Practice MCQ Questions and Answer on Volume and Surface Area

91.

A metallic sphere of radius 5 cm is melted to make a cone with base of the same radius. What is the height of the cone ?

• (A) 5 cm
• (B) 10 cm
• (C) 15 cm
• (D) 20 cm

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 Let the height of the cone be h cm Then, $$\eqalign{ & \frac{4}{3}\pi \times {\left( 5 \right)^3} = \frac{1}{3}\pi \times {\left( 5 \right)^2} \times h \cr & \Rightarrow h = 20\,cm \cr} $$

92.

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

93.

A hemispherical bowl is 176 cm round the brim. Supposing it to be half full, how many persons may be served from it in hemispherical glasses 4 cm in diameter at the top ?

• (A) 1172
• (B) 1272
• (C) 1372
• (D) 1472

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 Let the radius of the hemispherical bowl be r cm Then, $$\eqalign{ & 2\pi r = 176 \cr & \Rightarrow r = \frac{{176 \times 7}}{{2 \times 22}} \cr & \Rightarrow r = 28 \cr} $$ Volume of liquid in the bowl : $$\eqalign{ & = \frac{1}{2} \times \left( {\frac{2}{3} \times \pi \times 28 \times 28 \times 28} \right){\text{ c}}{{\text{m}}^3} \cr & = \left( {\frac{2}{3} \times \pi \times 14 \times 28 \times 28} \right){\text{ c}}{{\text{m}}^3} \cr} $$ Volume of 1 glass :$$ = \left( {\frac{2}{3} \times \pi \times 2 \times 2 \times 2} \right){\text{ c}}{{\text{m}}^3}$$ ∴ Required number of person : $$\eqalign{ & = \frac{{{\text{Volume of liquid in the bowl}}}}{{{\text{Volume of 1 glass}}}} \cr & = \left( {\frac{{14 \times 28 \times 28}}{{2 \times 2 \times 2}}} \right) \cr & = 1372 \cr} $$

94.

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 : $$\eqalign{ & = \frac{{\pi \times {{\left( {3x} \right)}^2} \times 4y}}{{\pi \times {{\left( {4x} \right)}^2} \times 3y}} \cr & = \frac{{36}}{{48}} \cr & = \frac{3}{4}\,Or\,3:4 \cr} $$

95.

The curved surface area of a sphere is 5544 sq.cm. Its volume is :

• (A) 22176 cm3
• (B) 33951 cm3
• (C) 38808 cm3
• (D) 42304 cm3

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 $$\eqalign{ & 4\pi {r^2} = 5544 \cr & \Rightarrow {r^2} = \left( {5544 \times \frac{1}{4} \times \frac{7}{{22}}} \right) \cr & \Rightarrow {r^2} = 441 \cr & \Rightarrow r = 21 \cr} $$ ∴ Volume : $$\eqalign{ & = \left( {\frac{4}{3} \times \frac{{22}}{7} \times 21 \times 21 \times 21} \right){\text{ c}}{{\text{m}}^3} \cr & = 38808{\text{ c}}{{\text{m}}^3} \cr} $$

96.

A hollow spherical metallic ball has an external diameter 6 cm and is $$\frac{1}{2}$$ cm thick. The volume of metal used in the ball is :

• (A) $$37\frac{2}{3}\text{ c}{\text{m}}^3$$
• (B) $$40\frac{2}{3}\text{ c}{\text{m}}^3$$
• (C) $$41\frac{2}{3}\text{ c}{\text{m}}^3$$
• (D) $$47\frac{2}{3}\text{ c}{\text{m}}^3$$

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 External radius = 3 cm Internal radius = (3 - 0.5) cm = 2.5 cm Volume of the metal : $$\eqalign{ & = \left[ {\frac{4}{3} \times \frac{{22}}{7} \times \left\{ {{{\left( 3 \right)}^3} - {{\left( {2.5} \right)}^3}} \right\}} \right]{\text{ c}}{{\text{m}}^3} \cr & = \left( {\frac{4}{3} \times \frac{{22}}{7} \times \frac{{91}}{8}} \right){\text{ c}}{{\text{m}}^3} \cr & = \left( {\frac{{143}}{3}} \right){\text{ c}}{{\text{m}}^3} \cr & = 47\frac{2}{3}{\text{ c}}{{\text{m}}^3} \cr} $$

97.

The volume of a sphere is $$2145\frac{11}{21}\text{c}{\text{m}}^3.$$   Its radius is equal to :

• (A) 7 cm
• (B) 8 cm
• (C) 9 cm
• (D) None of these

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 $$\eqalign{ & \frac{4}{3} \times \frac{{22}}{7} \times {r^3} = \frac{{45056}}{{21}} \cr & \Rightarrow {r^3} = \left( {\frac{{45056}}{{21}} \times \frac{3}{4} \times \frac{7}{{22}}} \right) \cr & \Rightarrow {r^3} = 512 \cr & \Rightarrow r = \root 3 \of {512} \cr & \Rightarrow r = 8\,cm \cr} $$

98.

What is the total surface area of a right circular cone of height 14 cm and base radius 7 cm?

• (A) 344.35 cm2
• (B) 462 cm2
• (C) 498.35 cm2
• (D) None of these

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 $$\eqalign{ & h = 14\,cm,\,r = 7\,cm \cr & {\text{So}},\,l = \sqrt {{{\left( 7 \right)}^2} + {{\left( {14} \right)}^2}} = \sqrt {245} = 7\sqrt 5 \,cm \cr & \therefore {\text{Total}}\,{\text{surface}}\,{\text{area}} \cr & = \pi \,rl + \pi \,{r^2} \cr & = \left( {\frac{{22}}{7} \times 7 \times 7\sqrt 5 + \frac{{22}}{7} \times 7 \times 7} \right)c{m^2} \cr & = \left[ {154\left( {\sqrt 5 + 1} \right)} \right]c{m^2} \cr & = \left( {154 \times 3.236} \right)c{m^2} \cr & = 498.35\,c{m^2} \cr} $$

99.

The radius of the base and height of a metallic solid cylinder are r cm and 6 cm respectively. It is melted and recast into a solid cone of the same radius of base. The height of the cone is :

• (A) 9 cm
• (B) 18 cm
• (C) 27 cm
• (D) 54 cm

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 Let the height of the cone be h cm Then, $$\eqalign{ & \pi \times {r^2} \times 6 = \frac{1}{3} \times \pi \times {r^2} \times h \cr & \Rightarrow h = 18\,cm \cr} $$

100.

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 : $$\eqalign{ & \frac{{dS}}{{dr}} = 0 \cr & \Rightarrow \frac{{ - 2V}}{{{r^2}}} + 4\pi r = 0 \cr & \Rightarrow V = 2\pi {r^3} \cr & \Rightarrow \pi {r^2}h = 2\pi {r^3} \cr & \Rightarrow h = 2r \cr} $$