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A hemispherical bowl is packed in a cuboidal box. The bowl just fits in the box. Inner radius of the bowl is $10 \text{ cm}$. Outer radius of the bowl is $10.5 \text{ cm}$.
Based on the above, answer the following questions :
(i) Find the dimensions of the cuboidal box.
(ii) Find the total outer surface area of the box.
(iii) (a) Find the difference between the capacity of the bowl and the volume of the box. (use $\pi = 3.14$)
OR
(iii) (b) The inner surface of the bowl and the thickness is to be painted. Find the area to be painted.
Based on the above, answer the following questions :
(i) Find the dimensions of the cuboidal box.
(ii) Find the total outer surface area of the box.
(iii) (a) Find the difference between the capacity of the bowl and the volume of the box. (use $\pi = 3.14$)
OR
(iii) (b) The inner surface of the bowl and the thickness is to be painted. Find the area to be painted.
Show SolutionHide Solution↓
(i) Diameter of bowl $= 21 \text{ cm}$. Dimensions of the box are $21 \text{ cm} \times 21 \text{ cm} \times 10.5 \text{ cm}$ (1 mark)
(ii) Total surface area of the box $= 2 \left( 441 + \frac{441}{2} + \frac{441}{2} \right) = 1764 \text{ sq. cm}$ ($\frac{1}{2} + \frac{1}{2}$ marks)
(iii) (a) Capacity of bowl $= \frac{2}{3} \times 3.14 \times 10^3 = \frac{6280}{3} \text{ cu. cm or } 2093.33 \text{ cu. cm}$ ($\frac{1}{2} + \frac{1}{2}$ marks)
Volume of box $= 21 \times 21 \times \frac{21}{2} = \frac{9261}{2} \text{ cu. cm. or } 4630.5 \text{ cu. cm}$ ($\frac{1}{2}$ mark)
Required difference $= \frac{15223}{6} \text{ cu. cm or } 2537.17 \text{ cu. cm}$ ($\frac{1}{2}$ mark)
OR
(iii) (b) Required area $= 2 \times \frac{22}{7} \times 10^2 + \frac{22}{7} \times (10.5^2 - 10^2) = \frac{4400}{7} + \frac{451}{14} = \frac{9251}{14} \text{ sq. cm or } 660.79 \text{ sq. cm}$ (1 mark)
(ii) Total surface area of the box $= 2 \left( 441 + \frac{441}{2} + \frac{441}{2} \right) = 1764 \text{ sq. cm}$ ($\frac{1}{2} + \frac{1}{2}$ marks)
(iii) (a) Capacity of bowl $= \frac{2}{3} \times 3.14 \times 10^3 = \frac{6280}{3} \text{ cu. cm or } 2093.33 \text{ cu. cm}$ ($\frac{1}{2} + \frac{1}{2}$ marks)
Volume of box $= 21 \times 21 \times \frac{21}{2} = \frac{9261}{2} \text{ cu. cm. or } 4630.5 \text{ cu. cm}$ ($\frac{1}{2}$ mark)
Required difference $= \frac{15223}{6} \text{ cu. cm or } 2537.17 \text{ cu. cm}$ ($\frac{1}{2}$ mark)
OR
(iii) (b) Required area $= 2 \times \frac{22}{7} \times 10^2 + \frac{22}{7} \times (10.5^2 - 10^2) = \frac{4400}{7} + \frac{451}{14} = \frac{9251}{14} \text{ sq. cm or } 660.79 \text{ sq. cm}$ (1 mark)