# SSC > Volume and Surface Area

Explore popular questions from Volume and Surface Area for SSC. This collection covers Volume and Surface Area previous year SSC questions hand picked by experienced teachers.

General Intelligence and Reasoning
General Awareness
Quantitative Aptitude
English Comprehension
Q 1.

Correct2

Incorrect-0.5

If the total surface area of a cube is 6 sq units, then what is the volume of the cube?

1 cu unit

B

2 cu units

C

4 cu units

D

6 cu units

##### Explanation

Total surface area of a cube = 6 {tex} a^{2} {/tex}

=> 6 = 6 {tex} a^{2} {/tex} = {tex} a^{2} {/tex} = 1

Therefore, a = 1

Now, volume of the cube = {tex} a^{3} {/tex} = {tex} 1^{3} {/tex} = 1 cu unit

Q 2.

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The surface area of a cube is 726 sq cm. Find the volume of the cube.

1331 {tex} cm^{3} {/tex}

B

1232 {tex} cm^{3} {/tex}

C

1626 {tex} cm^{3} {/tex}

D

1836 {tex} cm^{3} {/tex}

##### Explanation

According to the question,

{tex} 6a^{2} = 726 {/tex} [a = edge of the cube]

=>{tex} a^{2} = \frac{726}{6} = 121 {/tex}

Therefore, {tex} a = \sqrt{121} = 11 cm {/tex}

Therefore, Required volume =

{tex} a^{3} = 11^{3} = 1331 cm^{3} {/tex}

Q 3.

Correct2

Incorrect-0.5

If each side of a cube is decreased by 19%, then decrease in surface area is

A

40%

B

38.40%

C

35%

34.39%

##### Explanation

Here, x = y = -19%

According to the formula,

Percentage decrease in surface area

= {tex} \left[ x + y + \frac{xy}{100} \right] {/tex}%

= {tex} \left[ -19 - 19 = \frac{ \left(-19\right) \times \left(-19\right) }{100} \right] {/tex}%

= {tex} \left[ -38 + \frac{361}{100} \right] {/tex}%

= {tex} \left[-38 + 3.61 \right] \% {/tex} = - 34.39%

Q 4.

Correct2

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If the lateral surface area of a cylinder is 94.2 sq cm and its height is 5 cm, then find the radius of its base. (n = 3.14)

A

5 cm

B

8 cm

3 cm

D

4cm

##### Explanation

Given, lateral surface area = 94.2 sq. cm.

=> {tex} 2 \pi rh = 94.2 {/tex}

Therefore, {tex} r = \frac{94.2}{2 \times 3.14 \times 5} = 3 cm {/tex}

Q 5.

Correct2

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A rod of 2 cm diameter and 30 cm length is converted into a wire of 3 in length of uniform thickness. The diameter of the wire is

A

{tex} \frac{2}{10} {/tex} cm

{tex} \frac{2}{\sqrt{10}} {/tex} cm

C

{tex} \frac{1}{\sqrt{10}} {/tex} cm

D

{tex} \frac{1}{10} {/tex} cm

##### Explanation

Given,

{tex} r_{1} {/tex} = 1 cm, {tex} h_{1} {/tex} = 30 cm, {tex} h_{2} {/tex} = 300 cm

Volume of rod = Volume of wire

=> {tex} \pi r_{1}^{2}h_{1} = \pi r_{2}^{2}h_{2} {/tex}

=> {tex} \pi \times \left(1\right)^{2} \times 30 = \pi \times r_{2}^{2} \times 300 {/tex}

=> {tex} r_{2}^{2} = \frac{30}{300} {/tex}

Therefore, {tex} r_{2} = \frac{1}{\sqrt{10}}cm {/tex}

Therefore, diameter = 2{tex} r_{2} {/tex}

{tex} \frac{2}{\sqrt{10}} cm {/tex}

Q 6.

Correct2

Incorrect-0.5

What is the whole surface area of a cone of base radius 7 cm and height 24 cm?

A

654 sq cm

704 sq cm

C

724 sq cm

D

964 sq cm

##### Explanation

Slant height, {tex} l = \sqrt{h^{2} + r^{2}} {/tex}

= {tex} \sqrt{ \left(24\right)^{2} + \left(7\right)^{2} } = \sqrt{576 + 49} {/tex}

{tex} \sqrt{625} = 25 {/tex}

Total surface area = {tex} \pi r \left(l + r\right) {/tex}

= {tex} \frac{22}{7} \times 7 \left(25 + 7\right) {/tex}

= {tex} \frac{22}{7} \times 7 \times 32 = 704 sq.cm. {/tex}

Q 7.

Correct2

Incorrect-0.5

The volume of a right circular cone is 100 {tex} \pi cm^{3} {/tex} and its height is 12 cm. Find its slant height.

13 cm

B

16 cm

C

9 cm

D

26 cm

##### Explanation

Volume = {tex} \frac{1}{3} \pi r^{2}h {/tex}

According to the question,

{tex} \frac{1}{3} \pi r^{2}h = 100 \pi => \frac{1}{3} \pi r^{2} \times 12 = 100 \pi {/tex}

=> {tex} r^{2} = 25 {/tex}

Therefore, r = {tex} \sqrt{25} {/tex} = 5 cm

Therefore, Slant height (l) = {tex} \sqrt{h^{2}+r^{2}} = \sqrt{12^{2}+5^{2}} {/tex}

= {tex} \sqrt{169} {/tex} = 13 cm

Q 8.

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The radius of the base of a right circular cone is doubled. To keep the volume fixed, the height of the cone will be

A

half of the previous height

B

one-third of the previous height

one-fourth of the previous height

D

{tex} \frac{1}{\sqrt{2}} {/tex} times of the previous height

##### Explanation

In first situation,

Radius = {tex} r_{1} {/tex}, height = {tex} h_{1} {/tex} and volume = {tex} v_{1} {/tex}

In second situation,

Radius = 2 {tex} r_{1} {/tex}, height = {tex} h_{2} {/tex} and volume = {tex} v_{2} {/tex}

In the volume of fixed, then

{tex} v_{1} {/tex} = {tex} v_{2} {/tex}

{tex} \frac{1}{3} \pi r_{1}^{2}h_{1} = \frac{1}{3} \pi \left(2r_{1}\right)^{2}h_{2} {/tex}

{tex}h_{1} {/tex} = 4 {tex} h_{2} {/tex}

{tex} h_{2} = \frac{h_{1}}{4} {/tex}

Therefore, height of the cone will be one-fourth of the previous height.

Q 9.

Correct2

Incorrect-0.5

If the volumes of two right circular cones are in the ratio 1 : 3 and their diameters are in the ratio 3 : 5, then the ratio of their heights is

25 : 27

B

1 : 5

C

3 : 5

D

5 : 27

##### Explanation

Let diameter, radius and height of first cone are {tex} d_{1} {/tex}, {tex} r_{1} {/tex} and {tex} h_{1} {/tex} respectively and that of second cone are {tex} d_{2} {/tex}, {tex} r_{2} {/tex} and {tex} h_{2} {/tex} respectively.

{tex} \frac{r_{1}}{r_{2}} = \frac{d_{1}}{d_{2}} = \frac{3}{5}. \frac{h_{1}}{h_{2}} = ? {/tex}

Given, {tex} \frac{\frac{1}{3}\pi r_{1}^{2}h_{1}}{\frac{1}{3} \pi r_{2}^{2}h_{2}} = \frac{1}{3} => \left(\frac{r_{1}}{r_{2}}\right)^{2} \times \frac{h_{1}}{h_{2}} = \frac{1}{3} {/tex}

=> {tex} \left(\frac{3}{5}\right)^{2} \times \frac{h_{1}}{h_{2}} = \frac{1}{3} => \frac{h_{1}}{h_{2}} = \frac{1}{3} \times \frac{25}{9} = \frac{25}{27} {/tex}

Q 10.

Correct2

Incorrect-0.5

The diameters of two cones are equal, If their slant heights be in the ratio of 5 : 7 then find the ratio of their Curved surface areas.

A

25 : 7

B

25 : 49

C

5 : 49

5 : 7

##### Explanation

Given,

{tex} \frac{l_{1}}{l_{2}} = \frac{5}{7} {/tex}

Now, curved surface area of first cone

= {tex} \pi rl_{1} {/tex}

and curved surface area of second cone

= {tex} \pi rl_{2} {/tex}

Therefore, Ratio = {tex} \frac{ \pi rl_{1}}{ \pi rl_{2}} = \frac{l_{1}}{l_{2}} = 5 : 7 {/tex}

Q 11.

Correct2

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The curved surface area and the total surface area of a cylinder are in the ratio 1 : 2. If the total surface area of the right cylinder is 616 {tex} cm^{2} {/tex}, then its volume is

1632 {tex} cm^{3} {/tex}

B

1078 {tex} cm^{3} {/tex}

C

1232 {tex} cm^{3} {/tex}

D

1848 {tex} cm^{3} {/tex}

##### Explanation

Curved surface area of cylinder

= {tex} 2 \pi rh cm^{2} {/tex}

Total surface area of cylinder

= {tex} 2 \pi r \left(h + r\right) cm^{2} {/tex}

Now,

According to the question,

{tex} \frac{2 \pi rh}{2 \pi r \left(h + r\right) } = \frac{1}{2} {/tex}

=> {tex} \frac{h}{h+r} = \frac{1}{2} {/tex}

=> 2h = h + r

=> h = r

Now,

Total surface area = 616 {tex} cm^{2} {/tex}

=> {tex} 2 \pi r \left(h + r\right) = 616 {/tex}

=> {tex} 2 \pi r \left(r + r\right) = 616 {/tex}

{tex} \left(Because h = r \right) {/tex}

=> {tex} 2 \pi r \left(2r\right) = 616 {/tex}

=> {tex} 4 \pi r^{2} = 616{/tex}

=> {tex} r = \sqrt{\frac{616}{4 \pi }} = \sqrt{\frac{616 \times 7}{4 \times 22}} {/tex}

= {tex} \sqrt{7 \times 7} = 7 {/tex}

Therefore, r = h = 7 cm

Therefore, Volume = {tex} \pi r^{2}h {/tex}

= {tex} \frac{22}{7} \times \left(7 \times 7\right) \times 7 {/tex}

= 1078 {tex} cm^{3} {/tex}

Q 12.

Correct2

Incorrect-0.5

What is the volume of the frustum?

A

{tex} 3H \left(P+Q+\sqrt{PQ}\right) {/tex}

B

{tex} H \left(P+Q+\sqrt{PQ}\right) {/tex}

{tex} H \frac{ \left(P+Q+\sqrt{PQ}\right) }{3} {/tex}

D

{tex} H \frac{ \left(P+Q-\sqrt{PQ}\right) }{3} {/tex}

##### Explanation

We know that,

Volume of frustum

= {tex} \frac{\pi H}{3} \left(R^{2} + r^{2} + Rr\right) {/tex}

= {tex} \frac{ \pi }{3}H \{ \left(\sqrt{\frac{Q}{ \pi }}\right)^{2} + \left(\sqrt{\frac{P}{ \pi }}\right)^{2} + \sqrt{\frac{Q}{ \pi }} - \sqrt{\frac{P}{ \pi }} \} {/tex}

= {tex} \frac{ \pi H}{3} \{ \frac{Q}{ \pi } + \frac{P}{ \pi } + \frac{\sqrt{PQ}}{ \pi } \} {/tex}

= {tex} \frac{H}{3} \left(P + Q + \sqrt{PQ}\right) {/tex}

Q 13.

Correct2

Incorrect-0.5

The radius of the base of a right circular cone is increased by 15% keeping the height fixed. The volume of the cone will be increased by

30%

B

31%

C

32.25%

D

34.75%

##### Explanation

Let the fixed height of a right circular cone is h and initial radius is r.

Then, initial volume of cone, {tex} V_{1} = \frac{1}{3} \pi r^{2}h {/tex}

After increasing 15% radius of a cone

= {tex} \left(r + \frac{3r}{20}\right) = \frac{23}{20}r {/tex}

New volume becomes, {tex} V_{2} = \frac{1}{3} \pi \left(\frac{23}{20}\right)^{2}r^{2}h {/tex}

Therefore, Increasing percentage = {tex} \left(\frac{V_{2} - V_{1}}{V_{1}}\right) \times 100 {/tex}

= {tex} \frac{\frac{1}{3} \pi r^{2}h}{\frac{1}{3} \pi r^{2}h}\{ \left(\frac{23}{20}\right) ^{2} - 1 \} \times 100 {/tex}

= {tex} \left(\frac{23}{20}+1\right) \left(\frac{23}{20}-1\right) \times 100 {/tex}

= {tex} \frac{43}{20} \times \frac{3}{20} \times 100 {/tex}

= 32.25%

Q 14.

Correct2

Incorrect-0.5

The base of a right prism is a right angled angled isosceles triangle whose hypotenuse is 'a' cm. If the height of the prism is 'h' cm, then its volume is

{tex} \frac{a^{2}h}{4} cm^{3} {/tex}

B

{tex} \frac{a^{2}h}{6} cm^{3} {/tex}

C

{tex} \frac{a^{2}h}{8} cm^{3} {/tex}

D

{tex} \frac{a^{2}h}{12} cm^{3} {/tex}

##### Explanation

Let AB = BC = x

In triangle ABC,

{tex} AB^{2} + BC^{2} = AC^{2} {/tex}

=> {tex} x^{2} + x^{2} = a^{2} {/tex}

=> {tex} 2x^{2} = a^{2} {/tex}

Therefore, {tex} x = \frac{a}{\sqrt{2}} {/tex}

Therefore, Volume of prism = Area of base X Height

= {tex} \frac{1}{2}.\frac{a}{\sqrt{2}}.\frac{a}{\sqrt{2}} \times h = \frac{a^{2}h}{4} cm^{3} {/tex}

Q 15.

Correct2

Incorrect-0.5

The base of a cone and a cylinder have the same radius 6 cm. They have also the same height 8 cm. The ratio of the curved surfaces of the cylinder to that of the cone is

A

4 : 3

B

5 : 3

8 : 5

D

8 : 3

##### Explanation

Ratio of curved surface area of cylinder and cone

= {tex} \frac{2 \pi rh}{ \pi r\sqrt{h^{2}+r^{2}}}=\frac{2 \times 6 \times 8}{6 \times \sqrt{6^{2}+8^{2}}} {/tex}

= {tex} \frac{96}{6 \times 10} = \frac{96}{60} = \frac{8}{5} {/tex} = 8 : 5

Q 16.

Correct2

Incorrect-0.5

A solid metallic cylinder of base radius 3 cm and height 5 cm is melted to form cones, each of height 1 cm and base radius 1 mm. The number of cones is

A

7500

13500

C

3500

D

4500

##### Explanation

Let required number of coins be n.

Then,

{tex} n \times \frac{1}{3} \times \pi \left(\frac{1}{10}\right)^{2} \times 1 = \pi \times \left(3\right)^{2} \times 5 {/tex}

Therefore, {tex} n = 9 \times 5 \times 3 \times 100 {/tex} = 13500

Q 17.

Correct2

Incorrect-0.5

The volumes of a sphere and a right circular cylinder having the same radius are equal. The ratio of the diameter of the sphere to the height of the cylinder is

A

1 : 2

B

2 : 1

C

2 : 3

3 : 2

##### Explanation

Let r = Radius of cylinder = Radius of sphere, h = Height of the cylinder.

According to the question,

{tex} \frac{4}{3} \pi r^{3} = \pi r^{2}h {/tex}

=> {tex} h = \frac{4}{3}r => 4r = 3h {/tex}

=> {tex} 2r = \frac{3}{2}h => \frac{2r}{h} = \frac{3}{2}{/tex}

Therefore, Required ratio = 3 : 2

Q 18.

Correct2

Incorrect-0.5

Let A be a pyramid on a square base and B be a cube. It a, b and c denote the number of edges, number of faces and number of corners, respectively. Then, the result a = b + c is true for

A

only A

B

only 3

C

both A and B

neither A nor 8

##### Explanation

For cube figure,

Edges, a = 12 :

Faces, b = 6

Corners, c = 8

For pyramid figure,

Edges, a =8

Faces, b = 5

Corners, c = 5

So, a = b + c is neither true for cube nor for the pyramid.

Q 19.

Correct2

Incorrect-0.5

A hemispherical basin of 150 cm diameter holds water 120 times as much as a cylindrical tube. If the height of the tube is 15 cm, then the diameter of the tube is

A

27 cm

B

24 cm

25 cm

D

26 cm

##### Explanation

Given, diameter = 150 cm

Therefore, {tex} r = \frac{150}{2} cm {/tex}

{tex} \frac{2}{3} \pi \left(\frac{150}{2}\right)^{3} = 120 \pi r^{2} \times 15 {/tex}

=> {tex} \frac{2}{3} \times \frac{150 \times 150 \times 150}{8} = 120 \times 15 \times r^{2} {/tex}

{tex} r^{2} = \frac{150 \times 150 \times 150}{12 \times 120 \times 5} {/tex}

{tex} r^{2} = \frac{625}{4} => r = \sqrt{\frac{625}{4}} = \frac{25}{2} {/tex}

Therefore, Diameter = {tex} 2r = 2 \times \frac{25}{2} = 25 cm {/tex}

Q 20.

Correct2

Incorrect-0.5

A right circular metal cone (solid) is 8 cm high and the radius is 2 cm. It is melted and recast into a sphere. What is the radius of the sphere?

2 cm

B

3 cm

C

4 cm

D

5 cm

##### Explanation

Given that, the height and radius of a right circular metal cone (solid) are 8 cm and 2 cm, respectively.

i.e., h = 8 cm and r = 2cm

Let the radius of the sphere is R.

Then, by condition,

{tex} \frac{1}{3} \pi r^{2}h = \frac{4}{3} \pi R^{3} {/tex}

=> {tex} 4 \times 8 = 4 R^{3} {/tex}

=> {tex} R^{3} = \left(2\right)^{3} {/tex}

=> R = 2

Therefore, Radius of the sphere = 2 cm

Q 21.

Correct2

Incorrect-0.5

Let the largest possible right circular cone and largest possible sphere be fitted into two cubes of same length. Let C and S denote the volume of cone and volume of sphere respectively, then which one of the following is correct?

A

C = 2S

S = 2C

C

C = S

D

C = 3S

##### Explanation

Let the side of both cube is 'a' then height and radius of a cone

{tex} r = \frac{a}{2} \ and \ h = a {/tex}

{tex} R = \frac{a}{2} {/tex}

Therefore, Volume of cone (C) = {tex} \frac{1}{3} \pi r^{2}h {/tex}

= {tex} \frac{1}{3} \pi \left(\frac{a}{2}\right)^{2} \left(a\right) = \frac{ \pi a^{3}}{12} {/tex}

and volume of sphere (S)

= {tex} \frac{4}{3} \pi R^{3} = \frac{4}{3} \pi \left(\frac{a}{2}\right)^{3} = \frac{ \pi a^{3}}{6} {/tex}

From Eqs. (i) and (ii), we get

S = 2C

Q 22.

Correct2

Incorrect-0.5

10 circular plates each of thickness 3 cm, each are placed one above the other and a hemisphere of radius 6 cm is placed on the top just to cover the cylindrical solid. What is the volume of the solid so formed?

A

264 {tex} \pi cm^{3} {/tex}

B

252 {tex} \pi cm^{3} {/tex}

C

236 {tex} \pi cm^{3} {/tex}

None of these

##### Explanation

If 10 Circular plates each of thickness 3 cm, each are placed one above the other, then it forms a cylinder with height {tex} \left(3 \times 10 = 30 cm\right) {/tex} and a hemisphere of radius 6 cm is placed on the top just to cover the cylinder that means its radius is 6 cm also. Therefore, Radius of hemisphere (R) = 6 cm

Radius of cylinder (r) = 6 cm

and height of cylinder (h) = 30 cm

Therefore, Volume of the solid = Volume of cylinder + Volume of hemisphere

= {tex} \pi r^{2}h + \frac{2}{3} \pi R^{3} {/tex}

= {tex} \pi \left(6\right)^{2} \times 30 + \frac{2}{3} \pi \left(6\right)^{3} {/tex}

= {tex} \pi \times 36 \times 30 + \frac{2}{3} \pi \times 216 {/tex}

= {tex} 1080 \pi + 2 \pi \times 72 {/tex}

= {tex} 1080 \pi + 144 \pi {/tex}

= {tex} 1224 \pi cm^{3} {/tex}

which is the required volume of solid.

Q 23.

Correct2

Incorrect-0.5

A cylindrical box of radius 5 cm contains 10 solid spherical balls, each of radius 5 cm. If the top most ball touches the upper cover of the box, then volume of the empty space in the box is

{tex} \frac{2500}{3} \pi cm^{3} {/tex}

B

{tex} 5000 \pi cm^{3} {/tex}

C

{tex} 2500 \pi cm^{3} {/tex}

D

{tex} \frac{5000}{3} \pi cm^{3} {/tex}

##### Explanation

Height of the cylinder

= {tex} 10 \times Diameter \ of \ each \ ball {/tex}

= {tex} 10 \times 10 = 100 cm {/tex}

Therefore, Required empty space

= {tex} \pi \left(5\right)^{2} \left(100\right) - \frac{4}{3} \pi \left(5\right)^{3} \times 10 {/tex}

= {tex} \pi \left(5\right)^{2} \times 10 \left[ 10 - \frac{20}{3} \right] {/tex}

= {tex} \pi \left(5\right)^{2} \times 10 \left[ \frac{30 - 20}{3} \right] = \frac{2500}{3} \pi cm^{3} {/tex}

Q 24.

Correct2

Incorrect-0.5

What part of a ditch 48 m long, 16.5 m broad and 4 m deep can be filled by the earth got by digging a cylindrical tunnel of diameter 4 m and length 56 m?

A

{tex} \frac{1}{9} {/tex}

{tex} \frac{2}{9} {/tex}

C

{tex} \frac{7}{9} {/tex}

D

{tex} \frac{8}{9} {/tex}

##### Explanation

Volume of the earth dugout as a tunnel

= {tex} \pi r^{2}h = \frac{22}{7} \times 2 \times 2 \times 56 = 704 m^{3} {/tex}

Volume of the ditch = {tex} 48 \times \frac{33}{2} \times 4 {/tex}

= 24 X 33 X 4 = 3168

Therefore, Part required = {tex} \frac{704}{3168} = \frac{2}{9} {/tex}

Q 25.

Correct2

Incorrect-0.5

A cylinder is surmounted by a cone at one end, a hemisphere at the other end. The common radius is 3.5 cm, the height of the cylinder is 6.5 cm and the total height of the structure is 12.8 cm. The volume V of the structure lies between

370 {tex} cm^{3} {/tex} and 380 {tex} cm^{3} {/tex}

B

380 {tex} cm^{3} {/tex} and 390 {tex} cm^{3} {/tex}

C

390 {tex} cm^{3} {/tex} and 400 {tex} cm^{3} {/tex}

D

None of the above

##### Explanation

Let common radius be r cm.

Then, height of cylinder = h = 6.5 cm

and height of cone = h' = {tex} \left(12.8 - 6.5 - 3.5\right) {/tex} = 2.8 cm

Therefore, Volume of the complete structure

= (Volume of cylinder) + (Volume of cone) + (Volume of the hemisphere)

= {tex} \frac{1}{3} \pi r h + \pi r h +- \pi r {/tex}

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

= {tex} \pi \left(3.5\right)^{2} \left(\frac{2.8}{3}+6.5+\frac{2}{3} \times 3.5\right) {/tex}

= {tex} \frac{22}{7} \times 3.5 \times 3.5 \times 9.76 {/tex}

= 375.76 {tex} cm^{2} {/tex}

Hence, Volume (V) of the structure lies between 370 {tex} cm^{3} {/tex} and 380 {tex} cm^{3} {/tex}.