Question

The cost of painting outer curved surface of a hollow hemispherical iron ball at rate of Rs. 2 per cm square is Rs. 1256. If thickness of the iron is 1 cm, find the cost of painting the inner curved surface of the ball.(Take $\pi$ = 3.14).

• A. Rs. 1076.32
• B. Rs. 2034.72
• C. Rs. 1017.36
• D. Rs. 508.68

Answer 1

The correct option is C

Rs. 1017.36

The cost of painting outer curved surface of a hollow hemispherical iron ball at the rate of Rs. 2 per cm square is Rs. 1256. So, Outer curved surface area = $\frac{1256}{2}$ = 628 ${cm}^2$.

Then, curved surface area of hemisphere = 2$\pi$ $r^2$ = 628 ${cm}^2$

$\Rightarrow$$r^2$ = 100 ${cm}^2$

$\Rightarrow$ r = 10 cm

Now, thickness of iron = 1 cm. Therefore, inner radius = 9 cm.

Then, inner curved surface area = $2\times \frac{22}{7}\times 9\times 9$ = 508.68 ${cm}^2$.

So, cost of painting =$508.6\times 2$ = Rs. 1017.36 .