrt{5 g R}\)

Centrifugal Force NEET Physics Class 11 Solutions and Problems

2. Condition of leaving the circle \(\sqrt{2 \mathrm{gR}}<\mathrm{u}<\sqrt{5 \mathrm{gR}}\)

3. Condition of oscillation is 0 < u ≥ \(\sqrt{2 g R}\)

Question 1. A ball is released from height h. Find the condition for the particle to complete the circular path.
Answer:

According to law of conservation of energy (K.E. + P.E) at A = (K.E. + P.E) at B

⇒ \(0+m g h=\frac{1}{2} m v^2+0\)

⇒ \(v=\sqrt{2 g h}\)

But velocity at the lowest point of the circle,

⇒ \(v \geq \sqrt{5 g R} \Rightarrow \sqrt{2 g h} \geq \sqrt{5 g R} \Rightarrow h \geq \frac{5 R}{2}\)

Question 2. A body weighing 0.4 kg is whirled in a vertical circle making 2 revolutions per second. If the radius of the circle is 1.2 m, find the tension in the string, when the body is

1. At the top of the circle
2. At the bottom of the circle. Given : g = 9.8 ms^-2 and π = 1.2 m

Answer:

Mass m = 0.4 kg time period = \(\frac{1}{2}\) second and radius, r = 1.2 m

Angular velocity,\(\omega=\frac{2 \pi}{1 / 2}=4 \pi \mathrm{rad} \mathrm{s}^{-1}=12.56 \mathrm{rad} \mathrm{s}^{-1}\)

At the top of the circle, \(\mathrm{T}=\frac{\mathrm{m} \mathrm{v}^2}{\mathrm{r}}-\mathrm{mg}=\mathrm{mr} \omega^2-\mathrm{mg}=\mathrm{m}\left(\mathrm{r} \omega^2-\mathrm{g}\right)\)

= 0.4 (1.2 × 12.56 × 12.56 – 9.8) N = 71.8 N

At the lowest point, T = m(rω² + g) = 79.64 m

Question 3. In a circus a motorcyclist moves in a vertical loop inside a ‘death well’ (a hollow spherical chamber with holes, so that the spectators can watch from outside). Explain clearly why the motorcyclist does not drop down when the is at the uppermost point, with no support from below. What is the minimum speed required to perform a vertical loop if the radius of the chamber is 25 m?
Answer:

When the motorcyclist is at the highest point of the death well, the normal reaction R on the motorcyclist by the ceiling of the chamber acts downwards. His weight mg also acts downwards. These two forces are balanced by the outward centrifugal force acting on him.

∴ R + mg = \(\frac{m v^2}{r}\) …….(1) r = radius of the circle

Here v is the speed of the motorcyclist and m is the mass of the motorcyclist (including the mass of the motorcycle). Because of the balancing of the forces, the motorcyclist does not fall down.

The minimum speed required to perform a vertical loop is given by equation (i), when R = 0

∴ mg = \(\frac{m v_{\text {min }}^2}{r} \text { or } v_{\min }^2=\mathrm{gr}\)

or \(v_{\min }=\sqrt{\mathrm{gr}}=\sqrt{9.8 \times 25} \mathrm{~ms}^{-1}=15.65 \mathrm{~ms}^{-1}\)

So, the minimum speed at the top required to perform a vertical loop is 15.65 ms–1.

Question 4. A 4kg ball is swung in a vertical circle at the end of a cord 1 m long. What is the maximum speed at which it can swing if the cord can sustain a maximum tension of 163.6 N?
Answer:

Maximum tension = \(T=\frac{m v^2}{r}\) +mg(at lowest point)

∴ \(\frac{m v^2}{r}=T-m g\)

or \(\frac{4 v^2}{1}\) = 163.6 – 4 × 9.8

Solving we get v = 6 m/sec

Question 5. A small body of mass m = 0.1 kg swings in a vertical circle at the end of a chord of length 1 m. Its speed is 2 m/s when the chord makes an angle θ = 30º with the vertical. Find the tension in the chord.
Answer:

The equation of motion is

⇒ \(T-m g \cos \theta=\frac{m v^2}{r}\) or \(T=m g \cos \theta+\frac{m v^2}{r}\)

Substituting the given values, we get

T = 0.1 × 9.8 × cos 30 + \(\frac{0.1 \times(2)^2}{1}=0.98 \times\left(\frac{\sqrt{3}}{2}\right)+0.4\)

= 0.85 + 0.4 = 1.25 N

Special Notes Important Point:

If a particle of mass m is connected to a light rod and whirled in a vertical circle of radius R, then to complete the circle, the minimum velocity of the particle at the bottommost points is not. Because in this case, the velocity of the particle at the topmost point can be zero also. Using the conservation of mechanical energy between points A and B (1) we get

⇒ \(\frac{1}{2} m\left(u^2-v^2\right)=m g h\)

or \(\frac{1}{2} m u^2=m g(2 R)\)

∴ u = \(2 \sqrt{g R}\)

Therefore, the minimum value of u in this case is \(2 \sqrt{g R}\)

The same is the case when a particle is compelled to move inside a smooth vertical tube

NEET Class 11 Physics Centrifugal Force: Study Notes and Solutions

Particle Application Of Circular Motion

A Cyclist Making A Turn: Let a cyclist moving on a circular path of radius r bend away from the vertical by an angle θ. If R is the reaction of the ground, then R may be resolved into two components horizontal and vertical.

The vertical component R cos θ balances the weight mg of the cyclist and the horizontal component R sin θ provides the necessary centripetal force for circular motion.

R sin θ = \(\frac{m v^2}{r}\) ……. (1)

and R cos θ = mg ……. (2)

Dividing by (2), we get

⇒ \(\tan \theta \frac{\mathrm{v}^2}{\mathrm{rg}}\) ……. (3)

For less bedding of cyclists, his speed v should be smaller, and the radius r of a circular path should be greater. If μ is the coefficient of friction, then for no skidding of cycle (or overturning of cyclist)

⇒ \(\mu \geq \tan \theta\) …… (4)

⇒ \(\mu \geq \frac{v^2}{r g}\)

An Aeroplane Making A Turn

In order to make a circular turn, a plane must roll at some angle θ in such a manner that the horizontal component of the lift force L provides the necessary centripetal force for circular motion. The vertical component of the lift force balances the weight of the plane.

L sin θ = \(\frac{m v^2}{r}\)

and L cos θ = mg

or the angle θ should be such that tan θ = \(\frac{v^2}{rg}\)

Death Well And Rotor: Example of uniform circular motion In ‘Death Well’ a person drives a bicycle on the vertical surface of a large wooden well.

Centrifugal Force NEET Class 11: Practice Problems and Solutions

• In ‘Death Well’ walls are at rest while the person revolves.
• In a rotor, at a certain angular speed of the rotor, a person hangs resting against the wall without any floor.
• In the rotor, a person is at rest and the walls rotate.
• In both these cases friction balances the weight of a person while reaction provides the centripetal force necessary for circular motion i.e.

Force of fiction F_s= mg and Normal reaction \(F_N=\frac{m v^2}{r}\)

so \(\frac{F_N}{F_{\mathrm{s}}}=\frac{v^2}{r g}\) i.e., v = \(\sqrt{\frac{\mathrm{rgF}_{\mathrm{N}}}{\mathrm{F}_{\mathrm{s}}}}\)

Now for v to be minimum FS must be maximum, i.e., \(v_{\min }=\sqrt{\frac{\mathrm{gg}}{\mu}}\) [as F_SMax = μFN]

Question 6. A 62 kg woman is a passenger in a “rotor ride” at an amusement park. A drum of radius 5.0 m is spun with an angular velocity of 25 rpm. The woman is pressed against the wall of the rotating drum. Calculate the normal force of the drum of the woman (the centripetal force that prevents her from leaving her circular path).  While the drum rotates, the floor is lowered. A vertical static friction force supports the woman’s weight. What must the coefficient of friction be to support her weight? (ω = 25 rev/min, r = 5m)

Answer:

Normal force exerted by the drum on the woman towards the center

⇒ \(F_N=m a_c=m \omega^2 r=62 \mathrm{~kg} \times\left(25 \frac{\mathrm{rev}}{\mathrm{min}} \times \frac{2 \pi \mathrm{rad}}{1 \mathrm{rev}} \times \frac{1 \mathrm{~min}}{60 \mathrm{~s}}\right)^2 \times 5 \mathrm{~m}=2100 \mathrm{~N}\)

μ = FN = F mg N ………

dividing eqn. (2) be eq. (1)

⇒ \(\mu=\frac{g}{\omega^2 r}=\left(\frac{60}{2 \pi \times 25}\right)^2 \times \frac{10}{5}=0.292\)

Question 7. A 1.1 kg block slides on a horizontal frictionless surface in a circular path at the end of a 0.50 m long string.

1. Calculate the block’s speed if the tension in the string in 86 N.
2. By what percent does the tension change if the block speed decreases by 10 percent?

Answer:

1. Force diagram for the block. The upward normal force balances the block’s weight.

The tension force of the string on the block provides the centripetal force that keeps the 50 blocks moving in a circle. Newton’s second law for forces along the radial direction is ∑ F (in a radial direction) \(T=\frac{m v^2}{r}\)

or \(v=\sqrt{\frac{T r}{m}}=\sqrt{\frac{(80 \mathrm{~N})(0.50 \mathrm{~m})}{1.2 \mathrm{~kg}}}=5.0 \mathrm{~m} / \mathrm{s}\)

2. A 10 percent reduction in the speed results in a speed v’ = 5.4 m/s. The new tension is

⇒ \(T^{\prime}=\frac{m v^{\prime 2}}{\mathrm{r}}-\frac{(1.2 \mathrm{~kg})(5.4 \mathrm{~m} / \mathrm{s})^2}{0.50 \mathrm{~m}}=70 \mathrm{~N}\)

Thus, \(\frac{\mathrm{T}^{\prime}}{\mathrm{T}}=\frac{70 \mathrm{~N}}{86 \mathrm{~N}}=0.81\)

The percentage reduction in the tension is about 19%.

The same result is obtained using a proportionality method.

⇒ \(\frac{T^{\prime}}{T}=\frac{\left(\mathrm{mv}^{\prime 2} / \mathrm{r}\right)}{\left(\mathrm{mv}^2 / \mathrm{r}\right)}=\left(\frac{\mathrm{v}^{\prime}}{\mathrm{v}}\right)^2=\left(\frac{0.90}{\mathrm{v}}\right)^2=0.81\)

Looping The Loop: This is the best Question of nonuniform circular motion in a vertical plane.

For looping the pilot of the plane puts off the engine at the lowest point and traverses a vertical loop. (with variable velocity).

Question 8. An airplane moves at 64 m/s in a vertical loop of radius 120 m. Calculate the force of the plane’s seat on a 72 kg pilot while passing through the bottom part of the loop.

Answer:

Two forces act on the pilot his downward weight force w and the upward force of the aeroplane’s seat F_seat. Because the pilot moves in a circular path, these forces along the radial direction must, according to Newton’s second law (∑ F = ma), equal the pilot’s mass times his centripetal acceleration, where

⇒ \(a_c=v^2 /r\) we find F (in radial direction) = \(F_{\text {seat }}-w=\frac{m v^2}{r}\)

Remember that force pointing towards the center of the circle (Fseat) is positive and those pointing away from the center (w) are negative.

Substituting ω = mg and rearranging, we find that the force of the airplane seat on the pilot is

⇒ \(F_{\text {seat }}=m\left(\frac{\mathrm{v}^2}{\mathrm{r}}+\mathrm{g}\right)=72 \mathrm{~kg}\left[\frac{64(\mathrm{~m} / \mathrm{s})^2}{120 \mathrm{~m}}+9.8 \mathrm{~m} / \mathrm{s}^2\right]\)

= 72 kg (34.1 m/s² + 9.8m/s²) = 3160.8 N

The pilot in this Question feels very heavy. To keep him in the circular path, the seat must push the pilot upwards with a force of 3160 N, 4.5 times his normal weight. He experiences an acceleration of 4.5 g, that is, 4.5 times the acceleration of gravity.

A Car Taking A Turn On A Level Road: When a car takes a turn on a level road, the portion of the turn can be approximated by an arc of a circle of radius r.

If the car makes the turn at a constant speed v, then there must be some centripetal force acting on the car. This force is generated by the friction between the tire and the road. (A car tends to slip radially outward, so frictional force acts inwards)

μ_s is the coefficient of static friction

N = mg is the normal reaction of the surface

The maximum safe velocity v is –

⇒ \(\frac{m v^2}{r}-\mu_s N=\mu_s m g\)

or \(\mu_{\mathrm{s}}=\frac{v^2}{\mathrm{rg}}\)

or \(v=\sqrt{\mu_s r g}\)

It is independent of the mass of the car. The safe velocity is the same for all vehicles of larger and smaller mass.

Question 9. A car is traveling at 30 km/h in a circle of radius 60 m. What is the minimum value of μs for the car to make the turn without skidding?
Answer:

The minimum μSshould be that

⇒ \(\mu_{\mathrm{s}} \mathrm{mg}=\frac{\mathrm{mv}^2}{r}\) or \(\)

Here, \(v=30 \frac{\mathrm{km}}{\mathrm{h}}=\frac{30 \times 1000}{3600}=\frac{25}{3} \mathrm{~m} / \mathrm{s}\)

⇒ \(\mu_s=\frac{25}{3} \times \frac{25}{3} \times \frac{1}{60 \times 10}=0.115\)

For all values of μ_S greater than or equal to the above value, the car can make the turn without skidding. If the speed of the car is high so that the minimum μ_s is greater than the standard values (rubber tire on dry concrete μ_s= 1 and on wet concrete μ_s= 0.7), then the car will skid.

Banking Of Road: If a cyclist takes a turn, he can bend from his vertical position. This is not possible in the case of cars, trucks, or trains.

NEET Physics Class 11: Centrifugal Force Conceptual Solutions

The tilting of the vehicle is achieved by raising the outer edge of the circular track, slightly above the inner edge. This is known as the banking of the curved track. The angle of inclination with the horizontal is called the angle of banking. If the driver moves with a slow velocity friction does not play any role in negotiating the turn. The various forces acting on the vehicle are :

1. Weight of the vehicle (mg) in the downward direction.
2. Normal reaction (N) perpendicular to the inclined surface of the road.

Resolve N in two components.

N cosθ, vertically upwards which balances the weight of the vehicle.

∴ N cosθ = mg …….(1)

N sin θ, in the horizontal direction which provides the necessary centripetal force.

∴ N sin θ = \(\frac{m v^2}{r}\) …….(2)

on dividing eqn. (2) by eqn. (1)

⇒ \(\frac{N \sin \theta}{N \cos \theta}=\frac{\frac{m v^2}{r}}{m g}\)

or \(\tan \theta=\frac{\mathrm{v}^2}{\mathrm{rg}}\)

∴ \(\theta=\tan ^{-1}\left(\frac{\mathrm{v}^2}{\mathrm{rg}}\right)\)

Where m is the mass of the vehicle, r is the radius of curvature of the road, v is the speed of the vehicle and θ is the banking angle (sinθ =h/b).

Factors that decide the value of the angle of banking are as follows:

Thus, there is no need for the mass of the vehicle to express the value of the angle of banking i.e. angle of banking ⇒ is not dependent on the mass of the vehicle.

∵ v² = gr tanθ

∴ v = \(\sqrt{g r \tan \theta}\) (maximum safe speed)

This gives the maximum safe speed of the vehicle. In actual practice, some frictional forces are always present. So, the maximum safe velocity is always much greater than that given by the above equation. While constructing the curved track, the value of θ is calculated for fixed values of V_Max and r. This explains why along the curved roads, the speed limit at which the curve is to be negotiated is clearly incited on sign boards.

The outer side of the road is raised by h = b × θ.

When θ i small, then tan θ ≈ sin θ = \(\frac{\mathrm{h}}{\mathrm{b}}\)

Also tan θ = \(\frac{v^2}{\mathrm{rg}}\)

⇒ \(\frac{v^2}{\mathrm{rg}}=\frac{\mathrm{h}}{\mathrm{b}} \text { or } \mathrm{h}=\frac{\mathrm{v}^2}{\mathrm{rg}} \times \mathrm{b}\)

Question 10. At what should a highway be banked for cars traveling at a speed of 100 km/h if the radius of the road is 400 m and no frictional forces are involved?
Answer:

The banking should be done at an angle θ such that

tan θ = \(\frac{v^2}{r g}=\frac{\frac{250}{9} \times \frac{250}{9}}{400 \times 10}\)

or \(\tan \theta=\frac{625}{81 \times 40}=0.19\)

or θ = tan–1 0.19 ≈ 0.19 radian

≈ 0.19 × 57.3º ≈ 11º

Centrifugal Force in Circular Motion: NEET Physics Solutions

Question 11. The radius of curvature of a railway line at a place where the train is moving with a speed of 36 kmh^-1 is 1000 m, the distance between the two rails being 1.5 meters. Calculate the elevation of the outer rail above the inner rail so that there may be no side pressure on the rails.
Answer:

Velocity, \(v=36 \mathrm{~km} \mathrm{~h}^{-1}=\frac{36 \times 1000}{3600} \mathrm{~ms}^{-1}=10 \mathrm{~ms}^{-1}\)

radius, r = 1000 m ; tanθ = \(\frac{v^2}{\mathrm{rg}}=1000 \times 9.8=\frac{1}{9.8}\)

Let h be the height through which the outer rail is raised. Let l be the distance between the two rails.

Then, tan θ = \(\frac{h}{l}[latex]

[∵ θ is very small] or h = l tan θ

h = 1.5 × [latex]\frac{1}{98}[/frac] = 0.0153 m [∵ l = 1.5 m]

Question 12. An aircraft executes a horizontal loop at a speed of 720 km h–1 with its wing banked at 15º. Calculate the radius of the loop.
Answer:

Speed , v = [latex]720 \mathrm{~km} \mathrm{~h}^{-1}=\frac{720 \times 1000}{3600} \mathrm{~ms}^{-1}=200 \mathrm{~ms}^{-1}\)

and \(\tan \theta=\tan 15^{\circ}=0.2679\)

⇒ \(\tan \theta=\frac{\mathrm{v}^2}{\mathrm{rg}}\)

or \(r=\frac{v^2}{g \tan \theta}=\frac{200 \times 200}{9.8 \times 0.2679} \mathrm{~m}=1523.7 \mathrm{~m}=15.24 \mathrm{~km}\)

Question 13. A train rounds an unbanked circular bend of radius 30 m at a speed of 54 km h^-1. The mass of the train is 10⁶ kg. What provides the centripetal force required for this proposal? The engine or the rails? The outer or inner rails? Which rail will wear out faster, the outer or the inner rail? What is the angle of banking required to prevent wearing out of the rails?
Answer:

⇒ \(r=30 \mathrm{~m}, \mathrm{v}=54 \mathrm{~km} \mathrm{~h}^{-1}=\frac{54 \times 5}{18} \mathrm{~ms}^{-1}=15 \mathrm{~ms}^{-1} \mathrm{~m}=10^5 \mathrm{~kg}, \quad \theta=?\)

The centripetal force is provided by the lateral thrust by the outer rail on the flanges of the wheel of the train. The train causes an equal and opposite thrust on the outer rail (Newton’s third law of motion).

Thus, the outer rails wear out faster.

tan θ = \(\tan \theta=\frac{v^2}{\mathrm{rg}}=\frac{15 \times 15}{60 \times 9.8}=0.7653\)

∴ θ = tan–1(0.7653) = 37.43º 60 9.8

Special Points About Circular Motion

Centripetal force does not increase the kinetic energy of the particle moving in a circular path, hence the work done by the force is zero.

• Centrifuges are the apparatuses used to separate small and big particles from a liquid.
• The physical quantities that remain constant for a particle moving in the circular path are speed, kinetic energy, and angular momentum.
• If a body is moving on a curved road with a speed greater than the speed limit, the reaction at the inner wheel disappears and it will leave the ground first.
• On unbanked curved roads, the minimum radius of curvature of the curve for safe driving is r = v²/μg, where v is the speed of the vehicle and μ is the coefficient of friction.
• The skidding of a vehicle will occur if v²/r > μg i.e., skidding will take place if the speed is large, the curve is sharp and μ is small.
• If r is the radius of curvature of the speed breaker, then the maximum speed with which the vehicle can run on it without leaving contact with the ground is \(v=\sqrt{(g r)}\)
• While taking a turn on the level road sometimes vehicles overturn due to centrifugal force.

Points To Be Remember

Uniform Motion In A Circle –

Angular velocity = \(\omega=\frac{\mathrm{d} \theta}{\mathrm{dt}}=2 \pi \mathrm{n}=\frac{2 \pi}{\mathrm{T}}\)

Linear velocity \(v=\vec{\omega} \times \vec{r}\)

v = \(v=\omega r \text { when } \vec{\omega} \text { and } \vec{r}\) are perpendicular to each other.

Centripetal acceleration = \(a=\frac{v^2}{r}=\omega^2 r=\omega v=4 \pi^2 n^2 r\)

Equations Of Motion –

For constant angular acceleration –

1. \(\omega=\omega_0+\alpha \mathrm{t}\)
2. \(\theta=\omega_0 t+\frac{1}{2} \alpha t^2\)
3. \(\omega^2=\omega_0^2+2 \alpha \theta\)

Motion Of A Car On A Plane Circular Road –

For motion without skidding

⇒ \(\frac{M v_{\max }^2}{r}=\mu M_g, v_{\max } \sqrt{\mu \mathrm{g}}\)

Motion On A Banked Road – Angle of banking = θ

tan θ = \(\frac{h}{b}\)

Maximum safe speed at the bend \(\mathrm{v}_{\max }=\left[\frac{\mathrm{rg}(\mu+\tan \theta)}{1-(\mu \tan \theta)}\right]^{1 / 2}\)

If friction is negligible = \(\mathrm{v}_{\max }=\sqrt{\mathrm{rg} \tan \theta}=\sqrt{\frac{\mathrm{rhg}}{\mathrm{b}}} \text { and } \quad \tan \theta=\frac{\mathrm{v}^2 \max }{\mathrm{rg}}\)

Centrifugal Force Problems and Solutions for NEET Physics Class 11

Motion Of Cyclist On A Curve –

In equilibrium angle with vertical is θ then tan θ = \(\frac{v^2}{\mathrm{rg}}\)

Maximum safe speed = \(v_{\max }=\sqrt{\mu r g}\)

Motion In A Vertical Circle (particle tied to a string) –

At the top position – Tension \(T_A=m\left(\frac{v_A^2}{r}-g\right)\)

For T_A = 0, critical speed = \(\sqrt{\mathrm{gr}}\)

At the bottom – Tension \(T_B=m\left(\frac{v_B^2}{r}+g\right)\)

For completing the circular motion minimum speed at the bottom \(v_B=\sqrt{5 \mathrm{gr}}\)

Tension T_B = 6mg

Conical Pendulum (Motion in a horizontal circle)

Tension is string = \(\frac{\mathrm{mg} \ell}{\left(\ell^2-r^2\right)^{1 / 2}}\)

Angular velocity = \(\sqrt{\frac{\mathrm{g}}{\ell \cos \theta}}\)

Periodic time = \(2 \pi \sqrt{\frac{\ell \cos \theta}{g}}\)

⇒ \(2 \pi \sqrt{\frac{r}{g \tan \theta}}\)

NEET Physics Solutions For Class 11 Centripetal Force Centripetal Force

Centripetal Force: In uniform circular motion the force acting on the particle along the radius and towards the center keeps the body moving along the circular path. This force is called centripetal force.

Centripetal Force Explanation:

1. Centripetal force is necessary for uniform circular motion.
2. It is along the radius and towards the center.
3. Centripetal force = [mass] × [centripetal acceleration] = \(\frac{m v^2}{r}=m r \omega^2\)

4. Centripetal force is due to known interaction. Therefore it is a real force. If an object tied to a string is revolved uniformly in a horizontal circle, the centripetal force is due to the tension imparted to the string by the hand.

When a satellite revolves in a circular orbit around the Earth, the centripetal force is due to the gravitational force of attraction between the satellite and the Earth.

In an atom, an electron revolves in a circular orbit around the nucleus. The centripetal force is due to the electrostatic force of attraction between the positively charged nucleus and the negatively charged.

Centripetal Force NEET Physics Class 11: Solutions and Problems

Question 1. A stone of mass 1kg is whirled in a circular path of radius 1 m. Find out the tension in the string if the linear velocity is 10 m/s.
Answer: Tension

⇒ \(\frac{m v^2}{R}=\frac{1 \times(10)^2}{1}=100 \mathrm{~N}\)

Question 2. A satellite of mass 107 kg is revolving around the earth with a time period of 30 days at a height of 1600 km. Find out the force of attraction on satellite by Earth.
Answer:

Force = \(\mathrm{m} \omega^2 R \text { and } \frac{2 \pi}{\mathrm{T}}=\frac{2 \times 3.4}{30 \times 86400}=\frac{6.28}{2.59 \times 10^6}\)

Force = \(m \omega^2 r=\left(\frac{6.28}{2.59 \times 10^6}\right)^2 \times 10^7 \times(6400+1600) \times 10^3=2.34 \times 10^5 \mathrm{~N}\)

Centrifugal Force

The pseudo force experienced by a particle performing uniform circular motion due to an accelerated frame of reference which is along the radius and directed way from the center is called centrifugal force.

NEET Class 11 Physics Centripetal Force: Key Concepts and Solutions

Centrifugal Force Explanation:

1. Centrifugal force is a pseudo force as it is experienced due to accelerated frame of reference. The interaction of origin and away from the centre.
2. It is along the radius and away from the centre.
3. The centrifugal force in having the same magnitude as that of centripetal force. But, its direction is opposite to that of centripetal force . It is not due to reaction of centripetal force because without action, reaction not possible, but centrifugal force can exists without centripetal force.
4. Magnitude of the centrifugal force is \(\frac{\mathrm{mv}^2}{\mathrm{r}} \text { or } \mathrm{mr} \omega^2\)

Note: Pseudo force acts in a noninertial frame i.e. accelerated frame of reference in which Neutron’s laws of motion do not hold good.

• When a car moving along a horizontal curve takes a turn, the person in the car experiences a push in the outward direction.
• The coin placed slightly away from the center of a rotating gramophone disc slips towards the edge of the disc.
• A cyclist moving fast along a curved road has to lean inwards to keep his balance.

Difference Between Centripetal Force And Centrifugal Force:

Centripetal Force in Circular Motion: NEET Physics Solutions

Centripetal Force: Problems and Solutions for NEET Physics Class 11

Applications Of Centrifugal Force: The centrifugal pump used to lift the waterworks on the principle of centrifugal force.

A cream separator used in the diary work works on the principle of centrifugal force. Centrifuge used for the separation of suspended particles from the liquid, works on the principle of centrifugal force. Centrifugal drier.

• Calorimetry
• Calorimetry And Thermal Expansion
• Thermal Expansion
• Mathematical Tools

• Algebra
• Double Differentiation
• Multiplication Of Vectors
• Rules For Differentiation
• Rules For Integration
• Trigonometry
• Vector

NEET Physics Class 11 Chapter 10 Mathematical Tools

Mathematics is the language of physics. It becomes easier to describe, understand, and apply the physical principles if one has a good knowledge of mathematics.

Mathematical Tools

Mathematical Tools for NEET Physics Class 11: Key Concepts and Solutions

To solve the problems of physics Newton made significant contributions to Mathematics by inventing differentiation and integration.

Appropriate Choice Of Tool Is Very Important

What are mathematical tools?

This chapter includes all the necessary mathematics knowledge that we require to possess to study physics efficiently Importance of mathematical tools. This is the most important chapter of physics as it will be repeatedly used in all the upcoming chapters.

Importance of Mathematical Tools in Physics: This chapter is the foundation of physics that needs to be studied. Here we will learn about the mathematics that will be involved in Physics.

NEET Physics Chapter 10 Mathematical Tools: Study Notes and Solutions

NEET Physics Class 11 Chapter 10 Function

The function is a rule of relationship between two variables in which one is assumed to be dependent and the other independent variable, for example,

For example: The temperatures at which water boils depend on the elevation above sea level (the boiling point drops as you ascend). Here elevation above sea level is the independent and temperature is the dependent variable

For example:  The interest paid on a cash investment depends on the length of time the investment is held. Here time is the independent and interest is the dependent variable.

• In each of the above examples, the value of one variable quantity (dependent variable), which we might call y, depends on the value of another variable quantity (independent variable), which we might call x.
• Since the value of y is completely determined by the value of x, we say that y is a function of x and represent it mathematically as y = f(x).

Here f represents the function, x is the independent variable & y is the dependent variable.

• All possible values of independent variables (x) are called the domain of a function.
• All possible values of a dependent variable (y) are called a range of functions.
• Think of a function f as a kind of machine that produces an output value f(x) in its range whenever we feed it an input value x from its domain.

When we study circles, we usually call the area A and the radius r. Since area depends on radius, we say that A is a function of r, A = f(r). The equation A = πr² is a rule that tells how to calculate a unique (single) output value of A for each possible input value of the radius r.

A = f(r) = πr². (Here, the rule of relationship that describes the function may be square and multiplied by π).

If r = 1 A = π ; if r = 2A = 4π ; if r = 3 A = 9π

The set of all possible input values for the radius is called the domain of the function. The set of all output values of the area is the range of the function.

We usually denote functions in one of the two ways:

1. A formula such as y = x² uses a dependent variable, y, to denote the value of the function.
2. By giving a formula such as f(x) = x² that defines a function symbol f to name the function. Strictly speaking, we should call the function f and not f(x),

y = sin x. Here the function is sine, and x is the independent variable.

NEET Class 11 Mathematical Tools Chapter 10 Solutions

Question 1. The volume V of a ball (solid sphere) of radius r is given by the function V(r) = (4/3)π(r³). What is the volume of a ball with a radius of 3m?

Answer:

V= 4/3π(3)³ = 36π m³.

Function Of A Function:

Suppose we are given 2 functions, f (x) = x² and g (x) = x + 1

If we are required to find f(g(x))

i.e., value of f(x) at x = g(x)

f (g(x)) = (x + 1)² [put g (x) in place of x in f(x)]

g (f (x)) = x² + 1 [put f (x) in place of x in g (x)]

Question 2. Suppose that the function F is defined for all real numbers r by the formula F(r) = 2(r – 1) + 3. Evaluate F at the input values 0, 2, x + 2, and F(2).

Answer:

Given formula F(r) = 2(r – 1) + 3.

In each case, we substitute the given input value for r into the formula for F :

F(0) = 2(0 – 1) + 3 = – 2 + 3 = 1 ;

f(2) = 2(2 – 1) + 3 = 2(1) + 3 = 2 + 3 = 5

F(x + 2) = 2(x +2 – 1) + 3 = 2x + 5 ;

F(F(2)) = F(5) = 2(5 – 1) + 3 = 2(4) + 3 = 8 + 3 = 11.

Question 3. A function ƒ(x) is defined as ƒ(x) = x² + 3, Find ƒ(0), ƒ(1), ƒ(x²), ƒ(x+1) and ƒ(ƒ(1)).

Answer:

f(x) = x² + 3

We have to find f(0), f(1), f(x²), f(x + 2) and f(f(1))

ƒ(0) = 0² + 3 = 3 ;

ƒ(1)= 1² + 3 = 4 ;

ƒ(x²) = (x²)²+3 = x⁴+3

ƒ(x+1) = (x + 1)² + 3 = x² + 2x +1 + 3 = x² + 2x + 4;

ƒ(ƒ(1)) = ƒ(4)= 4²+3 = 16 + 3 = 19

Mathematical Tools in NEET Physics Class 11: Problems and Solutions

Question 4. If function F is defined for all real numbers x by the formula F(x) = x². Evaluate F at the input values 0, 2, x + 2, and F

Answer:

F(x) = x²

We have to find f(0), f(2), f(x + 2) and f(f(2)).

F(0) = 0 ;

F(2)= 2² = 4 ;

F(x+2) = (x+2)² = x² + 8x + 4;

F(F(2)) = F(4) = 4² = 16

NEET Physics Class 11 Chapter 10 Mathematical Tools – Geometry

Formulae For Determination Of Area:

1. Area of a square = (side)²
2. Area of rectangle = length × breadth
3. Area of a triangle = \(\frac{1}{2}\) (base × height)
4. Area of trapezoid = \(\frac{1}{2}\)(distance between parallel side) × (sum of parallel side)
5. Area enclosed by a circle = π r²(r = radius)
6. Surface area of a sphere = 4πr²(r = radius)
7. Area of a parallelogram = base × height
8. Area of curved surface of cylinder = 2πr l(r = radius and l = length)
9. Area of ellipse = π ab (a and b are semi-major and semi-minor axes respectively)
10. Surface area of a cube = 6(side)²
11. Total surface area of cone = \(\pi r^2+\pi r \ell\) where \(\pi r \ell=\pi r \sqrt{r^2+h^2}\)lateral area

Formulae For Determination Of Volume:

1. Volume of rectangular slab = length × breadth × height = abt
2. Volume of a cube = (side)³
3. Volume of a sphere = \(\frac{4}{3} \pi r^3 \ell\) (r = radius)
4. Volume of a cylinder = \(\pi r^2 \lambda \ell\) (r = radius and l is length)
5. Volume of a cone = \(\frac{1}{3} \pi r^2 h\) (r = radius and h is height) 3

Note:

⇒ \(\pi=\frac{22}{7}\)= 3.14, π² = 9.8776 ≈10

and \(\frac{1}{\pi}\)

= 0.3182 ≈0.3

NEET Physics Chapter 10: Mathematical Tools for Problem Solving

NEET Physics Class 11 Chapter 10 Mathematical Tools – Differentiation

Finite Difference

The finite difference between two values of a physical quantity is represented by Δ notation. For example:

The difference in two values of y is written as Δy as given in the table below.

Infinitely Small Difference

The infinitely small difference means a very small difference. And this difference is represented by ‘d’ notation instead of ‘Δ’.

For example, an infinitely small difference in the values of y is written as ‘dy’

if y₂= 100 and y₁= 99.99999999……..

then dy = 0.000000……………….00001

NEET Physics Class 11 Chapter 10 Definition Of Differentiation

Another name for differentiation is derivative. Suppose y is a function of x or y = f(x) Differentiation of y concerning x is denoted by the symbol f’(x)

where F(x) = \(\frac{d y}{d x}\)

dx is a very small change in x and dy corresponds very small change in y.

Notation: There are many ways to denote the derivative of a function y = f(x). Besides f ’(x), the most common notations are these:

Mathematical Tools NEET Physics Class 11: Conceptual Solutions

NEET Physics Class 11 Chapter 10 Mathematical Tools – Slope Of A Line

It is the tan of the angle made by a line with the positive direction of the x-axis, measured in an anticlockwise direction.

Slope = tan θ ( In 1^st quadrant tan θ is +ve and 2nd quadrant tan θ is –ve )

In Figure – 1 slope is positive In Figure – 2 slope is negative θ < 90° (1st quadrant) θ > 90° (2nd quadrant)

NEET Physics Class 11 Chapter 10 Mathematical Tools – Average Rates Of Change

Given an arbitrary function y = f(x) we calculate the average rate of change of y concerning x over the interval (x, x + Δx) by dividing the change in the value of y, i.e. Δy = f(x + Δx) – f(x), by the length of interval Δx over which the change occurred.

The average rate of change of y concerning x over the interval

[x, x + Δx] = \(\frac{\Delta y}{\Delta x}=\frac{f(x+\Delta x)-f(x)}{\Delta x}\)

Geometrically, \(\frac{\Delta y}{\Delta x}=\frac{Q R}{P R}\)= tan θ = Slope of the line PQ

therefore we can say that the average rate of change of y concerning x is equal to the slope of the line joining P and Q.

In triangle QPR tanθ = \(\frac{\Delta y}{\Delta x}\)

The Derivative Of A Function

We know that, average rate of change of y w.r.t. x is \(\frac{\Delta y}{\Delta x}=\frac{f(x+\Delta x)-f(x)}{\Delta x}\)

If the limit of this ratio exists as Δx → 0, then it is called the derivative of the given function f(x) and is denoted as

⇒ \(f^{\prime}(x)=\frac{d y}{d x}=\Delta x \rightarrow 0 \frac{\lim _x(x+\Delta x)-f(x)}{\Delta x}\)

NEET Physics Class 11 Chapter 10 Mathematical Tools with Solutions

NEET Physics Class 11 Chapter 10 Mathematical Tools – Geometrical Meaning Of Differentiation

The geometrical meaning of differentiation is very useful in the analysis of graphs in physics. To understand the geometrical meaning of derivatives we should know the secant and tangent to a curve

Secant And Tangent To A Curve

Secant: A secant to a curve is a straight line, which intersects the curve at any two points.

Tangent: A tangent is a straight line, which touches the curve at a particular point. Tangent is a limiting case of secant which intersects the curve at two overlapping points.

1. In the figure-1 shown, if the value of Δx is gradually reduced then the point Q will move nearer to the point P.
2. If the process is continuously repeated (Figure – 2) value of Δx will be infinitely small and the secant PQ to the given curve will become a tangent at point P.

Therefore \(_{\Delta x \rightarrow 0}\left(\frac{\Delta y}{\Delta x}\right)=\frac{d y}{d x}=\tan \theta\)= tan θ

we can say that differentiation of y with respect to x, i.e. \(\left(\frac{d y}{d x}\right)\) is equal to slope of the tangent at point P (x, y) or tanθ = \(\frac{d y}{d x}\)

(the average rate of change of y from x to x + Δx is identical to the slope of the second PQ.)

Mathematical Tools for NEET Physics: Practice Problems and Solutions

NEET Physics Class 11 Chapter 10 Integration

In mathematics, for each mathematical operation, there has been defined an inverse operation. For example- The inverse operation of addition is subtraction, the inverse operation of multiplication is division and the inverse operation of a square is a square root. Similarly, there is an inverse operation for differentiation which is known as integration

Antiderivatives Or Indefinite Integrals Definitions:

A function F(x) is an antiderivative of a function f(x) if F´(x) = f(x) for all x in the domain of f. The set of all antiderivatives of f is the indefinite integral of f concerning x, denoted by

The symbol ∫is an integral sign. The function f is the integrand of the integral and x is the variable of integration.

For example f(x) = x³ then f′(x) = 3x²

So the integral of 3x² is x³

Similarly if f(x) = x³ + 4 then f′(x) = 3x²

So the integral of 3x² is x³ + 4

there for general integral of 3x² is x³ + c where c is a constant

One antiderivative F of a function f, the other antiderivatives of f differ from F by a constant. We indicate this in integral notation in the following way:

∫f(x)dx F(x) C. = + ∫………….(1)

The constant C is the constant of integration or arbitrary constant, Equation (1) is read, “The indefinite integral of f concerning x is F(x) + C.” When we find F(x)+ C, we say that we have integrated f and evaluated the integral.

Example. Evaluate ∫2x dx.

Answer:

The formula x² + C generates all the antiderivatives of the function 2x. The function x² + 1, x² – π, and \(x^2+\sqrt{2}\) are all antiderivatives of the function 2x, as you can check by differentiation.

Many of the indefinite integrals needed in scientific work are found by reversing derivative formulas.

NEET Physics Class 11 Chapter 10: Mathematical Tools MCQs and Solutions

NEET Physics Class 11 Chapter 10 Integral Formulas

Indefinite Integral

⇒ \(\int x^n d x=\frac{x^{n+1}}{n+1}+C, n \neq-1, n \text { rational }\)

⇒ \(\int d x=\int 1 d x=x+C \text { (special case) }\)

⇒ \(\int \sin (A x+B) d x=\frac{-\cos (A x+B)}{A}+C\)

⇒ \(\int \cos k x d x=\frac{\sin k x}{k}+C\)

Reversed Derivation Formula

⇒ \(\frac{d}{d x}\left(\frac{x^{n+1}}{n+1}\right)=x^n\)

⇒ \(\frac{d}{d x}(x)=1\)

⇒ \(\frac{d}{d x}\left(-\frac{\cos k x}{k}\right)=\sin k x\)

⇒ \(\frac{d}{d x}\left(\frac{\sin k x}{k}\right)=\cos k x\)

Question 1. Examples based on the above formulas:

1. \(\int x^5 d x=\frac{x^6}{6}+C\) Formula 1 with n = 5
2. \(\int \frac{1}{\sqrt{x}} d x=\int x^{-1 / 2} d x=2 x^{1 / 2}+C=2 \sqrt{x}+C\) Formula 1 with n = –1/2
3. \(\int \sin 2 x d x=\frac{-\cos 2 x}{2}+C\) Formula 2 with k = 2
4. \(\int \cos \frac{x}{2} d x=\int \cos \frac{1}{2} x d x=\frac{\sin (1 / 2) x}{1 / 2}+C=2 \sin \frac{x}{2}+C\) Formula 3 with k = 1/2

Question 2. Right: ∫x cosx dx = x sin x + cos x + C

Reason: The derivative of the right-hand side is the integrand:

Check: \(\frac{\mathrm{d}}{\mathrm{dx}}\) (x sin x + cos x + C) = x cos x + sin x – sin x + 0 = x cos x

Wrong: ∫x cosx dx = x sin x + C

Reason: The derivative of the right-hand side is not the integrand:

Check: \(\frac{\mathrm{d}}{\mathrm{dx}}\)(x sin x + C) = x cos x + sin x + 0 x cos x.

NEET Physics Class 11 Chapter 10 Mathematical Tools – Algebra

Quadratic Equation And Its Solution:

An algebraic equation of second order (the highest power of the variable is equal to 2) is called a quadratic equation. Equation ax² + bx + c = 0 is the general quadratic equation.

The general solution of the above quadratic equation or value of variable x

⇒ \(x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

⇒ \(x_1=\frac{-b+\sqrt{b^2-4 a c}}{2 a}\)

and \(x_2=\frac{-b-\sqrt{b^2-4 a c}}{2 a}\)

Sum of roots = x₁ + x₂ = – \(\frac{\mathrm{b}}{\mathrm{a}}\) and product of roots = x₁ + x₂ = \(\frac{\mathrm{c}}{\mathrm{a}}\)

For real roots b²– 4ac ≥ 0 and for imaginary roots b² – 4ac < 0.

Algebra in Mathematical Tools NEET Physics Class 11 Solutions

Question 1. find roots of f (x) = x² – 4x + 3, f (x) = – x² + 3x – 5 Every quadratic equation has 2 roots (x₁ x₂) such that f (x₁) and f (x₂) is zero.

Graph of quadratic Equation: ax² + bx + c

Graph of a quadratic equation is of parabolic nature.

Points where the graph cuts the x-axis are roots of a quadratic equation.

Binomial Expression: An algebraic expression containing two terms is called a binomial expression.

For example (a + b), (a + b)³, (2x – 3y)^-1, \(\left(x+\frac{1}{y}\right)\),etc are binomial theorem

Binomial Theorem

⇒ \((a+b)^n=a^n+n a^{n-1} b^1+\frac{n(n-1)}{2 \times 1} a^{n-2} b^2+\ldots \ldots \ldots \ldots,(1+x)^n\)

⇒ \(1+n x+\frac{n(n-1)}{2 \times 1} x^2+\ldots \ldots \ldots\)

Binomial Approximation

If x is very small, then terms containing higher powers of x can be neglected so (1 + x)n = 1 + nx

Logarithm Definition: Every positive real number N can be expressed in exponential form as

N = a^x…… (1)for example.,  49 = 7²

where ‘a’ is also a positive real different than unity and is called the base and ‘x’ is called the exponent. We can write the relation (1) in logarithmic form as

loga N = x ……..  (2)

Hence the two relations and

NEET Class 11 Algebra in Mathematical Tools: Problems and Solutions

are identical where N > 0, a > 0, a ≠ 1

Hence logarithm of a number to some base is the exponent by which the base must be raised to get that number.

The logarithm of zero does not exist and the logarithm of (–) ve reals is not defined in the system of real numbers. a is raised what power to get N

Question 2. Find value of

1. \(\log _{81} 27\)
2. \(\log _{10} 100\)
3. \(\log 9 \sqrt{3}\)

Answer:

1. \(\log _{81} 27\)

⇒ \(3^3=3^{4 x}\)

gives x = 3/4

2. \(\log _{10} 100\)

⇒ \(10^2=10^x\)

gives x = 2

3. \(\log 9 \sqrt{3}\)

⇒ \(9 \sqrt{3}\left(\frac{1}{3}\right)^x\)

⇒ \(3^{5 / 2}=3^{-x}\)

gives x = -5/2

Note:

Unity has been excluded from the base of the logarithm as in this case, log1N will not be possible, and if N = 1 then log11 will have infinitely many solutions and will not be unique which is necessary in the functional notation.

a N log N a = is an identify for all N > 0 and a > 0, a ≠ 1 for example.,  \(2^{\log _2 5}\) = 5

The number N in (2) is called the antilog of ‘x’ to the base ‘a’. Hence If log₂512 is 9 then antilog₂9 is equal to 2² = 512

Using the basic definition of log we have 3 important deductions :

log_NN = 1 i.e. logarithm of the number to the same base is 1.

⇒ \(\log _{\frac{1}{N}}\) =–1 i.e. logarithm of a number to its reciprocal is –1.

log_a 1 = 0 i.e. logarithm of unity to any base is zero. (basic constraints on number and base must be observed.)

⇒ \(\log ^{\log _a n}\)= n is an identify for all N > 0 and a > 0; a ≠ 1 for example., \(\log ^{\log _a n}\)

Algebraic Methods in NEET Physics Class 11 Mathematical Tools

Whenever the number and base are on the same side of unity then the logarithm of that number to the base is (+ve), however, if the number and base are located on different sides of unity then the logarithm of that number to the base is (–ve) for example.,

log₁₀ 100 = 2

log_1/10 100 =– 2

For a non negative number ‘a’ and \(n \geq 2, n \in N \sqrt[n]{a}=a^{1 / n}\)

NEET Physics Chapter 10 Algebra: Study Notes and Solutions

Componend And Dividend Rule

If \(\frac{p}{q}=\frac{a}{b}\) then \(\frac{p+q}{p-q}=\frac{a+b}{a-b}\)

Arithmetic Progression (AP)

General from: a, a + d, a + 2d, ……………a + (n – 1)d

Here a = first term, d = common difference

Sum of n terms \(S_n=\frac{n}{2}[a+a+(n-1) d]=\frac{n}{2}\left[I^{\text {st }} \text { term }+n^{\text {th }} \text { term }\right]\)

NEET Physics Class 11 Chapter 10 Mathematical Tools – Trigonometry

Measurement Of Angle And Relationship Between Degrees And Radian

In navigation and astronomy, angles are measured in degrees, but in calculus, it is best to use units called radians because they simplify later calculations.

Let ACB be a central angle in a circle of radius r, as in the figure. Then the angle ACB or θ is defined in radius as –

θ = \(\frac{\text { Arc length }}{\text { Radius }}\)

θ = \(\frac{\widehat{A B}}{r}\)

If r = 1 then θ = AB

The radian measure for a circle of unit radius of angle ACB is defined as the length of the circular arc AB. Since the circumference of the circle is 2π and one complete revolution of a circle is 360º, the relation between radians and degrees is given by: π radians = 180º

Trigonometry Solutions for NEET Physics Class 11 Chapter 10

Angle Conversion Formulas

1 degree = \(\frac{\pi}{180}\) (≈ 0.02) radian

Degrees to radians: multiply by \(\frac{\pi}{180}\)

1 radian ≈ 57 degrees

Radians to degrees : multiply by \(\frac{180}{\pi}\)

Question 1.

1. Convert 45º to radians.
2. Convert \(\frac{\pi}{6}\) rad to degrees.

Answer:

1. \(\text { 45 – } \frac{\pi}{180}=\frac{\pi}{4} \mathrm{rad}\)
2. \(\frac{\pi}{6} \cdot \frac{180}{\pi}=30^{\circ}\)

Question 2. Convert 30º to radians.

Answer:

⇒ \(30^{\circ} \times \frac{\pi}{180}\)

⇒ \(\frac{\pi}{6} \mathrm{rad}\)

Question 3. Convert \(\frac{\pi}{3}\) rad to degrees.

Answer:

⇒ \(\frac{\pi}{3} \times \frac{180}{\pi}\)= 60º

Standard values

1. \(30^{\circ}=\frac{\pi}{6} \mathrm{rad}\)
2. \(45^{\circ}=\frac{\pi}{4} \mathrm{rad}\)
3. \(60^{\circ}=\frac{\pi}{3} \mathrm{rad}\)
4. \(90^{\circ}=\frac{\pi}{2} \mathrm{rad}\)
5. \(120^{\circ}=\frac{2 \pi}{3} \mathrm{rad}\)
6. \(135^{\circ}=\frac{3 \pi}{4} \mathrm{rad}\)
7. \(150^{\circ}=\frac{5 \pi}{6} \mathrm{rad}\)
8. \(180^{\circ}=\pi \mathrm{rad}\)
9. \(360^{\circ}=2 \pi \mathrm{rad}\)

(Check these values yourself to see that they satisfy the conversion formulae)

NEET Physics Class 11 Chapter 10 Mathematical Tools – Measurement Of Positive And Negative Angles

An angle in the xy-plane is said to be in standard position if its vertex lies at the origin and its initial ray lies along the positive x-axis.

Angles measured counterclockwise from the positive x-axis are assigned positive measures; angles measured clockwise are assigned negative measures.

NEET Physics Class 11 Chapter 10 Trigonometry: Problems and Solutions

NEET Physics Class 11 Chapter 10 Mathematical Tools – Six Basic Trigonometric Functions

The trigonometric function of a general angle θ is defined in terms of x, y, and r.

Sine: sinθ = \(\frac{\text { opp }}{\text { hyp }}=\frac{\mathrm{y}}{\mathrm{r}}\)

Cosecant: cosecθ = \(\frac{\text { hyp }}{\text { opp }}=\frac{\mathrm{r}}{\mathrm{y}}\)

Cosine: cosθ = \(\frac{\text { adj }}{\text { hyp }}=\frac{x}{r}\)

Secant: secθ = \(\frac{\text { hyp }}{\text { adj }}=\frac{r}{x}\)

Tangent: tanθ = \(\frac{\text { opp }}{\text { adj }}=\frac{y}{x}\)

Cotangent: cotθ = \(\frac{\text { adj }}{\text { opp }}=\frac{x}{y}\)

Values Of Trigonometric Functions

If the circle has radius r = 1, the equations defining sinθ and cos θ become Cos θ = x, sinθ = y

We can then calculate the values of the cosine and sine directly from the coordinates of P.

Question 1. Find the six trigonometric ratios from the given figure

Answer:

sinθ = \(\frac{\text { opp }}{\text { hyp }}=\frac{4}{5}\)

cosθ = \(\frac{\text { adj }}{\text { hyp }}=\frac{3}{5}\)

tan θ = \(\frac{\text { opp }}{\text { adj }}=\frac{4}{3}\)

cosec θ = \(\frac{\text { hyp }}{\text { opp }}=\frac{5}{4}\)

sec θ = \(\frac{\text { hyp }}{\text { adj }}=\frac{5}{3}\)

cotθ = \(\frac{\text { adj }}{\text { opp }}=\frac{3}{4}\)

Question 2. Find the sine and cosine of angle θ shown in the unit circle if the coordinates of point p are as shown.

Answer:

cos θ = x-coordinate of P = – \(\frac{1}{2}\)

2sin θ = y-coordinate of P = \(\frac{\sqrt{3}}{2}\)

Mathematical Tools – Trigonometry Solutions NEET Physics Class 11

NEET Physics Class 11 Chapter 10 Mathematical Tools – Rules For Finding Trigonometric Ratio Of Angles Greater Than 90°

Step 1 → Identify the quadrant in which the angle lies.

Step 2 → If angle = (nπ ± θ) where n is an integer. Then trigonometric function of (nπ ± θ)= same trigonometric function of θ and the sign will be decided by the CAST Rule.

If angle = \(\left[(2 n+1) \frac{\pi}{2} \pm \theta\right]\) where n is an integer. Then the trigonometric function of \(\left[(2 n+1) \frac{\pi}{2} \pm \theta\right]\)= complimentary trigonometric function of θ and sign will be decided by CAST Rule.

Values of sin θ, cos θ, and tan θ for some standard angles.

Question 1. Evaluate sin 120°

Answer:

sin 120°

= sin (90° + 30°)

= cos 30°

⇒ \(\frac{\sqrt{3}}{2}\)

Aliter sin 120°

= sin (180° – 60°)

= sin 60°

⇒ \(\frac{\sqrt{3}}{2}\)

Question 2. Evaluate cos 135°

Answer:

cos 135°

= cos (90° + 45°)

= – sin 45°

⇒ \(-\frac{1}{\sqrt{2}}\)

Question 3. Evaluate cos 210°

Answer:

cos 210°

= cos (180° + 30°)

= – cos30°

⇒ \(-\frac{\sqrt{3}}{2}\)

Trigonometry for NEET Physics Class 11: Formulas and Solutions

Question 4. Evaluate tan 210°

Answer:

tan 210°

= tan (180° + 30°)

= tan 30°

= \(\frac{1}{\sqrt{3}}\)

NEET Physics Class 11 Chapter 10 Mathematical Tools – General Trigonometric Formulas

Question 1. \(\cos ^2 \theta+\sin ^2 \theta=1\)

Answer:

⇒ \(1+\tan ^2 \theta=\sec ^2 \theta\)

⇒ \(1+\cot ^2 \theta={cosec}^2 \theta\)

Question 2. cos(A + B) = cos A cos B – sin A sin B

Answer:

sin( A + B) = sin A cos B + cos A sin B

⇒ \(\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}\)

NEET Class 11 Mathematical Tools – Trigonometry: Problems and Solutions

Question 3. sin 2θ = 2 sin θ cos θ; cos 2θ = cos² θ – sin²θ = 2cos² θ – 1 = 1 – 2sin² θ

Answer:

⇒ \(\cos ^2 \theta=\frac{1+\cos 2 \theta}{2}\)

⇒ \(\sin ^2 \theta=\frac{1-\cos 2 \theta}{2}\)

Question 4.

In Δ ABC, the sine rule

Answer:

ΔABC need not be right-angled, \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\)

Question 5. Cosine Rule:

Answer:

cosA = \(\frac{\mathrm{b}^2+\mathrm{C}^2-\mathrm{a}^2}{2 \mathrm{bc}}\)

cosB = \(\frac{a^2+C^2-b^2}{2 a c}\)

cosC = \(\frac{a^2+b^2-c^2}{2 a b}\)

NEET Physics Class 11 Chapter 10 Mathematical Tools – Coordinate Geometry

To specify the position of a point in space, we use a right-handed rectangular axes coordinate system. This system consists of

1. Origin
2. Axis or axes.

If a point is known to be on a given line or in a particular direction only one coordinate is necessary to specify its position, if it is in a plane, two coordinates are required, if it is in space three coordinates are needed.

Origin

This is any fixed point that is convenient to you. All measurements are taken w.r.t. this fixed point.

Axis or Axes

Any fixed direction passing through the origin and convenient to you can be taken as an axis.

• If the position of a point or the position of all the points under consideration always happens to be in a particular direction, then only one axis is required. This is generally called the x-axis.
• If the positions of all the points under consideration are always in a plane, two perpendicular axes are required.
• These are generally called the x and y-axis. If the points are distributed in space, three perpendicular axes are taken which are called the x, y, and z-axis.

Position Of A Point In xy Plane

The position of a point is specified by its distances from the origin along (or parallel to) the x and y-axis as shown in the figure.

Here x-coordinate and y-coordinate are called abscissa and ordinate respectively.

Distance Formula

The distance between two points (x₁, y₁) and (x₂, y₂) is given by

d = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)

Note: In space d = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2}\)

Slope Of A Line

The slope of a line joining two points A(x₁, y₁) and B(x₂, y₂) is denoted by
m and is given by

m = \(\frac{\Delta \mathrm{y}}{\Delta \mathrm{x}}=\frac{\mathrm{y}_2-\mathrm{y}_1}{\mathrm{x}_2-\mathrm{x}_1}=\tan \theta\) [If both axes have identical scales]

Here θ is the angle made by a line with a positive x-axis. The slope of a line is a quantitative measure of inclination.

Question 1. For points (2, 14) find abscissa and ordinate. Also, find the distance from the y and x-axis.

Answer:

Abscissa = x-coordinate = 2 = distance from y-axis.

Ordinate = y-coordinate = 14 = distance from the x-axis.

NEET Physics Chapter 10 Mathematical Tools: Trigonometry Solutions

Question 2. Find the value of a if distances between the points (–9 cm, a cm) and (3 cm, 3cm) is 13 cm.

Answer:

By using distance formula d = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)

⇒ \(13 \sqrt{[3-(-9)]^2+[3-a]^2}\)

⇒ 13² = 12² + (3 – a)²

⇒ (3 – a)² = 13² – 12² = 5²

⇒ (3 – a) = ± 5

⇒ a = 2 cm or 8 cm

Question 3. A dog wants to catch a cat. The dog follows the path whose equation is y-x = 0 while the cat follows the path whose equation is x² + y² = 8. The coordinates of possible points for catching the cat are.

1. (2, – 2)
2. (2, 2)
3. (–2, 2)
4. (–2, 2)
5. (2, 4)

Answer:

Let catching point be (x¹, y¹) then, y¹ – x¹ = 0 and x¹² + y¹² = 8

Therefore, 2x¹² = 8

⇒ x¹² = 4

⇒ x¹ = ± 2;

so possible ae (2, 2) and (–2, –2).

NEET Physics Class 11 Chapter 10 Mathematical Tools – Rules For Differentiation

Derivative Of A Constant:

The first rule of differentiation is that the derivative of every constant function is zero. If c is constant, then \(\frac{\mathrm{d}}{\mathrm{dx}} \mathrm{c}\) = 0

⇒ \(\frac{d}{d x}(8)=0\)

⇒ \(\frac{d}{d x}\left(-\frac{1}{2}\right)=0\)

Power Rule:

If n is a real number, then \(\frac{d}{d x} x^n=n x^{n-1}\)

To apply the power Rule, we subtract 1 from the original exponent (n) and multiply the result by n.

⇒ \(\begin{array}{l|l|l|l|l|l}\mathrm{f} & \mathrm{x} & \mathrm{x}^2 & x^3 & x^4 & \ldots \\\hline f^{\prime} & 1 & 2 x & 3 x^2 & 4 x^3 & \ldots\end{array}\)

Question 1. 

1. \(\frac{d}{d x}\left(\frac{1}{x}\right)\)

⇒ \(\frac{d}{d x}\left(x^{-1}\right)\)

⇒ \((-1) x^{-2}\)

⇒ \(-\frac{1}{x^2}\)

2. \(\frac{d}{d x}\left(\frac{4}{x^3}\right)\)

⇒ \(4 \frac{d}{d x} \quad\left(x^{-3}\right)\)

⇒ \(4(-3) x^{-4}\)

⇒ \(-\frac{12}{x^4}\)

The Constant Multiple Rule:

If u is a differentiable function of x, and c is a constant, then \(\frac{d}{d x}(c u)=c \frac{d u}{d x}\)

In particular, if n is a positive integer, then \(\frac{d}{d x}\left(c x^n\right)=c n x^{n-1}\)

Rules for Differentiation NEET Physics Class 11 Chapter 10

Question 2. The derivative formula

⇒ \(\frac{\mathrm{d}}{\mathrm{dx}}\left(3 \mathrm{x}^2\right)\)

= 3(2x)

= 6x

says that if we rescale the graph of y = x² by multiplying each y-coordinate by 3, then we multiply the slope at each point by 3.

Question 3. A useful special case

The derivative of the negative of a differentiable function is the negative of the function’s derivative. Rule 3 with c = –1 gives.

⇒ \(\frac{d}{d x}(-u)=\frac{d}{d x}(-1 \cdot u)=-1 \quad \frac{d}{d x}(u)=-\frac{d}{d x}\)

The Sum Rule

The derivative of the sum of two differentiable functions is the sum of their derivatives. If u and v are differentiable functions of x, then their sum u + v is differentiable at every point where u and v are both differentiable functions in their derivatives.

⇒ \(\frac{d}{d x}(u-v)=\frac{d}{d x}[u+(-1) v]\)

⇒ \(\frac{d u}{d x}+(-1) \frac{d v}{d x}=\frac{d u}{d x}-\frac{d v}{d x}\)

The Sum Rule also extends to sums of more than two functions, as long as there are only finitely many functions in the sum. If u₁, u₂,………u_n are differentiable at x, then so is u₁+ u₂+ ……..+ u_n, and

⇒ \(\frac{d}{d x}\left(u_1+u_2+\ldots . .+u_n\right)\)

⇒ \(\frac{d u_1}{d x}+\frac{d u_2}{d x}+\ldots \ldots .+\frac{d u_n}{d x}\)

Question 4.

1. \(y= x^4+12 x \\\)

⇒ \(\frac{d y}{d x}=\frac{d}{d x}\left(x^4\right)+\frac{d}{d x}(12 x)\)

⇒ \(4 x^3+12\)

2. \(y =x^3+\frac{4}{3} x^2-5 x+1\)

⇒ \(\frac{d y}{d x} =\frac{d}{d x}\left(x^3\right)+\frac{d}{d x}\left(\frac{4}{3} x^2\right)-\frac{d}{d x}(5 x)+\frac{d}{d x}(1)\)

⇒ \(3 x^2+\frac{4}{3} \cdot 2 x-5+0\)

⇒ \(3 x^2+\frac{8}{3} x-5\)

Notice that we can differentiate any polynomial term by term, the way we differentiated the polynomials in the above example.

The Product Rule

If u and v are differentiable at x, then so is their product uv, and \(\frac{d}{d x}(u v)=u \frac{d v}{d x}+v \frac{d u}{d x}\)

The derivative of the product uv is u times the derivative of v plus v times the derivative of u. In prime notation (uv)’ = uv’ + vu’.

While the derivative of the sum of two functions is the sum of their derivatives, the derivative of the product of two functions is not the product of their derivatives. For instance,

⇒ \(\frac{d}{d x}(x . x)=\frac{d}{d x}\left(x^2\right)=2 x,\),

while \(\frac{\mathrm{d}}{\mathrm{dx}} \text { (x) } \frac{\mathrm{d}}{\mathrm{dx}} \text {.(x) }\)= 1.1 = 1.

NEET Physics Class 11 Chapter 10 Differentiation Problems and Solutions

Question 5. Find the derivatives of y = (x² + 1) (x³ + 3).

Answer:

From the product Rule with u = x² + 1 and v = x³ + 3, we find

⇒ \(\frac{d}{d x}\left[\left(x^2+1\right)\left(x^3+3\right)\right]\)

= (x² + 1) (3x²) + (x³ + 3) (2x)

= 3x⁴ + 3x² + 2x⁴ + 6x

= 5x⁴ + 3x² + 6x.

For example can be done as well (perhaps better) by multiplying out the original expression for y and differentiating the resulting polynomial. We now check :

y=(x² + 1)(x³ + 3)=x⁵ +x³+3x² + 3

⇒ \(\frac{d y}{d x}=5 x^4+3 x^2+6 x\)

This is in agreement with our first calculation.

There are times, however, when the product Rule must be used. In the following examples. We have only numerical values to work with.

Question 6. Let y = uv be the product of the functions u and v. Find y’ (2)if u’(2)= 3, u’(2)= –4, v(2)= 1, and v’(2)= 2.

Answer:

From the Product Rule, in the form

y’ = (uv)’ = uv’ + vu’

we have y’(2) = u(2) vs(2) + v(2) up(2)

= (3)(2)+(1) (–4)

= 6 – 4 = 2.

The Quotient Rule

If u and v are differentiable at x, and v(x) ≠ 0, then the quotient u/v is differentiable at x, and

⇒ \(\frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^2}\)

Just as the derivative of the product of two differentiable functions is not the product of their derivatives, the derivative of the quotient of two functions is not the quotient of their derivatives.

Question 7. Find the derivative of y = \(\frac{t^2-1}{t^2+1}\)

Answer: We apply the Quotient Rule with u = t² – 1 and v = t² + 1:

⇒ \(\frac{d y}{d t}=\frac{\left(t^2+1\right) \cdot 2 t-\left(t^2-1\right) \cdot 2 t}{\left(t^2+1\right)^2}\)

⇒ \(\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{\mathrm{u}}{\mathrm{v}}\right)=\frac{\mathrm{v}(\mathrm{du} / \mathrm{dt})-\mathrm{u}(\mathrm{dv} / \mathrm{dt})}{\mathrm{v}^2}\)

⇒ \(\frac{2 t^3+2 t-2 t^3+2 t}{\left(t^2+1\right)^2}=\frac{4 t}{\left(t^2+1\right)^2}\)

Derivative Of Sine Function

⇒ \(\frac{\mathrm{d}}{\mathrm{dx}}\)(sinx) cos x

Question 8.

1. y = x² – sin x :

⇒ \(\frac{d y}{d x}=2 x-\frac{d}{d x}(\sin x)\) Difference Rule

= 2x -cos x

2. y = x² sin x :

⇒ \(\frac{d y}{d x}=x^2 \frac{d}{d x}(\sin x)+2 x \sin x\)Product Rule

⇒ \(x^2 \cos x+2 x \sin x\)

3. y = \(\frac{\sin x}{x}\)

⇒ \(\frac{d y}{d x}=\frac{x \cdot \frac{d}{d x}(\sin x)-\sin x .1}{x^2}\) Quotient rule

⇒ \(\frac{x \cos x-\sin x}{x^2}\)

Derivative Of Cosine Function

⇒ \(\frac{\mathrm{d}}{\mathrm{dx}}\)(cosx)= – sinx

Question 9.

1. y = 5x + cos x

⇒ \(\frac{d y}{d x}=\frac{d}{d x}(5 x)+\frac{d}{d x}(\cos x)\) Sum Rule

= 5 – sin x

2. y = sinx cosx

⇒ \(\frac{d y}{d x}=\sin x \frac{d}{d x}(\cos x)+\cos x \frac{d}{d x}(\sin x)\) Product Rule

= sinx (– sinx) + cosx (cosx)

= cos² x – sin² x

Derivatives Of Other Trigonometric Functions

Because sin x and cos x are differentiable functions of x, the related functions

tan x = \(\frac{\sin x}{\cos x}\)

cot x = \(\frac{\cos x}{\sin x}\)

sec x = \(\frac{1}{\cos x}\)

cosec x = \(\frac{1}{\sin x}\)

are differentiable at every value of x at which they are defined. There derivatives. Calculated from the Quotient Rule, are given by the following formulas.

⇒ \(\frac{d}{d x}\)(tan x) = sec² x ;

⇒ \(\frac{d}{d x}\)(sec x) = sec x tan

⇒ \(\frac{d}{d x}\)(cot x) = – cosec² x ;

⇒ \(\frac{d}{d x}\)(cosec x) = – cosec x cot x

Question 10. Find dy / dx if y = tan x.

Answer:

⇒ \(\frac{d}{d x}(\tan x)\)

⇒ \(\frac{d}{d x}\left(\frac{\sin x}{\cos x}\right)\)

⇒ \(\frac{\cos x \frac{d}{d x}(\sin x)-\sin x \frac{d}{d x}(\cos x)}{\cos ^2 x}\)

⇒ \(\frac{\cos x \cos x-\sin x(-\sin x)}{\cos ^2 x}\)

⇒ \(\frac{\cos ^2 x+\sin ^2 x}{\cos ^2 x}\)

⇒ \(\frac{1}{\cos ^2 x}=\sec ^2 x\)

Mathematical Tools – Differentiation Rules NEET Physics Class 11

Question 11.

1. \(\frac{d}{d x}(3 x+\cot x)\)

⇒ \(3+\frac{d}{d x}(\cot x)\)

⇒ \(3-{cosec}^2 x\)

2. \(\frac{d}{d x}\left(\frac{2}{\sin x}\right)\)

⇒ \(\frac{d}{d x}(2{cosec} x)\)

⇒ \(2 \frac{d}{d x}({cosec} x)\)

= 2 (–cosec x cot x)

= – 2 cosec x cot x

Derivative Of Logarithm And Exponential Functions

⇒ \(\frac{d}{d x}\left(\log _e x\right)=\frac{1}{x}\)

⇒ \(\frac{d}{d x}\left(e^x\right)=e^x\)

Question 12. y=\(e^x \cdot \log _e(x)\)

Answer:

⇒ \(\frac{d y}{d x}=\frac{d}{d x}\left(e^x\right) \cdot \log (x)+\frac{d}{d x}\left[\log _e(x)\right] e^x\)

⇒ \(\frac{d y}{d x}=e^x \cdot \log _e(x)+\frac{e^x}{x}\)

Question 13. \(\frac{d}{d t}\)

Answer: ω cos ωt

Question 14. \(\frac{d}{d t}\)(cos t)

Answer: −ω sin ωt

Question 15.

1. \(\frac{d}{d x} \cos 3 x \quad\)

⇒ \(-\sin 3 x \frac{d}{d x} 3 x\)

= – sin 3x

2. \(\frac{d}{d x} \sin 2 x \quad\)

⇒ \(\cos 2 x \frac{d}{d x}(2 x)\)= cos 2x.2

= 2 cos 2x

3. \(\frac{d}{d t}\)(A sin (ωt + φ)

= A cos (ωt + φ)\(\frac{d}{d t}\)(ωt + φ)

= A cos (ωt + φ). ω.

= A ω cos (ωt + φ)

Question 16. \(\frac{d}{d x}\left(\frac{1}{3 x-2}\right)\)

Answer:

⇒ \(\frac{d}{d x}(3 x-2)^{-1} \quad\)

⇒ \(-1(3 x-2)^{-2} \frac{d}{d x}(3 x-2)\)

⇒ \(-1(3 x-2)^{-2}(3)\)

⇒ \(-\frac{3}{(3 x-2)^2}\)

NEET Class 11 Differentiation Rules: Formulas and Solutions

Question 17. \(\frac{d}{d t}[A \cos (\omega t+\varphi)]\)

Answer:

= –Aω sin (ωt + φ)

Chain Rule

If f (x) is given as function of g(x) i.e., y = f(g(x)) and we are required to find \(\frac{d y}{d x}\) assume dx g(x)= u

⇒ y = f

⇒ \(\frac{d y}{d u}=f^{\prime}(4)\)

⇒ \(\frac{d u}{d x}=g^{\prime}(x)\)

Example: y = sin (x²)

y = log(x² + 5x)

y = sin (cos x)

y = A sin (ωt + φ), A,ω,φ, are constant

Radian Vs. Degrees

⇒ \(\frac{d}{d x} \sin \left(x^{\circ}\right)=\frac{d}{d x} \sin \left(\frac{\pi x}{180}\right)\)

⇒ \(\frac{\pi}{180} \cos \left(\frac{\pi x}{180}\right)=\frac{\pi}{180} \cos \left(x^{\circ}\right)\)

Rules For Integration

Constant Multiple Rule

A function is an antiderivative of a constant multiple of of a function f if and only if it is k times an antiderivative of f. ∫ k f(x)dx k = ∫f(x)dx; where k is a constant 

Question 1. \(\int 5 x^2 d x\)

Answer: \(\frac{5 x^3}{3}+C\)

Question 2. \(\int \frac{7}{x^2} d x\)

Answer:

⇒ \(\int 7 x^{-2} d x\)

⇒ \(-\frac{7 x^{-1}}{1}+C\)

⇒ \(\frac{-7}{x}+C\)

NEET Physics Class 11 Integration Problems and Solutions

Question 3. \(\int \frac{t}{\sqrt{t}} d t\)

⇒ \(\int t^{1 / 2} d t=\frac{t^{3 / 2}}{3 / 2}+C\)

⇒ \(\frac{2}{3} t^{3 / 2}+C\)

Sum And Difference Rule

A function is an antiderivative of a sum or difference f ± g if and only if it is the sum or difference of an antiderivative of f or an antiderivative of g.

⇒ \(\int[f(x) \pm g(x)] d x=\int f(x) d x \pm \int g(x) d x\)

Question 4. Term–by–term integration. Evaluate : ∫(x² – 2x + 5) dx.

Answer:

If we recognize that (x³ /3) – x² + 5x is an antiderivative of x² – 2x + 5, we can evaluate the integral as

If we do not recognize the antiderivative right away, we can generate it term by term with the sum and difference Rule:

⇒ \(\int\left(x^2-2 x+5\right) d x\)

⇒ \(\int x^2 d x-\int 2 x d x+\int 5 d x\)

⇒ \(\frac{x^3}{3}+C_1-x^2+C_2+5 x+C_3\)

This formula is more complicated than it needs to be. If we combine C₁,C₂ and C₃ into a single constant C = C₁+ C₂+ C₃, the formula simplifies to

⇒ \(\frac{x^3}{3}-x^2+5 x+C\)

and still gives all the antiderivatives there are. For this reason, we recommend that you go right to the final form even if you elect to integrate term by term. Write

⇒ \(\int\left(x^2-2 x+5\right) d x\)

⇒ \(\int x^2 d x-\int 2 x d x+\int 5 d x\)

⇒ \(\frac{x^3}{3}-x^2+5 x+C\)

Find the simplest antiderivative you can for each part and add the constant at the end.

Integration Rules NEET Physics Class 11 Chapter 10 Solutions

Question 5. Find a body’s velocity from its acceleration and initial velocity. The acceleration of gravity near the surface of the earth is 9.8 m/sec². This means that the velocity v of a body falling freely in a vacuum changes at the rate of \(\frac{\mathrm{dv}}{\mathrm{dt}}\) = 9.8 m/sec². If the body is dropped from rest, what will its velocity be t seconds after it is released?

Answer:

In mathematical terms, we want to solve the initial value problem that consists of

The differential condition: \(\frac{\mathrm{dv}}{\mathrm{dt}}\) = 9.8

The initial condition: v = 0 when t = 0 ( abbreviated as v (0) = 0 )

We first solve the differential equation by integrating both sides concerning t:

⇒ \(\frac{\mathrm{dv}}{\mathrm{dt}}\) = 9.8 The differential equation

⇒ \(\int \frac{d v}{d t} d t=\int 9.8 d t\) Integrate with respect to t.

v + C₁ = 9.8t + C₂ Integrals evaluated

v = 9.8t + C. Constants combined as one

This last equation tells us that the body’s velocity t seconds into the fall is 9.8t + C m/sec. For value of C: What value? We find out from the initial condition:

v = 9.8t + C

0 = 9.8(0) + C v( 0) = 0

C = 0.

Conclusion: The body’s velocity t seconds into the fall is

v = 9.8t + 0 = 9.8t m/sec.

• The indefinite integral F(x) + C of the function f(x) gives the general solution y = F(x) + C of the differential equation dy/dx = f(x).
• The general solution gives all the solutions of the equation (there are infinitely many, one for each value of C).
• We solve the differential equation by finding its general solution We then solve the initial value problem by finding the particular solution that satisfies the initial condition y(x_o) = y_o( y has the value yo when x = x_o.).

Definite Integration Or Integration With Limits

⇒ \(\int_a^b f(x) d x=[g(x)]_a^b=g(b)-g(a)\)

where g(x) is the antiderivative of f(x) i.e. g´(x) = f(x)

Question 6. \(\int_{-1}^4 3 d x\)

Answer:

⇒ \(3 \int_{-1}^4 d x\)

⇒ \(3[x]_{-1}^4\)

⇒ \(3[4-(-1)]\)

= (3)(5)=15

⇒ \(\int_0^{\pi / 2} \sin x d x\)

⇒ \([-\cos x]_0^{\pi / 2}\)

⇒ \(-\cos \left(\frac{\pi}{2}\right)+\cos (0)\)

= –0 + 1 = 1

NEET Physics Class 11 Chapter 10 Integration Techniques and Solutions

Application Of Definite Integral: Calculation Of Area Of A Curve

From the graph shown in the figure if we divide the whole area into infinitely small strips of dx width. We take a strip at x position of dx width.

A small area of this strip dA = f(x) dx

So, the total area between the curve and x-axis = sum of area of all strips = \(\int_a^b f(x) d x\)

Let f(x) ≥ 0 be continuous on [a,b]. The area of the region between the graph of f and the x-axis is

A = \(\int_a^b f(x) d x\)

Integration Methods for NEET Physics Class 11: Formulas and Solutions

Question 7. Find the area under the curve of y = x from x = 0 to x = a

Answer:

Multiplication Of Vectors

1. The Scalar Product:

The scalar product or dot product of any two vectors \(\vec{A} \text { and } \vec{B}\), denoted as \(\vec{A} \cdot \vec{B}\)(read \(\vec{A}\).\(\vec{B}\)) is defined as the product of their magnitude with cosine of angle between them. Thus, \(\overrightarrow{\mathrm{A}}\). \(\vec{B}\) = AB cos θ {here θ is the angle between the vectors}

The Scalar Product Properties:

It is always a scalar which is positive if the angle between the vectors is acute (i.e. < 90º) and negative if the angle between them is obtuse (i.e. 90º < θ ≤ 180º)

It is commutative, i.e., \(\vec{A} \cdot \vec{B}=\vec{A} \cdot \vec{B}\)

It is distributive, i.e., \(\vec{A} \cdot(\vec{B}+\vec{C})=\vec{A} \cdot \vec{B}+\vec{A} \cdot \vec{C}\)

As by definition \(\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}\)= AB cos θ.

The angle between the vectors θ = \(\cos ^{-1}\left[\frac{\vec{A} \cdot \vec{B}}{A B}\right]\)

⇒ \(\vec{A}.\vec{B}\) = A (B cos θ) = B (A cos θ)

Geometrically, B cos θ is the projection of \(\vec{B} \text { onto } \vec{A}\), and A cos θ is the projection of \(\vec{A} \text { onto } \vec{B}\) shown. So \(\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}\) is the product of the magnitude of \(\overrightarrow{\mathrm{A}}\) and the component of \(\overrightarrow{\mathrm{B}}\) along \(\overrightarrow{\mathrm{A}}\) and vice versa.

Component of \(\vec{B} \text { along } \overrightarrow{\mathrm{A}}\) = B cosθ = \(\frac{\vec{A} \cdot \vec{B}}{A}=\vec{A} \cdot \vec{B}\)

Component of \(\vec{A} \text { along } \overrightarrow{\mathrm{B}}\) along = \(\frac{\vec{A} \cdot \vec{B}}{B}=\vec{A} \cdot \vec{B}\)

Scalar product of two vectors will be maximum when cos θ = max = 1, i.e., θ = 0º, i.e., vectors are parallel

⇒ \((\vec{A} \cdot \vec{B})_{\max }=A B\)

If the scalar product of two nonzero vectors vanishes then the vectors are perpendicular. The scalar product of a vector by itself is termed as self dot product and is given by

⇒ \((\vec{A})^2=\vec{A} \cdot \vec{A}=A A \cos \theta=A A \cos 0^{\circ}=A^2\)

⇒ \(A=\sqrt{\vec{A} \cdot \vec{A}}\)

In case of unit vector \(\hat{n}\)

⇒ \(\hat{n} . \hat{n}=1 \times 1 \times \cos 0^{\circ}=1\)

⇒ \(\hat{n} \cdot \hat{n}=\hat{i} \cdot \hat{i}=\hat{j} \cdot \hat{j}=\hat{k} \cdot \hat{k}=1\)

In case of orthogonal unit vectors \(\hat{\mathrm{i}}, \hat{\mathrm{j}} \text { and } \hat{\mathrm{k}}\); \(\hat{i} \cdot \hat{j}=\hat{j} \cdot \hat{k}=\hat{k} \cdot \hat{i}=0\)

⇒ \(\vec{A} \cdot=\left(\hat{i} A_x+\hat{j} A_y+\hat{k} A_z\right) \cdot\left(\hat{i} B_x+\hat{j} B_y+\hat{k} B_z\right)\)

⇒ \(\left[A_x B_x+A_y B_y+A_z B_z\right]\)

Multiplication of Vectors NEET Physics Class 11 Solutions

Question 1. If the Vectors \(\vec{P}=a \hat{i}+a \hat{j}+3 \hat{k} \text { and } \vec{Q}=a \hat{i}-2 \hat{j}-\hat{k}\) are perpendicular to each other. Find the value of a?

Answer:

If vectors \(\vec{P} \text { and } \vec{Q}\) are perpendicular

⇒ \(\vec{P} \cdot \vec{Q}=0\)

⇒ \((a \hat{i}+a \hat{j}+3 \hat{k}) \cdot(a \hat{i}-2 \hat{j}-\hat{k})=0\)

⇒ \(a^2-2 a-3=0\)

⇒ \(a^2-3 a+a-3=0\)

⇒  a(a-3)+1(a-3) = 0

⇒ (a + 1)(a – 3) = 0

⇒ a + 1 = 0  or  a – 3 = 0

⇒ a = -1 or a = 3

⇒ a = -1,3

Question 2. Find the component of \(3 \hat{i}+4 \hat{j} \text { along } \hat{i}+\hat{j}\)?

Answer:

Component of \(\overrightarrow{\mathrm{A}}\)along \(\overrightarrow{\mathrm{B}}\) is given by \(\frac{\vec{A} \cdot \vec{B}}{B}\) hence required component

⇒ \(\frac{(3 \hat{i}+4 \hat{j}) \cdot(\hat{i}+\hat{j})}{\sqrt{2}}\)

⇒ \(\frac{7}{\sqrt{2}}\)

Question 3. Find angle between \(\vec{A}=3 \hat{i}+4 \hat{j} \text { and } \vec{B}=12 \hat{i}+5 \hat{j}\)

Answer:

We have cos θ = \(\frac{\vec{A} \cdot \vec{B}}{A B}\)

⇒ \(\frac{(3 \hat{i}+4 \hat{j}) \cdot(12 \hat{i}+5 \hat{j})}{\sqrt{3^2+4^2} \sqrt{12^2+5^2}}\)

⇒ \(\cos \theta=\frac{36+20}{5 \times 13}=\frac{56}{65}\)

⇒ \(\theta=\cos ^{-1} \frac{56}{65}\)

2. Vector Product

The vector product or cross product of any two vectors \(\overrightarrow{\mathrm{A}} \text { and } \overrightarrow{\mathrm{B}}\) A and B, denoted as \(\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}(\text { read } \overrightarrow{\mathrm{A}} \text { cross } \overrightarrow{\mathrm{B}})\) is defined as:

⇒ \(\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}=\mathrm{AB} \sin \theta \hat{n}\)

Here θ is the angle between the vectors and the direction \(\hat{n}\) is given by the right-hand-thumb rule.

Vector Multiplication Methods in NEET Physics Class 11

Right-Hand-Thumb Rule:

To find the direction of \(\hat{n}\) draw the two vectors \(\overrightarrow{\mathrm{A}} \text { and } \overrightarrow{\mathrm{B}}\) with both the tails coinciding.

Now place your stretched right palm perpendicular to the plane of \(\overrightarrow{\mathrm{A}} \text { and } \overrightarrow{\mathrm{B}}\) in such a way that the fingers are along the vector \(\overrightarrow{\mathrm{A}}\) and when the fingers are closed they go towards \(\overrightarrow{\mathrm{B}}\). The direction of the thumb gives the direction of \(\hat{n}\)

Vector Product Properties

The vector product of two vectors is always a vector perpendicular to the plane containing the two vectors i.e. orthogonal to both vectors.

A and B, though the vectors \(\vec{A} \text { and } \vec{B}\) though the vector \(\vec{A} \text { and } \vec{B}\) may or may not be orthogonal.

Vector product of two vectors is not commutative i.e. \(\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}} \neq \overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{A}}\)

But \(|\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}|=|\overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{A}}|\) = AB sin θ

The vector product is distributive when the order of the vectors is strictly maintained i.e.

⇒ \(\mathrm{A} \times(\overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{C}})=\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{C}}\)

The magnitude of the vector product of two vectors will be maximum when sinθ = max = 1, i.e, θ = 90º

⇒ \(|\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}|_{\max }=\mathrm{AB}\) i.e., the magnitude of the vector product is maximum if the vectors are orthogonal.

The magnitude of the vector product of two non–zero vectors will be minimum when |sinθ| = minimum = 0,i.e., θ = 0º or 180º, and vectors are collinear.
min | A B | 0 × =i.e., if the vector product of two non–zero vectors vanishes, the vectors are collinear.

Note: When θ = 0º then vectors may be called like vectors or parallel vectors and when θ = 180º then vectors may be called unlike vectors or antiparallel vectors.

The self-cross product i.e. product of a vector by itself vanishes i.e. is a null vector.

Note: Null vector or zero vector: A vector of zero magnitude is called a zero vector. The direction of a zero vector is determinate (unspecified).

⇒ \(\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{A}}=\mathrm{AA} \sin 0^{\circ} \hat{\mathrm{n}}=\overrightarrow{0}\)

Note: Geometrical meaning of vector product of two vectors

1. Consider two vectors \(\overrightarrow{\mathrm{A}} \text { and } \overrightarrow{\mathrm{B}}\) which are represented by \(\overrightarrow{\mathrm{OP}} \text { and } \overrightarrow{\mathrm{QP}} \text { and } \angle \mathrm{POQ}=\theta\)
2. Complete the parallelogram OPRQ. Join P with Q. Here OP = A and OQ = B. Draw QN ⊥OP
3. Magnitude of the cross product of \(\vec{A} \text { and } \vec{B}\)

⇒ \(|\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}|=\mathrm{AB} \sin \theta\)

= (OP)(OQ sin θ)

= (OP)(NQ) (∵ NQ = OQ sin θ)

= base × height

= Area of parallelogram OPRQ

Area of ΔPOQ = ΔPOQ = \(\frac{\text { base } \times \text { height }}{2}=\frac{(\mathrm{OP})(\mathrm{NQ})}{2}\)

⇒ \(\frac{1}{2}|\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}|\)

∴ Area of parallelogram OPRQ = 2 [area of Δ OPQ] = \(|\vec{A} \times \vec{B}|\)

Formulae to find Area

If \(\vec{A} \text { and } \vec{B}\) are two adjacent sides of a triangle, then its area = \(\frac{1}{2}|\vec{A} \times \vec{B}|\)

If \(\vec{A} \text { and } \vec{B}\) are two adjacent sides of a parallelogram, then its area = \(|\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}|\)

If \(\vec{A} \text { and } \vec{B}\) are diagonals of a parallelogram then its area = \(\frac{1}{2}|\vec{A} \times \vec{B}|\)

In case of unit vector \(\hat{n}, \quad \hat{n} \times \hat{n}=\overrightarrow{0} \Rightarrow \hat{i} \times \hat{i}=\hat{j} \times \hat{j}=\hat{k} \times \hat{k}=\overrightarrow{0}\)

In case of orthogonal unit vectors \(\hat{i}, \hat{j} \text { and } \hat{k}\) in accordance with right-hand-thumb-rule,

In terms of compound, \(\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}\)

⇒ \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
A_x & A_y & A_z \\
B_x & B_y & B_z
\end{array}\right|=\hat{i}\left|\begin{array}{cc}
A_y & A_z \\
B_y & B_z
\end{array}\right|-\hat{j}\left|\begin{array}{cc}
A_x & A_z \\
B_x & B_z
\end{array}\right|+\hat{k}\left|\begin{array}{cc}
A_x & A_y \\
B_x & B_y
\end{array}\right|\)

The magnitude of the area of the parallelogram formed by the adjacent sides of vectors \(\vec{A} \text { and } \vec{B}\) equal to

NEET Class 11 Chapter 10 Vector Multiplication: Cross and Dot Products

Question 4. \(\overrightarrow{\mathrm{A}}\) is Eastwards and \(\overrightarrow{\mathrm{B}}\) is downwards. Find the direction of \(\overrightarrow{\mathrm{A}} \times \vec{B}\)?

Answer:

Applying the right-hand thumb rule we find that \(\overrightarrow{\mathrm{A}} \times \vec{B}\) is along North.

Question 5. If \(\vec{A} \cdot \vec{B}=|\vec{A} \times \vec{B}|\), find angle between \(\overrightarrow{\mathrm{A}} \text { and } \overrightarrow{\mathrm{B}}\)

Answer:

⇒ \(\vec{A} \cdot \vec{B}=|\vec{A} \times \vec{B}|\)

AB cos θ = AB sin θ

tan θ = 1

⇒ θ = 45º

Question 6. Two vectors \(\vec{A} \text { and } \vec{B}\) are inclined to each other at an angle θ. Find a unit vector that is perpendicular to both \(\vec{A} \text { and } \vec{B}\)

Answer:

⇒ \(\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}=\mathrm{AB} \sin \theta \hat{n}\)

⇒ \(\hat{n}=\frac{\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}}{\mathrm{AB} \sin \theta}\) here \(\hat{n}\) is perpendicular to both \(\vec{A} \text { and } \vec{B}\)

Question 7. Find \(\vec{A} \times \vec{B}\) if \(\vec{A}=\hat{i}-2 \hat{j}+4 \hat{k}\) and \(\vec{B}=2 \hat{i}-\hat{j}+2 \hat{k}\)

Answer:

⇒ \(\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}\)

⇒ \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & -2 & 4 \\
3 & -1 & 2
\end{array}\right|\)

⇒ \(\hat{i}(-4-\hat{j}(-4))-(2-12)+\hat{k}(-1-(-6))=10 \hat{j}+5 \hat{k}\)

Question 8. Find the value of

1. sin (− θ)
2. cos (− θ)
3. tan (− θ)
4. cos (\(\frac{\pi}{2}\)− θ)
5. sin (\(\frac{\pi}{2}\)+ θ)
6. cos (\(\frac{\pi}{2}\)+ θ) 2
7. sin (π − θ)
8. cos (π − θ)
9. sin (\(\frac{3 \pi}{2}\)− θ)
10. cos (\(\frac{3 \pi}{2}\)− θ)
11. sin (\(\frac{3 \pi}{2}\)+ θ)
12. cos (\(\frac{3 \pi}{2}\)+ θ)
13. tan (\(\frac{\pi}{2}\)− θ)
14. cot (\(\frac{\pi}{2}\)− θ) 2

Answers :

1. – sin θ
2. cos θ
3. – tan θ
4. sin θ
5. cos θ
6. – sin θ
7. sin θ
8. – cos θ
9. – cos θ
10. – sin θ
11. – cos θ
12. sin θ
13. cot θ
14. tan θ

Multiplication of Vectors for NEET Physics Class 11

Question 9.

1. For what value of m the vector \(\vec{A}=2 \hat{i}+3 \hat{j}-6 \hat{k}\) is perpendicular to \(\vec{B}=3 \hat{i}-m \hat{j}+6 \hat{k}\)
2. Find the components of vector \(\vec{A}=2 \hat{i}+3 \hat{j}\) along the direction of \(\hat{i}+\hat{j}\)?

Answers :

1. m = –10
2. \(\frac{5}{\sqrt{2}}\)

Question 10.

1. \(\overrightarrow{\mathrm{A}}\) is North–East and \(\overrightarrow{\mathrm{B}}\) is downwards, find the direction of \(\vec{A} \times \vec{B}\)
2. Find \(\vec{B} \times \vec{A} \text { if } \vec{A}=3 \hat{i}-2 \hat{j}+6 \hat{k}\) and \(\vec{B}=\hat{i}-\hat{j}+\hat{k}\)

Answers :

1. North-West.
2. \(-4 \hat{i}-3 \hat{j}+\hat{k}\)

NEET Physics Class 11 Chapter 10 Mathematical Tools Multiple Choice Question And Answers

Question 1. The surface area of a sphere as a function of its radius is A(r) = 4πr₂ the value of A(10) will be :

1. 1358 m²
2. 324 m²
3. 314 m²
4. 1256 m²

Answer: 4. 1256 m²

Question 2. If f (x) = x² –1

1. 5
2. 6
3. 7
4. 8

Answer: 4. 8

Question 3. If f(x) = \(x+\frac{1}{x}\), then the value of f(1) will be

1. 2
2. – 2
3. 1
4. – 1

Answer: 1. 2

NEET Physics Class 11 Chapter 10 Mathematical Tools MCQs

Question 4. Find v (0), where v (t) = 3 + 2t

1. 5
2. 6
3. 3
4. None

Answer: 3. 3

Question 5. If f(θ) = sin θ, find \(f\left(\frac{\pi}{6}\right)\)

1. \(\frac{\pi}{6}\)
2. \(\frac{1}{2}\)
3. 2
4. \(\frac{\pi}{3}\)

Answer: 2. \(\frac{1}{2}\)

Question 6. If f (x) = 5, then the value of f (10) will be

1. 10
2. 5
3. 15
4. None

Answer: 2. 5

Question 7. tan15° is equivalent to :

1. \((2-\sqrt{3})\)
2. \((5+\sqrt{3})\)
3. \(\left(\frac{5-\sqrt{3}}{2}\right)\)
4. \(\left(\frac{5+\sqrt{3}}{2}\right)\)

Answer: 1. \((2-\sqrt{3})\)

Question 8. sin²θ is equivalent to:

1. \(\left(\frac{1+\cos \theta}{2}\right)\)
2. \(\left(\frac{1+\cos 2 \theta}{2}\right)\)
3. \(\left(\frac{1-\cos 2 \theta}{2}\right)\)
4. \(\left(\frac{\cos 2 \theta-1}{2}\right)\)

Answer: 3. \(\left(\frac{1-\cos 2 \theta}{2}\right)\)

Question 9. sinA. sin(A + B) is equal to

1. cos²A . cosB + sinA sin²B
2. \(\sin ^2 A \cdot \cos B+\frac{1}{2} \cos 2 A \cdot \sin B\)
3. \(\sin ^2 A \cdot \cos B+\frac{1}{2} \sin 2 A \cdot \sin B\)
4. sin²A . sinB + cosA cos²B

Answer: 3. \(\sin ^2 A \cdot \cos B+\frac{1}{2} \sin 2 A \cdot \sin B\)

Question 10. –sinθ is equivalent to :

1. \(\cos \left(\frac{\pi}{2}+\theta\right)\)
2. \(\cos \left(\frac{\pi}{2}-\theta\right)\)
3. \(\sin (\theta-\pi)\)
4. \(\sin (\pi+\theta)\)

Answer: (1,2,4)

Question 11. θ is the angle between the side CA and CB of a triangle, shown in the figure then θ is given by :

1. \(\cos \theta=\frac{2}{3}\)
2. \(\sin \theta=\frac{\sqrt{5}}{3}\)
3. \(\tan \theta=\frac{\sqrt{5}}{2}\)
4. \(\tan \theta=\frac{2}{3}\)

Answer: 2. \(\sin \theta=\frac{\sqrt{5}}{3}\)

Question 12. If tan θ = \(\frac{1}{\sqrt{5}}\) and θ lies in the first quadrant, the value of cos θ is :

1. \(\sqrt{\frac{5}{6}}\)
2. \(-\sqrt{\frac{5}{6}}\)
3. \(\frac{1}{\sqrt{6}}\)
4. \(-\frac{1}{\sqrt{6}}\)

Answer: 1. \(\sqrt{\frac{5}{6}}\)

Question 13. Calculate the slope of a shown line

1. 2/3
2. – 2/3
3. 3/2
4. –3/2

Answer: 2. – 2/3

NEET Physics Class 11 Chapter 10 Mathematical Tools MCQs Practice

Question 14. The speed (v) of a particle moving along a straight line is given by v = t² + 3t – 4 where v is in m/s and t in second. Find a time t at which the particle will momentarily come to rest.

1. 3
2. 4
3. 2
4. 1

Answer: 4. 1

Find the derivative of given functions w.r.t. corresponding independent variable.

Question 15. y = x² + x + 8

1. \(\frac{d y}{d x}=2 x+1\)
2. \(\frac{d y}{d x}=2+1\)
3. \(\frac{d y}{d x}=2 x-1\)
4. \(\frac{d y}{d x}=x+1\)

Answer: 1. \(\frac{d y}{d x}=2 x+1\)

Question 16. y = tan x + cot x

1. tan² x + cosec² x
2. cot² x – sin² x
3. sec² x – cosec² x
4. sec x + cosec x 2

Answer: 3. sec2 x – cosec2 x

Question 17. y = lnx + ex, then \(\frac{d^2 y}{d x^2}\) is equal to

1. \(\frac{1}{x^2}-e^x\)
2. \(\frac{1}{\mathrm{x}^2}+\mathrm{e}^{\mathrm{x}}\)
3. \(\frac{1}{x}+e^x\)
4. \(-\frac{1}{x^2}+e^x\)

Answer: 4. \(-\frac{1}{x^2}+e^x\)

Question 18. y = \(\mathrm{e}^{\mathrm{x}} \ell \mathrm{n} \mathrm{x}\)

1. \(\mathrm{e}^{\mathrm{x}} \ell \mathrm{n} x+\frac{\mathrm{e}^{\mathrm{x}}}{\mathrm{x}}\)
2. \(e^x \ell n x-\frac{e^x}{x}\)
3. \(\mathrm{e}^{\mathrm{x}} \ln x-\frac{e}{\mathrm{x}} \)
4. None of these

Answer: 1. \(\mathrm{e}^{\mathrm{x}} \ell \mathrm{n} x+\frac{\mathrm{e}^{\mathrm{x}}}{\mathrm{x}}\)

Question 19. y = sin 5 x

1. 5 cos 5 x
2. 3 cos 3 x
3. 5 cos 5x
4. 2 cos 2x

Answer: 1. 5 cos 5 x

Question 20. (x + y)² = 4

1. \(\frac{d y}{d x}=+1\)
2. \(\frac{d y}{d x}=-1\)
3. \(\frac{d}{d x}=-1\)
4. \(\frac{d y}{d}=-1\)

Answer: 2. \(\frac{d y}{d x}=-1\)

Question 21. y = 2u³, u = 8x – 1

1. \(\frac{d y}{d x}=48(8 x-1)^2\)
2. \(\frac{d y}{d x}=58(5 x-1)^2\)
3. \(\frac{d y}{d x}=48(8 x-1)^2\)
4. \(\frac{d y}{d x}=28(8 x-1)\)

Answer: 1. \(\frac{d y}{d x}=48(8 x-1)^2\)

Question 22. Given s = t² + 5t + 3, find \(\frac{\mathrm{ds}}{\mathrm{dt}}\) at t = 1

1. 7
2. 9
3. 12
4. 15

Answer: 1. 7

Question 23. If s = ut + \(s=u t+\frac{1}{2} a t^2\), where u and a are constants. Obtain the value of \(\frac{\mathrm{ds}}{\mathrm{dt}}\)

1. u – at
2. u + at
3. 2u + at
4. None of these

Answer: 2. u + at

Question 24. The minimum value of y = 5x² – 2x + 1 is

1. \(\frac{1}{5}\)
2. \(\frac{2}{5}\)
3. \(\frac{4}{5}\)
4. \(\frac{3}{5}\)

Answer: 3. \(\frac{4}{5}\)

Question 25. y = \(\frac{2 x+5}{3 x-2}\)

1. \(y^{\prime}=\frac{-19}{(3 x-2)^2}\)
2. \(y^{\prime}=\frac{19}{(3 x-2)}\)
3. \(y^{\prime}=\frac{-19}{(3 x+2)}\)
4. \(y^{\prime}=\frac{-19}{(3 x+2)^2}\)

Answer: 4. \(y^{\prime}=\frac{-19}{(3 x+2)^2}\)

Question 26. A uniform metallic solid sphere is heated uniformly. Due to thermal expansion, its radius increases at the rate of 0.05 mm/second. Find its rate of change of volume concerning the time when its radius becomes 10 mm. (take p = 3.14)

1. 31.4 mm³/second
2. 62.8 mm³/second
3. 3.14 mm³/second
4. 6.28 mm³/second

Answer: 2. 62.8 mm³/second

Question 27. If y = 3t² – 4t; then the minima of y will be at :

1. 3/2
2. 3/4
3. 2/3
4. 4/3

Answer: 3. 2/3

Question 28. If y = sin(t²) ,then \(\frac{d^2 y}{d t^2}\) will be –

1. 2t cos(t²)
2. 2 cos (t²) – 4t² sin (t²)
3. 4t² sin (t²)
4. 2 cos (t²)

Answer: 2. 2 cos (t²) – 4t² sin (t²)

Class 11 NEET Physics Chapter 10 Mathematical Tools MCQs

Question 29. The displacement of a body at any time t after starting is given by s = 15t – 0.4t². The velocity of the body will be 7 ms^-1 after time :

1. 20 s
2. 15 s
3. 10 s
4. 5 s

Answer: 3. 10 s

Question 30. For the previous question, the acceleration of the particle at any time t is :

1. –0.8 m/s²
2. 0.8 m/s²
3. –0.6 m/s²
4. 0.5 m/s²

Answer: 1. –0.8 m/s²

Question 31. If the velocity of a particle is given by v = 2t⁴ then its acceleration (dv/dt) at any time t will be given by :

1. 8t³
2. 8t
3. –8t³
4. t²

Answer: 1. 8t³

Question 32. The maximum value of xy subject to x + y = 8, is :

1. 8
2. 16
3. 20
4. 24

Answer: 2. 16

Question 33. If y = 3t² – 4t; then the minima of y will be at :

1. 3/2
2. 3/4
3. 2/3
4. 4/3

Answer: 3. 2/3

Question 34. The slope of the graph as shown in the figure at points 1, 2 and is m₁, m₂, and m₃ respectively then

1. m₁ > m₂ > m₃
2. m₁ < m₂ < m₃
3. m₁ = m₂ = m₃
4. m₁ = m₂ > m₃

Answer: 2. m₁ < m₂ < m₃

Question 35. The magnitude of the slope of the shown graph.

1. First increases then decreases
2. First decrease then increases
3. Increase
4. Decrease

Answer: 2. First decrease then increases

Question 36. y = – x² + 3

1. \(\frac{d y}{d x}=-2 x, \frac{d^2 y}{d x^2}=-2\)
2. \(\frac{d y}{d x}=2 x, \frac{d^2 y}{d x^2}=-2\)
3. \(\frac{d y}{d x}=-2 x, \frac{d^2 y}{d x^2}=2\)
4. None of these

Answer: 1. \(\frac{d y}{d x}=-2 x, \frac{d^2 y}{d x^2}=-2\)

Question 37. y = \(\frac{x^3}{3}+\frac{x^3}{2}+\frac{x}{4}\)

1. \(\frac{d y}{d x}=x^2-x+\frac{1}{4}, \frac{d^2 y}{d x^2}=2 x+3\)
2. \(\frac{d y}{d x}=x^2+x-\frac{1}{4}, \frac{d^2 y}{d x^2}=2 x+1\)
3. \(\frac{d y}{d x}=x^2+x+\frac{1}{4}, \frac{d^2 y}{d x^2}=2 x+1\)
4. \(\frac{d y}{d x}=x^2+x+\frac{1}{4}, \frac{d^2 y}{d x^2}=2 x-1\)

Answer: 3. \(\frac{d y}{d x}=x^2+x+\frac{1}{4}, \frac{d^2 y}{d x^2}=2 x+1\)

Question 38. y = 4 – 2x – x^-3

1. \(\frac{d y}{d x}=2+3 x^{-4}, \frac{d^2 y}{d x^2}=-12 x^{-5}\)
2. \(\frac{d y}{d x}=-2+3 x^{-4}, \frac{d^2 y}{d x^2}=-12 x^{-5}\)
3. \(\frac{d y}{d x}=-2+3 x^{-4}, \frac{d^2 y}{d x^2}=12 x^{-5}\)
4. \(\frac{d y}{d x}=-2-3 x^{-4}, \frac{d^2 y}{d x^2}=-12 x^{-5}\)

Answer: 2. \(\frac{d y}{d x}=-2+3 x^{-4}, \frac{d^2 y}{d x^2}=-12 x^{-5}\)

Question 39. y = – 10x + 3 cos x

1. 10 – 3 sin x
2. – 10 + 3 sin x
3. – 10 + 5 sin x
4. – 10 – 3 sin x

Answer: 4. – 10 – 3 sin x

Question 40. y = \(\frac{3}{x}+5 \sin x\)

1. \(-\frac{3}{x^2}+5 \cos x\)
2. \(\frac{3}{x^2}+5 \cos \mathrm{x}\)
3. \(-\frac{3}{x^2}-\cos x\)
4. \(-\frac{3}{x^2}-5 \cos x\)

Answer: 1. \(-\frac{3}{x^2}+5 \cos x\)

Question 41. y = cosec \(x-4 \sqrt{x}+7\)

1. \(-\csc x \cot x-\frac{2}{\sqrt{x}}\)
2. \(\csc x \cot x+\frac{2}{\sqrt{x}}\)
3. \(-\csc x \cot x+\frac{2}{\sqrt{x}}\)
4. \(\csc x \cot x+\frac{2}{\sqrt{x}}\)

Find \(\frac{\mathrm{ds}}{\mathrm{dt}}\)

Answer: 1. \(-\csc x \cot x-\frac{2}{\sqrt{x}}\)

Question 42. s = tan t – t

1. sec²t + t
2. sec²t
3. sec t – 1
4. sec²t – 1

Answer: 4. sec²t – 1

Question 43. s = t² – sec t + t

1. 2t + sec t tan t + 1
2. 2t – sec t tan t + 1
3. 2t – sec t tan t –1
4. 2t + sec²tan t – 1

Answer: 2. 2t – sec t tan t + 1

Question 44. p = \(5+\frac{1}{\cot q}\) find \(\frac{d p}{d q}\)

1. sec² q
2. sec³ q
3. sec q
4. tan² q

Answer: 1. sec² q

Question 45. p = (1 + cosec q) cos q, find \(\frac{d p}{d q}\)

1. sin q – cosec² q
2. – sin q – cosec² q
3. – sin q – cos² q
4. sec q – cosec² q

Answer: 2. – sin q – cosec² q

Question 46. y = sin³ x , find the \(\frac{d y}{d x}\)

1. 3 sin² x (cosx)
2. 3 sin³ x (cosx)
3. 3 sin x (cos x)²
4. sin x (cos x)

Answer: 1. 3 sin² x (cosx)

Question 47. y = 5 cos^-4 x, find \(\frac{d y}{d x}\)

1. 20 in x cos^-5 x
2. 10 in x cos^-5 x
3. 20 in x cos^-3 x
4. 20 in x sin^-5 x

Answer: 1. 20 sin x cos^-5 x

Find the derivatives of the functions

Question 48. s = \(\frac{4}{3 \pi} \sin 3 t+\frac{4}{5 \pi} \cos 5 t\)

1. \(\frac{4}{\pi}\)(cos 3t – sin 5t)
2. \(\frac{4}{\pi}\) 4(cos 3t + sin 5t)
3. \(\frac{4}{\pi}\)(cos t – sin t)
4. \(\frac{4}{\pi}\)(cot 3t – sec 5t)

Answer: 1. \(\frac{4}{\pi}\)(cos 3t – sin 5t)

Question 49. s = \(\sin \left(\frac{3 \pi t}{2}\right)+\cos \left(\frac{3 \pi t}{2}\right)\)

1. \(\frac{3 \pi}{2}\left[\cos \left(\frac{3 \pi t}{2}\right)-\sin \left(\frac{3 \pi t}{2}\right)\right]\)
2. \(\frac{3 \pi}{2}\left[\cos \left(\frac{3 \pi t}{2}\right)+\sin \left(\frac{3 \pi t}{2}\right)\right]\)
3. \(\frac{3 \pi}{2}\left[\cot \left(\frac{3 \pi t}{2}\right)+\sin \left(\frac{3 \pi t}{2}\right)\right]\)
4. None of these

Answer: 1. \(\frac{3 \pi}{2}\left[\cos \left(\frac{3 \pi t}{2}\right)-\sin \left(\frac{3 \pi t}{2}\right)\right]\)

Find integrals of given functions

Question 50. \(\int\left(x^2-2 x+1\right) d x\)

1. \(\frac{x^3}{3}+x^2-x-c\)
2. \(\frac{x^3}{3}+x+x+c\)
3. \(\frac{x}{3}+x^2+x-c\)
4. \(\frac{x^3}{3}-x^2+x+c\)

Answer: 4. \(\frac{x^3}{3}-x^2+x+c\)

Question 51. \(\int\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right) d x\)

1. \(\frac{2 \sqrt{x}}{3}+2 \sqrt{x}-c\)
2. \(\frac{2 \sqrt{x^2}}{3}-2 \sqrt{x}+c\)
3. \(\frac{2 \sqrt{x^3}}{3}+2 \sqrt{x}+c\)
4. \(\frac{2 \sqrt{x}}{2}+2 \sqrt{x}-c\)

Answer: 3. \(\frac{2 \sqrt{x^3}}{3}+2 \sqrt{x}+c\)

Question 52. \(\int \frac{1}{3 x} d x\)

1. \(\frac{1}{3} \ln x+x\)
2. \(\frac{1}{3} \ln x\)
3. \(\frac{1}{2} \ln x+x\)
4. \(\frac{1}{3} \ln x+x\)

Answer: 4. \(\frac{1}{3} \ln x+x\)

Question 53. \(\int x \sin \left(2 x^2\right) d x\), (use,u = 2x²)

1. \(-\frac{1}{4} \cos \left(2 x^2\right)+C\)
2. \(\frac{1}{4} \cos \left(2 x^2\right)+C\)
3. \(-\frac{1}{2} \cos (2 x)+C\)
4. \(-\frac{1}{3} \cos \left(3 x^2\right)+C\)

Answer: 1. \(-\frac{1}{4} \cos \left(2 x^2\right)+C\)

Question 54. \(\int \frac{3}{(2-x)^2} d x\)

1. \(\frac{3}{2-x}+C\)
2. \(\frac{2}{2-x}+C\)
3. \(\frac{3}{2-x}+C\)
4. \(\frac{3}{2+x}+C\)

Answer: 1. \(\frac{3}{2-x}+C\)

NEET Physics Class 11 Mathematical Tools Multiple Choice Questions

Question 55. \(\int_{-4}^{-1} \frac{\pi}{2} d \theta\)

1. \(\frac{3 \pi}{3}\)
2. \(\frac{3 \pi}{2}\)
3. \(\frac{2 \pi}{3}\)
4. \(\frac{\pi}{2}\)

Answer: 1. \(\frac{3 \pi}{3}\)

Question 56. \(\int_0^1 e^x d x\)

1. e – 1 m
2. e + 1
3. e – 2
4. None of these

Answer: 1. e – 1 m

Question 57. y = 2x, the area under the curve from x = 0 to x = b will be

1. b²/2 units
2. b² units
3. 2b² units
4. b/2 units

Answer: 2. b2 units

Question58. y = \(\int_0^\pi \sin x d x\)

1. 2 units
2. 3 units
3. 4 units
4. 5 units

Answer: 1. 2 units

Question 59. The integral \(\int_1^5 x^2 d x\) is equal to

1. \(\frac{125}{3}\)
2. \(\frac{124}{3}\)
3. \(\frac{1}{3}\)
4. 45

Answer: 2. \(\frac{124}{3}\)

Question 60. \(\int x^{-\frac{3}{2}} d x\) is equal to :

1. \(\frac{-2}{\sqrt{x}}+C\)
2. \(\frac{2}{\sqrt{x}}+C\)
3. \(2 \sqrt{x}+C\)
4. \(-2 \sqrt{x}+C\)

Answer: 1. \(\frac{-2}{\sqrt{x}}+C\)

Question 61. \(\int x^{-\frac{5}{3}} d x\) is equal to :

1. \(\frac{3}{2} x^{\frac{2}{3}}+C\)
2. \(-\frac{3}{2} x^{\frac{2}{3}}+C\)
3. \(\frac{3}{2} x^{-\frac{2}{3}}+C\)
4. \(-\frac{3}{2} x^{-\frac{2}{3}}+C\)

Answer: 4. \(-\frac{3}{2} x^{-\frac{2}{3}}+C\)

Question 62. \(\int x^{2019} d x\) dx ∫is equal to :

1. \(\frac{x^{2020}}{2020}+C\)
2. \(\frac{x^{2018}}{2018}+C\)
3. \(2019 \mathrm{X}^{2018}+\mathrm{C}\)
4. \(-2012 X^{2011}+C\)

Answer: 1. \(\frac{x^{2020}}{2020}+C\)

Question 63. ∫2sin(x)dx is equal to :

1. –2cos x + C
2. 2 cosx + C
3. –2 cos x
4. 2 cox

Answer: 1. –2cos x + C

Question 64. \(\int(\sin x+\cos x) d x\) is equal to :

1. –cox + sinx
2. – cox + sinx + C
3. cosx – sinx + C
4. – cosx – sinx + C

Answer: 2. – cost + sinx + C

Question 65. \(\int\left(x+x^2+x^3+x^4\right) d x\) is equal to :

1. 1+2x+3x²+4x³ + C
2. 1+2x+3x²+4x³
3. \(\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\frac{x^5}{5}+C\)
4. \(\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\frac{x^5}{5}\)

Answer: 3. \(\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\frac{x^5}{5}+C\)

Question 66. If y = sin(ax+b), ∫y dx will be :

1. \(\frac{\cos (a x+b)}{a}+C\)
2. \(-\frac{\cos (a x+b)}{a}+c\)
3. a cos(ax+b)+C
4. – a cos(ax+b)+C

Answer: 2. \(-\frac{\cos (a x+b)}{a}+c\)

Question 67. If y = x²sin(x³), then ∫y dx will be :

1. \(-\cos \left(x^3\right)+C\)
2. \(\left(-\frac{\cos x^3}{3}\right)+C\)
3. \(\cos \left(x^3\right)+C\)
4. \(\frac{\cos x^3}{3}+C\)

Answer: 2. \(\left(-\frac{\cos x^3}{3}\right)+C\)

Question 68. If y = x², then area of curve y v/s x from x = 0 to 2 will be :

1. 1/3
2. 8/3
3. 4/3
4. 2/3

Answer: 2. 8/3

Question 69. If y = t sin (t²) then ∫ydt will be : 

1. \(\frac{\cos \left(\mathrm{t}^2\right)}{2}+\mathrm{c}\)
2. \(\frac{\cos \left(\mathrm{t}^2\right)}{2}+\mathrm{c}\)
3. \(\frac{-\cos \left(\mathrm{t}^2\right)}{2}+\mathrm{c}\)
4. \(\cos \left(t^2\right)\)

Answer: 3. \(\frac{-\cos \left(\mathrm{t}^2\right)}{2}+\mathrm{c}\)

Question 70. If x = (6y + 4) (3y² + 4y + 3) then ∫x day will be :

1. \(\frac{1}{3 y^2+4 y+3}\)
2. \(\frac{\left(3 y^2+4 y+3\right)^2}{2}+C\)
3. \(\left(3 y^2+4 y+3\right)\)
4. \(\frac{(6 y+4)}{\left(3 y^2+4 y+3\right)}\)

Answer: 2. \(\frac{\left(3 y^2+4 y+3\right)^2}{2}+C\)

Question 71. Value of \(\int_0^{\pi / 2} \cos 3 t\) it is

1. \(\frac{2}{3}\)
2. \(-\frac{1}{3}\)
3. \(-\frac{2}{3}\)
4. \(\frac{1}{3}\)

Answer: 2. \(-\frac{1}{3}\)

Question 72. \(\int_0^1\left(t^2+9 t+c\right) d t=\frac{9}{2}\). Then the value of ‘c’.

1. \(\frac{2}{3}\)
2. \(-\frac{1}{3}\)
3. \(-\frac{2}{3}\)
4. \(\frac{1}{3}\)

Answer: 2. \(-\frac{1}{3}\)

Question 73. Find the value of the following integration. \(\int_0^{2 \pi} \sin ^2 \theta d \theta\) Here c,a are constants.

1. π
2. 2π
3. 3 π
4. 4 π

Answer: 1. π

NEET Physics Chapter 10: Mathematical Tools MCQs and Solutions

Question 74. If y = \(\frac{1}{a x+b}\), then ∫y dx will be :

1. \(\frac{1}{(a x+b)^2}+C\)
2. \(a x+b+C\)
3. \(a \ln (a x+b)+C\)
4. \(\frac{\ln (a x+b)}{a}+C\)

Answer: 4. \(\frac{\ln (a x+b)}{a}+C\)

Question 75. \(\int_\pi^{2 \pi} \theta d \theta\)

1. \(\frac{3 \pi^2}{2}\)
2. \(\frac{3 \pi^3}{2}\)
3. \(\frac{\pi^3}{2}\)
4. π

Answer: 1. \(\frac{3 \pi^2}{2}\)

Question 76. \(\int_0^{\sqrt[3]{7}} x^2 d x\)

1. \(\frac{7}{3}\)
2. \(\frac{7}{4}\)
3. \(\frac{5}{4}\)
4. 0

Answer: 1. \(\frac{7}{3}\)

Question 77. \(\int_0^1 \frac{d x}{3 x+2}\)

1. \(\ln \left(\frac{5}{2}\right)^{1 / 3}\)
2. \(\ln \left(\frac{5}{2}\right)^{1 / 2}\)
3. \(\ln \left(\frac{5}{2}\right)^{1 / 4}\)
4. None of these

Answer: 1. \(\ln \left(\frac{5}{2}\right)^{1 / 3}\)

Question 78. \(\int(x+1) d x\)

1. \(\frac{x^2}{2}+2 x-C\)
2. \(\frac{x^2}{2}+x+C\)
3. \(\frac{x^2}{2}-x+C\)
4. \(\frac{x^2}{2}-x-C\)

Answer: 2. \(\frac{x^2}{2}+2 x-C\)

Question 79. ∫ (5–6x) dx

1. 5x – x² + C
2. x – 3x²– C
3. 5x + 3x² + C
4. 5x – 3x² + C

Answer: 4. 5x – 3x² + C

Question 80. \(\int\left(3 t^2+\frac{t}{2}\right) d t\)

1. \(t^2+\frac{t^2}{4}-C\)
2. \(t^2+\frac{t^2}{4}+C\)
3. \(t^3-\frac{t^2}{4}-C\)
4. \(\frac{t^3}{6}+t^4+C\)

Answer: 2. \(t^2+\frac{t^2}{4}+C\)

Question 81. \(\int\left(\frac{t^2}{2}+4 t^3\right) d t\)

1. \(\frac{t^3}{6}+t^2+C\)
2. \(\frac{t^3}{6}+t+C\)
3. \(\frac{t^3}{6}-t+C\)
4. \(\frac{t^3}{6}+t^4+C\)

Answer: 4. \(\frac{t^3}{6}+t^4+C\)

Question 82. \(\int x^{-1 / 3} d x\)

1. \(\frac{3}{2} x^{2 / 3}+C\)
2. \(\frac{3}{2} x^{2 / 5}+C\)
3. \(\frac{3}{2} x^{1 / 3}+C\)
4. \(\frac{3}{2} x^{2 / 7}+C\)

Answer: 1. \(\frac{3}{2} x^{2 / 3}+C\)

Question 83. \(\int\left(\frac{\sqrt{x}}{2}+\frac{2}{\sqrt{x}}\right) d x\)

1. \(\frac{x^{3 / 2}}{3}+4 x^{1 / 2}+C\)
2. \(\frac{x^{3 / 2}}{3}+x^{1 / 2}+C\)
3. \(\frac{x^{3 / 2}}{3}+4 x^{2 / 5}+C\)
4. \(\frac{x^{3 / 2}}{3}+4 x^2+C\)

Answer: 1. \(\frac{x^{3 / 2}}{3}+4 x^{1 / 2}+C\)

Question 84. \(\int\left(8 y-\frac{2}{y^{1 / 4}}\right) d y\)

1. \(4 y^2-\frac{8}{3} y^{3 / 4}+C\)
2. \(4 y^2+\frac{8}{3} y^{3 / 4}+C\)
3. \(y^2-\frac{8}{3} y^{3 / 4}+C\)
4. \(4 y^2-\frac{8}{3} y^{1 / 3}+C\)

Answer: 1. \(4 y^2-\frac{8}{3} y^{3 / 4}+C\)

Question 85. \(\int 2 x\left(1-x^{-3}\right) d x\)

1. \(x+\frac{2}{x}-C\)
2. \(x^2+\frac{2}{x}+C\)
3. \(2 x^2+\frac{2}{x}+C\)
4. \(5 x^2+\frac{2}{x}+C\)

Answer: 2. \(x^2+\frac{2}{x}+C\)

Question 86. ∫(– 2cost) dt

1. – 2 sin t + C
2. – 3 sin t + C
3. – 5 sin t + C
4. – 7 sin t + C

Answer: 1. – 2 sin t + C

Question 87. ∫(– 5 sint) dt

1. 5 cos t + C
2. 2 cos t – C
3. 5 cosec t + C
4. 5 tan t + C

Answer: 1. 5 cos t + C

Question 88. \(\int 7 \sin \frac{\theta}{3} d \theta\)

1. \(-21 \cos \frac{\theta}{3}+\mathrm{C}\)
2. \(-14 \cos \frac{\theta}{3}+C\)
3. \(-42 \cos \frac{\theta}{3}\)
4. \(-7 \cos \frac{\theta}{3}+C\)

Answer: 1. \(-21 \cos \frac{\theta}{3}+\mathrm{C}\)

Question 89. ∫3cos 5θ+C

1. \(\frac{3}{5} \sin 5 \theta+C\)
2. \(\frac{3}{5} \sin 3 \theta+C\)
3. \(\frac{3}{5} \cos 5 \theta+C\)
4. \(\frac{3}{5} \sec 5 \theta+C\)

Answer: 1. \(\frac{3}{5} \sin 5 \theta+C\)

Question 90. \(\int\left(-3 \csc ^2 x\right) d x\) dx

1. 3 cot x + C
2. cot x + C
3. 3 tan x + C
4. 5 cot x + C

Answer: 1. 3 cot x + C

Question 91. \(\int\left(-\frac{\sec ^2 x}{3}\right) d x\)

1. \(\frac{-\tan x}{3}+x\)
2. \(\frac{-\tan x}{3}+C\)
3. \(\frac{-\tan x}{3}+C\)
4. None

Answer: 2. \(\frac{-\tan x}{3}+C\)

Question 92. \(\int \frac{\csc \theta \cot \theta}{2} d \theta\)

1. \(-\frac{1}{2} \csc \theta+C\)
2. \(-\frac{1}{2} \tan \theta+C\)
3. \(-\frac{1}{2} \cot \theta+C\)
4. \(-\frac{1}{2} \sec \theta+C\)

Answer: 1. \(-\frac{1}{2} \csc \theta+C\)

NEET Physics Mathematical Tools MCQs for Chapter 10

Question 93. \(\int \frac{2}{5} \sec \theta \tan \theta \mathrm{d} \theta\)

1. \(\frac{2}{5} \sec \theta+C\)
2. \(\frac{2}{5} \cos \theta+\mathrm{C}\)
3. \(\frac{2}{5} \tan \theta+C\)
4. \(\frac{2}{5}{cosec} \theta+C\)

Answer: 1. \(\frac{2}{5} \sec \theta+C\)

Question 94. \(\int\left(4 \sec x \tan -2 \sec ^2 x\right) d x\)

1. 4 sec x – 2 tan x + C
2. 2 sec x – 2 tan x + C
3. 4 sec x – 3 tan x + C
4. 4 sec x–5 tan x + C

Answer: 1. 4 sec x – 2 tan x + C

Question 95. \(\int \frac{1}{2}\left(\csc ^2 x-{cxc} x \cot x\right) d x\)

1. \(-\frac{1}{2} \cot x+\frac{1}{2} \csc x+C\)
2. \(\frac{1}{2} \tan x+\frac{1}{2} \csc x+C\)
3. \(-\frac{1}{2} \sec x+\frac{1}{2} \csc x+C\)
4. \(-\frac{1}{2} \sin x+\frac{1}{2} \csc x+C\)

Answer: 1. \(-\frac{1}{2} \cot x+\frac{1}{2} \csc x+C\)

Question 96. \(\int\left(\sin 2 x-\csc ^2 x\right) d x\)

1. \(-\frac{1}{2} \cos 2 x-\cot \mathrm{x}+C\)
2. \(-\frac{1}{2} \cos 2 x+\cot \mathrm{x}+\mathrm{C}\)
3. \(-\frac{1}{2} \cos 3 x-\cot x+C\)
4. \(-\frac{1}{2} \cos 2 x+\tan x+C\)

Answer: 2. \(-\frac{1}{2} \cos 2 x+\cot \mathrm{x}+\mathrm{C}\)

Question 97. \(\int(2 \cos 2 x-3 \sin 3 x) d x\)

1. sin 2x + cos 3x + C
2. sin 2x + cos 5x + C
3. sin 2x + cot 3x + C
4. sin 3x + cos 3x + C

Answer: 1. sin 2x + cos 3x + C

Question 98. \(\int \frac{1+\cos 4 t}{2} d t\)

1. \(\frac{t}{2}+\frac{\sin 4 t}{8}+C\)
2. \(\frac{t}{2}-\frac{\sin 4 t}{8}-C\)
3. \(\frac{t}{3}+\frac{\sin 4 t}{8}+C\)
4. All of these

Answer: 1. \(\frac{t}{2}+\frac{\sin 4 t}{8}+C\)

Question 99. \(\int \frac{1-\cos 6 \mathrm{t}}{2} \mathrm{dt}\)

1. \(\frac{t}{2}+\frac{\sin 6 t}{12}+C\)
2. \(\frac{t}{2}-\frac{\sin 6 t}{12}+C\)
3. \(2 \times \frac{t}{2}-\frac{\sin 6 t}{12}+C\)
4. \(\frac{t}{2}-\frac{\sin 6 t}{12}+C\)

Answer: 2. \(\frac{t}{2}-\frac{\sin 6 t}{12}+C\)

Question 100. \(\int\left(1+\tan ^2 \theta\right) d \theta\)

1. tan θ + C
2. cot θ + C
3. sec θ + C
4. cosec θ + C

Answer: 1. tan θ + C

Question 101. \(\int_{1 / 2}^{3 / 2}(-2 x+4) d x\)

1. 2 square units
2. 4 square units
3. 6 square units
4. 8 square units

Answer: 1. 2 square units

Evaluate definite integrals of the following functions

Question 102. \(\int_0^{\pi / 2} \theta^2 d \theta\)

1. \(\frac{\pi^3}{24}\)
2. \(\frac{\pi^2}{24}\)
3. \(\frac{\pi^2}{36}\)
4. \(\frac{\pi^2}{48}\)

Answer: 1. \(\frac{\pi^3}{24}\)

Question 103. \(\int_0^{3 b} x^2 d x\)

1. 9b³
2. 3b³
3. 27b³
4. 81 b³

Answer: 1. 9b³

Question 104. The forces, each numerically equal to 5N, are acting as shown in the Figure. Find the angle between forces.

1. 90º
2. 180º
3. 120º
4. 160º

Answer: 3. 120º

Question 105. The vector joining the points A (1, 1, –1) and B (2, –3, 4) and pointing from A to B is –

1. \(-\hat{i}+4 \hat{j}-5 \hat{k}\)
2. \(\hat{i}+4 \hat{j}+5 \hat{k}\)
3. \(\hat{i}-4 \hat{j}+5 \hat{k}\)
4. \(-\hat{i}-4 \hat{j}-5 \hat{k}\)

Answer: 4. \(-\hat{i}-4 \hat{j}-5 \hat{k}\)

Question 106. A vector of magnitude 30 and direction eastwards is added with another vector of magnitude 40 and direction Northwards. Find the magnitude and direction of the resultant with the east.

1. 45, 50º with East
2. 53, 75 with East
3. 53, 50º with East
4. 50, 53º with East

Answer: 4. 50, 53º with East

Question 107. The vector sum of the forces of 10 N and 6 N can be

1. 2 N
2. 8 N
3. 18 N
4. 20 N.

Answer: 2. 8 N

NEET Physics Class 11 Chapter 10: Key Concepts and MCQs

Question 108. The vector sum of two forces P and Q is minimum when the angle θ between their positive directions, is

1. \(\frac{\pi}{4}\)
2. \(\frac{\pi}{3}\)
3. \(\frac{\pi}{2}\)
4. \(\pi\)

Answer: 4. \(\pi\)

Question 109. The vector sum of two vectors \(\overrightarrow{\mathrm{A}} \text { and } \overrightarrow{\mathrm{B}}\) is maximum, then the angle θ between two vectors is –

1. 0º
2. 30°
3. 45°
4. 60°

Answer: 1. 0º

Question 110. Find the magnitude of \(3 \hat{i}+2 \hat{j}+\hat{k}\)?

1. \(\sqrt{10}\)
2. \(\sqrt{11}\)
3. \(\sqrt{13}\)
4. \(\sqrt{14}\)

Answer: 4. \(\sqrt{14}\)

Question 111. If \(\overrightarrow{\mathrm{A}}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}\) then find \(\hat{A}\)

1. \(\frac{3 \hat{i}+4 \hat{j}}{5}\)
2. \(\frac{2 \hat{i}+3 \hat{j}}{5}\)
3. \(\frac{2 \hat{i}+4 \hat{j}}{5}\)
4. \(\frac{3 \hat{i}-2 \hat{j}}{5}\)

Answer: 1. \(\frac{3 \hat{i}+4 \hat{j}}{5}\)

Question 112. One of the rectangular components of a velocity of 60 km h–1 is 30 km h–1. Find another rectangular component.

1. \(30 \sqrt{3} \mathrm{~km} \mathrm{~h}^{-1}\)
2. \(20 \sqrt{3} \mathrm{~km} \mathrm{~h}^{-1}\)
3. \(30 \sqrt{2} \mathrm{~km} \mathrm{~h}^{-1}\)
4. \(30 \sqrt{2} \mathrm{~km} \mathrm{~h}^{-1}\)

Answer: 1. \(30 \sqrt{3} \mathrm{~km} \mathrm{~h}^{-1}\)

Question 113. The x and y components of a force are 2 N and – 3N. The force is

1. \(2 \hat{i}-3 \hat{j}\)
2. \(2 \hat{i}+3 \hat{j}\)
3. \(-2 \hat{i}-3 \hat{j}\)
4. \(3 \hat{i}+2 \hat{j}\)

Answer: 1. \(2 \hat{i}-3 \hat{j}\)

Question 114. A force of 30 N is inclined at an angle θ to the horizontal. If its vertical component is 18 N, find the horizontal component & the value of θ

1. 24 N ; 370 approx
2. 20 N ; 470 approx
3. 25 N ; 350 approx
4. 37 N ; 240 approx

Answer: 1. 24 N; 370 approx

Question 115. The angle θ between directions of forces \(\vec{A} \text { and } \vec{B}\) is 90º where A = 8 dyne and B = 6 dyne. If the resultant \(\vec{R}\) makes an angle α with \(\vec{A}\) then find the value of ‘α’?

1. 47º
2. 37º
3. 75º
4. 120º

Answer: 2. 37º

Question 116. If \(\vec{A}=3 \hat{i}+4 \hat{j} \text { and } \vec{B}=\hat{i}+\hat{j}+2 \hat{k}\) then find out unit vector along \(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}\)

1. \(\frac{4 \hat{i}+5 \hat{j}-2 \hat{k}}{\sqrt{45}}\)
2. \(\frac{2 \hat{i}-5 \hat{j}-2 \hat{k}}{\sqrt{45}}\)
3. \(\frac{4 \hat{i}-2 \hat{j}+2 \hat{k}}{\sqrt{45}}\)
4. \(\frac{4 \hat{i}+5 \hat{j}+2 \hat{k}}{\sqrt{45}}\)

Answer: 4. \(\frac{4 \hat{i}+5 \hat{j}+2 \hat{k}}{\sqrt{45}}\)

Question 117. The x and y components of the vector \(\overrightarrow{\mathrm{A}}\) are 4m and 6m respectively. The x,y components of vector are \(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}\) 10m and 9m respectively. Find the length of \(\overrightarrow{\mathrm{B}}\) and the angle that \(\overrightarrow{\mathrm{B}}\) makes with the x-axis.

1. \(3 \sqrt{3}, \tan ^{-1} \frac{1}{2}\)
2. \(3 \sqrt{5}, \tan ^{-1} \frac{1}{2}\)
3. \(3 \sqrt{5}, \tan \frac{1}{3}\)
4. \(2 \sqrt{3}, \tan ^{-1} \frac{1}{2}\)

Answer: 2. \(3 \sqrt{5}, \tan ^{-1} \frac{1}{2}\)

Question 118. A vector is not changed if

1. It is displaced parallel to itself
2. It is rotated through an arbitrary angle
3. It is cross-multiplied by a unit vector
4. It is multiplied by an arbitrary scalar.

Answer: 1. It is displaced parallel to itself

Question 119. If the angle between two forces increases, the magnitude of their resultant

1. Decreases
2. Increases
3. Remains unchanged
4. First decreases and then increases

Answer: 1. Decreases

Question 120. Which of the following sets of displacements might be capable of bringing a car to its returning point?

1. 5, 10, 30 and 50 km
2. 5, 9, 9 and 16 km
3. 40, 40, 90 and 200 km
4. 10, 20, 40 and 90 km

Answer: 21. 5, 9, 9 and 16 km

Question 121. When two vectors \(\overrightarrow{\mathrm{a}} \text { and } \overrightarrow{\mathrm{b}}\) are added, the magnitude of the resultant vector is always

1. Greater than (a + b)
2. Less than or equal to (a + b)
3. Less than (a + b)
4. Equal to (a + b)

Answer: 2. Less than or equal to (a + b)

Question 122. If \(|\vec{A}+\vec{B}|=|\vec{A}|=|\vec{B}|\), then the angle between \(\overrightarrow{\mathrm{A}} \text { and } \overrightarrow{\mathrm{B}}\) is

1. 0º
2. 60º
3. 90º
4. 120º

Answer: 4. 120º

Question 123. Vector \(\overrightarrow{\mathrm{A}}\) is of length 2 cm and is 60º above the x-axis in the first quadrant. Vector \(\overrightarrow{\mathrm{B}}\) is of length 2 cm and 60º below the x-axis in the fourth quadrant. The sum \(\) is a vector of magnitude –

1. 2 along + y-axis
2. 2 along + x-axis
3. 1 along – x-axis
4. 2 along – x-axis

Answer: 2. 2 along + x-axis

Question 124. Which of the following is a true statement?

1. A vector cannot be divided by another vector
2. Angular displacement can either be a scalar or a vector.
3. Since the addition of vectors is commutative therefore vector subtraction is also commutative.
4. The resultant of two equal forces of magnitude F acting at a point is F if the angle between the two forces is 120º.

Answer: 1. A vector cannot be divided by another vector

Question 125. In the Figure which of the ways indicated for combining the x and y components of vector a is proper to determine that vector?

Answer: 1

Question 126. Two vectors having an equal magnitude of 5 units have an angle of 60º between them. Find the magnitude of their resultant vector and its angle from one of the vectors.

1. 5, 20º
2. \(5 \sqrt{3}\), 30º
3. 3, 40º
4. 3, 50º

Answer: 2. \(5 \sqrt{3}\), 30º

Mathematical Tools MCQs for NEET Physics Chapter 10

Question 127. Two forces each numerically equal to 10 dynes are acting as shown in the figure, then find the resultant of these two vectors.

1. 5 dyne
2. 10 dyne
3. 15 dyne
4. 25 dyne

Answer: 2. 10 dyne

Question 128. The magnitude of pairs of displacement vectors is given. Which pairs of displacement vectors cannot be added to give a resultant vector of magnitude 13 cm?

1. 4 cm, 16 cm
2. 20 cm, 7 cm
3. 1 cm, 15 cm
4. 6 cm, 8 cm

Answer: 3. 1 cm, 15 cm

Question 129. If \(\overrightarrow{\mathrm{A}}=3 \hat{\mathrm{i}}+2 \hat{\mathrm{i}} \text { and } \overrightarrow{\mathrm{B}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{i}}-\hat{\mathrm{k}}\), then find a unit vector along \((\vec{A}-\vec{B})\)

1. \(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}}{\sqrt{3}}\)
2. \(\frac{\hat{i}-\hat{j}-\hat{k}}{\sqrt{3}}\)
3. \(\frac{\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{3}}\)
4. \(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{3}}\)

Answer: 3. \(\frac{\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{3}}\)

Question 130. If \(\hat{n}\) is a unit vector in the direction of the vector \(\overrightarrow{\mathrm{A}}\), then –

1. \(\hat{n}=\frac{\vec{A}}{|A|}\)
2. \(\hat{n}=\vec{A}|\vec{A}|\)
3. \(\hat{n}=\frac{|\vec{A}|}{\vec{A}}\)
4. \(\hat{n}=\hat{n} \times \vec{A}\)

Answer: 1. \(\hat{n}=\frac{\vec{A}}{|A|}\)

Question 131. The resultant of \(\vec{A} \text { and } \vec{B}\) makes an angle α with \(\vec{A} \text { and } \vec{B}\), then :

1. α < β
2. α < β
3. α < β if A > B
4. α < β if A = B

Answer: 3. α < β if A > B

Question 132. If \(\vec{P}+\vec{Q}=\vec{P}-\vec{Q}\) and θ is the angle between \(\overrightarrow{\mathrm{P}} \text { and } \overrightarrow{\mathrm{Q}}\), then

1. θ = 0º
2. θ = 90º
3. P =0
4. Q = 0

Answer: 4. Q = 0

Question 133. The magnitudes of the sum and difference of two vectors are the same, then the angle between them is

1. 90º
2. 40º
3. 45º
4. 60º

Answer: 1. 90º

Question 134. The projection of a vector \(3 \hat{i}+4 \hat{k}\) on the y-axis is:

1. 5
2. 4
3. 3
4. 3

Answer: 4. 3

Question 135. Two forces of 12N and 8N act up the n body. The resultant force on the body has a maximum value of-

1. 4N
2. 0N
3. 20 N
4. 8 N

Answer: 3. 20 N

Question 136. In figure, \(\vec{E}\) equals

1. \(\vec{A}\)
2. \(\vec{B}\)
3. \(\vec{A}+\vec{B}\)
4. \(-(\vec{A}+\vec{B})\)

Answer: 4. \(-(\vec{A}+\vec{B})\)

Question 137. In figure, \(\overrightarrow{\mathrm{D}}-\overrightarrow{\mathrm{C}}\) equals

1. \(\vec{A}\)
2. \(-\overrightarrow{\mathrm{A}}\)
3. \(\vec{B}\)
4. \(-\vec{B}\)

Answer: 1. \(\vec{A}\)

Question 138. In the figure, \(\vec{E}+\vec{D}-\vec{C}\) equals

1. \(\vec{A}\)
2. \(-\overrightarrow{\mathrm{A}}\)
3. \(\vec{B}\)
4. \(-\vec{B}\)

Answer: 4. \(-\vec{B}\)

Question 139. Forces proportional to AB, BC, and 2CA act along the sides of triangle ABC in order. They’re resultant represented in magnitude and direction as

1. CA
2. AC
3. BC
4. CB

Answer: 1. CA

Question 140. A given force is resolved into components P and Q equally inclined to it. Then :

1. P = 2Q
2. 2P = Q
3. P = Q
4. None of these

Answer: 3. P = Q

Question 141. A particle starting from the origin (0,0) moves in a straight line in the (x, y) plane. Its coordinates at a later time are (\(\sqrt{3}\), 3). The path of the particle makes with the x-axis an angle of :

1. 30º
2. 45º
3. 60º
4. 0º

Answer: 3. 60º

Question 142. If \(\overrightarrow{\mathrm{A}}=\hat{\mathrm{i}}+\hat{\mathrm{J}}+\hat{\mathrm{k}} \text { and } \overrightarrow{\mathrm{B}}=2 \hat{\mathrm{i}}+\hat{\mathrm{J}}\) find (a) \(\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}\) (b) \(\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}\)

1. 3 and \(-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}\)
2. 5 and \(-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}\)
3. 1 and \(-\hat{i}+2 \hat{j}+\hat{k}\)
4. 3 and \(-\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}\)

Answer: 1. 3 and \(-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}\)

Question 143. If \(|\vec{A}|=4,|\vec{B}|\) = 3 and θ = 60º in figure, Find (a) \(\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}\) (b) \(\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}\)

1. 3 and \(6 \sqrt{3}\)
2. 6 and \(3 \sqrt{3}\)
3. 6 and \(3 \sqrt{6}\)
4. 6 and \(6 \sqrt{3}\)

Answer: 4. 6 and \(6 \sqrt{3}\)

NEET Class 11 Physics Chapter 10 Mathematical Tools MCQs with Answers

Question 144. Three non zero vector \(\overrightarrow{\mathrm{A}}, \overrightarrow{\mathrm{B}} and \overrightarrow{\mathrm{C}}\) satisfy the relation \(\vec{A} \cdot \vec{B}=0 \text { and } \vec{A}. \vec{C}=0\). Then \(\overrightarrow{\mathrm{A}}\)= 0 can be parallel to :

1. \(\vec{B}\)
2. \(\overrightarrow{\mathrm{C}}\)
3. \(\vec{B} \cdot \vec{C}\)
4. \(\overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{C}}\)

Answer: 4. \(\overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{C}}\)

Question 145. If \(\vec{A}=4 \hat{i}+n \hat{J}-2 \hat{k} \text { and } \vec{B}=2 \hat{i}+3 \hat{j}+\hat{k}\), then find the value of n so that \(\overrightarrow{\mathrm{A}} \perp \overrightarrow{\mathrm{B}}\)

1. n = 2
2. n = – 1
3. n + 1
4. n = – 2

Answer: 4. n = – 2

Question 146. If \(\vec{F}=(4 \hat{i}-10 \hat{j}) \text { and } \vec{r}=(5 \hat{i}-3 \hat{j})\), then calculate torque \((\vec{\tau}=\vec{r} \times \vec{F})\)

1. \(-38 \hat{k}\)
2. \(-35 \hat{k}\)
3. \(-55 \hat{k}\)
4. \(-28 \hat{k}\)

Answer: 1. \(-38 \hat{k}\)

Question 147. Find a unit vector perpendicular to both the vectors) \((2 \hat{i}+3 \hat{j}+\hat{k}) \text { and }(\hat{i}-\hat{j}+2 \hat{k})\)

1. \(\hat{n}= \pm \frac{1}{\sqrt{83}}(7 \hat{i}+3 \hat{j}+5 \hat{k})\)
2. \(\hat{n}= \pm \frac{1}{\sqrt{83}}(-7 \hat{i}-3 \hat{j}+5 \hat{k})\)
3. \(\hat{n}= \pm \frac{1}{\sqrt{83}}(7 \hat{i}-3 \hat{j}-5 \hat{k})\)
4. \(\hat{n}= \pm \frac{1}{\sqrt{58}}(7 \hat{i}-3 \hat{j}-5 \hat{k})\)

Answer: 3. \(\hat{n}= \pm \frac{1}{\sqrt{83}}(7 \hat{i}-3 \hat{j}-5 \hat{k})\)

Question 148. Which of the following vector identities is false?

1. \(\vec{P}+\vec{Q}=\vec{Q}+\vec{P}\)
2. \(\vec{P}+\vec{Q}=\vec{Q} \times \vec{P}\)
3. \(\vec{P} \cdot \vec{Q}=\vec{Q} \cdot \vec{P}\)
4. \(\vec{P} \times \vec{Q} \neq \vec{Q} \times \vec{P}\)

Answer: 2. \(\vec{P}+\vec{Q}=\vec{Q} \times \vec{P}\)

Question 149. The area of a parallelogram, whose diagonals are \(\) will be

1. 14 unit
2. \(5 \sqrt{3}\)
3. \(10 \sqrt{3}\)
4. \(20 \sqrt{3}\)

Answer: 3. \(10 \sqrt{3}\)

Question 150. If \(\overrightarrow{\mathrm{A}}=\hat{\mathrm{i}}+\hat{\mathrm{j}} \) and \(\overrightarrow{\mathrm{B}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}\) The value of \((\vec{A}+\vec{B}) \cdot(\vec{A}-\vec{B})\) is :

1. \(\sqrt{2}\)
2. 0
3. \(\frac{1}{2}\)
4. 2

Answer: 2. 0

Question 151. Vectors \(\vec{A}=\hat{i}+\hat{j}-2 \hat{k} \text { and } \vec{B}=3 \hat{i}+3 \hat{j}-6 \hat{k}\) are:

1. Parallel
2. Antiparallel
3. Perpendicular
4. At acute angle with each other

Answer: 3. Perpendicular

Question 152. If two vectors are given as : \(\overrightarrow{\mathrm{A}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}\) and \( \overrightarrow{\mathrm{B}}=-\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}\) then the vector is not perpendicular to \(\) is:

1. \(-2 \hat{i}+4 \hat{j}+2 \hat{k}\)
2. \(\hat{i}+\hat{j}+\hat{k}\)
3. \(25 \hat{i}-625 \hat{j}-25 \hat{k}\)
4. \(3 \hat{i}-2 \hat{j}-3 \hat{k}\)

Answer: 1. \(-2 \hat{i}+4 \hat{j}+2 \hat{k}\)

Question 153. If a vector \(2 \hat{i}+3 \hat{j}+8 \hat{k}\) is perpendicular to the vector \(4 \hat{i}-4 \hat{j}+\alpha \hat{k}\), then the value of α is :

1. –1
2. \(\frac{1}{2}\)
3. \(-\frac{1}{2}\)
4. 1

Answer: 3. \(-\frac{1}{2}\)

Question 154. If the angle between the vectors \(\overrightarrow{\mathrm{A}} \text { and } \overrightarrow{\mathrm{B}}\)is θ, the value of the product \((\overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{A}}) \cdot \overrightarrow{\mathrm{A}}\) is equal to :

1. BA² cos θ
2. BA² sin θ
3. BA² sin θ cos θ
4. Zero

Answer: 4. Zero

Question 155. \(\overrightarrow{\mathrm{A}} \text { and } \overrightarrow{\mathrm{B}}\) are two vectors and θ is the angle between them, if \(|\vec{A} \times \vec{B}|=\sqrt{3}(\vec{A} \cdot \vec{B})\) the value of θ is :

1. 60º
2. 45º
3. 30º
4. 90º

Answer: 1. 60º

Question 156. Two forces P and Q acting at a point are such that if P is reversed, the direction of the resultant is turned through 90º. Then

1. P = Q
2. P =2Q
3. P = \(\frac{Q}{2}\)
4. No relation between P and Q

Answer: 1. P = Q

Question 157. The vector sum of two forces is perpendicular to their vector differences. In that case, the forces :

1. Are not equal to each other in magnitude
2. Cannot be predicted
3. Are equal to each other
4. Are equal to each other in magnitude

Answer: 4. Are equal to each other in magnitude

Question 158. If \(|\vec{A} \times \vec{B}|=\sqrt{3} \cdot \vec{A}, \vec{B}\) then the value of \(|\vec{A}+\vec{B}|\) is :

1. \(\left(\mathrm{A}^2+\mathrm{B}^2+\mathrm{AB}\right)^{1 / 2}\)
2. \(\left(A^2+B^2+\frac{A B}{\sqrt{3}}\right)^{1 / 2}\)
3. \(A+B\)
4. \(\left(\mathrm{A}^2+\mathrm{B}^2+\sqrt{3} A B\right)^{1 / 2}\)

Answer: 1. \(\left(\mathrm{A}^2+\mathrm{B}^2+\mathrm{AB}\right)^{1 / 2}\)

Question 159. Velocity as a function of time is V(t) = sin²t – cos(2t). Then the value of \(\left(\frac{\pi}{3}\right)\) will be :

1. \(\frac{2}{3}\)
2. \(\frac{1}{4}\)
3. \(\frac{1}{4}\)
4. \(\frac{5}{4}\)

Answer: 4. \(\frac{5}{4}\)

Question 160. If f = \(2 \pi \frac{x^3 y^5}{\sqrt{z}}\) then log f is equal to :

1. \(\log 2 \pi+3 \log x+5 \log y+\frac{1}{2} \log z\)
2. \(\log 2 \pi+3 \log x+5 \log y-\frac{1}{2} \log z\)
3. \(\log 2 \pi-3 \log x+5 \log y+\frac{1}{2} \log z\)
4. \(\log 2 \pi+3 \log x+5 \log y+\log z\)

Answer: 2. \(\log 2 \pi+3 \log x+5 \log y-\frac{1}{2} \log z\)

Question 161. Which of the following is true

1. sin37° + cos37° = sin53° + cos53°
2. sin37° – cos37° = cos53° – sin53°
3. tan37° + 1 = tan 53° – 1
4. tan37° × tan53° = 1

Answer: 1. sin37° + cos37° = sin53° + cos53°

Question 162. If y₁ = A sinθ₁ and y₂ = A sin θ₂ then

1. \(y_1+y_2=2 A \sin \left(\frac{\theta_1+\theta_2}{2}\right) \cos \left(\frac{\theta_1-\theta_2}{2}\right)\)
2. \(y_1+y_2=2 A \sin \theta_1 \sin \theta_1\)
3. \(y_1-y_2=2 A \sin \left(\frac{\theta_1-\theta_2}{2}\right) \cos \left(\frac{\theta_1+\theta_2}{2}\right)\)
4. \(y_1, y_2=-2 A^2 \cos \left(\frac{\pi}{2}+\theta_1\right) \cdot \cos \left(\frac{\pi}{2}-\theta_2\right)\)

Answer: (1,3)

Question 163. If R² = A² + B² + 2AB cosθ , if |A| = |B| then value of magnitude of R is equivalent to :

1. 2Acosθ
2. \(A \cos \frac{\theta}{2}\)
3. \(2 A \cos \frac{\theta}{2}\)
4. \(2 B \cos \frac{\theta}{2}\)

Answer: (3,4)

Question 164. A particle starting from the origin (0, 0) moves in a straight line in the (x, y) plane. Its coordinates at a later time are (\(\sqrt{3}\), 3). The path of the particle makes with the x-axis an angle of :

1. 30º
2. 45º
3. 60º
4. 0º

Answer: 3. 60º

Question 165. Find the value of an if the distance between the point (–9cm, a cm) and (3cm, 3 cm) is 13 cm.

1. 6 cm
2. 8 cm
3. 10 cm
4. 12 cm

Answer: 2. 8 cm

Question 166. y = lnx² + sin x

1. \(\frac{d y}{d x}=\frac{2}{x}+\cos x, \frac{d^2 y}{d x^2}=\frac{-2}{x^2}-\sin x\)
2. \(\frac{d y}{d x}=\frac{2}{x}-\cos x, \frac{d^2 y}{d x^2}=\frac{-2}{x^2}+\sin x\)
3. \(\frac{d y}{d x}=-\frac{2}{x}+\cos x, \frac{d^2 y}{d x^2}=\frac{-2}{x^2}-\sin x\)
4. \(\frac{d y}{d x}=-\frac{2}{x}-\cos x, \frac{d^2 y}{d x^2}=\frac{2}{x^2}-\sin x\)

Answer: 1. \(\frac{d y}{d x}=\frac{2}{x}+\cos x, \frac{d^2 y}{d x^2}=\frac{-2}{x^2}-\sin x\)

Question 167. y = \(\sqrt[7]{x}+\tan x\)

1. \(\frac{d y}{d x}=-\frac{x^{-\frac{6}{7}}}{7}+\sec ^2 x, \frac{d^2 y}{d x^2}=\frac{-6}{49} x^{\frac{-13}{7}}-2 \tan x \sec ^2 x\)
2. \(\frac{d y}{d x}=\frac{x^{-\frac{6}{7}}}{7}+\sec ^2 x, \frac{d^2 y}{d x^2}=\frac{-6}{49} x^{\frac{-13}{7}}+2 \tan x \sec ^2 x\)
3. \(\frac{d y}{d x}=\frac{x^{-\frac{6}{7}}}{7}-\sec ^2 x, \frac{d^2 y}{d x^2}=\frac{-6}{49} x^{\frac{-13}{7}}-2 \tan x \sec ^2 x\)
4. \(\frac{d y}{d x}=\frac{x^{-\frac{6}{7}}}{7}+\sec ^2 x, \frac{d^2 y}{d x^2}=\frac{6}{49} x^{\frac{-13}{7}}+2 \tan x \sec ^2 x\)

Answer: 2. \(\frac{d y}{d x}=\frac{x^{-\frac{6}{7}}}{7}+\sec ^2 x, \frac{d^2 y}{d x^2}=\frac{-6}{49} x^{\frac{-13}{7}}+2 \tan x \sec ^2 x\)

NEET Physics Chapter 10: Mathematical Tools MCQs and Solutions

Find the derivative of given functions w.r.t. the corresponding independent variable.

Question 168. y = \(\left(x+\frac{1}{x}\right)\left(x-\frac{1}{x}+1\right)\)

1. \(\frac{d y}{d x}=1+2 x+\frac{2}{x^3}-\frac{1}{x^2}\)
2. \(\frac{d y}{d x}=1+2 x-\frac{2}{x^3}-\frac{1}{x^2}\)
3. \(\frac{d y}{d x}=1+2 x+\frac{2}{x^3}-\frac{1}{x^2}\)
4. \(\frac{d y}{d x}=1+2 x+\frac{2}{x^3}+\frac{1}{x^2}\)

Answer: 3. \(\frac{d y}{d x}=1+2 x+\frac{2}{x^3}-\frac{1}{x^2}\)

Question 169. r = (1 + sec θ) sin θ, r′ is

1. \(\frac{d r}{d \theta}=\cos \theta+\sec ^2 \theta\)
2. \(\frac{d r}{d \theta}=\cos \theta-\sec ^2 \theta\)
3. \(\frac{d r}{d \theta}=\cos \theta+\tan ^2 \theta\)
4. \(\frac{d r}{d \theta}=\cos \theta+\sec ^2 \theta\)

Answer: 1. \(\frac{d r}{d \theta}=\cos \theta+\sec ^2 \theta\)

Question 170. q = \(\sqrt{2 r-r^2}\), find \(\frac{\mathrm{dq}}{\mathrm{dr}}\)

1. \(\frac{1-r}{\sqrt{2 r-r^2}}\)
2. \(\frac{1+r}{\sqrt{2 r+r^2}}\)
3. \(\frac{1-r}{\sqrt{3 r+r}}\)
4. \(\frac{1-r}{\sqrt{2 r-r^2}}\)

Answer: 1. \(\frac{1-r}{\sqrt{2 r-r^2}}\)

Find \(\frac{d y}{d x}\)

Question 171. y = \(\frac{\cot x}{1+\cot x}\)

1. \(\frac{-\csc ^2 x}{(1+\cot x)^2}\)
2. \(\frac{-\csc ^2 x}{(1-\cot x)^2}\)
3. \(\frac{-\csc ^2 x}{(1+\cot x)^2}\)
4. \(\frac{-\csc ^2 x}{(1+\tan x)^2}\)

Answer: 1. \(\frac{-\csc ^2 x}{(1+\cot x)^2}\)

Question 172. y = \(\frac{\ell \mathrm{nx}+\mathrm{e}^{\mathrm{x}}}{\tan \mathrm{x}}\), then \(\frac{d y}{d x}\) is

1. \(\frac{\tan x\left(e^x+\frac{1}{x}\right)+\sec ^2 x\left(e^x+\ell \ln x\right)}{\tan ^2 x}\)
2. \(\frac{\tan x\left(e^x+\frac{1}{x}\right)-\sec ^2 x\left(e^x+\ell \ln x\right)}{\tan ^2 x}\)
3. \(\frac{\tan x\left(e^x-\frac{1}{x}\right)-\sec ^2 x\left(e^x+\ell n x\right)}{\tan ^2 x}\)
4. \(\frac{\tan x\left(e^x+\frac{1}{x}\right)-\sec ^2 x\left(e^x-\ell n x\right)}{\tan ^2 x}\)

Answer: 1. \(\frac{\tan x\left(e^x+\frac{1}{x}\right)+\sec ^2 x\left(e^x+\ell \ln x\right)}{\tan ^2 x}\)

Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) as a function of x

Question 173. x³ + y³ = 18 xy

1. \(\frac{d y}{d x}=\frac{18 y+3 x^2}{3 y^2+18 x}\)
2. \(\frac{d y}{d x}=\frac{15 y+3 x^2}{3 y^2-18 x}\)
3. \(\frac{d y}{d x}=\frac{18 y-3 x^2}{3 y^2-18 x}\)
4. \(\frac{d y}{d x}=\frac{18 y-3 x^2}{3 y^2+12 x}\)

Answer: 3. \(\frac{d y}{d x}=\frac{18 y-3 x^2}{3 y^2-18 x}\)

Question 174. Find two positive numbers x & y such that x + y = 60 and xy is maximum –

1. 15, 45
2. 30, 30
3. 20, 40
4. 10, 50

Answer: 2. 30, 30

Find integrals of given functions.

Question 175. \(\int x^{-3}(x+1) d x\)

1. \(-\frac{1}{x}-\frac{1}{2 x^2}+C\)
2. \(\frac{1}{x}+\frac{1}{2 x^2}+C\)
3. \(3-\frac{1}{2 x^2}+C\)
4. \(-\frac{1}{x}+\frac{1}{2 x^2}+C\)

Answer: 1. \(-\frac{1}{x}-\frac{1}{2 x^2}+C\)

Question 176. \(\int\left(1-\cot ^2 x\right) d x\)

1. 2x + cot x + C
2. x + cot x + C
3. 2x – cot x + C
4. 2x + tan x + C

Answer: 1. 2x + cot x + C

Question 177. ∫cos (tanθ + secθ)dθ

1. – cos θ + θ + C
2. – cos θ – θ + C
3. – cosec θ + θ + C
4. – 2cos θ + θ + C

Answer: 1. – cos θ + θ + C

Question 178. \(\int \sqrt{3-2 s} d s\)

1. \(-\frac{1}{3}(3-2 s)^{2 / 3}+C\)
2. \(-\frac{1}{3}(3-2 s)^{3 / 2}+C\)
3. \(-\frac{1}{3}(3+2 s)^{3 / 2}+C\)
4. None of these

Answer: 2. \(-\frac{1}{3}(3-2 s)^{3 / 2}+C\)

Question 179. \(\int \frac{d x}{\sqrt{5 x+8}}\)

1. \(\left[\frac{2}{5} \sqrt{5 x+8}\right]+C\)
2. \(\left[\frac{2}{5} \sqrt{3 x-8}\right]+C\)
3. \(\left[\frac{2}{5} \sqrt{5 x-4}\right]-C\)
4. \(\left[\frac{2}{5} \sqrt{5 x-4}\right]\)

Answer: 1. \(\left[\frac{2}{5} \sqrt{5 x+8}\right]+C\)

Question 180. \(\int_0^{\sqrt{\pi}} x \sin x^2 d x\)

1. 1
2. 2
3. 3
4. 1

Answer: 1. 1

Use a definite integral to find the area of the region between the given curve and the x-axis on the interval [0,b],

Question 181. y = 3x²

1. b³
2. b²
3. b
4. b⁵

Answer: 1. b³

NEET Physics Mathematical Tools MCQs for Chapter 10

Question 182. Two vectors \(\overrightarrow{\mathrm{a}} \text { and } \overrightarrow{\mathrm{b}}\) inclined at an angle θ w.r.t. each other have a resultant \(\overrightarrow{\mathrm{c}}\) which makes an angle β with \(\overrightarrow{\mathrm{a}}\). If the directions of \(\overrightarrow{\mathrm{a}} \text { and } \overrightarrow{\mathrm{b}}\) are interchanged, then the resultant will have the same

1. Magnitude
2. Direction
3. Magnitude as well as direction
4. Neither magnitude nor direction.

Answer: 1. Magnitude

Question 183. Two vectors \(\overrightarrow{\mathrm{A}} \text { and } \overrightarrow{\mathrm{B}}\) lie in a plane. Another vector \(\overrightarrow{\mathrm{C}}\) lies outside this plane. The resultant \(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{C}}\) of these three vectors

1. Can be zero
2. Cannot be zero
3. Lies in the plane of \(\vec{A} and \vec{B}\)
4. Lies in the plane of \(\overrightarrow{\mathrm{A}} and \overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}\)

Answer: 2. Cannot be zero

Question 184. The rectangular components of a vector are (2, 2). The corresponding rectangular components of another vector are (1, \(\sqrt{3}\)). Find the angle between the two vectors

1. 10º
2. 15º
3. 20º
4. 25º

Answer: 2. 15º

Question 185. Given : \(\vec{a}+\vec{b}+\vec{c}\)= 0. Out of the three vectors \(\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}} \text { and } \overrightarrow{\mathrm{c}}\) two are equal in magnitude. The magnitude of the vector \(\sqrt{2}\) times that of either of the two having equal magnitude. The angles between vectors are:

1. 90º, 135º,. 135º
2. 30º, 60º, 90º
3. 45º, 45º, 90º
4. 45º, 60º, 90º

Answer: 1. 90º, 135º,. 135º

Question 186. Let \(\vec{a} \text { and } \vec{b}\) be two non-null vectors such that \(|\vec{a}+\vec{b}|=|\vec{a}-2 \vec{b}|\). Then the value of \(\frac{|\vec{a}|}{|\vec{b}|}\) may be :

1. \(\frac{1}{4}\)
2. \(\frac{1}{8}\)
3. 1
4. 2

Answer: (3,4)

Question 187. A truck traveling due north at 20 ms^-1 turns west and travels with the same speed. What is the change in velocity?

1. \(20 \sqrt{2} \mathrm{~ms}^{-1}\) south-west
2. \(40 \mathrm{~ms}^{-1}\) south-west
3. \(20 \sqrt{2} \mathrm{~ms}^{-1}\) north-west
4. \(40 \mathrm{~ms}^{-1}\) north-west

Answer: 1. \(20 \sqrt{2} \mathrm{~ms}^{-1}\) south-west

Question 188. Determine that vector which when added to the resultant of \(\overrightarrow{\mathrm{P}}=2 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}-10 \hat{\mathrm{k}}\) and \(\vec{Q}=\hat{i}+2 \hat{j}+3 \hat{k}\) a unit vector along X-axis.

1. \(-2 \hat{i}-9 \hat{j}+7 \hat{k}\)
2. \(+2 \hat{i}+9 \hat{j}-7 \hat{k}\)
3. \(-2 \hat{i}+7 \hat{j}+9 \hat{k}\)
4. \(+2 \hat{i}-5 \hat{j}+3 \hat{k}\)

Answer: 1. \(-2 \hat{i}-9 \hat{j}+7 \hat{k}\)

Question 189. Two vectors acting in opposite directions have a resultant of 10 units. If they act at right angles to each other, then the result is 50 units. Calculate the magnitude of two vectors.

1. P = 40 ; Q = 30
2. P = 30 ; Q = 40
3. P = 80 ; Q = 50
4. P = 30 ; Q = 40

Answer: 1. P = 40 ; Q = 30

Question 160. Find the resultant of the three vectors \(\overrightarrow{\mathrm{OA}}, \overrightarrow{\mathrm{OB}} \text { and } \overrightarrow{\mathrm{OC}}\) each of magnitude r as shown in the figure?

1. \(r(1+\sqrt{2})\)
2. \(r(1-\sqrt{2})\)
3. \((1+\sqrt{2})\)
4. \(r(1+\sqrt{2})^2\)

Answer: 1. \(r(1+\sqrt{2})\)

Question 161. A car is moving on a straight road due north with a uniform speed of 50 km h^-1 when it turns left through 90º. If the speed remains unchanged after turning, the change in the velocity of the car in the turning process is

1. Zero
2. \(50 \sqrt{2} \mathrm{~km} \mathrm{~h}^{-1}\) S-W direction
3. \(50 \sqrt{2} \mathrm{~km} \mathrm{~h}^{-1}\) N-W direction
4. 50 km h^-1 due west.

Answer: 2. \(50 \sqrt{2} \mathrm{~km} \mathrm{~h}^{-1}\) S-W direction

Question 162. Six forces, 9.81 N each, acting at a point are coplanar. If the angles between neighboring forces are equal, then the resultant is

1. 0 N
2. 9.81 N
3. 2 × 9.81 N
4. 3 × 9.81 N.

Answer: 1. 0 N

Question 163. At what angle must the two forces (x + y) and (x – y) act so that the resultant may be \(\sqrt{\left(x^2+y^2\right)}\)?

1. \(\cos ^{-1}\left[\frac{-\left(x^2+y^2\right)}{2\left(x^2-y^2\right)}\right]\)
2. \(\cos ^{-1}\left[\frac{-2\left(x^2-y^2\right)}{x^2+y^2}\right]\)
3. \(\cos ^{-1}\left[\frac{-\left(x^2+y^2\right)}{x^2-y^2}\right]\)
4. \(\cos ^{-1}\left[\frac{\left(x^2-y^2\right)}{x^2+y^2}\right]\)

Answer: 1. \(\cos ^{-1}\left[\frac{-\left(x^2+y^2\right)}{2\left(x^2-y^2\right)}\right]\)

Question 164. The magnitude of the scalar product of two vectors is 8 and that of the vector product is \(8 \sqrt{3}\). The angle between them is :

1. 30º
2. 60º
3. 120º
4. 150º

Answer: (2,3)

Question 165. A vector \(\vec{A}\) points vertically downward and \(\vec{B}\) points towards the east, then the vector product \(\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}\) is

1. Along west
2. Along east
3. Zero
4. Along south

Answer: 4. Along south

Question 166. Which of the arrangement of axes in a figure? can be labeled “right-handed coordinate system”? As usual, each axis label indicates the positive side of the axis.

1. (1), (2)
2. (3), (4)
3. (6)
4. (5)

Answer: (1,2,3)

NEET Physics Mathematical Tools MCQs for Chapter 10

Question 167. The unit vector perpendicular to each of the vectors \(3 \hat{i}+\hat{j}+2 \hat{k}\) and \(2 \hat{i}-2 \hat{j}+\hat{k}\) is given by

1. \(\frac{1}{\sqrt{3}}(\hat{i}-\hat{j}-\hat{k})\)
2. \(\frac{1}{\sqrt{3}}(\hat{\mathbf{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})\)
3. \(\frac{5 \hat{i}+\hat{j}+4 \hat{k}}{\sqrt{46}}\)
4. \(\pm \frac{5 \hat{i}+\hat{j}-4 \hat{k}}{\sqrt{42}}\)

Answer: 4. \(\pm \frac{5 \hat{i}+\hat{j}-4 \hat{k}}{\sqrt{42}}\)

Question 168. Three vectors \(\overrightarrow{\mathrm{A}}, \overrightarrow{\mathrm{B}} \text { and } \overrightarrow{\mathrm{C}}\) are such that \(\overrightarrow{\mathrm{A}}=\overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{C}}\) and their magnitudes are in ratio 5: 4 : 3 respectively. Find the angle between vector \(\overrightarrow{\mathrm{A}} \text { and } \overrightarrow{\mathrm{C}}\)

1. 35º
2. 53º
3. 60º
4. 75º

Answer: 2. 53º

Question 169. A car travels 6 km towards the north at an angle of 45° to the east and then travels a distance of 4 km towards the north at an angle of 135° to the east. How far is the point from the starting point? What angle does the straight line joining its initial and final position make with the east?

1. \(\sqrt{50} \mathrm{~km} and \tan ^{-1}(5)\)
2. \(10 \mathrm{~km} and \tan ^{-1}(\sqrt{5})\)
3. \(\sqrt{52} \mathrm{~km} and \tan ^{-1}(5)\)
4. \(\sqrt{52} \mathrm{~km} and \tan ^{-1}(\sqrt{5})\)

Answer: 3. \(\sqrt{52} \mathrm{~km} and \tan ^{-1}(5)\)

Question 170. The vectors \(\vec{A} \text { and } \vec{B}\) are such that: 

⇒\(|\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}|=|\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}|\)

The angle between the two vectors is :

1. 90°
2. 60°
3. 75°
4. 45°

Answer: 1. 90°

Question 171. Six vectors, \(\vec{a}\) through \(\vec{f}\) have the magnitudes and directions indicated in figure e. Which of the following statements is true?

1. \(\vec{b}+\vec{c}=\vec{f}\)
2. \(\vec{d}+\vec{c}=\vec{f}\)
3. \(\vec{d}+\vec{e}=\vec{f}\)
4. \(\vec{b}+\vec{e}=\vec{f}\)

Answer: 3. \(\vec{d}+\vec{e}=\vec{f}\)

Question 172. If dimensions of critical velocity υc of a liquid flowing through a tube are expressed as \(\left[\eta^{\mathrm{x}} \rho^{\mathrm{y}} \mathrm{r}^{\mathrm{x}}\right]\), where η, ρ, and r are the coefficient of viscosity of the liquid, density of a liquid, and radius of the tube respectively, then the values of x, y, and z are given by :

1. –1, –1, 1
2. –1, –1, –1
3. 1,1,1
4. 1, –1, –1

Answer: 4. 1, –1, –1

Question 173. If vectors \(\vec{A}=\cos \omega t \hat{i}+\sin \omega t \hat{j} \text { and } \vec{B}=\cos \frac{\omega t}{2} \hat{i}+\sin \frac{\omega t}{2} \hat{j}\) are functions of time, then the value of t at which they are orthogonal to each other is :

1. \(t=\frac{\pi}{2 \omega}\)
2. \(t=\frac{\pi}{\omega}\)
3. \(t=0\)
4. \(t=\frac{\pi}{4 \omega}\)

Answer: 2. \(t=\frac{\pi}{\omega}\)

Question 174. If the magnitude of the sum of two vectors is equal to the magnitude of the difference of the two vectors, the angle between these vectors is :

1. 180°
2. 0°
3. 90°
4. 45°

Answer: 3. 90°

Question 175. A particle moves so that its position vector is given by \(\overrightarrow{\mathrm{r}}=\cos \omega t \hat{x}+\sin \omega t \hat{y}\). Where ω is a constant. Which of the following is true?

1. Velocity and acceleration both are perpendicular to \(\overrightarrow{\mathrm{r}}\)
2. Velocity and acceleration both are parallel to \(\overrightarrow{\mathrm{r}}\)
3. Velocity is perpendicular to \(\overrightarrow{\mathrm{r}}\) and acceleration is directed towards the origin
4. Velocity is perpendicular to \(\overrightarrow{\mathrm{r}}\) and acceleration is directed away from the origin

Answer: 3. Velocity is perpendicular to \(\overrightarrow{\mathrm{r}}\) and acceleration is directed towards the origin

Question 176. The x and y coordinates of the particles at any time are x = 5t -2t² and y = 10t respectively, where x and y are in meters and t in seconds. The acceleration of the particle at t = 2s is

1. 5m/s²
2. -4 m/s²
3. – 8 m/s²
4. 0

Answer: 2. -4 m/s²

Question 177. The moment of the force, \(\overrightarrow{\mathrm{F}}=4 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}\) at (2, 0, –3), about the point (2, –2, –2), is given by

1. \(-8 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}-7 \hat{\mathrm{k}}\)
2. \(-4 \hat{\mathrm{i}}-\hat{\mathrm{j}}-8 \hat{\mathrm{k}}\)
3. \(-7 \hat{i}-4 \hat{j}-8 \hat{k}\)
4. \(-7 \hat{i}-8 \hat{j}-4 \hat{k}\)

Answer: 3. \(-7 \hat{i}-4 \hat{j}-8 \hat{k}\)

Question 178. In the cube of side ‘as shown in the figure, the vector from the central point of the face ABOD to the central point of the face BEFO will be :

1. \(\frac{1}{2} a(\hat{j}-\hat{k})\)
2. \(\frac{1}{2} a(\hat{j}-\hat{i})\)
3. \(\frac{1}{2} \mathrm{a}(\hat{\mathrm{k}}-\hat{\mathrm{i}})\)
4. \(\frac{1}{2} a(\hat{i}-\hat{k})\)

Answer: 2. \(\frac{1}{2} a(\hat{j}-\hat{i})\)

NEET Physics Class 11 Chapter 10 Mathematical Tools MCQs Practice

Question 179. Two forces P and Q, of magnitude 2F and 3F, respectively are at an angle θ with each other. If the force Q is doubled, then their resultant also gets doubled. Then, the angle θ is :

1. 30°
2. 60°
3. 90°
4. 120°

Answer: 4. 120°

Question 180. Two vectors \(\vec{A} \text { and } \vec{B}\) have equal magnitudes. The magnitude of \((\vec{A}+\vec{B})\) is ‘n’ times the magnitude of \((\vec{A}-\vec{B})\). The angle between \(\vec{A} \text { and } \vec{B}\) is

1. \(\sin ^{-1}\left[\frac{n-1}{n+1}\right]\)
2. \(\cos ^{-1}\left[\frac{n^2-1}{n^2+1}\right]\)
3. \(\sin ^{-1}\left[\frac{n^2-1}{n^2+1}\right]\)
4. \(\cos ^{-1}\left[\frac{n-1}{n+1}\right]\)

Answer: 2. \(\cos ^{-1}\left[\frac{n^2-1}{n^2+1}\right]\)