## Volume and Surface Area Questions and Answers Part-3

1. A cube of length 1 cm is taken out from a cube of length 8 cm. What is the weight of the remaining portion ?
a) $$\frac{7}{8}$$ of the weight of the original cube
b) $$\frac{8}{9}$$ of the weight of the original cube
c) $$\frac{63}{64}$$ of the weight of the original cube
d) $$\frac{511}{512}$$ of the weight of the original cube

Explanation: Volume of the bigger cube = (83) cm3 = 512 cm3
Volume of the cut-out cube = (13) cm3 = 1 cm3
Volume of the remaining portion = (512 - 1) cm3 = 511 cm3

$$\frac{{{\text{Weight of the remaining portion}}}}{{{\text{Weight of the original cube}}}}$$      $$= \frac{{511}}{{512}}$$

2.Capacity of a cylindrical vessel is 25.872 litres. If the height of the cylinder is three times the radius of its base, what is the area of the base ?
a) 336 cm2
b) 616 cm2
c) 1232 cm2
d) None of these

Explanation: Volume of cylinder :
\eqalign{ & = 25.872\,{\text{litres}} \cr & = \left( {25.872 \times 1000} \right){\text{c}}{{\text{m}}^{\text{3}}} \cr & = 25872\,{\text{ c}}{{\text{m}}^3} \cr}
Let the radius of the base of the cylinder be r cm
Then, height = (3r) cm
\eqalign{ & \frac{{22}}{7} \times {r^2} \times \left( {3r} \right) = 25872 \cr & {r^3} = \frac{{25872 \times 7}}{{66}} \cr & {r^3} = 2744 \cr & r = \root 3 \of {2744} \cr & r = 14 \cr}
Hence, area of the base :
\eqalign{ & = \pi {r^2} \cr & = \left( {\frac{{22}}{7} \times 14 \times 14} \right){\text{ c}}{{\text{m}}^2} \cr & = 616\,{\text{ c}}{{\text{m}}^2} \cr}

3. What part of a ditch, 48 metres long, 16.5 metres board and 4 metres deep can be filled by the earth got by digging a cylindrical tunnel of diameter 4 metres and length 56 metres ?
a) $$\frac{1}{9}$$
b) $$\frac{2}{9}$$
c) $$\frac{7}{9}$$
d) $$\frac{8}{9}$$

Explanation: Volume of earth dug :
\eqalign{ & = \left( {\frac{{22}}{7} \times 2 \times 2 \times 56} \right){{\text{m}}^3} \cr & = 704\,{{\text{m}}^3} \cr}
Volume of ditch :
\eqalign{ & = \left( {48 \times 16.5 \times 4} \right){{\text{m}}^3} \cr & = 3168\,{{\text{m}}^3} \cr}
Required fraction :
\eqalign{ & = \frac{{704}}{{3168}} \cr & = \frac{2}{9} \cr}

4. 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

Explanation:
\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 & l = \sqrt {1225 + 7056} \cr & l = \sqrt {8281} \cr & 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}

5. Ice cream completely filled in a cylinder of diameter 35 cm and height 32 cm is to be served by completely filling identical disposable cones of diameter 4 cm and height 7 cm. The maximum number of persons that can be served this way is :
a) 950
b) 1000
c) 1050
d) 1100

Explanation: Volume of cylinder :
\eqalign{ & = \left( {\pi \times \frac{{35}}{2} \times \frac{{35}}{2} \times 32} \right){\text{ c}}{{\text{m}}^3} \cr & = 9800\pi {\text{ c}}{{\text{m}}^3} \cr}
Volume of 1 cone :
\eqalign{ & = \left( {\frac{1}{3} \times \pi \times 2 \times 2 \times 7} \right){\text{ c}}{{\text{m}}^3} \cr & = \frac{{28\pi }}{3}{\text{ c}}{{\text{m}}^3} \cr}
Number of persons that can be served :
\eqalign{ & = \left( {9800\pi \times \frac{3}{{28\pi }}} \right) \cr & = 1050 \cr}

6.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) $$47\frac{2}{3}{\text{ c}}{{\text{m}}^3}$$

Explanation: 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}

7.If a hollow sphere of internal and external diameters 4 cm and 8 cm respectively is melted into a cylinder of base diameter 8 cm, then the height of the cylinder is :
a) 4 cm
b) $$\frac{13}{3}$$ cm
c) $$\frac{14}{3}$$ cm
d) 5 cm

Explanation: Let the height of the cylinder be h cm
\eqalign{ & \frac{4}{3}\pi \left[ {{{\left( 4 \right)}^3} - {{\left( 2 \right)}^3}} \right] = \pi \times {4^2} \times h \cr & \frac{4}{3} \times \pi \times 56 = \pi \times 16h \cr & h = \frac{{4 \times 56}}{{3 \times 16}} \cr & h = \frac{{14}}{3}\,cm \cr}

8. The ratio of the volume of a hemisphere and a cylinder circumscribing this hemisphere and having a common base is :
a) 1 : 2
b) 2 : 3
c) 3 : 4
d) 4 : 5

Explanation: Let the radius of the hemisphere be be r cm
Then, radius of the cylinder = r cm
Height of the cylinder = r cm
Required ratio :
\eqalign{ & = \frac{{{\text{Volume of hemisphere}}}}{{{\text{Volume of cylinder}}}} \cr & = \frac{{\frac{2}{3}\pi {r^3}}}{{\pi {r^2} \times r}} \cr & = \frac{2}{3}\, Or\,2:3 \cr}

9.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 dm
b) 194.04 dm
c) 19.404 dm
d) 1940.4 dm

Explanation: 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}

10. If the diameter of a sphere is 6 m, its hemisphere will have a volume of :
a) $$18\pi$$ m3
b) $$36\pi$$ m3
c) $$72\pi$$ m3
d) None of these

\eqalign{ & = \left( {\frac{2}{3}\pi \times 3 \times 3 \times 3} \right){m^3} \cr & = \left( {18\pi } \right){m^3} \cr}