Practice MCQ Questions and Answer on Volume and Surface Area

51.

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 : $$\eqalign{ & = \left( {\frac{{\frac{{18}}{5}}}{{0.1}}} \right) \cr & = \left( {\frac{{18}}{5} \times 10} \right) \cr & = 36 \cr} $$

52.

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 : $$\eqalign{ & = \frac{1}{3}\pi {r^2}h \cr & = \left( {\frac{1}{3} \times \pi \times {3^2} \times 4} \right)\pi {r^3} \cr & = 12\pi {\text{ c}}{{\text{m}}^3} \cr} $$

53.

A 4 cm cube is cut into 1 cm cubes. The total surface area of all the small cubes is :

• (A) 24 cm2
• (B) 96 cm2
• (C) 384 cm2
• (D) None of these

Show Answer

 Number of small cube formed : $$\eqalign{ & = \left( {\frac{{4 \times 4 \times 4}}{{1 \times 1 \times 1}}} \right) \cr & = 64 \cr} $$ Total surface area of the small cubes : $$\eqalign{ & = \left[ {64 \times \left( {6 \times {1^2}} \right)} \right]{\text{c}}{{\text{m}}^2} \cr & = 384{\text{ c}}{{\text{m}}^2} \cr} $$

54.

The radius of a cylinder is 5 m more than its height. If the curved surface area of the cylinder is 792 m², what is the volume of the cylinder ?

• (A) 5712 m3
• (B) 5244 m3
• (C) 5544 m3
• (D) 5306 m3

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 Let the height of the cylinder be x cm Then, radius = (x + 5) m Curved surface area of the cylinder = $$2\pi rh$$ Now, $$\eqalign{ & 2\pi \left( {x + 5} \right) \times x = 792 \cr & \Rightarrow 2 \times \frac{{22}}{7} \times \left( {{x^2} + 5x} \right) = 792 \cr & \Rightarrow {x^2} + 5x = \frac{{792 \times 7}}{{44}} = 126 \cr & \Rightarrow {x^2} + 5x - 126 = 0 \cr & \Rightarrow {x^2} + 14x - 9x - 126 = 0 \cr & \Rightarrow x\left( {x + 14} \right) - 9\left( {x + 14} \right) = 0 \cr & \Rightarrow \left( {x - 9} \right)\left( {x + 14} \right) = 0 \cr & \therefore x = 9, - 14{\text{(neglect negative value)}} \cr} $$ ∴ Height of cylinder = 9 m ∴ Radius of cylinder = 9 + 5 = 14 m Volume of cylinder : $$\eqalign{ & = \pi {r^2}h \cr & = \frac{{22}}{7} \times 14 \times 14 \times 9 \cr & = 5544\,{m^3} \cr} $$

55.

The slant height of a right circular cone is 10 m and its height is 8 m. Find the area of its curved surface.

• (A) 30π m2
• (B) 40π m2
• (C) 60π m2
• (D) 80π m2

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 $$\eqalign{ & l = 10\,m,h = 8\,m \cr & So, \cr & r = \sqrt {{l^2} + {h^2}} \cr & \,\,\,\, = \sqrt {{{\left( {10} \right)}^2} - {{\left( 8 \right)}^2}} \cr & \,\,\,\, = 6\,m \cr} $$ ∴ Curved surface area : $$\eqalign{ & = \pi rl \cr & = \left( {\pi \times 6 \times 10} \right){m^2} \cr & = 60\pi \,{m^2} \cr} $$

56.

Find the number of coins 1.5 cm in diameter and 0.2 cm thick, to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.

• (A) 430
• (B) 440
• (C) 450
• (D) 460

Show Answer

 Volume one coin : $$\eqalign{ & = \left( {\frac{{22}}{7} \times \frac{{75}}{{100}} \times \frac{{75}}{{100}} \times \frac{2}{{10}}} \right){\text{c}}{{\text{m}}^3} \cr & = \frac{{99}}{{280}}{\text{ c}}{{\text{m}}^3} \cr} $$ Volume of larger cylinder : $$ = \left( {\frac{{22}}{7} \times \frac{9}{4} \times \frac{9}{4} \times 10} \right){\text{ c}}{{\text{m}}^3}$$ Number of coins : $$\eqalign{ & = \left( {\frac{{22}}{7} \times \frac{9}{4} \times \frac{9}{4} \times 10 \times \frac{{280}}{{99}}} \right) \cr & = 450 \cr} $$

57.

A cylindrical tank of diameter 35 cm is full of water. If 11 litres of water is drawn off, the water level in the tank will drop by :

• (A) $$10\frac{1}{2}\text{ cm}$$
• (B) $$11\frac{3}{7}\text{ cm}$$
• (C) $$12\frac{6}{7}\text{ cm}$$
• (D) $$14\text{ cm}$$

Show Answer

 Let the drop in the water level be h cm Then, $$\eqalign{ & \Rightarrow \frac{{22}}{7} \times \frac{{35}}{2} \times \frac{{35}}{2} \times h = 11000 \cr & \Rightarrow h = \left( {\frac{{11000 \times 7 \times 4}}{{22 \times 35 \times 35}}} \right){\text{ cm}} \cr & \Rightarrow h = \frac{{80}}{7}{\text{ cm}} \cr & \Rightarrow h = 11\frac{3}{7}{\text{ cm}} \cr} $$

58.

A sphere and a cube have equal surface area. The ratio of the volume of the sphere to that of the cube is :

• (A) $$\sqrt{\pi }:\sqrt{6}$$
• (B) $$\sqrt{2}:\sqrt{\pi }$$
• (C) $$\sqrt{\pi }:\sqrt{3}$$
• (D) $$\sqrt{6}:\sqrt{\pi }$$

Show Answer

 $$\eqalign{ & 4\pi {R^2} = 6{a^2} \cr & \Rightarrow \frac{{{R^2}}}{{{a^2}}} = \frac{3}{{2\pi }} \cr & \Rightarrow \frac{R}{a} = \frac{{\sqrt 3 }}{{\sqrt {2\pi } }} \cr} $$ $$\eqalign{ & \therefore \frac{{{\text{Volume of spere}}}}{{{\text{Volume of cube}}}} \cr & = \frac{{\frac{4}{3}\pi {R^3}}}{{{a^3}}} \cr & = \frac{4}{3}\pi {\left( {\frac{R}{a}} \right)^3} \cr & = \frac{4}{3}\pi \frac{{3\sqrt 3 }}{{2\pi \sqrt {2\pi } }} \cr & = \frac{{2\sqrt 3 }}{{\sqrt {2\pi } }} \cr & = \frac{{\sqrt {12} }}{{\sqrt {2\pi } }} \cr & = \frac{{\sqrt 6 }}{{\sqrt \pi }} \cr & \text{or, }\sqrt 6 :\sqrt \pi \cr} $$

59.

Capacity of a cylindrical vessel is 25.872 litres. If the height of the cylinder is three times the radius of its base, what is the area of the base ?

• (A) 336 cm2
• (B) 616 cm2
• (C) 1232 cm2
• (D) Cannot be determined

Show Answer

 Volume of cylinder : $$\eqalign{ & = 25.872\,{\text{litres}} \cr & = \left( {25.872 \times 1000} \right){\text{c}}{{\text{m}}^{\text{3}}} \cr & = 25872\,{\text{ c}}{{\text{m}}^3} \cr} $$ Let the radius of the base of the cylinder be r cm Then, height = (3r) cm $$\eqalign{ & \therefore \frac{{22}}{7} \times {r^2} \times \left( {3r} \right) = 25872 \cr & \Rightarrow {r^3} = \frac{{25872 \times 7}}{{66}} \cr & \Rightarrow {r^3} = 2744 \cr & \Rightarrow r = \root 3 \of {2744} \cr & \Rightarrow r = 14 \cr} $$ Hence, area of the base : $$\eqalign{ & = \pi {r^2} \cr & = \left( {\frac{{22}}{7} \times 14 \times 14} \right){\text{ c}}{{\text{m}}^2} \cr & = 616\,{\text{ c}}{{\text{m}}^2} \cr} $$

60.

The radii of the bases of two cylinders are in the ratio 3 : 4 and their height are in the ratio 4 : 3. The ratio of their volume is :

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

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 Let their radii be 3x, 4x and heights be 4y, 3y Ratio of their volumes : $$\eqalign{ & = \frac{{\pi \times {{\left( {3x} \right)}^2} \times 4y}}{{\pi \times {{\left( {4x} \right)}^2} \times 3y}} \cr & = \frac{{36}}{{48}} \cr & = \frac{3}{4}\,Or\,3:4 \cr} $$