Question

A company deals in casting and moulding of metal on orders received from its clients. In one such order, the company is supposed to make 50 toys in the form of a hemisphere surmounted by a right circular cone of the same base radius as that of the hemisphere. The radius of the base of the cone is 21 cm and height is 28 cm.

Based on the above information, answer the following questions.

1. Find the volume of 50 toys.
2. Find the ratio of the volume of hemisphere to the volume of cone.
3. Find the cost of painting the hemispherical part of the 50 toys at the rate of ₹10 per cm².
4. If one such toy is melted and recast into a solid cylinder of diameter 28 cm then find the height of the cylinder.

Answer 1

i. Volume of 50 toys

Given: Radius of cone and hemisphere, \[r = 21\text{ cm}\]

Height of cone, \[h = 28\text{ cm}\]

Formula: \[\text{Volume of 1 toy} = \text{Volume of cone} + \text{Volume of hemisphere}\]

\[\text{Volume of 1 toy} = \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3\]

Calculation: \[\text{Volume of 1 toy} = \left(\frac{1}{3} \times \frac{22}{7} \times 21^2 \times 28\right) + \left(\frac{2}{3} \times \frac{22}{7} \times 21^3\right)\]

\[\text{Volume of 1 toy} = 12936 + 19404 = 32340\text{ cm}^3\]

\[\text{Volume of 50 toys} = 50 \times 32340 = 1617000\text{ cm}^3\]

ii. Ratio of the volume of hemisphere to the volume of cone

Calculation: \[\text{Ratio} = \frac{\text{Volume of hemisphere}}{\text{Volume of cone}} = \frac{\frac{2}{3}\pi r^3}{\frac{1}{3}\pi r^2 h} = \frac{2r}{h}\]

\[\text{Ratio} = \frac{2 \times 21}{28} = \frac{42}{28} = \frac{3}{2}\]

Thus, the ratio is 3 : 2.

iii. Cost of painting the hemispherical part of 50 toys

Formula: \[\text{Curved surface area of hemisphere} = 2\pi r^2\]

Calculation: \[\text{Area of 1 toy} = 2 \times \frac{22}{7} \times 21 \times 21 = 2772\text{ cm}^2\]

\[\text{Total area for 50 toys} = 2772 \times 50 = 138600\text{ cm}^2\]

\[\text{Total Cost} = 138600 \times ₹10 = ₹1386000\]

iv. Height of the recast cylinder

Given: Diameter of cylinder, \[d = 28\text{ cm} \implies R = 14\text{ cm}\]

\[\text{Volume of cylinder} = \text{Volume of 1 toy}\]

\[\pi R^2 H = 32340\]

\[\frac{22}{7} \times 14 \times 14 \times H = 32340\]

\[616 \times H = 32340\]

\[H = \frac{32340}{616} = 52.5\text{ cm}\]