Practice MCQ Questions and Answer on Volume and Surface Area

31.

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

Show Answer

 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 : $$\eqalign{ & = \pi {r^2}h \cr & = \frac{{22}}{7} \times 14 \times 14 \times 9 \cr & = 5544\,{m^3} \cr} $$

32.

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

Show Answer

 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

33.

The diameter of the base of a cylindrical drum is 35 dm and the height is 24 dm. It is full of kerosene. How many tins each of size 25 cm × 22 cm × 35 cm can be filled with kerosene from the drum ?

• (A) 120
• (B) 600
• (C) 1020
• (D) 1200

Show Answer

 $$\eqalign{ & {\text{Number of tins}} \cr & = \frac{{{\text{Voulme of the drum}}}}{{{\text{Volume of each tin}}}} \cr & = \frac{{\left( {\frac{{22}}{7} \times \frac{{35}}{2} \times \frac{{35}}{2} \times 24} \right)}}{{\left( {\frac{{25}}{{10}} \times \frac{{22}}{{10}} \times \frac{{35}}{{10}}} \right)}} \cr & = 1200 \cr} $$

34.

If the volume of a sphere is divided by its surface area, the result is 27 cm. The radius of the sphere is :

• (A) 9 cm
• (B) 36 cm
• (C) 54 cm
• (D) 81 cm

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 $$\eqalign{ & \frac{{\frac{4}{3}\pi {r^3}}}{{4\pi {r^2}}} = 27 \cr & \Rightarrow r = 81\,cm \cr} $$

35.

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

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

36.

The length of the longest rod that can be placed in a room of dimensions 10 m × 10 m × 5 m is :

• (A) 15$$\sqrt{3}$$ m
• (B) 15 m
• (C) 10$$\sqrt{2}$$ m
• (D) 5$$\sqrt{3}$$ m

Show Answer

 Required length : $$\eqalign{ & = \sqrt {{{\left( {10} \right)}^2} + {{\left( {10} \right)}^2} + {{\left( 5 \right)}^2}} \,m \cr & = \sqrt {225} \,m \cr & = 15\,m \cr} $$

37.

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) $$\sqrt{6}:\sqrt{\pi }$$

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

38.

A hemisphere and a cone have equal bases. If their heights are also equal, then the ratio of their curved surface will be :

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

Show Answer

 Let, $$\eqalign{ & OP = OQ = OR = r \cr & \therefore OR = h = r \cr} $$ ∴ Curved surface area of the hemisphere = $$2\pi {r^2}$$ Curved surface area of a cone = $$\pi rl$$ Where, $$\eqalign{ & l = \sqrt {{h^2} + {r^2}} \cr & \,\,\,\,\, = \sqrt {{r^2} + {r^2}} \cr & \,\,\,\,\, = r\sqrt 2 \cr} $$ ∴ Required ratio : $$\eqalign{ & = \frac{{2\pi {r^2}}}{{\pi rl}} \cr & = \frac{{2\pi {r^2}}}{{\pi r \times r\sqrt 2 }} \cr & = \frac{2}{{\sqrt 2 }} \cr & = \frac{{2 \times \sqrt 2 }}{{\sqrt 2 \times \sqrt 2 }} \cr & = \frac{{2\sqrt 2 }}{2} \cr & = \frac{{\sqrt 2 }}{1}\,Or\,\sqrt 2 :1 \cr} $$

39.

The base of a pyramid is an equilateral triangle of side 1 m. If the height of the pyramid is 4 metres, then the volume is :

• (A) 0.550 m3
• (B) 0.577 m3
• (C) 0.678 m3
• (D) 0.750 m3

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 Area of the base : $$\eqalign{ & = \left( {\frac{{\sqrt 3 }}{4} \times {1^2}} \right){m^2} \cr & = \frac{{\sqrt 3 }}{4}{m^2} \cr} $$ ∴ Volume of pyramid : $$\eqalign{ & = \left( {\frac{1}{3} \times \frac{{\sqrt 3 }}{4} \times 4} \right){m^3} \cr & = \left( {\frac{{\sqrt 3 }}{3}} \right){m^3} \cr & = \left( {\frac{{1.732}}{3}} \right){m^3} \cr & = 0.577\,{m^3} \cr} $$

40.

A solid is in the form of a right circular cylinder with hemispherical ends. The total length of the solid is 35 cm. The diameter of the cylinder is $$\frac{1}{4}$$ of its height. The surface area of the solid is :

• (A) 462 cm2
• (B) 693 cm2
• (C) 750 cm2
• (D) 770 cm2

Show Answer

 Let the radius of the cylinder and the hemisphere be r cm Diameter of the cylinder = (2r) cm Height of the cylinder = (4 × 2r) cm = 8r cm Total length of the solid : = (8r + r + r) cm = 10r cm ⇒ 10r = 35 ⇒ r = 3.5 ∴ Surface area of the solid : = Curved surface area of the cylinder + 2 × (Curved surface area of the hemisphere) $$ = \left( {2 \times \frac{{22}}{7} \times 3.5 \times 28 + 2 \times 2 \times \frac{{22}}{7} \times 3.5 \times 3.5} \right){\text{ c}}{{\text{m}}^2}$$ $$\eqalign{ & = \left( {616 + 154} \right){\text{ c}}{{\text{m}}^2} \cr & = 770{\text{ c}}{{\text{m}}^2} \cr} $$