## Volume And Surface Area Of Solids

The objects which occupy space have three dimensions) arc called solids.

## Volume And Surface Area Of Solids Cuboid And Cube

A cuboid is a rectangular solid having six faces all of which are rectangles. The cuboid has six faces. The opposite faces are all congruent. Two adjacent faces meet in a line.

This line is called the edge of the cuboid. There are 12 edges of a cuboid. The point where three adjacent edges meet is called the vertex of the cuboid. There are 8 vertices of a cuboid.

The three edges that meet on the vertex of a cuboid are called its length, breadth and height. A cube is a cuboid in which all the edges are of equal length and every one of the six faces is a square.

**Total Surface (Surface Area) of a Cuboid and a Cube**

The sum of the area of the faces of the cuboid is called its surface. The areas of opposite faces are equal, so the total surface of the A cuboid is

2lb + 2bh + 2hl = 2 (bh+ hl +lb)

For a cube, l = b = h = a (say)

Surface of a cube = 2(a.a + a.a + a.a) = 6a^{2}

**The Length of the Diagonal of a Cuboid and a Cube**

Diagonal of a floor \(D^{\prime} B^{\prime}=\sqrt{l^2+b^2}\). So, if we want to find the length of the longest rod DB’, then we make a new right-angled A treating, DB’ as hypotenuse, base as D’B’ and height as DD’ which is h. So, the length of the diagonal of a cuboid DB’.

= \(\sqrt{\left(D^{\prime} B^{\prime}\right)^2+\left(D D^{\prime}\right)^2}=\sqrt{l^2+b^2+h^2}\)

For a cube, l = b = h = a (say)

∴ The length of the diagonal of a cube = \(\sqrt{a^2+a^2+a^2}=a \sqrt{3}\)

**Lateral Surface Area of a Cuboid and a Cube**

- Lateral surface area of a cuboid = 2(l + b)h. [CSA = TSA- (area of top and area of bottom)]
- Lateral surface area of a cube = 4a
^{2}, where a = edge of the cube. It is also called the curved surface area or area of 4 walls.

**The volume of a Cuboid and a Cube**

The volume of any solid figure is the amount of space enclosed within its bounding faces.

The volume of a cuboid of length l units, breadth b units and height h units are lbh cubic units.

= (length x breadth x height) cubic units

= (area of the base x height) cubic units.

For a cube,

Volume of cube = a^{3} cubic units,

where, the length of the edge of the cube is a unit.

Note: The measurements of volume are as follows:

1 cm^{3} = 10^{3} cubic mm

1 m^{3} = 1000 litre = 1 kl

## Volume And Surface Area Of Solids Cylinder

If a rectangle is revolved about one of its sides as its axis, the solid so formed is called a right circular cylinder.

The side AD about which the rectangle ABCD revolves is called the height of the cylinder.

The line CD is called the generating line because when it revolves round AB, it generates the cylinder.

If you cut the hollow cylinder along AD and spread the piece, it becomes a rectangle whose one side is AD and the other side is AB which is equal to the circumference of a circle whose radius is the radius of the cylinder.

Hence, Side AB = 2πr

If h is the height of the cylinder, the area of ABCD

= AB x AD = 2πrh

Thus, the area of the curved surface of the cylinder

= 2πrh = perimeter of base x height

**Whole Surface of a Cylinder**

The whole surface area of a cylinder

= Curved surface + Area of the base + Area of the top

= 2πrh + πr2 + nr^{2} = 2πrh + 2πr^{2} = 2πr(h + r)

**Volume of a Cylinder**

The volume of a cylinder = Area of the base x Height

= nr2 x h = nr2h

**Note:** If the top and bottom are removed from the total surface area, then only the curved surface area remains.

## Volume And Surface Area Of Solids Hollow Cylinder

Solids like iron pipes, rubber tubes, etc., are in the shape of hollow cylinders.

For a hollow cylinder of height h and with external and internal radii R and r respectively, we have

Volume of the material = Exterior volume- Interior volume

= πR^{2}h- πr^{2}h = πh (R^{2}– r^{2})

The curved surface of the hollow cylinder

= External surface + Internal surface = 2πRh + 2πrh = 2πh(R + r)

The total surface area of the hollow cylinder

= Curved surface + 2(Area of base rings)

= (2πRh + 2πrh) + 2(πR^{2}– πr^{2}) = 2πh(R + r) + 2π(R^{2 }– r^{2})

= 2π(R + r) (h+R-r)

## Volume And Surface Area Of Solids Right Circular Cone

A right circular cone is a solid generated by the revolution of a right-angled triangle about sides containing the right angle as the axis.

Let CAB be a right-angled triangle, right-angled at A. The hypotenuse CB revolves around the side CA as the fixed axis. The hypotenuse CB will generate the curved surface of a cone.

The radius of the circular base is called the radius of the cone. It is usually denoted by ‘r’.

The point C is called the vertex of the cone.

The length CA of the axis is called the height of the cone. It is usually ‘denoted by ‘h’.

The hypotenuse CB is called the generating line of the cone and its length is called the slant height. It is usually denoted by l.

**The volume of Right Circular Cone**

Volume of a cone = \(\frac{1}{3}\) (area of the base) x height’

i.e., \(V=\frac{1}{3} \pi r^2 h\)

where r = radius of the base and h = height.

Slant height of right circular cone = \(l=\sqrt{h^2+r^2}\)

**The curved surface of the Right Circular Cone**

Curved surface of a cone = \(\pi r l=\pi r \sqrt{h^2+r^2}\)

where r = radius of the base, l = slant height of the cone.

**Total Surface of Right Circular Cone**

Total surface of the cone = Area of the base + Area of the curved surface

i.e., total surface of the cone = πr^{2} + πrl = πr(r + l).

## Volume And Surface Area Of Solids Sphere

When a circle is revolved about its diameter, the solid thus formed is a sphere. Let AB be the diameter of a circle ADB and 0 be its centre.

If the circle ADB revolves around AB, point D takes different positions and a closed solid is formed.

A sphere may also be defined as a solid bounded by a closed surface, all points on which are at a constant distance from a fixed point.

**Curved Surface Area and Volume of a Sphere**

The curved surface area of a sphere = 4πr^{2}, where r is the radius of the sphere.

The volume of sphere = \(\frac{4}{3} \pi r^3\)

## Volume And Surface Area Of Solids Hemisphere

A plane through the centre of a sphere divides the sphere into two equal parts. Each part is called a hemisphere.

If r is the radius of the hemisphere, then

Volume of hemisphere = \(\frac{2}{3} \pi r^3\)

Curved surface area = 2πr^{2}

Total surface area = 2πr^{2} + πr^{2} = 3πr^{2}.

## Volume And Surface Area Of Solids Spherical Shell

A spherical shell having external radius R and internal radius r, then we have

Volume of material = \(\frac{4}{3} \pi R^3-\frac{4}{3} \pi r^3=\frac{4}{3} \pi\left(R^3-r^3\right)\)

**Volume And Surface Area Of Solids Solved Examples**

**Question 1. A tent of cloth is cylindrical upto 1 m in height and conical above it of the same radius of base. If the diameter of the tent is 6 m and the slant height of the conical part is 5 m, find the cloth required to make this tent.**

**Solution:**

Diameter of base 2r = 6 m

⇒ \(r=\frac{6}{2}=3 \mathrm{~m}\)

Height of cylindrical path h = 1 m

The slant height of conical part l = 5 m

Cloth required in tent = 2πrh + πrl

= πr (2h + l)

= \(\frac{22}{7} \times 3(2 \times 1+5)=66 \mathrm{~m}^2\)

**Question 2. The base and top of a right circular cylindrical drum are hemispherical. The diameter of the cylindrical part is 14 cm and the total height is 30 cm. Find the total surface area of the drum.**

**Solution:**

Here 2r = 14

⇒ r = 7 cm

Height of cylindrical part h = 30 – 2 x 7 = 16 cm

The total surface area of the drum = 2 x the Curved surface of the hemisphere + the Curved surface of the cylinder

= 2 x 2πr^{2} + 2πrh

= 2πr(2r + h)

= \(2 \times \frac{22}{7} \times 7(2 \times 7+16)\)

= 44 x 30 = 1320 cm^{2}

**Question 3. Three cubes, each with an 8 cm edge, are joined end to end. Find the total surface area of the resulting cuboid.**

**Solution:**

As is clear from the adjoining figure;

the length of the resulting cuboid = 3 x 8 cm = 24 cm

Its width = 8 cm and its height = 8 cm

i.e., l = 24 cm, b = 8 cm and h = 8 cm

∴ The total surface area of the resulting cuboid

= 2(1 x b + b x h + h x l)

= 2 (24 x 8 + 8 x 8 + 8 x 24) cm^{2} = 896 cm^{2}

**Question 4. A rectangular sheet of tin 58 cm x 44 cm is to be made into an open box by cutting off equal squares from the corners and folding up the flaps. What should be the volume of box if the surface area of box is 2452 cm ^{2}?**

**Solution:**

Let a square of x cm from each corner be removed from a rectangular sheet.

So, length of box = (58 – 2x) cm

breadth of box = (44 – 2x) cm

and height of box = x cm

∴ Surface area of open box = 2 (lb + bh + hl)- lb

⇒ 2[(58 – 2x) (44 – 2x) + x (44 – 2x) + x (58 – 2x)]- (58 – 2x)(44 – lx) = 2452

⇒ (58- 2x) (44- 2x) + 2x(44- 2x) + 2x (58 – 2x) = 2452

⇒ (29 – x) (22 – x) + x (22 – x) + x (29 – x) = 613

⇒ 638 – 51x + x2 + 22x – x2 + 29x – x2 = 613

⇒ x^{2} = 25

⇒ x = 5 cm (x = – 5 is not admissible)

∴ Volume of box = x(58 – 2x) (44- 2x) = 5(58- 10)(44 – 10)

= 5 x 48 x 34 cm^{3} = 8160 cm^{3}

**Question 5. The radii of two right circular cylinders are in the ratio 2 : 3 and their heights are in the ratio 5:4. Calculate the ratio of their curved surface areas and also the ratio of their volumes.**

**Solution:**

Let the radii of the two cylinders be 2r and 3r respectively and their heights be 5h and 4h.

∴ \(\frac{\text { Curved surface area of 1st cylinder }\left(S_1\right)}{\text { Curved surface area of 2nd cylinder }\left(S_2\right)}=\frac{2 \pi \times 2 r \times 5 h}{2 \pi \times 3 r \times 4 h}\)

i.e., \(\frac{S_1}{S_2}=\frac{5}{6}\) or S1: S2 = 5:6

⇒ \(\frac{\text { Volume of 1st cylinder }\left(V_1\right)}{\text { Volume of 2nd cylinder }\left(V_2\right)}=\frac{\pi \times(2 r)^2 \times 5 h}{\pi \times(3 r)^2 \times 4 h}\)

i.e., \(\frac{V_1}{V_2}=\frac{5}{9}\) or V_{1}:V_{2}= 5:9

**Question 6. The volume and surface area of a solid hemisphere are numerically equal. What is the diameter of the hemisphere?**

**Solution:**

We have,

Volume of hemisphere = Surface area of hemisphere

⇒ \(\frac{2}{3} \pi r^3=3 \pi r^2\) = 2r = 9

Hence, the diameter of the hemisphere = 9 units

**Question 7. The curved surface area of a cone of height 8 m is 1 88.4 m ^{2}. Find the volume of a cone.**

**Solution:**

πrl = 188.4

⇒ \(r l=\frac{188.4}{3.14}=60\)

⇒ r^{2}l^{2} = 3600

⇒ r^{2}(h^{2} + r^{2}) = 3600

⇒ r^{2}(64 + r^{2}) = 3600

⇒ r^{4} + 64r^{2} – 3600 = 0

⇒ (r^{2}+ 100) (r^{2}– 36) = 0

∴ r^{2} = -100 or r^{2} = 36

⇒ r = 6 (∵ r^{2} = -100 is not possible)

∴ Volume of a cone = \(\frac{1}{3} \pi r^2 h=\frac{1}{3} \times 3.14 \times 36 \times 8=301.44 \mathrm{~m}^3\)

**Question 8. A conical tent is required to accommodate 157 persons, each person must have 2 m ^{2} of space on the ground and 15 m^{3} of air to breathe. Find the height of the tent; also calculate the slant height.**

**Solution:**

1 person needs 2 m^{2} of space.

∴ 157 persons needs 2 x 157 m space on the ground

∴ r^{2} = 2 x 157

∴ \(r^2=\frac{2 \times 157}{3.14}=100\) = r = 100

Also, 1 person needs 15 m of air.

∴ 157 persons need 15 x 157 m3 of air.

∴ \(\frac{1}{3} \pi r^2 h=15 \times 157\)

⇒ \(h=\frac{15 \times 157 \times 3}{3.14 \times 100}=22.5 \mathrm{~m}\)

∴ l^{2} =h^{2} + r^{2} = (22.5)^{2} + (10)^{2} = 606.25

∴ \(l=\sqrt{606.25}=24.62 \mathrm{~m}\)

**Question 9. The radius and height of a solid right circular cone are in the ratio of 5:12. If its volume is 314 cm find its total surface area. [Take π = 3.14]**

**Solution:**

Let the radius of the cone = 5x

∴ Height of cone = 12r

∴ l^{2} = (5r)^{2} + (12x)^{2} = 169 x^{2}

∴ \(l=\sqrt{169 x^2}=13 x\)

It is given that volume = 314 cm2

∴ \(\frac{1}{3} \pi(5 x)^2(12 x)=314\)

⇒ \(\frac{1}{3} \times 3.14 \times 25 \times 12 \times x^3=314\)

⇒ \(x^3=\frac{314 \times 3}{3.14 \times 25 \times 12}=1\)

∴ x = 1

∴ Radius r = 5 x 1 = 5 cm

Height h = 1 2 x 1 = 1 2 cm

and slant height l= 13 x I = 13 cm

Now, total surface area of cone = πr(l + r) = 3.14 x 5 (13 + 5) = 3.14 x 5 x 18

= 282.60 cm^{2}

Hence, the total surface area of a cone is 282.60 cm.

**Question 10. If h, C, and V respectively are the height, the curved surface area and the volume of a cone. Prove that 3πVh ^{3} – C^{2}h^{2} + 9V^{2} = 0.**

**Solution:**

⇒ \(C=\pi r l, V=\frac{1}{3} \pi r^2 h, l^2=h^2+r^2\)

L.H.S. = 3πVh^{3} – C^{2}h^{2} + 9V^{2}

= \(3 \pi\left(\frac{1}{3} \pi r^2 h\right) h^3-(\pi r l)^2 h^2+9\left(\frac{1}{3} \pi r^2 h\right)^2\)

= \(\pi^2 r^2 h^4-\pi^2 r^2 h^2\left(h^2+r^2\right)+9 \times \frac{1}{9} \pi^2 r^4 h^2\)

= π^{2}r^{2}h^{4} – π^{2}r^{2}h^{4} – π^{2}r^{4}h^{2} + π^{2}r^{4}h^{2} = 0 = R.H.S. Hence Proved.

**Question 11. A cone of equal height and equal base is cut off from a cylinder of height 24 cm and base radius 7 cm. Find the total surface and volume of the remaining solid.**

**Solution:**

Here r = 7 cm, h = 24 cm

The volume of remaining solid = Volume of the cylinder – Volume of a cone

= \(\pi r^2 h-\frac{1}{3} \pi r^2 h=\frac{2}{3} \pi r^2 h\)

= \(\frac{2}{3} \times \frac{22}{7} \times 7 \times 7 \times 24=2464 \mathrm{~cm}^3\)

Now, l^{2} = h^{2} + r^{2} = (24)^{2} + (7)^{2}

= 576 + 49 = 625

⇒ l = 25 cm

∴ Total surface area of remaining solid

= Curved surface of cylinder + Curved surface of cone + Area of top

= 2πrh +πrI + πr^{2}

= πr(2h + l + r)

= \(\frac{22}{7} \times 7(2 \times 24+25+7)=22 \times 80=1760 \mathrm{~cm}^2\)

**Question 12: From a wooden cubical block of edge 7 cm, the largest possible right conical piece is cut out whose base is on one of the faces of the cube. Calculate:**

- The volume of the wood left in the block, and
- the total surface area of the block left. (\(\text { Take } \pi=\frac{22}{7}\))

**Solution:**

Initial volume of the block = a^{3} = 7^{3} cm^{3}

The base of the largest cone will touch the sides of the base of the cube, and the height will be equal to the length of the edge of the cube.

∴ For the cone, r = 3.5 cm and h = 7 cm.

1. The volume of the cone = \(\frac{1}{3} \pi r^2 h=\frac{\pi}{3} \cdot\left(\frac{7}{2}\right)^2 \cdot 7 \mathrm{~cm}^3\)

∴ volume of the wood left = \(\left\{7^3-\frac{1}{3} \cdot \frac{22}{7} \cdot\left(\frac{7}{2}\right)^2 \cdot 7\right\} \mathrm{cm}^3\)

= \(\left(7^3-\frac{11 \times 7^2}{3 \times 2}\right) \mathrm{cm}^3=7^2\left(7-\frac{11}{6}\right) \mathrm{cm}^3\)

= \(\frac{49 \times 31}{6} \mathrm{~cm}^3=\frac{1519}{6} \mathrm{~cm}^3=253 \frac{1}{6} \mathrm{~cm}^3\)

2. The total surface area of the wood left

= Total surface area of the cube- Area of the base of the cone + Curved surface area of the cone.

= 6a^{2} – πr^{2} + πrl

= \(\left\{6 \times 7^2-\frac{22}{7} \cdot\left(\frac{7}{2}\right)^2+\frac{22}{7} \cdot \frac{7}{2} \cdot \sqrt{7^2+\left(\frac{7}{2}\right)^2}\right\} \mathrm{cm}^2\)

= \(\left(6 \times 49-\frac{11 \times 7}{2}+11 \cdot 7 \cdot \frac{\sqrt{5}}{2}\right) \mathrm{cm}^2=\left\{294+\frac{77}{2}(\sqrt{5}-1)\right\} \mathrm{cm}^2\)

= \(\left(294+\frac{77}{2} \times 1.24\right) \mathrm{cm}^2=341.74 \mathrm{~cm}^2\)

**Question 13. A sphere is inscribed in a cylinder such that the sphere touches the cylinder. Show that the curved surface is equal to the curved surface of the cylinder.**

**Solution:**

A cylinder circumscribed a sphere is shown.

Here, the radius of the sphere = radius of the cylinder = r

Height of cylinder h = 2r

Now, the curved surface of the sphere = 4πr^{2}

and the curved surface of sphere of cylinder = 2πrh = 2πr(2r) = 4πr^{2}

Therefore, the curved surface of the sphere and the curved surface of the cylinder are equal.

Hence proved.

**Question 14. The largest possible cube Is made from a wooden sphere of radius 6√3 cm. Find the surface area of the cube.**

**Solution:**

Here, a diagonal of the cube will be the diameter of the sphere

∴ length of a diagonal of the cube = 2 x 6√3 cm = 12√3 cm

If an edge of the cube is a then a diagonal of the cube = √3 a

= 12√3 cm

∴ a = 12 cm.

∴ The surface area of the cube = 6a^{2} = 6.(12)^{2} cm^{2} = 864 cm^{2}

**Question 15. A hemisphere is inscribed in a cylinder and a cone is inscribed in the hemisphere. The cone vertex lies in the centre of the upper circular part of the cylinder. Show that: \(\frac{1}{3}\) x Volume of cylinder = \(\frac{1}{2}\)x Volume of hemisphere = Volume of cone**

**Solution:**

Let radius of cone = radius of cylinder = radius of hemisphere = r

∴ Height of cone = Height of cylinder = r

Now, the volume of cone = \(\frac{1}{3} \pi r^2 \cdot(r)=\frac{1}{3} \pi r^3\) → (1)

Volume of hemisphere = \(\frac{2}{3} \pi r^3\)

⇒ \(\frac{1}{2}\)x Volume of hemisphere = \(\frac{1}{3} \pi r^3\) → (2)

and Volume of cylinder = πr^{2}(r) = r^{3}

⇒ \(\frac{1}{3}\) x Volume of cylinder = \(\frac{1}{3} \pi r^3\) → (3)

From equations (1), (2) and (3)

⇒ \(\frac{1}{3}\) x Volume of cylinder = \(\frac{1}{2}\)x Volume of hemisphere

= Volume of cone

Hence proved.

**Question 16. Water in a canal, 5.4 m wide and 1.8 m deep is flowing at a speed of 25 km/hr. How much area can it irrigate in 40 minutes, if 10 cm of standing water is required for irrigation?**

**Solution:**

Width of canal = 5.4 m

Height of canal = 1.8 m

In one hour, water is moving upto 25 km = 25000 m

∴ We treat it as the length of the canal.

∴ Volume of water in canal in 1 hour = (25000 x 5.4 x 1.8)m^{3}

⇒ Volume of water in the canal in 40 minutes = 243000 m^{3}

∴ Volume of water in canal in 60 minutes = \(\frac{243000}{60} \times 40 \mathrm{~m}^3\)

= 162000 m^{3}

This water can irrigate a field upto the height of 10 cm = 0. 1 m

∴ Volume = Area = Area x height

⇒ 162000 = Area x 0.1

⇒ Area of field = \(\frac{162000}{0.1} \mathrm{~m}^2=1620000 \mathrm{~m}^2\) (**∵Note:** 1 Hectare = 1000 m)

= \(\frac{1620000}{10000} \text { hectare }\)

= 162 hectare.

So, 162 hectare area can be irrigated.

## Conversion Of Solid Form One Shape To Another And Mixed Problems

When a solid is converted into another without any loss of material then its volume remains the same.

If a larger solid is converted into smaller solids then the number of smaller solids

= \(\frac{\text { Volume of larger solid }}{\text { Volume of } 1 \text { smaller solid }}\)

**Conversion Of Solid Form One Shape To Another And Mixed Problems Solved Examples**

**Question 1. A solid metallic cuboid of dimensions 9 m x 8 m x 2 m is melted and recast into solid cubes of edge 2m. Find the number of cubes so formed.**

**Solution:**

Let x cubes each of edge 2 m are formed to melt a cuboid of dimensions 9 x 8 x 2.

So, \(\text { number of cubes }=\frac{\text { Volume of cuboid }}{\text { Volume of } 1 \text { cube }}\)

∴ \(x=\frac{9 \times 8 \times 2}{2}=72\)

So, 72 cubes will be formed

**Question 2. Three cubes of metal whose edges are in the ratio 3:4 are melted down into a single cube whose diagonal is 12√3 cm, Find the edges of the three cubes.**

**Solution:**

The ratio in the edges = 3:4:5

Let edges be 3x, 4x and 5x respectively.

∴ Volumes of three cubes will be 27x^{3}, 64x^{3} and 125x^{3} in cm^{3} respectively.

Now, sum of the volumes of these three cubes = 27x^{3} + 64x^{3} + 125 x^{3} = 216 x^{3} cm^{3}

Let the edge of the new cube be a cm.

Diagonal of new cube = a√3 cm

a√3 = 12√3

Volume of new cube = (12)^{3} = 1728 cm^{3}

Now by the given condition

216x^{3} = 1728

⇒ x^{3} = 8

⇒ x = 2

∴ \(\left.\begin{array}{rl}

\text { Edge of 1 cube } & =3 \times 2=6 \mathrm{~cm} \\

\text { Edge of 2 cube } & =4 \times 2=8 \mathrm{~cm} \\

\text { Edge of 3 cube } & =5 \times 2=10 \mathrm{~cm}

\end{array}\right\}\).

**Question 3. A cube of metal with a 2.5 cm edge is melted and cast into a rectangular solid whose base is 1.25 cm by 0.25 cm. Assuming no loss in melting find the height of the solid. Also, find the gain in the surface area.**

**Solution:**

The volume of the cube = (edge)^{3} = (2.5)^{3} cu. cm

Area of the base of rectangular solid = 1 .25 x 0.25 sq. cm

∴ Height of solid = \(\frac{\text { Volume }}{\text { Area of base }}=\frac{(2.5)^3}{1.25 \times 0.25}=50 \mathrm{~cm}\)

Surface area of the cube = 6 x (edge)^{2} = 6 x (2.5)^{2} = 37.5 sq. cm

Surface area of the solid = 2(lb + bh + hl) = 2(50 x 1.25 + 1.25 x 0.25 + 50 x 0.25)

= 2(62.5 + 0.3125 + 12.5) = 150.625 sq. cm

∴ Gain in surface area = 150.625 – 37.50 = 113.125 sq. cm

**Question 4. A granary is in the shape of a cuboid of size 8m x 6m x 3m. If a bag of grain occupies a space of 0.65 m3, how many bags can be stored in the granary?**

**Solution:**

The size of the granary is 8 m x 6 m x 3 m,

Volume of granary = 8 x 6 x 3 = 144m^{3}

The volume of one bag of grain = 0.65 m^{3}

The number of bags which can be stored in the granary

= \(\frac{\text { Volume of granary }}{\text { Volume of each bag }}=\frac{144}{0.65}=221.54 \text { or } 221 \text { bags. }\)

**Question 5. A cylinderical bucket 28 cm in diameter and 72 cm high is full of water. The water is emptied into a rectangular tank 66 cm long and 28 cm wide. Find the height of the water level in the tank.**

**Solution:**

Let the height of the water level in the tank = x m,

then according to the problem

πr^{2}h = l x b x x

or \(\frac{22}{7} \times 14 \times 14 \times 72=66 \times 28 \times x\)

or \(x=\frac{\frac{22}{7} \times 14 \times 14 \times 72}{66 \times 28}=24 \mathrm{~cm}\)

**Question 6. A rectangular container whose base is a square of side 15 cm stands on a horizontal table and holds water upto 3 cm from the top. When a cube is placed in the water and is completely submerged, the water rises to the top and 54 cm of water overflows. Calculate the volume of the cube and its surface area.**

**Solution:**

Volume of the cube submerged = Volume of water that fills 3 cm height of the container + Volume of water that overflows

= 15 x 15 x 3 + 54 = 729 cm^{3}

If the side of the cube submerged = x cm

Its volume = x^{3} cm^{3}

∴ x^{3} = 729 = 9 x 9 x 9

⇒ x = 9 cm.

∴ The side of the cube = 9 cm

And its surface area = 6 x (side) =6 x 9 x 9 = 486 cm^{2}

**Question 7. A solid spherical ball of iron with a radius of 6 cm is melted and recast into three solid spherical balls. The radii of the two of balls are 3 cm and 4 cm respectively, determine the diameter of the third ball.**

**Solution:**

Let the radius of the third ball = r cm

∴ The volume of three balls formed = Volume of the ball melted

⇒ \(\frac{4}{3} \pi(3)^3+\frac{4}{3} \pi(4)^3+\frac{4}{3} \pi(r)^3=\frac{4}{3} \pi(6)^3\)

⇒ 27 + 64 + r^{3} = 216

⇒ r^{3} = 125, i.e., r = 5 cm

∴ The diameter of the third ball = 2 x 5 cm = 10 cm

**Question 8. 50 circular plates each of radius 7 cm and thickness 0.5 cm are placed one above the other to form a solid right circular cylinder. Find**

- The total surface area and
- The volume of the cylinder so formed

**Solution:**

The height of the cylinder formed by placing 50 plates = 50 x 0.5 = 25 cm

Radius of cylinder formed = Radius of plate = 7 cm

1. Total surface area of cylinder = 2πrh + 2πr^{2}

= \(\left[2 \times \frac{22}{7} \times 7 \times 25+2 \times \frac{22}{7} \times(7)^2\right] \mathrm{cm}^2\)

= (1100 + 308) cm^{2} = 1408 cm^{2}

2. Volume of cylinder = \(\pi r^2 h=\frac{22}{7} \times(7)^2 \times 25=3850 \mathrm{~cm}^3\)

**Question 9. A rectangular paper of 22 cm x 12 cm is folded in two different ways and forms two cylinders.**

- Find the ratio of the volumes of two cylinders.
- Find the difference in the volumes of the two cylinders.

**Solution:**

**When the paper is folded along a 12 cm side,**

then height of cylinder h_{1} = 22 cm and 2πr_{1} = 12

⇒ \(r_1=\frac{12}{2 \pi}=\frac{6}{\pi} \mathrm{cm}\)

∴ Volume = \(V_1=\pi r_1^2 h_1=\pi \times\left(\frac{6}{\pi}\right)^2 \times 22=\frac{792}{\pi}=\frac{792 \times 7}{22}=252 \mathrm{~cm}^3\)

When the paper is folded along a 22 cm side,

then height of cylinder h_{2} = 12 cm and 2πr_{2} = 22

⇒ \(2 \times \frac{22}{7} \times r_2=22\)

⇒ \(r_2=\frac{7}{2} \mathrm{~cm}\)

∴ Volume V_{2} = r_{22}h_{2}

= \(\frac{22}{7} \times\left(\frac{7}{2}\right)^2 \times 12\)

= \(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 12=462 \mathrm{~cm}^3\)

- Ratio of volumes, V
_{1}:V_{2}= 252 : 462 = 6: 11 - Difference in volume = (462 – 252) cm
^{3}= 210 cm^{3}

**Question 10. A semicircle of radius 17.5 cm is rotated about its diameter. Find the curved surface of the generated solid.**

**Solution:**

The solid generated by a circle rotated about its diameter is a sphere.

Now, the radius of the sphere r = 17.5 cm.

and its curved surface = \(4 \pi r^2=4 \times \frac{22}{7} \times 17.5 \times 17.5=3850 \mathrm{~cm}^2\)

**Question 11. A sphere of radius 6 cm is melted and recast into a cone of height 6 cm. Find the radius of the cone.**

**Solution:**

Radius of sphere = 6 cm

∴ Volume of sphere = \(\frac{4}{3} \pi(6)^3=288 \pi \mathrm{cm}^3\)

Let the radius of cone = r

Height of cone = 6 cm

∴ Volume of cone = \(\frac{1}{3} \pi r^2 h=\frac{1}{3} \pi r^2 \times 6=2 \pi r^2\)

Given that, Volume of cone = Volume of a sphere

⇒ 2nr^{2} – 288 7r

⇒ r^{2}= 144

⇒ r = 12

Therefore, the radius of the cone = 12 cm

**Question 12. The height and radius of the base of a metallic cone are 27 cm and 16 cm respectively. It is melted and recast into a sphere. Find the radius and curved surface of the sphere.**

**Solution:**

Height of cone h = 27 cm

The radius of cone r = 16 cm

∴ Volume of cone = \(\frac{1}{3} \pi r^2 h=\frac{1}{3} \pi(16)^2 \times 27 \mathrm{~cm}^3\)

Let the radius of the sphere = R

∴ Volume of sphere = \(\frac{4}{3} \pi R^3\)

Now, Volume of sphere = Volume of cone

⇒ \(\frac{4}{3} \pi R^3=\frac{1}{3} \pi(16)^2 \times 27\)

⇒ \(R^3=\frac{16^2 \times 27}{4}=4^3 \times 3^3\)

R = 4 x 3 = 12cm

and curved surface of sphere = 4R^{2} = 4(12)^{2} = 576 cm^{2}

**Question 13. The radius of a metallic sphere is 60 mm. It is melted and recast into a wire of diameter 0.8 mm. Find the length of the wire.**

**Solution:**

Radius of sphere = 60 mm = 6 cm

∴ Volume of sphere = \(\frac{4}{3} \pi(6)^3=288 \pi \mathrm{cm}^3\)

Let, the length of wire = l cm

Radius of wire \(r=\frac{0.8}{2} \mathrm{~mm}=0.4 \mathrm{~mm}=\frac{0.4}{10} \mathrm{~cm}=\frac{4}{100} \mathrm{~cm}\)

∴ Volume of wire = \(\pi r^2 l=\pi\left(\frac{4}{100}\right)^2 \cdot l\)

Now, Volume of wire = Volume of a sphere

⇒ \(\pi \times \frac{4}{100} \times \frac{4}{100} \times l=288 \pi\)

⇒ \(l=\frac{288 \times 100 \times 100}{4 \times 4}=180000 \mathrm{~cm}=1800 \mathrm{~m}\)

**Question 14. A wire of diameter 3 mm is wound around a cylinder whose height is 1 2 cm and radius 5 cm so as to cover the curved surface of the cylinder completely. Find**

- Length of the wire.
- Mass of the wire, assuming the density of copper to be 8.88 g per cm
^{3}.

**Solution:**

1. Let the wire wounded around the cylinder complete n revolutions.

Diameter (width) of wire = 3 m- 0.3 cm

So, the whole height of the cylinder = 0.3 n cm

But the whole height of the cylinder = 12 cm

∴ \(0.3 n=12 \Rightarrow n=\frac{12}{0.3}=\frac{120}{3}=40\)

So, 40 revolutions are completed to wound the wire completely on the cylinder.

In 1 revolution, the length of the wire = 2 πr

∴ In 40 revolutions, the length of the wire = 40 x 2πr

= \(80 \times \frac{22}{7} \times 5 \mathrm{~cm}=1257.14 \mathrm{~cm}\)

= 12.57 m.

2. Now, radius of wire = \(\frac{0.3}{2}=0.15 \mathrm{~cm}\)

Volume of wire = area of cross section x length of wire

= π(0.15)^{2} x 1257.14 = 88.898 cm^{3}

∴ Mass of wire = volume x density

= 88.898 x 8.88 g= 789.41 g

Hence, the length of the wire = 12.57 m

and mass of wire = 789.41 g

**Question 15. A metallic cylinder of diameter 16 cm and height 9 cm is melted and recast into a sphere of diameter 6 cm. How many such spheres can be formed?**

**Solution:**

For the cylinder,

Radius = \(\frac{16}{2} = 8cm[latex]

Height = 9cm

∴ Volume of cylinder = JI(8)2(9) = 576K cm3

Diameter of sphere = 6 cm

∴ Radius of sphere = [latex]\frac{6}{2}=3 cm\)

Now, volume of one sphere = \(\frac{4}{3} \pi(3)^3=36 \pi \mathrm{cm}^3\)

∴ Number of spheres formed = \(\frac{\text { Volume of cylinder }}{\text { Volume of one sphere }}=\frac{576 \pi}{36 \pi}=16\)

**Question 16. A solid metallic right circular cone of height 6.75 cm and radius of the base 12 cm is melted and two solid spheres formed from it. If the volume of one of the spheres is 8 times that of the other, find the radius of the smaller sphere.**

**Solution:**

Let the radius of the smaller sphere be r and the radius of the larger sphere be R.

image

So, \(\frac{1}{3} \pi(12)^2 \times 6.75=\frac{4}{3} \pi r^3+\frac{4}{3} \pi R^3\)

⇒ (12)^{2} x 6.75 = 4(r^{3} + R^{3}) → (1)

Now, the volume of the larger sphere = 8 x the Volume of the smaller sphere

⇒ \(\frac{4}{3} \pi R^3=8 \times \frac{4}{3} \pi r^3\)

R^{3} = 8r^{3} = R = 2r → (2)

From equations (1) and (2), we get

12 x 12 x 6.75 = 4(r^{3} + 8r^{3})

⇒ \(9 r^3=\frac{12 \times 12 \times 6.75}{4}\)

⇒ \(r^3=\frac{12 \times 12 \times 6.15}{4 \times 9}=27=(3)^3\)

r = 3 cm.

Hence, the radius of a smaller sphere = 3 cm.

**Question 17. Water is flowing at the rate of 3 km an hour through a circular pipe of 20 cm internal diameter into a circular cistern of diameter 10 m and depth 2m. In how much time will the cistern be filled?**

**Solution:**

Internal diameter of pipe = 20 cm

∴ Internal radius of pipe = 10 cm = \(\frac{1}{10}m\)

Rate of flow of water = 3 km/h

⇒ Rate of flow of water = \(\frac{3 \times 1000}{60 \times 60} \mathrm{~m} / \mathrm{sec}\)

Diameter of cistern = 10 m

Radius of cistern = 5 m

Height of cistern = 2 m

Water discharge by pipe in 1 sec

= π x r^{2} x flow of water

= \(\pi \times\left(\frac{1}{10}\right)^2 \times \frac{3 \times 1000}{60 \times 60} \mathrm{~m}^3 / \mathrm{sec}\)

Volume of cistern = π x 5^{2} x 2 m^{3}

Time taken to fill the cistern = \(\frac{\text { Volume of cistern }}{\text { Volume of water discharge in }1 \mathrm{sec}}\)

= \(\frac{\pi \times 25 \times 2}{\pi \times \frac{1}{100} \times \frac{3 \times 1000}{60 \times 60}} \mathrm{sec}=\frac{25 \times 2 \times 100 \times 60 \times 60}{3 \times 1000} \mathrm{sec}\)

= \(\frac{5 \times 60 \times 60}{3} \sec =\frac{5 \times 60 \times 60}{3 \times 60 \times 60} \text { hours }=\frac{5}{3} \text { hours }\)

**Question 18. The volume of a sphere is 288 π cm ^{3}. 27 small spheres can be formed with this sphere. Find the radius of a small sphere.**

**Solution:**

Volume of 27 small spheres = Volume of one big sphere = 288π

Volume of 1 small sphere = \(\frac{288}{27} \pi\)

\(\frac{4}{3} \pi r^3=\frac{32}{3} \pi\) where r = radius of small sphere

r^{3} = 8

r = 2 cm

**Question 19. Find:**

- The lateral or curved surface area of a closed cylindrical petrol storage tank is 4.2 m in diameter and 4.5 m high.
- How much steel was actually used, if 1/12 of the steel actually used was wasted in making the tank?

**Solution:**

1. Curved surface area = \(2 \pi r h=2 \times \frac{22}{7} \times \frac{4.2}{2} \times 4.5=59.4 \mathrm{~m}^2\)

2. Total steel used = Total surface area (Assuming thickness = 0 m)

= 2πr(r + h)

= \(2 \times \frac{22}{7} \times \frac{4.2}{2}\left(\frac{4.2}{2}+4.5\right)\)

= \(\frac{44}{7} \times 2.1 \times \frac{13.2}{2}=22 \times 0.3 \times 13.2=87.12 \mathrm{~m}^2\)

Let the actual steel used be x m^{2}

= \(\frac{11}{12} x=87.12\)

= \(x=\frac{87.12 \times 12}{11}=95.04 \mathrm{~m}^2\)

**Question 20. In one fortnight (15 days) of a given month, there was a rainfall of 11.10 cm in a river valley. If the area of the valley is 97280 km ^{2}, show that the total rainfall of a day was approximately equivalent to the addition (sum) to the normal water of three rivers earth 1072 km long, 7.5 m wide and 3m deep. (Round off the volumes upto one place of decimal).**

**Solution:**

Area of the valley = 97280 km^{2} = 9.728 x 10^{4} km^{2}

= 9.728 x 10^{10} m^{2}

∴ Volume (amount) of rainfall in the valley in one fortnight (15 days)

= \(9.728 \times 10^{10} \times \frac{11.10}{100} \mathrm{~m}^3=107.98 \times 10^8 \mathrm{~m}^3\)

∴ Amount of rainfall in 1 day = \(\frac{107.98 \times 10^8}{15} \mathrm{~m}^3\)

= 7.198 x 10^{8} m^{3} = 7.2 x 10^{8}m^{3} → (1)

Now, length of each river = 1072 km = 1072000 m

breadth of each river = 75 m

and depth of each river = 3 m

∴ Amount (volume) of water in each river = 1072000 x 75 x 3 m^{3}

∴ Volume of water in 3 such rivers = 3 x 1.072 x 10^{6} x 75 x 3 m^{3}

= 7.236 x 10^{8} m^{3} = 7.2 x 10^{8} m^{3} → (2)

From (1 ) and (2), we can say that total rainfall in a day is approximately the same as the volume of 3 rivers.

**Question 21. The dimensions of a solid iron cuboid are 4.4 m x 2.6 m x 1.0 m. It is melted and recast into a hollow cylindrical pipe of 30 cm inner radius and thickness of 5 cm. Find the length of the pipe.**

**Answer:**

Length of solid iron cuboid = 4.4 m

Breadth of solid iron cuboid = 2.6 m

Height of solid iron cuboid = 1.0 m

∴ Volume of solid cuboid = 4.4 x 2.6 x 1 = 11.44 m^{3}

= 11.44 x 100 x 100 x 100 cm^{3}

This cuboid is melted and recast into a hollow cylinder.

∴ The volume of a cuboid = Volume of a hollow cylinder

⇒ 11.44 x 10^{6} = πR^{2}h – πr^{2}h

⇒ 11.44 x 10^{6} = πh(R^{2}– r^{2}) [∵ r = 30 cm, width = 5 cm, R = 35 cm]

⇒ \(11.44 \times 10^6=\frac{22}{7} \times h(R-r)(R+r)\)

⇒ 11 .44 x 106 x 7 = 22 x h (35 – 30) (35 + 30)

⇒ \(h=\frac{11.44 \times 10^6 \times 7}{22 \times 5 \times 65} \mathrm{~cm}=\frac{11440000 \times 7}{22 \times 5 \times 65} \mathrm{~cm}\)

= 11200 cm = 112 m

Hence, the length of the pipe = 112 m

**Question 22. A tank measures 2 m long, 1.6 m wide and 1 m depth. Water is there upto 0.4 m in height. Brides measuring 25 cm x 14 cm x 10 cm are put into the tank so that water may come upto the top. Each brick absorbs water equal \(\frac{1}{7} \text { th }\) to that of its own volume. How many bricks will be needed?**

**Solution:**

Let x bricks be needed

∴ Volume of x bricks = x X 0.25 X 0.1 4 X 0.1 m^{3}

The volume of water absorbed by bricks

= \(=\frac{1}{7} x \times 0.25 \times 0.14 \times 0.1\)

∴ Remaining water in the tank = \(2 \times 1.6 \times 0.4-\frac{1}{7} \times x \times 0.25 \times 0.14 \times 0.1\)

Now, the Volume of water in the tank + Volume of x bricks

= Volume of tank

⇒ \(\left(2 \times 1.6 \times 0.4-\frac{x}{7} \times 0.25 \times 0.14 \times 0.1\right)+x \times 0.25 \times 0.14 \times 0.1\) = 2 x 1.6 x 1

⇒ \(\frac{6 x}{7} \times 0.25 \times 0.14 \times 0.1=2 \times 1.6 \times 0.6\)

⇒ \(x=\frac{2 \times 1.6 \times 0.6 \times 7}{6 \times 0.25 \times 0.14 \times 0.1}=\frac{2 \times 16 \times 6 \times 7 \times 1000}{6 \times 25 \times 14 \times 1}=640\)

**Question 23. A metallic solid sphere of radius 10.5 cm is melted and recast into smaller solid cones, each of radius 3.5 cm and height 3 cm. How many cones will be made?**

**Solution:**

Let n cones be recast from the sphere.

∴ The sum of volumes of n cones = volume of a sphere

⇒ \(n\left[\frac{1}{3} \pi(3.5)^2 \times 3\right]=\frac{4}{3} \pi(10.5)^3\)

⇒ \(n=\frac{4 \times 10.5 \times 10.5 \times 10.5}{3.5 \times 3.5 \times 3}=126\)

Hence, 126 cones will be made.

## Frustum Of A Right Circular Cone

**Frustum:** If a right circular cone is cut by a plane parallel to the base of the cone then the portion between the plane and base is called the frustum of the cone.

**How To Find The Volume And Surface Area Of A Bucket**

Let a bucket of height h and radii of upper and lower ends be r_{1 }and r_{2} respectively.

Now we shall find three parts:

- The slant is the height of the bucket.
- Curved surface area and total surface area of the bucket.
- The volume of the bucket or capacity in litres.

**Proof :**

1. Let the slant height of the bucket be l

Now, draw DC ⊥ AB

l^{2}=h^{2 + }(r_{1 }– r_{2})^{2}

⇒ \(l=\sqrt{h^2+\left(r_1-r_2\right)^2}\)

2. Let EO = H and DO = L

ΔABO ~ ΔDEO

∴ \(\frac{A B}{D E}=\frac{B O}{E O}=\frac{A O}{D O}\)

⇒ \(\frac{r_1}{r_2}=\frac{h+H}{H}=\frac{l+L}{L}\)

⇒ \(\frac{r_1}{r_2}=\frac{h+H}{H} \quad \text { and } \quad \frac{r_1}{r_2}=\frac{l+L}{L}\)

⇒ Hr_{1} = hr_{2} + Hr_{2} and Lr_{1} = lr_{2} + Lr_{2}

⇒ H(r_{1}– r_{2}) = hr_{2} ⇒ L(r_{1} – r_{2}) = lr_{2}

⇒ \(H=\frac{h r_2}{r_1-r_2} \quad \text { and } \quad L=\frac{l r_2}{r_1-r_2}\)

Curved surface area of bucket = C.S.A of larger cone- C.S.A of smaller cone

= \(\pi r_1(L+l)-\pi r_2 L=\pi r_1\left[\frac{l r_2}{r_1-r_2}+l\right]-\pi r_2\left(\frac{l r_2}{r_1-r_2}\right)\)

= \(\pi r_1 l\left(\frac{r_2+r_1-r_2}{r_1-r_2}\right)-\frac{\pi r_2^2 l}{r_1-r_2}\)

= \(\frac{\pi l}{r_1-r_2}\left(r_1^2-r_2^2\right)=\frac{\pi l\left(r_1+r_2\right)\left(r_1-r_2\right)}{\left(r_1-r_2\right)}\)

= \(\pi l\left(r_1+r_2\right) \text { where } l=\sqrt{h^2+\left(r_1-r_2\right)^2}\)

Total surface area = C.S.A. of bucket + Area of the smaller circle

= πl(r_{1} + r_{2}) + πr_{2}^{2}

3. Volume of bucket = Volume of larger cone – Volume of smaller cone

= \(\frac{1}{3} \pi l_l\left(r_1^2+r_1 r_2+r_2^2\right)\)

= \(\frac{1}{3} \pi r_1^2(H+h)-\frac{1}{3} \pi r_2^2 \cdot H\)

= \(\frac{1}{3} \pi r_1^2\left[\frac{h r_2}{r_1-r_2}+h\right]-\frac{1}{3} \pi r_2^2\left(\frac{h r_2}{r_1-r_2}\right)\)

= \(\frac{1}{3} \pi r_1^2 h\left(\frac{r_2+r_1-r_2}{r_1-r_2}\right)-\frac{1}{3} \frac{\pi h r_2^3}{r_1-r_2}\)

= \(\frac{1}{3} \frac{\pi h}{r_1-r_2}\left(r_1^3-r_2^3\right)\)

= \(\frac{1}{3} \frac{\pi h}{r_1-r_2}\left(r_1-r_2\right)\left(r_1^2+r_1 r_2+r_2^2\right)\)

= \(\frac{1}{3} \pi h\left(r_1^2+r_2^2+r_1 r_2\right)\)

**Note:** Actually, the bucket is a frustum cone made of cutting the bucket by a plane parallel to the base.

## Frustum Of A Right Circular Cone Solved Examples

**Question 1. A cone is divided into two parts by drawing a plane through the mid-point of its axis, parallel to its base, Compare the volume of the two parts.**

**Solution:**

We can solve this using similarity.

Let r and li be the radius and height of a cone OAB

Let OE = \(\frac{h}{2}\)

As OED and OFB are similar

∴ \(\frac{O E}{O F}=\frac{E D}{F B} \frac{h / 2}{h}=\frac{E D}{r}\)

⇒ \(E D=\frac{r}{2}\)

Now volume of cone OCD = \(\frac{1}{3} \pi r^2 h=\frac{1}{3} \times \pi \times\left(\frac{r}{2}\right)^2 \times \frac{h}{2}=\frac{\pi r^2 h}{24}\)

and Volume of cone OAB = \(\frac{1}{3} \times \pi \times r^2 \times h=\frac{\pi r^2 h}{3}\)

∴ \(\frac{\text { Volume of part } O C D}{\text { Volume of part } C D A B}=\frac{\frac{\pi r^2 h}{24}}{\frac{\pi r^2 h}{3}-\frac{\pi r^2 h}{24}}=\frac{\frac{1}{24}}{\frac{1}{3}-\frac{1}{24}}=\frac{\frac{1}{24}}{\frac{8-1}{24}}=\frac{1}{7}\)

**Question 2. The height of a right circular cone is trisected by two planes drawn parallel to the base. Show that the ratio of volumes of the three portions starting from the top is in the ratio 1:7:19.**

**Solution:**

Since height is trisected, therefore by basic proportionality theorem, base radii of three cones VCD, VA’B’ and VAB are also in the ratio 1: 2 : 3.

( ∵ VL : VM : VN = r_{2 }: r_{1} : r = 1:2:3)

Let volume of cone VCD = V = \(\frac{1}{3} \pi r_2{ }^2 h\)

∴ Volume of cone VA’B’ = \(\frac{1}{3} \pi r_1^2(2 h)=\frac{1}{3} \pi\left(2 r_2\right)^2(2 h)=8 \times \frac{1}{3} \pi r_2{ }^2 h=8 V\)

and volume of cone VAB = \(\frac{1}{3} \pi r^2(3 h)=\frac{1}{3} \pi\left(3 r_2\right)^2(3 h)=27 \times \frac{1}{3} \pi r_2^2 h=27 \mathrm{~V}\)

∴ Ratio of volumes of 3 portions

= Volume(VCD) : Volume(CD£’A’) : Volume(A’B’BA)

= V: 8V – V: 27V – 8V = 1: 7: 19

Hence Proved.

**Question 3. The radii of the faces of a frustum of a cone are 3 cm and 4 cm and its height is 5cm. Find its volume.**

**Solution:**

Here r = 3 cm, R = 4 cm and h = 5 cm

Volume of frustum of cone = \(\frac{1}{3} \pi l\left(R^2+r^2+R r\right) \text { cu. units }\)

= \(\frac{1}{3} \times \frac{22}{7} \times 5\left(4^2+3^2+4 \times 3\right)\)

= \(\frac{1}{3} \times \frac{110}{7}(16+9+12)\)

= \(\frac{110}{21} \times 37=\frac{4070}{21} \mathrm{~cm}^3\)

**Question 4. The height of a cone is 30 cm. A small cone is cut off at the top by a plane parallel to the base. If its volume is \(\frac{1}{27}\) of the volume of the given cone, at what height above the base, the section has been made?**

**Solution:**

Let OA – h ⇒ AB = 30- h

and let AC = r, BD = R

ΔOAC ~ ΔOBD,

∴ \(\frac{h}{30}=\frac{r}{R} \Rightarrow r=\frac{h R}{30}\)

Now, Volume of smaller cone = \(\frac{1}{27}\) x Volume of larger cone

⇒ \(\frac{1}{3} \pi r^2 h=\frac{1}{27} \times \frac{1}{3} \pi R^2 \times 30\)

⇒ \(\frac{1}{3} \pi\left(\frac{h R}{30}\right)^2 \cdot h=\frac{1}{27} \times \frac{1}{3} \pi R^2 \times 30\)

⇒ \(\frac{h^3}{30^2}=\frac{30}{27} \Rightarrow h^3=\frac{30^3}{3^3} \Rightarrow h=10\)

∴ Reqired height = 30 – 10 = 20 cm.

1 kl = 1 m^{3}

1000 l = 100 x 100 x 100 cm^{3}

1 litre = 1000 cm^{3}

= \(1 \mathrm{~cm}^3=\frac{1}{1000} \text { litre }\)

**Question 5. The radii of the ends of a bucket 30 cm high are 21 cm and 7 cm. Find its capacity in litres and the amount of sheets required to make this bucket. (\(\text { Take } \pi=\frac{22}{7}\))**

**Solution:**

Although we can solve this problem by using similar triangles, here we are using the direct formula.

Volume of bucket = \(\frac{1}{3} \pi h\left(r_1^2+r_1 r_2+r_2^2\right)\)

= \(\frac{1}{3} \times \frac{22}{7} \times 30\left(21^2+21 \times 7+7^2\right)\)

= \(\frac{1}{3} \times \frac{22}{7} \times 30 \times 637=20020 \mathrm{~cm}^3=\frac{20020}{1000} \text { lit. }\)

⇒ Capacity = 20.02 lit

Area of sheet = C.S.A. of bucket + Area of base

= πl(r_{1} + r_{2}) + r_{2}^{2} = π[l(r_{1} + r_{2}) + r2^{2} (∵ \(l=\sqrt{h^2+\left(n_1-r_2\right)^2}\) = \(\sqrt{900+14^2}\) = \(\sqrt{1096}\) = 33.106)

= \(\frac{22}{7}[33.106(28)+49]\)

= 3067.32 = 3067cm^{2} (approx.)

**Question 6. A bucket is 32 cm in diameter at the top and 20 cm in diameter at the bottom. Find the capacity of the bucket in litres if it is 21 cm deep. Also, find the cost of the tin sheet used in making the bucket at the rate of ₹ 1.50 per sq dm.**

**Solution:**

Here R = 16 cm, r = 10 cm and h = 21 cm

Volume of frustum of a cone = \(\frac{\pi h}{3}\left(R^2+r^2+R r\right)\)

= \(\frac{22}{7} \times \frac{1}{3} \times 21\left(16^2+10^2+16 \times 10\right)\)

= 22(256 + 100 + 160)

= 22 x 516 = 11352cm^{3}

= \(\frac{11352}{1000} \text { litres }=11.352 \text { litres }\)

Now for slant height l of frustum

l^{2} = h^{2} + (R-r)^{2}

l^{2} = 212 + (16 – 10)^{2}

l^{2} = 441 + 36

l^{2} = 477

∴ \(l=\sqrt{477}=21.84 \mathrm{~cm}\)

Now S.A of bucket = πl(R + r) + πr^{2}

= \(\frac{22}{7} \times 21.84 \times(16+10)+\frac{22}{7} \times 10^2\)

= \(\frac{22}{7}(21.84 \times 26+100)\)

= \(\frac{22}{7} \times 667.84=2098.92 \mathrm{~cm}^2=20.99 \mathrm{dm}^2\)

Cost of sheet @ ₹ 1 .50 per sq. dm. = 20.99 x 1.50 = ₹ 31.49

## Volume And Surface Area Of Solids Exercise 13.1

**Question 1. 2 cubes each of volume 64 cm ^{3} are joined end to end. Find the surface area of the resulting cuboid.**

**Solution:**

Given,

volume of cube = 64 cm^{3}

(side)^{3} = 64

(side)^{3} = 4^{3}

side = 4 cm

Side of cube = 4 cm

A cuboid is formed by joining two cubes together as shown.

∴ For cuboid

length l = 4 + 4 = 8 cm

height h = 4 cm

Now, the total surface area of a cuboid

= 2(l.b + b.h + l.h)

= 2 (8 x 4 + 4 x 4 + 8 x 4)

= 2 (32 + 16 + 32) = 160 cm^{2}

**Question 2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.**

**Solution:**

In the adjoining figure, a cylinder is surmounted on the hemisphere.

Diameter of hemisphere 2r = 14 cm

⇒ Radius of hemisphere r = 7 cm

The radius of cylinder r = 1 cm

Now, the total height of the vessel = 13 cm

⇒ h + r = 13 cm

⇒ h = 13 – 7 = 6 cm

Height of cylinder, h = 6 cm

The inner surface area of the cylinder = 2πrh

Inner curved surface area of hemisphere = 2πr^{2}

Inner surface area of vessel = 2πrh + 2πr^{2}

= 2πr(h + r)

= \(2 \times \frac{22}{7} \times 7 \times(6+7)\)

= 44 x 13 = 572 cm^{2}

**Question 3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of the same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.**

**Solution:**

In the adjoining figure, a cone of the same cross-section is surmounted on the hemisphere.

The radius of base of cone r = 3.5 cm

⇒ Radius of hemisphere r = 3.5 cm

Total height of toy = 15.5 cm

⇒ h + r = 15.5

⇒ h = 15.5 – 3.5 = 12 cm

∴ Height of cone h = 12 cm

Now, from l^{2} = h^{2} + r^{2}

l^{2} = (12)^{2} + (3.5)^{2}

= 144 + 12.25

= 156.25

⇒ \(l=\sqrt{156.25}=12.5 \mathrm{~cm}\)

∴ The curved surface area of the cone = πrl

and curved surface area of hemisphere = 2πr^{2}

So, the total surface area of the toy = πrl + 2πr^{2}

= πr (l + 2r)

= \(\frac{22}{7} \times 3.5 \times(12.5+2 \times 3.5)\)

= 11 x 19.5 = 214.5cm^{2}

**Question 4. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.**

**Solution:**

The base of the hemisphere is on the upper face of a cube of edge 7 cm,

∴ Maximum diameter of hemisphere = edge of the cube

= 7 cm

⇒ 2r = 7 ⇒ \(r=\frac{7}{2} \mathrm{~cm}\)

Now, the total surface area of the solid

= total surface area of cube + curved surface of the hemisphere – an area of the base of the hemisphere

= 6 x (side)^{2} + 2πr^{2}– πr^{2}

= 6 x (side)^{2} + πr^{2}

= \(=6 \times 7 \times 7+\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\)

= 294 + 38.5 = 332.5 cm^{2}

**Question 5. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.**

**Solution:**

Let the side of the cube = a

Diameter of hemisphere = side of the cube

⇒ 2r = a = \(r=\frac{a}{2}\)

Now, the surface area of remaining solid = total surface of the hemisphere – an area of the base of the hemisphere

= 6a^{2} + 2πr^{2}– πr^{2}

= 6a^{2} + πr^{2}

= \(6 a^2+\pi\left(\frac{a}{2}\right)^2=\frac{24 a^2+\pi a^2}{4}\)

= \(\frac{a^2(24+\pi)}{4} \text { square units }\)

**Question 6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.**

**Solution:**

Diameter of capsule = 5 mm

2r = 5 mm

r = 2.5 mm

∴ The radius of the cylindrical part = radius of the hemisphere = 2.5 mm

Length of capsule = 14 mm

⇒ h + 2r = 14 mm

⇒ h = 14 – 2r = 14 – 5 = 9 mm

Now the surface area of the capsule = 2 x curved surface of hemisphere + curved surface of the cylinder

= 2 x 2πr^{2} + 2πrh = 2πr (2r + h)

= \(2 \times \frac{22}{7} \times 2.5 \times(5+9)\)

= \(\frac{110}{7} \times 14=220 \mathrm{~mm}^2\)

**Question 7. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of ₹ 500 per m ^{2}.**

**Solution:**

In the adjoining, a tent is shown in which a cone is surmounted by a cylinder.

Diameter of cylindrical part 2r = 2. 1 m

⇒ \(r=\frac{2.1}{2} \mathrm{~m}\)

Height of cylindrical part h = 4 m

Radius of conical part \(r=\frac{2.1}{2} \mathrm{~m}\)

The slant height of conical part l = 2.8 m

Now, the area of canvas required to form a tent = curved surface of the cylindrical part + curved surface of the conical part

= 2πrh + πrl

= πr (2h + l)

= \(\frac{22}{7} \times \frac{2.1}{2} \times(2 \times 4+2.8)\)

= 3.3 x 10.8 = 35.64 m^{2}

**Question 8. From a solid cylinder whose height is 2.4 cm and diameter is 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm ^{2}.**

**Solution:**

Diameter of cylinder 2r = 1.4 cm

⇒ r = 0.7 cm

∴ The radius of the cylinder = radius of the cone = r = 0.7 cm

Height of cylinder = height of cone = h = 2.4 cm

If the slant height of the cone is l, then

l^{2} = h^{2} + r^{2}= (2.4)^{2} + (0.7)^{2}

= 5.76 + 0.49 = 6.25

⇒ \(l=\sqrt{6.25}=2.5 \mathrm{~cm}\)

Surface area of remaining solid

= area of base of cylinder + curved surface of cylinder + curved surface of cone

= πr^{2} + 2πrh + πrl

= πr (r + 2h + l)

= \(\frac{22}{7} \times 0.7 \times(0.7+2 \times 2.4+2.5)\)

= 2.2 x 8 = 17.6 cm^{2}

**Question 9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.**

**Solution:**

The total surface area of the article

= curved surface of cylinder + 2 x curved surface of a hemisphere

= 2πrh + 2 x 2πr^{2}

= 2πr (h+2r)

= \(2 \times \frac{22}{7} \times 3.5 \times(10+2 \times 3.5)\)

= 22 x 17 = 374 cm^{2}

## Volume And Surface Area Of Solids Exercise 13.2

**Question 1. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π.**

**Solution:**

Radius of hemisphere = radius of cone = r = 1 cm

Height of cone h = radius of cone = 1 cm

Volume of hemisphere = \(\frac{2}{3} \pi r^3\)

Volume of cone = \(\frac{1}{3} \pi r^2 h\)

∴ The volume of solid = volume of hemisphere + volume of a cone

= \(\frac{2}{3} \pi r^3+\frac{1}{3} \pi r^2 h\)

= \(\frac{2}{3} \pi(1)^3+\frac{1}{3} \pi(1)^2(1)\)

= \(\frac{2}{3} \pi+\frac{1}{3} \pi=\pi \mathrm{cm}^3\)

**Question 2. Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)**

**Solution:**

The model is shown in the given below.

Diameter 2r = 3 cm

⇒ \(r=\frac{3}{2} \mathrm{~cm}\)

Height of each cone h = 2 cm

Let the height of the cylinder = H

∴ H + h + h = 12 cm

⇒ H + 2 + 2 = 12

⇒ H = 12 – 4 = 8 cm

The volume of model = volume of cylinder + 2 x volume of a cone

= \(\pi r^2 H+2 \times \frac{1}{3} \pi r^2 h\)

= \(\pi r^2\left(H+\frac{2 h}{3}\right)\)

= \(\frac{22}{7} \times \frac{3}{2} \times \frac{3}{2} \times\left(8+\frac{2 \times 2}{3}\right)\)

= \(\frac{22}{7} \times \frac{3}{2} \times \frac{3}{2} \times \frac{28}{3}=66 \mathrm{~cm}^3\)

Air contained in model = 66 cm^{3}

**Question 3. A gulab jamun contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with a length of 5 cm and a diameter of 2.8 cm.**

**Solution:**

For one gulab jamun,

Diameter 2r = 2.8 cm

⇒ r = 1.4 cm

Height of cylindrical part h = 5 – r – r

= 5-1.4 – 1.4 = 2.2 cm

Volume of one gulab jamun

= volume of cylindrical part + 2 x volume of hemispherical part

= \(\pi r^2 h+2 \times \frac{2}{3} \pi r^3\)

= \(\pi r^2\left(h+\frac{4 r}{3}\right)\)

= \(\frac{22}{7} \times 1.4 \times 1.4 \times\left(2.2+\frac{4 \times 1.4}{3}\right)\)

= 25.051 cm^{3}

⇒ Volume of 45 gulab jamuns = 45 x 25.051

= 1127.295 cm^{3}

Volume of sugar syrup in 45 gulab jamuns

= 30% of 1127.295

= \(1127.295 \times \frac{30}{100} \mathrm{~cm}^3\)

= 338.1885 cm3 ≈ 338 cm^{3}

**Question 4. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand.**

**Solution:**

Volume of cuboid = 15 x 10 x 3.5 cm^{3}

= 525 cm^{3}

For conical depression

r = 0.5 cm and h = 1.4 cm

∴ The volume of one cavity

= \(\frac{1}{3} \pi r^2 h=\frac{1}{3} \times \frac{22}{7} \times 0.5 \times 0.5 \times 1.4\)

= \(\frac{1.1}{3} \mathrm{~cm}^3\)

⇒ Volume of four depression= \(4 \times \frac{1.1}{3} \mathrm{~cm}^3\)

= 1.467 cm^{3}

Now, the volume of wood used in the pen stand

= volume of cuboid – volume of four depression

= (525 – 1 .467) cm^{3} = 523.533 cm^{3}

**Question 5. A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.**

**Solution:**

The radius of cone r = 5 cm

and height h = 8 cm

⇒ Volume of cone = \(\frac{1}{3} \pi r^2 h\)

The volume of water filled in the cone

= \(\frac{1}{3} \pi r^2 h=\frac{1}{3} \pi \times(5)^2 \times 8\)

= \(\frac{200}{3} \pi \mathrm{cm}^3\)

The volume of water flows out on dropping lead shots in it.

= \(\frac{1}{4} \times \frac{200 \pi}{3}=\frac{50 \pi}{3} \mathrm{~cm}^3\)

The radius of shot R = 0.5 cm

Volume of one lead shot = \(\frac{4}{3} \pi R^3=\frac{4}{3} \pi(0.5)^3 \mathrm{~cm}^3\)

= \(\frac{\pi}{6} \mathrm{~cm}^3\)

Now, number of shots

= \(\frac{\text { volume of water flows out }}{\text { volume of one shot }}\)

= \(\frac{50 \pi / 3}{\pi / 6}=\frac{50 \pi}{3} \times \frac{6}{\pi}=100\)

**Question 6. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm ^{3} of iron has approximately 8g mass. (Use π = 3.14)**

**Solution:**

For the first cylinder,

Diameter 2r = 24 cm

⇒ r = 12 cm

Height h = 220 cm

∴ Volume = πr^{2}h = π x 12 x 12 x 220 cm^{3}

= 31680 π cm^{3}

For the second cylinder,

Radius R = 8 cm

Height H = 60 cm

Volume = πr^{2}H = π x 8 x 8 x 60

= 3840π cm^{3}

Volume of the pole = (31680π + 3840π) cm^{3}

= 35520π cm^{3}

∴ Weight of pole = 35520π x 8 g

= 35520 x 3.14 x 8 g

= 892262.4 g = 892.2624 kg

= 892.26 kg

**Question 7. A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.**

**Solution:**

The radius of cylinder r = 60 cm and

height h = 1 80 cm

∴ Volume of cylinder = πr^{2}h

= π x 60 x 60 x 180 cm^{3}

= 648000π cm3

The radius of cone R = 60 cm

and height H = 120 cm

∴ Volume of cone = \(\frac{1}{3} \pi R^2 H\)

= \(\frac{1}{3} \pi \times 60 \times 60 \times 120 \mathrm{~cm}^3\)

= 144000 cm^{3}

The volume of solid formed from the cone and hemisphere

= (144000π + 144000π) cm^{3}

= 288000π cm^{3}

⇒ Volume of water displaced by this solid

= 288000π cm^{3}

∴ The volume of the remaining water in a cylinder

= (648000π – 288000π) cm^{3}

= 360000π cm^{3}

= \(360000 \times \frac{22}{7} \mathrm{~cm}^3\)

= \(1131428.57 \mathrm{~cm}^3=\frac{1131428.57}{1000000} \mathrm{~m}^3\)

= 1.131 m^{3} (approx.)

**Question 8. A spherical glass vessel has a cylindrical neck 8 cm long, and 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm ^{3}. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.**

**Solution:**

Radius of cylindrical part \(r=\frac{2}{2}=1 \mathrm{~cm}\)

and height h = 8 cm

∴ The volume of the cylindrical part = nr2h

= π(1)^{2}(8) = 8πcm^{3}

Radius of spherical part, \(R=\frac{8.5}{2}=\frac{17}{4} \mathrm{~cm}\)

∴ The volume of the spherical part,

= \(\frac{4}{3} \pi R^3=\frac{4 \pi}{3} \times \frac{17}{4} \times \frac{17}{4} \times \frac{17}{4} \mathrm{~cm}^3\)

= \(\frac{4913}{48} \pi \mathrm{cm}^3\)

∴ Volume of cylinder = \(\left(8 \pi+\frac{4913 \pi}{48}\right) \mathrm{cm}^3\)

= \(\frac{384 \pi+4913 \pi}{48} \mathrm{~cm}^3[/latex

= [latex]\frac{5297 \pi}{48} \mathrm{~cm}^3=\frac{5297}{48} \times 3.14 \mathrm{~cm}^3\)

= 346.51 cm^{3}

So the answer 345 cm^{3} of a child is not correct.

∴ Correct volume of cylinder = 346.51 cm^{3}

## Volume And Surface Area Of Solids Exercise 13.3

**Question 1. A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.**

**Solution:**

The radius of sphere R = 4.2 cm

The radius of cylinder r = 6 cm

Let the height of the cylinder = h

Now, the volume of the cylinder = volume of a sphere

⇒ \(\pi r^2 h=\frac{4}{3} \pi R^3\)

⇒ \(h=\frac{4 R^3}{3 r^2}=\frac{4 \times 4.2 \times 4.2 \times 4.2}{3 \times 6 \times 6} \mathrm{~cm}\)

= 2.744 cm

∴ Height of cylinder = 27.44 cm

**Question 2. Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.**

**Solution:**

Let r_{1} = 6 cm, r_{2} = 8 cm and r_{3} = 10 cm

Let the radius of a bigger solid sphere = R

The volume of a bigger solid volume

= sum of volumes of three given spheres

⇒ \(\frac{4}{3} \pi R^3=\frac{4}{3} \pi r_1^3+\frac{4}{3} \pi r_2^3+\frac{4}{3} \pi r_3^3\)

⇒ \(R^3=r_1^3+r_2^3+r_3^3\)

= 63 + 83+ 10^{2}

= 216 + 512 + 1000 = 1728 = 12^{2}

⇒ R = 12 cm

∴ The radius of the new solid sphere = 12 cm

**Question 3. A 20 m deep well with a diameter of 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.**

**Solution:**

Radius of well, \(r=\frac{7}{2} m\)

and depth h = 20 m

Let the height of the platform be H metre.

∴ The volume of the platform = the volume of the well.

⇒ 22 x 14 x H = πr^{2}h

⇒ \(22 \times 14 \times H=\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 20\)

⇒ \(H=\frac{22 \times 7 \times 7 \times 20}{7 \times 2 \times 2 \times 22 \times 14}=2.5 \mathrm{~m}\)

∴ Height of platform = 2.5 m

**Question 4. A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.**

**Solution:**

Diameter of well 2r = 3 m

⇒ \(r=\frac{3}{2}=1.5 \mathrm{~m}\)

and depth h = 14 m

∴ The volume of earth taken out from the well = nr2h

= \(\frac{22}{7} \times 1.5 \times 1.5 \times 14=99 \mathrm{~m}^3\)

Now, the outer radius of the well, R = 1.5 + 4 = 5.5m

∴ Area of the ring of platform = π (R^{2} – r^{2})

= \(\frac{22}{7}\left[(5.5)^2-(1.5)^2\right]\)

= \(\frac{22}{7} \times 7 \times 4=88 \mathrm{~m}^2\)

Let the height of the embankment = H

∴ 88 x H = 99

⇒ \(H=\frac{99}{88}=\frac{9}{8}=1.125 \mathrm{~m}\)

Height of embankment = 1.1 25 m

**Question 5. A container shaped like a right circular cylinder having a diameter of 12 cm and a height of 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.**

**Answer:**

The radius of the cylindrical container

∴ \(r=\frac{12}{2} \mathrm{~cm}=6 \mathrm{~cm}\)

and height h= 15 cm

∴ Volume = πr^{2}h = 71 x 6 x 6 x 15

= 540 π cm^{3}

⇒ The total volume of ice cream = 540cm^{3}

The radius of cone = radius of hemisphere = R

⇒ \(R=\frac{6}{2}=3 \mathrm{~cm}\)

Height of cone, H = 12 cm

The volume of ice cream in one cone + hemisphere

= \(\frac{1}{3} \pi R^2 H+\frac{2}{3} \pi R^3=\frac{1}{3} \pi R^2(H+2 R)\)

= \(\frac{1}{3} \pi \times 3 \times 3 \times(12+2 \times 3)=54 \pi\)

Now, number of cones = \(\frac{\text { Total volume of ice cream }}{\text { Volume of ice cream in one cone }}\)

= \(\frac{540 \pi}{54 \pi}=10\)

**Question 6. How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm x 10 cm x 3.5 cm?**

**Solution:**

Let the number of silver coins = n

Radius of coin \(r=\frac{1.75}{2} \mathrm{~cm}=\frac{7}{8} \mathrm{~cm}\)

Height h = 2 mm = \(\frac{2}{10} \mathrm{~cm}=\frac{1}{5} \mathrm{~cm}\)

The volume of one coin = πr^{2}h

= \(\frac{22}{7} \times \frac{7}{8} \times \frac{7}{8} \times \frac{1}{5}=\frac{77}{160} \mathrm{~cm}^3\)

∴ Volume of n coins = \(\frac{77 n}{160} \mathrm{~cm}^3\)

Volume of cuboid = 5.5 x 10 x 3.5 = 192.5 cm^{3}

Now, the volume of n coins = volume of a cuboid

⇒ \(\frac{77 n}{160}=192.5 \Rightarrow n=\frac{192.5 \times 160}{77}\)

⇒ n = 400

∴ Number of silver coins = 400

**Question 7. A cylinder bucket, 32 cm high and with radius of base of 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.**

**Solution:**

For cylindrical buckets,

Radius r = 18 cm

Height h = 32 cm

∴ The volume of sand = volume of the bucket

= πr^{2}h = π x 18 x 18 x 32 cm^{3}

= 10368π cm^{3}

For conical heap,

Let Raidus = R

Height H = 24 cm

∴ Volume of conical heap = \(\frac{1}{3} \pi R^2 H=\frac{1}{3} \pi R^2 \times 24=8 \pi R^2\)

Now, the volume of conical heap = volume of sand

⇒ 8πR^{2} = 10368π ⇒ R^{2} = 1296

⇒ R = 36 cm

∴ l^{2} = H^{2} + R^{2} = 24^{2} + 36^{2}

= 576 + 1296 = 1872

⇒ \(l=\sqrt{1872}=12 \sqrt{13} \mathrm{~cm}\)

**Question 8. Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?**

**Solution:**

Speed of water in canal = 10 km/h

= \(\frac{10 \times 1000}{60} \mathrm{~m} / \mathrm{min}\)

= \(\frac{500}{3} \mathrm{~m} / \mathrm{min}\)

Width of canal = 6 m and depth = 1.5 m

Now, the canal will transfer the water equal to the volume of a cuboid of dimensions.

⇒ \(6 \mathrm{~m} \times 1.5 \mathrm{~m} \times \frac{500}{3} m\) in 1 minute.

∴ Volume of water transfer in 30 minutes

= \(30 \times 6 \times 1.5 \times \frac{500}{3}=45000 \mathrm{~m}^3\)

If the depth of the irrigating region = 8 cm

= \(\frac{8}{100} \mathrm{~m}\) then

area x depth = 45000

⇒ \({Area} \times \frac{8}{100}=45000\)

⇒ \(\text { Area }=\frac{45000 \times 100}{8}=562500 \mathrm{~m}^2\)

Therefore, the area of the region irrigated by the canal in 30 minutes = 562500 m^{2}

**Question 9. A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?**

**Solution:**

Diameter of tank 2r = 10 m

⇒ r = 5 m

and depth h = 2m

∴ Volume of tank = r2h = (5)2 x 2 = 50 m3

Again, diameter of pipe = 2R = 20 cm

⇒ \(R=10 \mathrm{~cm}=\frac{10}{100} \mathrm{~m}=\frac{1}{10} \mathrm{~m}\)

Speed of water in pipe = 3 km/h

= \(\frac{3 \times 1000}{60} \mathrm{~m} / \mathrm{min}\)

= 50m/min

Now the pipe will transfer the water into the tank in 1 minute equal to the volume of a cylinder of radius \(\frac{1}{10}\) m and length 50 m.

∴ Time taken to fill the tank = \(\frac{\text { volume of cylindrical tank }}{\text { volume of water transfer in tank in 1 minute }}\)

= \(\frac{50 \pi}{\pi \times\left(\frac{1}{10}\right)^2 \times 50}=100 \text { minutes }\)

∴ Time taken to fill the tank completely = 100 minutes

## Volume And Surface Area Of Solids Exercise 13.4

**Question 1. A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.**

**Solution:**

The diameters of the frustum of the cone are 4 cm and 2 cm.

∴ Radius r_{1} = 2 cm and r_{2} = 1 cm

∴ Height of glass h = 14 cm

∴ The volume of a glass of the shape of a frustum of a cone

= \(\frac{1}{3} \pi h\left(r_1^2+r_1 r_2+r_2^2\right)\)

= \(\frac{1}{3} \times \frac{22}{7} \times 14\left[(2)^2+2 \times 1+(1)^2\right]\)

= \(\frac{44}{3}[4+2+1]\)

= \(\frac{44 \times 7}{3}=\frac{308}{3} \mathrm{~cm}^3\)

= \(=102 \frac{2}{3} \mathrm{~cm}^3\)

So, capacity of glass = = \(102 \frac{2}{3} \mathrm{~cm}^3\)

**Question 2. The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.**

**Solution :**

The slant height of the frustum of cone l = 4 cm.

Circumference of one end 2πr_{1} = 18 cm

∴ πr_{1} = 9 cm

Circumference of other ends 2πr_{2} = 6 cm

∴ πr_{2} = 3 cm

Curved surface area of frustum = π(r_{1} + r_{2}) l

= (πr_{1} + πr_{2}) l

= (9 + 3) x 4

= 48 cm^{2}

Therefore, the curved surface area of the frustum of the cone = 48 cm^{2}.

**Question 3. A fez, tire cap used by the Turks, is shaped like the frustum of a cone. If its radius on the open side is 10 cm, its radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it.**

**Solution:**

The cap is in the form of a frustum of a cone whose slant height is l = 15 cm.

Radius r_{1} = 10 cm and radius r_{2}= 4 cm

∴ Curved surface of cap = π(r_{1} + r_{2}) l

= \(\frac{22}{7}(10+4) \times 15=660 \mathrm{~cm}^2\)

Area of the closed end of the cap

= \(\pi r_2^2=\frac{22}{7} \times(4)^2 \mathrm{~cm}^2\)

= \(\frac{352}{7} \mathrm{~cm}^2=50 \frac{2}{7} \mathrm{~cm}^2\)

∴ Total canvas used in cap = Curved surface of cap + area of closed-end

= \(\left(660+50 \frac{2}{7}\right) \mathrm{cm}^2\)

= \(710 \frac{2}{7} \mathrm{~cm}^2\)

Therefore, the area of material used for making cap

∴ \(710 \frac{2}{7} \mathrm{~cm}^2\)

**Question 4. A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 1 6 cm with radii of its lower arid upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of ₹ 20 per litre. Also find the cost of the metal sheet used to make the container, if it costs ₹ 8 per 100 cm ^{2}. (Take π = 3.14)**

**Solution:**

The vessel is in the shape of a frustum of a cone whose height is h = 16 cm.

And radius of upper end r_{1} = 20 cm and radius of lower end r_{2} = 8 cm

Then, the volume of the vessel = volume of the frustum

= \(\frac{1}{3} \pi h\left(r_1^2+r_1 r_2+r_2^2\right)\)

= \(\frac{1}{3} \pi \times 16 \times\left[(20)^2+20 \times 8+(8)^2\right]\)

= \(\frac{16}{3} \pi \times 624 \mathrm{~cm}^3=3328 \pi \mathrm{cm}^3\)

= 3328 x 3.14 cm^{3} = 10449.92 cm^{3} [∴ π = 3.14]

The milk required to fill the vessel is 10449.92 cm^{3} or 10.450 litre.

Then, cost of milk at ₹ 20 per litre = 20 x ₹ 10.45 = ₹ 209

The sheet will be used to make the curved surface and base of the vessel.

Then, the area of the base of the vessel = nr_{2}^{2}

= 3.14 x (8)^{2} = 3.14 x 64 = 200.96 cm^{2}

The slant height of the vessel

⇒ \(l=\sqrt{h^2+\left(r_1-r_2\right)^2}=\sqrt{(16)^2+(20-8)^2}\)

= \(\sqrt{256+144}=\sqrt{400}=20 \mathrm{~cm}\)

Then, curved surface of vessel = π (r_{1} + r_{2}) l

= 3.14 (20 + 8) x 20 cm^{2}

= 3.14 x 28 x 20 cm^{2}

= 1758.4 cm^{2}

∴ Area of sheet used in vessel

= (1758.4 + 200.96) cm^{2}

= 1959.36 cm^{2}

Cost of the sheet at the rate of ₹ 8 per 1 00 cm^{2}

= \(₹ \frac{8}{100} \times ₹ 1959.36=₹ 156.7488\)

= ₹ 156.75

∴ Cost of milk = ₹ 209

and cost of sheet = ₹ 156.75

**Question 5. A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter \({1}{16}\) cm, find the length of the wire.**

**Solution:**

In the given, the diameter of the base of the cone is A’OA and the vertex is V. The angle of the vertex is A’VA = 60°, and then the semi-vertical angle of the cone is α = 30°.

Height of cone = 20 cm

Then, in right ΔOAV,

⇒ \(\tan \alpha=\frac{O A}{O V} \Rightarrow \tan 30^{\circ}=\frac{r_1}{20} \Rightarrow \frac{1}{\sqrt{3}}=\frac{r_1}{20}\)

⇒ \(r_1=\frac{20}{\sqrt{3}} \mathrm{~cm}\)

∵ ΔVO’B and ΔVOA are similar.

∴ \(\frac{V O^{\prime}}{V O}=\frac{O^{\prime} B}{O A} \Rightarrow \frac{10}{20}=\frac{r_2}{r_1} \Rightarrow \frac{r_2}{r_1}=\frac{1}{2}\)

⇒ \(r_1=2 r_2 \Rightarrow r_2=\frac{1}{2} r_1=\frac{1}{2} \times \frac{20}{\sqrt{3}}=\frac{10}{\sqrt{3}} \mathrm{~cm}\)

and height of frustum \(\dot{h}=\frac{1}{2}\) x height of cone

= 10 cm

Then, volume of frustum = \(\frac{1}{3} \pi h\left(r_1^2+r_1 r_2+r_2^2\right)\)

⇒ \(\frac{1}{3} \pi(10)\left[\left(\frac{20}{\sqrt{3}}\right)^2+\frac{20}{\sqrt{3}} \times \frac{10}{\sqrt{3}}+\left(\frac{10}{\sqrt{3}}\right)^2\right]\)

⇒ \(\frac{1}{3} \pi 10\left[\frac{400}{3}+\frac{200}{3}+\frac{100}{3}\right]\)

⇒ \(\frac{7000}{9} \pi \mathrm{cm}^3\)

Diameter of cylindrical wire = \(\frac{1}{16}\) cm

Radius of wire \(r=\frac{1}{32} \mathrm{~cm}\) cm

Let the length of the wire drawn be l cm.

Then, the volume of wire = πr^{2}l

⇒ \(\pi \times \frac{1}{32} \times \frac{1}{32} \times l=\frac{\pi}{1024} l \mathrm{~cm}^3\)

∵ The wire is drawn from the frustum of the cone.

∴ The volume of wire = volume of the frustum

⇒ \(\frac{\pi}{1024} l=\frac{7000}{9} \pi\)

⇒ \(l=\frac{7000 \pi}{9} \times \frac{1024}{\pi} \mathrm{cm}^3\)

= \(\frac{70}{9} \times 1024 \mathrm{~m}\)

= \(\frac{71680}{9} \mathrm{~m}=7964.44 \mathrm{~m}\)

Therefore, the length of the wire = 7964.44 m

## Volume and Surface Area of Solids Exercise 13.5

**Question 1. A copper wire, 3 mm in diameter, is wound about a cylinder whose length is 12 cm, and diameter 10 cm, to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per cm3.**

**Solution:**

The diameter of the cylinder = 10 cm and

height of cylinder = 12 cm

∴ Circumference of cylinder

= π x diameter = π x 10 = 10π cm

∴ Length of wire used in one round about a cylinder

= 10 cm

∵ The length of the cylinder is 12 cm or 120 mm. When one round of wire is wound on the cylinder then it covers 3mm length of the cylinder.

When two rounds of wire are wound on the cylinder then it covers the (2 x 3) mm length of the cylinder.

When three rounds of wire are wound on the cylinder then it covers (3×3) mm length of the cylinder.

When four rounds of wire are wound on the cylinder then it covers the (4×3) mm length of the cylinder.

The number of wounds of wire to cover the cylinder = \(\frac{120}{3} = 40\)

Length of required wire to wound 40 rounds on cylinder

= 40 x 10 π = 400 π cm

= 400 x 3.14 cm = 1256 cm(approx.)

= 12.56 m

So, the required length of wire = 12.56 m

The volume of wire = length x area of the wire

= \(1256 \times \pi \frac{d^2}{4} \quad\left[d=3 \mathrm{~mm}=\frac{3}{10} \mathrm{~cm}\right]\)

= \(1256 \times 3.14 \times \frac{9}{100 \times 4}\)

= \(\frac{314 \times 3.14 \times 9}{100}=88.74 \mathrm{~cm}^3\)

and mass of wire = 88.74 x 8.88 g

= 788.01 g = 0.788 kg

**Question 2. A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose the value of π as found appropriate.)**

**Solution:**

In right ΔABC, ∠B = 90°, AB = 4 cm,

BC = 3 cm

Then, area of \(\triangle A B C=\frac{3 \times 4}{2}=6 \mathrm{~cm}^2\)

Hypotenuse \(A C=\sqrt{A B^2+B C^2}\)

= \(\sqrt{(4)^2+(3)^2}=\sqrt{25}=5\)

BOB’ is perpendicular to AC, if BO = r, then area of \(\triangle A B C=\frac{A C \times B O}{2}=\frac{5}{2} B O=\frac{5}{2} r\)

Then, \(\triangle A B C=\frac{A C \times B O}{2}=\frac{5}{2} B O=\frac{5}{2} r\)

Then, \(\frac{5}{2} r=6\) (∵\(\frac{5}{2} r\)and 6 both are area of AABC)

∴ \(r=\frac{6 \times 2}{5}=2.4 \mathrm{~cm}\)

Now, the radius r = 2.4 cm of the double cone formed by rotating the right ΔABC.

Then, volume of double cone (two cones) = volume of cone (A, BB’) + volume of cone (C, BB’)

= \(\frac{1}{3} \pi r^2(A O)+\frac{1}{3} \pi r^2(O C)\)

= \(\frac{1}{3} \pi r^2(A O+O C)\)

= \(\frac{1}{3} \pi r^2(A C)\) [ AO + OC = AC]

= \(\frac{1}{3} \pi \times(2.4)^2 \times 5=9.6 \pi \mathrm{cm}^3\)

= 9.6 x 3.14 cm3 (π = 3.14)

= 30.144 cm^{3}

and surface area of the double cone (both cones)

= curved surface of the cone (A, BB’) + curved surface of the cone (C, BB’)

= πr (AB) + πr (BC) = nr (AB +BC)

= 3.14 x 2.4 x (4 +3) =3.14 x 2.4 x 7

= 52.75 cm^{2}

Therefore, the volume of the double cone = 30.144 cm^{3} and surface area = 52.75 .cm^{2} (approximately).

**Question 3. A cistern, internally measuring 150cm x 120cm x 110cm, has 129600 cm ^{3} of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water. How many bricks can be put in without overflowing the water, each brick being 22.5 cm x 7.5 cm x 6.5 cm?**

**Solution:**

Volume of cistern

= 150 x 120 x 110 cm^{3}

= 1980000 cm^{3}

The volume of water filled in cistern = 129600 cm^{3}

Volume of each brick = 22.5 x 7.5 x 6.5 cm^{3}

= 1096.875 cm^{3}

Let on placing x bricks, the water rises upto the brim in the cistern.

Then, volume of x bricks = 1096.875 x cm^{3}

Then, volume of absorbs water by bricks = \(1096.875 x \times \frac{1}{17}=\frac{1096.875 x}{17} \mathrm{~cm}^3\)

Then, the volume of remaining water in cistern = \(\left(129600-\frac{1096.875 x}{17}\right) \mathrm{cm}^3\)

Now, volume of x bricks + volume of water in cistern = volume of cistern

∴ \(1096.875 x+129600-\frac{1096.875 x}{17}\) = 1980000

or \(1096.875 x-\frac{1096.875 x}{17}\) = 1980000 – 129600

or \(1096.875 x\left(1-\frac{1}{17}\right)=1850400\)

or \(1096.875 x=\frac{1850400 \times 17}{16}\)

or \(x=\frac{1850400 \times 17}{16 \times 1096.875}\)

= 1792.4 = 1792 (approximately)

Therefore, the number of bricks placed in the cistern is 1792 (approximately).

**Question 4. In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is 7280 km ^{2}, show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each 1072 km long, 75 m wide and 3 m deep.**

**Solution:**

The volume of each river

= 10721cm x 75 m x 3 m

= 1072 x 75 x 3 x 1000 m^{3}

= 241200000 m^{3}

∴ The volume of total water in three rivers

= 3 x 241200000 m^{3}

∴ Total water of rivers = 723600000 m^{3}

∴ Area of valley = 7280 km^{2}

= 7280 x (1000)^{2} m^{2}

= 7280000000 m

∴ Volume of rainwater

= \(7280000000 \times \frac{10}{100} \mathrm{~m}^3\) (∵\(10 \mathrm{~cm}=\frac{10}{100} \mathrm{~m}\))

= 728000000 m^{3}

These two volumes are not equal.

So, it is clear that the given data given in the question are incorrect.

**Question 5. An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, the diameter of the cylindrical portion is 8 cm and the diameter portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel.**

**Solution:**

Height of cylindrical part h = 10 cm

Total height of funnel = 22 cm

∴ Height of frustum of cone (H) = 22 – 10 = 12 cm

Upper radius of frustum of cone = \(R_1=\frac{18}{2}=9 \mathrm{~cm}\)

Lower radius of frustum of cone = \(R_2=\frac{8}{2}=4 \mathrm{~cm}\)

The radius of cylindrical part r = 4 cm

The curved surface of the cylindrical part = 2 πrh

= 2π x 4 x 10 = 80 π cm^{2}

The slant height of the frustum of a cone

⇒ \(l=\sqrt{H^2+\left(R_1-R_2\right)^2}\)

= \(\sqrt{(12)^2+(9-4)^2}=\sqrt{144+25}\)

= \(\sqrt{169}=13 \mathrm{~cm}\)

So, the total surface area of the funnel

∴ The curved surface of the frustum of a cone

= π (R_{1}+ R_{2}) l

= π (9 + 4) x 13 = 169 π cm^{2}

∴ The curved surface area of the cylindrical part + curved surface area of the frustum of a cone

= 80 π + 169π = 249π cm^{2}

= \(249 \times \frac{22}{7} \mathrm{~cm}^2\)

= \(\frac{5478}{7}=782 \frac{4}{7} \mathrm{~cm}^2\)

Therefore, area of tin sheet used in funnel = \(782 \frac{4}{7} \mathrm{~cm}^2\)

**Question 6. Derive the formula for the curved surface area and total surface area of the frustum of a cone.**

**Solution:**

Let for the (V, AB), V is the vertex, r_{2} the base radius and l_{2} the slant height. A cone (V, CD) is cut off from this cone from a point O’ below h_{1} from the vertex V of this cone, parallel to the base whose’ radius is r_{1} and slant height is l_{1}.

Draw the perpendicular DE from point D to the base.

In ΔVO’D and ΔDEB,

∠VO’D = ∠DEB (VO and DE both are perpendicular to the base)

∠VDO’ = ∠DBE (the bases of two cones are parallel to each other)

∴ ΔVO’D and ΔDEB are similar

\(\frac{V D}{B D}=\frac{O^{\prime} D}{E B}\)or \(\frac{l_1}{l}=\frac{O^{\prime} D}{O B-O E}=\frac{O^{\prime} D}{O B-O^{\prime} D}\)

while BD = l = slant height of the frustum

⇒ \(\frac{l_1}{l}=\frac{r_1}{r_2-r_1} \Rightarrow l_1=\left(\frac{r_1}{r_2-r_1}\right) l\) → (1)

The curved surface area of a frustum

= curved surface area of cone (V, AB) – curved surface area of cone ( V, CD)

= πr_{2}l_{2} -πr_{1}l_{1 }= πr_{2}(l_{1} + BD) – πr_{1}l_{1}

= πr_{2}l_{1} +πr_{2} (BD) – πr_{1}l_{1}

= π(r_{2}-r_{1}) l_{1} + πr2 l

= \(\pi\left(r_2-r_1\right)\left(\frac{r_1}{r_2-r_1}\right) l+\pi r_2 l\) [from eqn. (1)]

= π r_{1}l + πr_{2}l

So, curved surface area of frustum = π (r_{1} + r_{2}) l

Hence Proved

And total surface area of a frustum

= curved surface + area of first end + area of second end

= π(r_{1} + r_{2}) / + πr_{1}^{2} + πr_{2}^{2}

= π(r_{1} + r_{2})l + π (r_{12} + r_{22})

Hence Proved.

**Question 7. Derive the formula for the volume of the frustum of a cone.**

**Solution:**

From the last question, for the cone (V, AB), height = h_{2} and radius = r_{2}

∴ Volume of cone (V, AB) = \(\frac{1}{3} \pi r_2^2 h_2\)

and volume of cone (V, CD) = \(\frac{1}{3} \pi r_1^2 h_1\)

∴ Volume of frustum = volume of cone (V, AB) – volume of cone (V, CD)

∴ Volume of frustum (V)

= \(\frac{1}{3} \pi r_2^2 h_2-\frac{1}{3} \pi r_1^2 h_1\) → (1)

∴ h_{2} = VO’= VO’ + O’O = h_{1} + h

∴ Put h_{2} = h_{1} +h in eqn. (1),

Volume of frustum V = \(\frac{1}{3} \pi r_2^2\left(h_1+h\right)-\frac{1}{3} \pi r_1^2 h_1\)

Volume of frustum V = \(\frac{1}{3} \pi\left(r_2^2-r_1^2\right) h_1+\frac{1}{3} \pi r_2^2 h\) → (2)

In similar ΔVO’D and ΔDEB,

\(\frac{h_1}{h}=\frac{r_1}{r_2-r_1} \quad \Rightarrow \quad h_1=\left(\frac{r_1}{r_2-r_1}\right) h\)Put \(h_1=\left(\frac{r_1}{r_2-r_1}\right) h\) in eqn. (2),

⇒ \(V=\frac{1}{3} \pi\left(r_2^2-r_1^2\right) \frac{\eta_1}{\left(r_2-r_1\right)} h+\frac{1}{3} \pi r_2^2 h\)

= \(\frac{1}{3} \pi\left(r_2+r_1\right) r_1 h+\frac{1}{3} \pi r_2^2 h\)

= \(\frac{1}{3} \pi\left(r_1^2+r_1 r_2\right) h+\frac{1}{3} \pi r_2^2 h\)

= \(\frac{1}{3} \pi\left(r_1^2+r_1 r_2+r_2^2\right) h\)

Therefore, the volume of the frustum of the cone

= \(\frac{1}{3} \pi\left(r_1^2+r_1 r_2+r_2^2\right) h\) Hence Proved.

## Volume And Surface Area Of Solids Multiple Choice Questions

**Question 1. A surah is a combination of:**

- A sphere and a cylinder
- A hemisphere and a cylinder
- Two hemispheres
- A cylinder and a cone

**Answer:** 1. A sphere and a cylinder

**Question 2. A glass is generally of the shape of:**

- A cone
- A frustum of a cone
- A cylinder
- A sphere

**Answer:** 2. A frustum of a cone

**Question 3. A plummet is a combination of:**

- A cone and a cylinder
- A hemisphere and a cone
- A frustum of a cone and a cylinder
- A sphere and a cylinder

**Answer:** 2. A hemisphere and a cone

**Question 4. An iron piece in the shape of a cuboid of dimensions 49 cm x 33 cm x 24 cm is melted and recast into a solid sphere. The radius of the sphere is:**

- 21 cm
- 23 cm
- 25 cm
- 19 cm

**Answer:** 1. 21 cm

**Question 5. While converting a shape of a solid into another shape, the volume of the new shape:**

- Increases
- Decreases
- Remains same
- Becomes twice.

**Answer:** 3. Remains same

**Question 6. The ratio of the surface of two spheres is 16:9. The ratio of their volumes is:**

- 3:4
- 64: 27
- 27: 64
- 4:3

**Answer:** 2. 64: 27

**Question 7. The diameter of a sphere exactly inscribed in a right circular cylinder of radius r cm and height h cm (h > 2r) is:**

- r cm
- 2r cm
- h cm
- 2h cm

**Answer:** 2. 2r cm