Practice MCQ Questions and Answer on Volume and Surface Area

1.

The radius of a cylindrical cistern is 10 metres and its height is 15 metres. Initially the cistern is empty. We start filling the cistern with water through a pipe whose diameter is 50 cm. Water is coming out of the pipe with a velocity of 5 m/sec. How many minutes will it take in filling the cistern with water ?

• (A) 20 min
• (B) 40 min
• (C) 60 min
• (D) 80 min

Show Answer

 Volume of cistern : $$\eqalign{ & = \left( {\pi \times {{10}^2} \times 15} \right){m^3} \cr & = 1500\pi {m^3} \cr} $$ Volume of water flowing through the pipe in 1 sec : $$\eqalign{ & = \left( {\pi \times 0.25 \times 0.25 \times 5} \right){m^3} \cr & = 0.3125\pi {m^3} \cr} $$ ∴ Time taken to fill the cistern : $$\eqalign{ & = \left( {\frac{{1500\pi }}{{0.3125\pi }}} \right) \cr & = \left( {\frac{{1500 \times 10000}}{{3125}}} \right) \cr & = 4800\,\sec \cr & = \left( {\frac{{4800}}{{60}}} \right)\min \cr & = 80\,\min \cr} $$

2.

If the total length of diagonals of a cube is 12 cm, then what is the total length of the edges of the cube ?

• (A) 6$$\sqrt{3}$$ cm
• (B) 12 cm
• (C) 15 cm
• (D) 12$$\sqrt{3}$$ cm

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 Since a cube has 4 diagonals, we have : Length of a diagonal $$\eqalign{ & = \left( {\frac{{12}}{4}} \right)cm \cr & = 3\,cm \cr} $$ Let the length of each edge of the cube be a cm Then, $$\eqalign{ & \sqrt 3 a = 3 \cr & or,a = \sqrt 3 \cr} $$ ∴ Total length of the edges of the cube = $$12\sqrt 3\, $$ cm

3.

A spherical ball of lead, 3 cm in diameter is melted and recast into three spherical ball. The diameter of two of these are 1.5 cm and 2 cm respectively. The diameter of the third ball is :

• (A) 2.5 cm
• (B) 2.66 cm
• (C) 3 cm
• (D) 3.5 cm

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 Let the radius of the third ball be R cm Then, $$\frac{4}{3}\pi \times {\left( {\frac{3}{4}} \right)^3} + \frac{4}{3}\pi \times {\left( 1 \right)^3}$$   $$ + \frac{4}{3}\pi \times {R^3}$$   $$ = \frac{4}{3}\pi \times {\left( {\frac{3}{2}} \right)^3}$$ $$\eqalign{ & \Rightarrow \frac{{27}}{{64}} + 1 + {R^3} = \frac{{27}}{8} \cr & \Rightarrow {R^3} = \frac{{125}}{{64}} = \frac{{{{\left( 5 \right)}^3}}}{{{{\left( 4 \right)}^3}}} \cr & \Rightarrow R = \frac{5}{4} \cr} $$ ∴ Diameter of the third ball : $$ = 2R = \frac{5}{2}cm = 2.5\,cm$$

4.

The radius of a sphere is equal to the radius of the base of a right circular cone, and the volume of the sphere is double the volume of the cone. The ratio of the height of the cone to the radius of its base is :

• (A) 2 : 1
• (B) 1 : 2
• (C) 2 : 3
• (D) 3 : 2

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 Let the radius of cone and the sphere be R and the height of the cone be H Volume of sphere $$ = \frac{4}{3}\pi {r^3}$$ Volume of cone $$ = \frac{1}{3}\pi {r^2}h$$ According to given information : $$\eqalign{ & \Rightarrow \frac{4}{3}\pi {R^3} = 2 \times \frac{1}{3}\pi {R^2}H \cr & \Rightarrow 4R = 2H \cr & \Rightarrow \frac{H}{R} = \frac{4}{2}\,Or\,2:1 \cr} $$

5.

The truck of a tree is a right cylinder 1.5 m in radius and 10 m high. The volume of the timber which remains when the truck is trimmed just enough to reduce it to a rectangular parallelopiped on a square base is :

• (A) 44 m3
• (B) 45 m3
• (C) 46 m3
• (D) 47 m3

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 Let the length of each side of the square base be x metres Then, $$\eqalign{ & {x^2} + {x^2} = {\left( 3 \right)^2} \cr & \Rightarrow 2{x^2} = 9 \cr & \Rightarrow {x^2} = \frac{9}{2} \cr & \Rightarrow x = \frac{3}{{\sqrt 2 }} \cr} $$ ∴ Volume of parallelopiped : $$\eqalign{ & = \left( {\frac{3}{{\sqrt 2 }} \times \frac{3}{{\sqrt 2 }} \times 10} \right){{\text{m}}^{\text{3}}} \cr & = \frac{{90}}{2}{{\text{m}}^{\text{3}}} \cr & = 45\,{{\text{m}}^{\text{3}}} \cr} $$

6.

It is required to fix a pipe such that water flowing through it at a speed of 7 metres per minute fills a tank of capacity 440 cubic metres in 10 minutes. The inner radius of the pipe should be :

• (A) $$\sqrt{2}m$$
• (B) $$2m$$
• (C) $$\frac{1}{2}m$$
• (D) $$\frac{1}{\sqrt{2}}m$$

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 Let the inner radius of the pipe be r metres Then, Volume of water flowing through the pipe in 10 minutes : $$\eqalign{ & = \left[ {\left( {\frac{{22}}{7} \times {r^2} \times 7} \right) \times 10} \right]{m^3} \cr & = \left( {220{r^2}} \right){m^3} \cr & \therefore 220{r^2} = 440 \cr & \Rightarrow {r^2} = 2 \cr & \Rightarrow r = \sqrt 2 \, m \cr } $$

7.

The dimensions of a cuboid are 7 cm, 11 cm and 13 cm. The total surface area is :

• (A) 311 $$\text{ c}{\text{m}}^2$$
• (B) 622 $$\text{ c}{\text{m}}^2$$
• (C) 1001 $$\text{ c}{\text{m}}^2$$
• (D) 2002 $$\text{ c}{\text{m}}^2$$

Show Answer

 Surface area : $$\eqalign{ & = {2\left( {7 \times 11 + 11 \times 13 + 7 \times 13} \right)} \cr & = {2 \times 311} \cr & = 622{\text{ c}}{{\text{m}}^2} \cr} $$

8.

A large cube is formed from the material obtained by melting three smaller cubes of 3, 4 and 5 cm side. What is the ratio of the total surface areas of the smaller cubes and the large cube?

• (A) 2 : 1
• (B) 3 : 2
• (C) 25 : 18
• (D) 27 : 20

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 $$\eqalign{ & {\text{Volume}}\,{\text{of}}\,{\text{the}}\,{\text{large}}\,{\text{cube}} \cr & = \left( {{3^3} + {4^3} + {5^3}} \right) = 216\,c{m^3} \cr & {\text{Let}}\,{\text{the}}\,{\text{edge}}\,{\text{of}}\,{\text{the}}\,{\text{large}}\,{\text{cube}}\,{\text{be}}\,a \cr & So,\,{a^3} = 216\,\,\,\,\, \Rightarrow \,\,\,\,\,a = 6\,cm \cr & \therefore {\text{Required}}\,{\text{ratio}} \cr & = {\frac{{6 \times \left( {{3^2} + {4^2} + {5^2}} \right)}}{{6 \times {6^2}}}} \cr & = \frac{{50}}{{36}} \cr & = 25:18 \cr} $$

9.

The ratio of the volume of a hemisphere and a cylinder circumscribing this hemisphere and having a common base is :

• (A) 1 : 2
• (B) 2 : 3
• (C) 3 : 4
• (D) 4 : 5

Show Answer

 Let the radius of the hemisphere be be r cm Then, radius of the cylinder = r cm Height of the cylinder = r cm ∴ Required ratio : $$\eqalign{ & = \frac{{{\text{Volume of hemisphere}}}}{{{\text{Volume of cylinder}}}} \cr & = \frac{{\frac{2}{3}\pi {r^3}}}{{\pi {r^2} \times r}} \cr & = \frac{2}{3}\, Or\,2:3 \cr} $$

10.

The length, breadth and height of a cuboid are in the ratio 1 : 2 : 3. The length, breadth and height of the cuboid are increased by 100%, 200% and 200% respectively. Then the increase in the volume of the cuboid is :

• (A) 5 Times
• (B) 6 Times
• (C) 12 Times
• (D) 17 Times

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 Let the original length, breadth and height of the cuboid be x, 2x and 3x units respectively Then, original volume = (x × 2x × 3x) cu.units = 6x3 cu.units New length = 200% of x = 2x New breadth = 300% of 2x = 6x New height = 300% of 3x = 9x ∴ New volume : = (2x × 6x × 9x) cu.units = 108x3 cu.units Increase in volume : = (108x3 - 6x3) cu.units = (102x3) cu.units ∴ Required ratio : $$\eqalign{ & = \frac{{102{{\text{x}}^3}}}{{6{{\text{x}}^3}}} \cr & = 17{\text{ }}\left( {{\text{Times}}} \right) \cr} $$