Practice MCQ Questions and Answer on Volume and Surface Area

91.

If the areas of three adjacent faces of a cuboid are x, y, z respectively, then the volume of the cuboid is :

• (A) $$xyz$$
• (B) $$2xyz$$
• (C) $$\sqrt {xyz} $$
• (D) $$3\sqrt {xyz} $$

Show Answer

 Let, length = l, breadth = b, height = h Then, x = lb, y = bh, z = lh Let, V be the volume of the cuboid Then, V = lbh $$\eqalign{ & \therefore xyz = lb \times bh \times lh \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = {\left( {lbh} \right)^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = {V^2} \cr & \,\,\,\,\,\,\,\,\,\,\,or\,V = \sqrt {xyz} \cr} $$

92.

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, $$\eqalign{ & \frac{{\frac{1}{3} \times \pi \times {x^2} \times h}}{{\frac{1}{3} \times \pi \times {{\left( {2x} \right)}^2} \times H}} = \frac{2}{3} \cr & \Leftrightarrow \frac{h}{H} = \frac{2}{3} \times 4 \cr & \Leftrightarrow \frac{h}{H} = \frac{8}{3}\,Or\,8:3 \cr} $$

93.

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

94.

Some solid metallic right circular cones, each with radius of the base 3 cm and height 4 cm, are melted to form a solid sphere of radius 6 cm. The number of right circular cones is :

• (A) 6
• (B) 12
• (C) 24
• (D) 48

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 Volume of sphere : $$\eqalign{ & = \left( {\frac{4}{3}\pi \times {6^3}} \right){\text{c}}{{\text{m}}^{\text{3}}} \cr & = \left( {288\pi } \right){\text{c}}{{\text{m}}^{\text{3}}} \cr} $$ Volume of each cone : $$\eqalign{ & = \left( {\frac{1}{3}\pi \times {3^2} \times 4} \right){\text{c}}{{\text{m}}^{\text{3}}} \cr & = \left( {12\pi } \right){\text{c}}{{\text{m}}^{\text{3}}} \cr} $$ ∴ Number of cone : $$\eqalign{ & = \frac{{288\pi }}{{12\pi }} \cr & = 24 \cr} $$

95.

A pyramid has an equilateral triangle as its base of which each side is 1 m. Its slant edge is 3 m. The whole surface are of the pyramid is equal to :

• (A) $$\frac{{\sqrt 3 + 2\sqrt {13} }}{4}sq.m$$
• (B) $$\frac{{\sqrt 3 + 3\sqrt {13} }}{4}sq.m$$
• (C) $$\frac{{\sqrt 3 + 3\sqrt {35} }}{4}sq.m$$
• (D) $$\frac{{\sqrt 3 + 2\sqrt {35} }}{4}sq.m$$

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 Area of base : $$\eqalign{ & = \left( {\frac{{\sqrt 3 }}{4} \times {1^2}} \right){m^2} \cr & = \frac{{\sqrt 3 }}{4}{m^2} \cr} $$ Clearly, the pyramid has 3 triangular faces each with sides 3m, 3m and 1 m So, area of each lateral face : $$\eqalign{ & = \sqrt {\frac{7}{2} \times \left( {\frac{7}{2} - 3} \right)\left( {\frac{7}{2} - 3} \right)\left( {\frac{7}{2} - 1} \right)} {m^2} \cr & \left[ {\because s = \frac{{3 + 3 + 1}}{2} = \frac{7}{2}} \right] \cr & = \sqrt {\frac{7}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{5}{2}} {m^2} \cr & = \frac{{\sqrt {35} }}{4}{m^2} \cr} $$ ∴ Whole surface area of the pyramid : $$\eqalign{ & = \left( {\frac{{\sqrt 3 }}{4} + 3 \times \frac{{\sqrt {35} }}{4}} \right){m^2} \cr & = \frac{{\sqrt 3 + 3\sqrt {35} }}{4}{m^2} \cr} $$

96.

If the height of a right circular cone is increased by 200% and the radius of the base is reduced by 50%, then the volume of the cone :

• (A) Remains unaltered
• (B) Decrease by 25%
• (C) Increase by 25%
• (D) Increase by 50%

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 Let the radius of a right circular cine be R cm and height be H cm Volume of right circular cone $$ = \frac{1}{3}\pi {R^2}H{\text{ cu}}{\text{.cm}}$$ When height of right circular cone is increased by 200% and radius of the base is reduce by 50% New volume :$$\eqalign{ & {\text{ = }}\frac{1}{3}\pi {\left( {\frac{R}{2}} \right)^2}.3H \cr & = \frac{1}{3}\pi \frac{{{R^2}4}}{4}.3H \cr & = \frac{{\pi {R^2}H}}{4} \cr} $$ Difference : $$\eqalign{ & = \pi {R^2}H\left( {\frac{1}{3} - \frac{1}{4}} \right) \cr & = \frac{1}{{12}}\pi {R^2}H \cr} $$ Decrease percentage : $$\eqalign{ & = \frac{{\frac{1}{{12}}\pi {R^2}H}}{{\frac{1}{3}\pi {R^2}H}} \times 100 \cr & = 25\% \cr} $$

97.

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

98.

A rectangular water tank is 8 m high, 6 m long and 2.5 m wide. How many litres of water can it hold ?

• (A) 120 litres
• (B) 1200 litres
• (C) 12000 litres
• (D) 120000 litres

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 Volume of the tank : $$\eqalign{ & = \left( {8 \times 100 \times 6 \times 100 \times 2.5 \times 100} \right)c{m^3} \cr & = 120000000\,c{m^3} \cr & = \left( {\frac{{120000000}}{{1000}}} \right){\text{litres}} \cr & = 120000{\text{ litres}} \cr} $$

99.

A bucket is in the from of a frustum of a cone and holds 28.490 litres of water. The radii of the top and bottom are 28 cm and 21 cm respectively. Find the height of the bucket.

• (A) 15 cm
• (B) 20 cm
• (C) 25 cm
• (D) 30 cm

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 Volume of bucket : = 28.490 litres = (28.490 ×1000) cm3 = 28490 cm3 Let the height of the bucket be h cm We have : r = 21 cm. and R = 28 cm $$\therefore \frac{\pi }{3}h\left[ {{{\left( {28} \right)}^2} + {{\left( {21} \right)}^2} + 28 \times 21} \right]$$     $$ = 28490$$ $$ \Rightarrow h\left( {784 + 441 + 588} \right) = $$     $$\frac{{28490 \times 21}}{{22}}$$ $$\eqalign{ & \Rightarrow 1813h = 27195 \cr & \Rightarrow h = \frac{{27195}}{{1813}} \cr & \Rightarrow h = 15\,cm \cr} $$

100.

A well has to be dug out that is to be 22.5 m deep and of diameter 7 m. Find the cost of plastering the inner curved surface at Rs. 3 per sq.meter :

• (A) Rs. 1465
• (B) Rs. 1475
• (C) Rs. 1485
• (D) Rs. 1495

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 Curved surface area : $$\eqalign{ & = 2\pi rh \cr & = \left( {2 \times \frac{{22}}{7} \times \frac{7}{2} \times 22.5} \right){{\text{m}}^2} \cr & = 495{\text{ }}{{\text{m}}^2} \cr} $$ ∴ Cost of plastering : = Rs. (495 × 3) = Rs. 1485