Practice MCQ Questions and Answer on Volume and Surface Area

161.

A hall is 15 m long and 12 m broad. If the sum of the areas of the floor and the ceiling is equal to the sum of the areas of four walls, the volume of the hall is:

• (A) 720 m3
• (B) 900 m3
• (C) 1200 m3
• (D) 1800 m3

Show Answer

 $$\eqalign{ & 2\left( {15 + 12} \right) \times h = 2\left( {15 \times 12} \right) \cr & \Rightarrow h = \frac{{180}}{{27}}m = \frac{{20}}{3}m \cr & \therefore {\text{Volume}} = \left( {15 \times 12 \times \frac{{20}}{3}} \right){m^3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1200\,{m^3} \cr} $$

162.

A hemispherical bowl has 3.5 cm radius. It is to be painted inside as well as outside. The cost of painting it at the rate of Rs. 5 per 10 sq. cm will be :

• (A) Rs. 77
• (B) Rs. 100
• (C) Rs. 175
• (D) Rs. 50

Show Answer

 Radius of a hemispherical bowl = 3.5 cm Inner and outer surface area of the bowl : $$\eqalign{ & = 4\pi {r^2} \cr & = 4 \times \frac{{22}}{7} \times 3.5 \times 3.5 \cr & = 154{\text{ sq}}{\text{.cm}} \cr} $$ Total cost of painting at the rate of Rs. 5 per 10 sq.cm : $$\eqalign{ & = 154 \times \frac{5}{{10}} \cr & = {\text{Rs}}{\text{. 77}} \cr} $$

163.

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 : $$\eqalign{ & = \left( {\frac{3}{{\sqrt 2 }} \times \frac{3}{{\sqrt 2 }} \times 10} \right){{\text{m}}^{\text{3}}} \cr & = \frac{{90}}{2}{{\text{m}}^{\text{3}}} \cr & = 45\,{{\text{m}}^{\text{3}}} \cr} $$

164.

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

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 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 : $$\eqalign{ & = \frac{{{\text{Volume of hemisphere}}}}{{{\text{Volume of cylinder}}}} \cr & = \frac{{\frac{2}{3}\pi {r^3}}}{{\pi {r^2} \times r}} \cr & = \frac{2}{3}\, Or\,2:3 \cr} $$

165.

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 : $$\eqalign{ & = \left[ {\left( {\frac{{22}}{7} \times {r^2} \times 7} \right) \times 10} \right]{m^3} \cr & = \left( {220{r^2}} \right){m^3} \cr & \therefore 220{r^2} = 440 \cr & \Rightarrow {r^2} = 2 \cr & \Rightarrow r = \sqrt 2 \, m \cr } $$

166.

When a ball bounces, it rises to $$\frac{2}{3}$$ of the height from which it fell. If the ball is dropped from a height of 36 m, how high will it rise at the third bounce ?

• (A) $$10\frac{1}{3}$$ m
• (B) $$10\frac{2}{3}$$ m
• (C) $$12\frac{1}{3}$$ m
• (D) $$12\frac{2}{3}$$ m

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 Ball is dropped from the height of 36 m when the ball will rise at the third bounce ∴ Required height : $$\eqalign{ & = \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} \times 36 \cr & = \frac{{32}}{3} \cr & = 10\frac{2}{3}\,m \cr} $$

167.

A closed box made of wood of uniform thickness has length, breadth and height 12 cm, 10 cm and 8 cm respectively. If the thickness of the wood is 1 cm, the inner surface area is :

• (A) 264 cm2
• (B) 376 cm2
• (C) 456 cm2
• (D) 696 cm2

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 Internal length = (12 - 2) cm = 10 cm Internal breadth = (10 - 2) cm = 8 cm Internal height = (8 - 2) cm = 6 cm Inner surface area : = 2 [10 × 8 + 8 × 6 + 10 × 6] cm2 = (2 × 188) cm2 = 376 cm2

168.

A cistern, open at the top, is to be lined with sheet of lead which weights 27 kg/m². The cistern is 4.5 m long and 3 m wide and holds 50 m³. The weight of lead required is :

• (A) 1660.5 kg
• (B) 1764.5 kg
• (C) 1860.5 kg
• (D) 1864.5 kg

Show Answer

 Let the depth of the cistern be h metres Then, $$\eqalign{ & 4.5 \times 3 \times h = 50 \cr & \Rightarrow h = \frac{{50}}{{13.5}} \cr & \Rightarrow h = \frac{{100}}{{27}} \cr} $$ Area of sheet required : $$\eqalign{ & = lb + 2\left( {bh + lh} \right) \cr & = lb + 2h\left( {l + b} \right) \cr & = \left[ {4.5 \times 3 + 2 \times \frac{{100}}{{27}} \times \left( {4.5 + 3} \right)} \right]{{\text{m}}^2} \cr & = \left( {13.5 + \frac{{200}}{{27}} \times 7.5} \right){\text{ }}{{\text{m}}^2} \cr & = \left( {\frac{{27}}{2} + \frac{{500}}{9}} \right){{\text{m}}^2} \cr & = \frac{{1243}}{{18}}{\text{ }}{{\text{m}}^2} \cr} $$ ∴ Weight of lead : $$\eqalign{ & = \left( {27 \times \frac{{1243}}{{18}}} \right)kg \cr & = \left( {\frac{{3729}}{2}} \right)kg \cr & = 1864.5\,kg \cr} $$

169.

A school room is be built to accommodate 70 children so as to allow 2.2 m² of floor and 11 m³ of space for each child. If the room be 14 metres long, what must be its breadth and height ?

• (A) 11 m, 4 m
• (B) 11 m, 5 m
• (C) 12 m, 5.5 m
• (D) 13 m, 6 m

Show Answer

 Let the breadth and height of the room be b and h metres respectively. Then, Area of the floor $$ = \left( {14b} \right)\,{m^2}$$ $$\eqalign{ & \therefore 14b = 2.2 \times 70 \cr & \Rightarrow b = \frac{{2.2 \times 70}}{{14}} \cr & \Rightarrow b = 11 \cr} $$ Volume of the room : $$\eqalign{ & = \left( {14 \times 11 \times h} \right){m^3} \cr & = \left( {154h} \right){m^3} \cr} $$ $$\eqalign{ & \therefore 154h = 11 \times 70 \cr & \Rightarrow h = \frac{{11 \times 70}}{{154}} \cr & \Rightarrow h = 5 \cr} $$

170.

The sum of perimeters of the six faces of a cuboid is 72 cm and the total surface area of the cuboid is 16 cm². Find the longest possible length that can be kept inside the cuboid :

• (A) 5.2 cm
• (B) 7.8 cm
• (C) 8.05 cm
• (D) 8.36 cm

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 Sum of perimeters of the six faces : $$\eqalign{ & = 2\left[ {2\left( {l + b} \right) + 2\left( {b + h} \right) + 2\left( {l + h} \right)} \right] \cr & = 4\left( {2l + 2b + 2h} \right) \cr & = 8\left( {l + b + h} \right) \cr} $$ Total surface area $$ = 2\left( {lb + bh + lh} \right)$$ $$\eqalign{ & \therefore 8\left( {l + b + h} \right) = 72 \cr & \Rightarrow l + b + h = 9 \cr & 2\left( {lb + bh + lh} \right) = 16 \cr & \Rightarrow lb + bh + lh = 8 \cr} $$ Now, $${\left( {l + b + h} \right)^2} = {l^2} + {b^2} + {h^2} + 2$$     $$\left( {lb + bh + lh} \right)$$ $$\eqalign{ & \Rightarrow {\left( 9 \right)^2} = {l^2} + {b^2} + {h^2} + 16 \cr & \Rightarrow {l^2} + {b^2} + {h^2} = 81 - 16 \cr & \Rightarrow {l^2} + {b^2} + {h^2} = 65 \cr} $$ Required length : $$\eqalign{ & = \sqrt {{l^2} + {b^2} + {h^2}} \cr & = \sqrt {65} \cr & = 8.05\,cm \cr} $$