Practice MCQ Questions and Answer on Volume and Surface Area

111.

The sum of the radius and the height of a cylinder is 19 m. The total surface area of the cylinder is 1672 m², what is the volume of the cylinder ?

• (A) 3080 m3
• (B) 2940 m3
• (C) 3220 m3
• (D) 2660 m3

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 Let the radius of the cylinder be r and height be h Then, r + h = 19 Again, total surface area of cylinder = $$\left( {2\pi rh + 2\pi {r^2}} \right)$$ Now, $$\eqalign{ & 2\pi r\left( {h + r} \right) = 1672 \cr & \Rightarrow 2\pi r \times 19 = 1672 \cr & \Rightarrow 38\pi r = 1672 \cr & \therefore \pi r = \frac{{1672}}{{38}} = 44\,m \cr & \therefore r = \frac{{44 \times 7}}{{22}} = 14\,m \cr} $$ ∴ Height = 19 - 14 = 5 m Volume of cylinder : $$\eqalign{ & = \pi {r^2}h \cr & = \frac{{22}}{7} \times 14 \times 14 \times 5 \cr & = 22 \times 2 \times 14 \times 5 \cr & = 3080\,{m^3} \cr} $$

112.

A metallic sheet is of rectangular shape with dimensions 48 m x 36 m. From each of its corners, a square is cut off so as to make an open box. If the length of the square is 8 m, the volume of the box (in m³) is:

• (A) 4830
• (B) 5120
• (C) 6420
• (D) 8960

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 Clearly, l = (48 - 16)m = 32 m, b = (36 -16)m = 20 m, h = 8 m. ∴ Volume of the box = (32 x 20 x 8) m3 = 5120 m3

113.

The volume of a right circular cone which is obtained from a wooden cube of edge 4.2 dm wasting minimum amount of wood is :

• (A) 19404 dm3
• (B) 194.04 dm3
• (C) 19.404 dm3
• (D) 1940.4 dm3

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 The volume of cone should be maximum ∴ Radius of the base of cone : $$\eqalign{ & = \frac{{{\text{Edge of cube}}}}{2} \cr & = \frac{{4.2}}{2} \cr & = 2.1\,{\text{dm.}} \cr} $$ Height of cone = Edge of cube = 4.2 dm. ∴ Volume of cone : $$\eqalign{ & = \frac{1}{3}\pi {r^2}h \cr & = \left( {\frac{1}{3} \times \frac{{22}}{7} \times 2.1 \times 2.1 \times 4.2} \right){\text{cu}}{\text{.dm}}{\text{.}} \cr & = 19.404\,{\text{cu}}{\text{.dm}}{\text{.}} \cr} $$

114.

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

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 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 : $$\eqalign{ & = \frac{{2\pi {r^2}}}{{\pi rl}} \cr & = \frac{{2\pi {r^2}}}{{\pi r \times r\sqrt 2 }} \cr & = \frac{2}{{\sqrt 2 }} \cr & = \frac{{2 \times \sqrt 2 }}{{\sqrt 2 \times \sqrt 2 }} \cr & = \frac{{2\sqrt 2 }}{2} \cr & = \frac{{\sqrt 2 }}{1}\,Or\,\sqrt 2 :1 \cr} $$

115.

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

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 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 : $$\eqalign{ & = \left( {\frac{{22}}{7} \times \frac{9}{4} \times \frac{9}{4} \times 10 \times \frac{{280}}{{99}}} \right) \cr & = 450 \cr} $$

116.

If the area of the base of a right circular cone is 3850 cm² and its height is 84 cm, then the curved surface area of the cone is :

• (A) 10001 cm2
• (B) 10010 cm2
• (C) 10100 cm2
• (D) 11000 cm2

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 $$\eqalign{ & \pi {r^2} = 3850 \cr & \Rightarrow {r^2} = \left( {\frac{{3850 \times 7}}{{22}}} \right) = 1225 \cr & \Rightarrow r = 35 \cr & {\text{Now, }}r = 35{\text{ cm, }}h = 84{\text{ cm}} \cr & {\text{So,}} \cr & l = \sqrt {{{\left( {35} \right)}^2} + {{\left( {84} \right)}^2}} \cr & \Rightarrow l = \sqrt {1225 + 7056} \cr & \Rightarrow l = \sqrt {8281} \cr & \Rightarrow l = 91{\text{ cm}} \cr} $$ ∴ Curved surface area : $$\eqalign{ & = \left( {\frac{{22}}{7} \times 35 \times 91} \right){\text{ c}}{{\text{m}}^2} \cr & = 10010{\text{ c}}{{\text{m}}^2} \cr} $$

117.

If three metallic spheres of radii 6 cm, 8 cm and 10 cm are melted to from a single sphere, the diameter of the new sphere will be :

• (A) 12 cm
• (B) 24 cm
• (C) 30 cm
• (D) 36 cm

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 Volume of new sphere : $$\eqalign{ & = \left[ {\frac{4}{3}\pi \times {{\left( 6 \right)}^3} + \frac{4}{3}\pi \times {{\left( 8 \right)}^3} + \frac{4}{3}\pi \times {{\left( {10} \right)}^3}} \right]{\text{ c}}{{\text{m}}^3} \cr & = \left[ {\frac{4}{3}\pi \left\{ {{{\left( 6 \right)}^3} + {{\left( 8 \right)}^3} + {{\left( {10} \right)}^3}} \right\}} \right]{\text{ c}}{{\text{m}}^3} \cr & = \left( {\frac{4}{3}\pi \times 1728} \right){\text{c}}{{\text{m}}^3} \cr & = \left[ {\frac{4}{3}\pi \times {{\left( {12} \right)}^3}} \right]{\text{ c}}{{\text{m}}^3} \cr} $$ Let the radius of the new sphere be R Then, $$\eqalign{ & \frac{4}{3}\pi {R^3} = \frac{4}{3}\pi \times {\left( {12} \right)^3} \cr & \Rightarrow R = 12\,cm \cr} $$ ∴ Diameter : $$\eqalign{ & = 2R \cr & = 2 \times 12 \cr & = 24\,cm \cr} $$

118.

A hollow iron pipe is 21 cm long and its external diameter is 8 cm. If the thickness of the pipe is 1 cm and iron weights 8 g/cm³, then the weight of the pipe is:

• (A) 3.6 kg
• (B) 3.696 kg
• (C) 36 kg
• (D) 36.9 kg

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 $$\eqalign{ & {\text{External}}\,{\text{radius}} = 4\,cm \cr & {\text{Internal}}\,{\text{radius}} = 3\,cm \cr & {\text{Volume}}\,{\text{of}}\,{\text{iron}} \cr & = \left( {\frac{{22}}{7} \times \left[ {{{\left( 4 \right)}^2} - {{\left( 3 \right)}^2}} \right] \times 21} \right)c{m^3} \cr & = \left( {\frac{{22}}{7} \times 7 \times 1 \times 21} \right)c{m^3} \cr & = 462\,c{m^3} \cr & \therefore {\text{Weight}}\,{\text{of}}\,{\text{iron}} = \left( {462 \times 8} \right)gm \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 3696\,gm \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 3.696\,kg \cr} $$

119.

A water tank is 30 m long, 20 m wide and 12 m deep. It is made of iron sheet which is 3 m wide. The tank is open at the top. If the cost of the iron sheet is Rs. 10 per metre, then the total cost of the iron sheet required to build the tank is :

• (A) Rs. 6000
• (B) Rs. 8000
• (C) Rs. 9000
• (D) Rs. 10000

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 Since the tank is open at the top, we have : Area of sheet required = Surface area of the tank $$\eqalign{ & = lb + 2\left( {bh + lh} \right) \cr & = \left[ {30 \times 20 + 2\left( {20 \times 12 + 30 \times 12} \right)} \right]{{\text{m}}^2} \cr & = \left( {600 + 1200} \right){{\text{m}}^2} \cr & = 1800{\text{ }}{{\text{m}}^2} \cr} $$ Length of sheet required : $$\eqalign{ & = \left( {\frac{{{\text{Area}}}}{{{\text{Width}}}}} \right) \cr & = \frac{{1800}}{3}m \cr & = 600\,m \cr} $$ ∴ Cost of the sheet = Rs. (600 × 10) = Rs. 6000

120.

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

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 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 : $$\eqalign{ & = \left[ {64 \times \left( {6 \times {1^2}} \right)} \right]{\text{c}}{{\text{m}}^2} \cr & = 384{\text{ c}}{{\text{m}}^2} \cr} $$