Practice MCQ Questions and Answer on Volume and Surface Area

171.

If the ratio of volumes of two cones is 2 : 3 and the ratio of the radii of their bases is 1 : 2, then the ratio of their height will be :

• (A) 3 : 4
• (B) 4 : 3
• (C) 3 : 8
• (D) 8 : 3

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 let their radii be x and 2x, and their heights be h and H respectively Then, $$\eqalign{ & \frac{{\frac{1}{3} \times \pi \times {x^2} \times h}}{{\frac{1}{3} \times \pi \times {{\left( {2x} \right)}^2} \times H}} = \frac{2}{3} \cr & \Leftrightarrow \frac{h}{H} = \frac{2}{3} \times 4 \cr & \Leftrightarrow \frac{h}{H} = \frac{8}{3}\,Or\,8:3 \cr} $$

172.

For a sphere of radius 10 cm, What percent of the numerical value of its volume would be the numerical value of the surface area ?

• (A) 24%
• (B) 26.5%
• (C) 30%
• (D) 45%

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 Volume of the sphere : $$ = \left[ {\frac{4}{3}\pi {{\left( {10} \right)}^3}} \right]{\text{ c}}{{\text{m}}^3}$$ Surface area of the sphere : $$ = \left[ {4\pi {{\left( {10} \right)}^2}} \right]{\text{ c}}{{\text{m}}^2}$$ ∴ Required percentage : $$\eqalign{ & = \left[ {\frac{{4\pi {{\left( {10} \right)}^2}}}{{\frac{4}{3}\pi {{\left( {10} \right)}^3}}}} \times 100 \right]\% \cr & = 30\% \cr} $$

173.

The radius of the base and height of a metallic solid cylinder are r cm and 6 cm respectively. It is melted and recast into a solid cone of the same radius of base. The height of the cone is :

• (A) 9 cm
• (B) 18 cm
• (C) 27 cm
• (D) 54 cm

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 Let the height of the cone be h cm Then, $$\eqalign{ & \pi \times {r^2} \times 6 = \frac{1}{3} \times \pi \times {r^2} \times h \cr & \Rightarrow h = 18\,cm \cr} $$

174.

How many small cubes, each of 96 cm surface area, can be formed from the material obtained by melting a larger cube of 384 cm surface area ?

• (A) 5
• (B) 8
• (C) 800
• (D) 8000

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 Let the length of each edge of small cube be a1 and that of larger cube be a2 Then, $$\eqalign{ & 6a_1^2 = 96\, \cr & \Rightarrow a_1^2 = 16 \cr & \Rightarrow {a_1} = 4\,and \cr & 6a_2^2 = 384 \cr & \Rightarrow a_2^2 = 64 \cr & \Rightarrow {a_2} = 8 \cr} $$ ∴ Number of cubes formed : $$\eqalign{ & = \frac{{{\text{Volume of larger cube}}}}{{{\text{Volume of smaller cube}}}} \cr & = \left( {\frac{{8 \times 8 \times 8}}{{4 \times 4 \times 4}}} \right) \cr & = 8 \cr} $$

175.

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 : $$\eqalign{ & = \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} \times 36 \cr & = \frac{{32}}{3} \cr & = 10\frac{2}{3}\,m \cr} $$

176.

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

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

177.

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 $${\text{ c}}{{\text{m}}^2}$$

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

178.

Base of a right prism is a rectangle, the ratio of whose length and breadth is 3 : 2. If the height of the prism is 12 cm and total surface area is 288 sq.cm the volume of the prism is :

• (A) 291 cm3
• (B) 288 cm3
• (C) 290 cm3
• (D) 286 cm3

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 Let the length of base be 3a cm and breadth be 2a cm Total surface area of prism : = [Perimeter of base × height] + [2 × Area of base] = [2 (3a + 2a) × 12 + 2 × 3a × 2a] sq.cm = (120a + 12a2) sq.cm According to the question, 120a + 12a2 = 288 ⇒ a2 + 10a = 24 ⇒ a2 + 10a - 24 = 0 ⇒ a2 + 12a - 2a - 24 = 0 ⇒ a (a + 12) - 2 (a + 12) = 0 ⇒ (a - 2)(a + 12) = 0 ⇒ a = 2 because a $$ \ne $$ -12 ∴ Volume of prism : = Area of base × Height = (3a × 2a × 12)cu.cm = 72a2 cu.cm = (72 × 2 × 2)cu.cm = 288 cu.cm

179.

50 men took a dip in a water tank 40 m long and 20 m broad on a religious day. If the average displacement of water by a man is 4 m³, then the rise in the water level in the tank will be:

• (A) 20 cm
• (B) 25 cm
• (C) 35 cm
• (D) 50 cm

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 $$\eqalign{ & {\text{Total}}\,{\text{Volume}}\,{\text{of}}\,{\text{water}}\,{\text{displaced}} \cr & = \left( {4 \times 50} \right){m^3} = 200\,{m^3} \cr & \therefore {\text{Rise}}\,{\text{in}}\,{\text{water}}\,{\text{level}} \cr & = \left( {\frac{{200}}{{40 \times 20}}} \right)m \cr & = 0.25\,m \cr & = 25\,cm \cr} $$

180.

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 : $$ = 2R = \frac{5}{2}cm = 2.5\,cm$$