Question

A hemispherical bowl made of brass has inner radius of 3.5 cm. Find the cost of tin-plating the inner part of the bowl as well as tin plating a plate that exactly fits as a lid on bowl. Tin plating is done at the rate of ₹10 per 100 sq.cm.

• A. $=Rs.17.55$
• B. $=Rs.31.55$
• C. $=Rs.21.55$
• D. $=Rs.11.55$

Answer 1

The correct option is D

$=Rs.11.55$

Surface area that has to be plated = Curved surface area of the hemisphere bowl + surface area of the plate

$=2\pi r^2+\pi r^2$

$=2\times \frac{22}{7}\times {3.5}^2+\frac{22}{7}\times {3.5}^2$

$=3\times \frac{22}{7}\times {3.5}^2$

$=115.5{cm}^2$

Hence cost of tin plating

$=115.5\times \frac{10}{100}$

= ₹$11.55$