## Volume and Surface Area Questions and Answers Part-4

1. A cuboidal water tank contains 216 litres of water. Its depth is $$\frac{1}{3}$$ of its length and breadth is $$\frac{1}{2}$$ of $$\frac{1}{3}$$ of the difference between length and depth. The length of the tank is :
a) 2 dm
b) 6 dm
c) 18 dm
d) 72 dm

Explanation:Let the length of the tank be x dm
Then, depth of the tank = $$\frac{x}{3}$$ dm
\eqalign{ & = \left[ {\frac{1}{2}{\text{ of }}\frac{1}{3}{\text{ of }}\left( {x - \frac{x}{3}} \right)} \right]{\text{dm}} \cr & = \left( {\frac{1}{2} \times \frac{1}{3} \times \frac{{2x}}{3}} \right){\text{dm}} \cr & = \frac{x}{9}\,{\text{dm}} \cr}
\eqalign{ & x \times \frac{x}{9} \times \frac{x}{3} = 216 \cr & {x^3} = 216 \times 27 \cr & x = 6 \times 3 \cr & x = 18\,dm \cr}

2. A covered wooden box has the inner measures as 115 cm, 75 cm and 35 cm and the thickness of wood is 2.5 cm. Find the volume of the wood :
a) 81000 cu.cm
b) 81775 cu.cm
c) 82125 cu.cm
d) None of these

Explanation: The external measures of the box are (115 + 5) cm, (75 + 5) cm, and (35 + 5) cm i.e., 120 cm, 80 cm and 40 cm
Volume of the wood :
= External volume - Internal volume
= [(120 × 80 × 40) - (115 × 75 × 35)] cm3
= (384000 - 301875) cm3
= 82125 cm3

3. Find the cost of a cylinder of radius 14 m and height 3.5 m when the cost of its metal is Rs. 50 per cubic meter :
a) Rs. 100208
b) Rs. 107800
c) Rs. 10800
d) Rs. 109800

Explanation: Volume :
\eqalign{ & = \pi {r^2}h \cr & = \left( {\frac{{22}}{7} \times 14 \times 14 \times 3.5} \right){m^3} \cr & = 2156\,{m^3} \cr}
Cost of the cylinder :
\eqalign{ & = {\text{Rs}}{\text{.}}\left( {2156 \times 50} \right) \cr & = {\text{Rs}}{\text{. 107800}} \cr}

4.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

Explanation: 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}

5.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

Explanation: 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}

6. For a sphere of radius 10 cm, What percent of the numerical value of its volume would be the numerical value of the surface area ?
a) 24%
b) 26.5%
c) 30%
d) 45%

Explanation: Volume of the sphere :
$$= \left[ {\frac{4}{3}\pi {{\left( {10} \right)}^3}} \right]{\text{ c}}{{\text{m}}^3}$$
Surface area of the sphere :
$$= \left[ {4\pi {{\left( {10} \right)}^2}} \right]{\text{ c}}{{\text{m}}^2}$$
Required percentage :
\eqalign{ & = \left[ {\frac{{4\pi {{\left( {10} \right)}^2}}}{{\frac{4}{3}\pi {{\left( {10} \right)}^3}}}} \times 100 \right]\% \cr & = 30\% \cr}

7. How many lead shots each 3 mm in diameter can be made from a cuboid of dimensions 9 cm × 11 cm × 12 cm ?
a) 7200
b) 8400
c) 72000
d) 84000

Explanation: Volume of each lead shot :
\eqalign{ & = \left[ {\frac{4}{3}\pi \times {{\left( {\frac{{0.3}}{2}} \right)}^3}} \right]{\text{ c}}{{\text{m}}^3} \cr & = \left( {\frac{4}{3} \times \frac{{22}}{7} \times \frac{{27}}{{8000}}} \right){\text{ c}}{{\text{m}}^3} \cr & = \frac{{99}}{{7000}}{\text{ c}}{{\text{m}}^3} \cr}
\eqalign{ & = \left( {9 \times 11 \times 12 \times \frac{{7000}}{{99}}} \right) \cr & = 84000 \cr}

8. 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

Explanation: Let the height of the cone be h cm
\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}

9. A hemisphere of lead of radius 6 cm is cast into a right circular cone of height 75 cm. The radius of the base of the cone is :
a) 1.4 cm
b) 2 cm
c) 2.4 cm
d) 4.2 cm

Explanation: Let the radius of the cone be R cm
\eqalign{ & \frac{1}{3}\pi \times {R^2} \times 75 = \frac{2}{3}\pi \times 6 \times 6 \times 6 \cr & {R^2} = \left( {\frac{{2 \times 6 \times 6 \times 6}}{{75}}} \right) \cr & {R^2} = \frac{{144}}{{25}} \cr & {R^2} = \frac{{{{\left( {12} \right)}^2}}}{{{{\left( 5 \right)}^2}}} \cr & R = \frac{{12}}{5} \cr & R = 2.4\,cm \cr}

10. 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 cm
b) 288 cm
c) 290 cm
d) 286 cm

Explanation: 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