Practice MCQ Questions and Answer on Volume and Surface Area

171.

The diameter of the iron ball used for the shot-put game is 14 cm. It is melted and then a solid cylinder of height $$2\frac{1}{3}$$ cm is made. What will be the diameter of the base of the cylinder ?

• (A) 14 cm
• (B) $$\frac{14}{3}$$ cm
• (C) 28 cm
• (D) $$\frac{28}{3}$$ cm

Show Answer

 Let the radius of the cylinder be R Then, $$\eqalign{ & \pi \times {R^2} \times \frac{7}{3} = \frac{4}{3}\pi \times 7 \times 7 \times 7 \cr & \Rightarrow {R^2} = \left( {\frac{{4 \times 7 \times 7 \times 7}}{3} \times \frac{3}{7}} \right) \cr & \Rightarrow {R^2} = 196 \cr & \Rightarrow {R^2} = {\left( {14} \right)^2} \cr & \Rightarrow {R^2} = 14{\text{ cm}} \cr} $$ ∴ Diameter = 2R = 28

172.

A right triangle with sides 3 cm, 4 cm and 5 cm is rotated about the side of 3 cm to form a cone. The volume of the cone so formed is :

• (A) 12π cm3
• (B) 15π cm3
• (C) 16π cm3
• (D) 20π cm3

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 Clearly, we have r = 3 cm and h = 4 cm ∴ Volume : $$\eqalign{ & = \frac{1}{3}\pi {r^2}h \cr & = \left( {\frac{1}{3} \times \pi \times {3^2} \times 4} \right)\pi {r^3} \cr & = 12\pi {\text{ c}}{{\text{m}}^3} \cr} $$

173.

A swimming bath is 24 m long and 15 m broad. When a number of men dive into the bath, the height of the water rises by 1 cm. If the average amount of water displaced by one of the men be 0.1 cu.m, how many men are there in the bath ?

• (A) 32
• (B) 36
• (C) 42
• (D) 46

Show Answer

 Volume of water displaced : $$\eqalign{ & = \left( {24 \times 15 \times \frac{1}{{100}}} \right){m^3} \cr & = \frac{{18}}{5}{m^3} \cr} $$ Volume of water displaced by 1 man = 0.1 m3 ∴ Number of men : $$\eqalign{ & = \left( {\frac{{\frac{{18}}{5}}}{{0.1}}} \right) \cr & = \left( {\frac{{18}}{5} \times 10} \right) \cr & = 36 \cr} $$

174.

The height of a closed cylinder of given volume and the minimum surface area is :

• (A) Equal to its diameter
• (B) Half of its diameter
• (C) Double of its diameter
• (D) None of these

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 $$\eqalign{ & V = \pi {r^2}h{\text{ and }} \cr & S = 2\pi rh + 2\pi {r^2} \cr & \,\,\,\,\,\,\, = 2\pi r\left( {h + r} \right) \cr & {\text{Where, }}h = \frac{V}{{\pi {r^2}}} \cr & \Rightarrow S = 2\pi r\left( {\frac{V}{{\pi {r^2}}} + r} \right) \cr & \Rightarrow S = \frac{{2V}}{r} + 2\pi {r^2} \cr & \Rightarrow \frac{{dS}}{{dr}} = \frac{{ - 2V}}{{{r^2}}} + 4\pi r{\text{ and}} \cr & \frac{{{d^2}S}}{{d{r^2}}} = \left( {\frac{{4V}}{{{r^3}}} + 4\pi } \right){\text{ > 0}} \cr} $$ ∴ S is minimum when : $$\eqalign{ & \frac{{dS}}{{dr}} = 0 \cr & \Rightarrow \frac{{ - 2V}}{{{r^2}}} + 4\pi r = 0 \cr & \Rightarrow V = 2\pi {r^3} \cr & \Rightarrow \pi {r^2}h = 2\pi {r^3} \cr & \Rightarrow h = 2r \cr} $$

175.

How many bricks, each measuring 25 cm x 11.25 cm x 6 cm, will be needed to build a wall of 8 m x 6 m x 22.5 cm?

• (A) 5600
• (B) 6000
• (C) 6400
• (D) 7200

Show Answer

 $$\eqalign{ & {\text{Number}}\,{\text{of}}\,{\text{bricks}} \cr & = \frac{{{\text{Volume}}\,{\text{of}}\,{\text{the}}\,{\text{wall}}}}{{{\text{Volume}}\,{\text{of}}\,{\text{1}}\,{\text{brick}}}} \cr & = {\frac{{800 \times 600 \times 22.5}}{{25 \times 11.25 \times 6}}} \cr & = 6400 \cr} $$

176.

The radius of base and curved surface area of a right cylinder is 'r' units and 4πrh square units respectively. The height of the cylinder is :

• (A) $$\frac{\text{h}}{2}$$ units
• (B) 1h units
• (C) 2h units
• (D) 4h units

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 Radius of the base = r units Curved surface area of a right cylinder = $$4\pi {\text{r}}h$$ Curved surface area of cylinder = $$2\pi {\text{RH}}$$ ∴ According to the question, $$2\pi {\text{rH = }}4\pi {\text{rh}}$$ ⇒ Height of cylinder = 2h units

177.

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

Show Answer

 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

178.

From a cube of side 8 m, a square hole of 3 m side is hollowed from end to end. What is the volume of the remaining solid ?

• (A) 440 m3
• (B) 480 m3
• (C) 508 m3
• (D) 520 m3

Show Answer

 Volume of the remaining solid : = Volume of the cube - Volume of the cuboid cut out from it = [(8 × 8 × 8) - (3 × 3 × 8)] m3 = (512 - 72) m3 = 440 m3

179.

Given that 1 cu. cm of marble weights 25 gms, the weight of a marble block 28 cm in width and 5 cm thick is 112 kg. The length of the block is :

• (A) 26.5 cm
• (B) 32 cm
• (C) 36 cm
• (D) 37.5 cm

Show Answer

 Let length = x cm Then, $$\eqalign{ & x \times 28 \times 5 \times \frac{{25}}{{1000}} = 112 \cr & \Rightarrow x = \left( {112 \times \frac{{1000}}{{25}} \times \frac{1}{{28}} \times \frac{1}{5}} \right) \cr & \Rightarrow x = 32\,cm \cr} $$

180.

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 : $$\eqalign{ & = \sqrt {{l^2} + {b^2} + {h^2}} \cr & = \sqrt {65} \cr & = 8.05\,cm \cr} $$