Question

A cubical block of side $10cm$ is surmounted by a hemisphere. What is the largest diameter that the hemisphere can have? Find the cost of painting the total surface area of the solid so formed, at the rate of $Rs.5$ per $100sq.cm$. [Use $\pi 3.14$]

Answer 1

The total surface area of the solid $=$Total surface area of the cube$+$Curved surface area of the hemisphere$-$Area of the base of the hemisphere.

$=6a^2+2\pi r^2-\pi r^2$

$=[6\times {10}^2+2\times 3.14\times 5^2-3.14\times 5^2]{cm}^2$

$=600+157-78.5=678.5{cm}^2$

Cost of painting$=$Rs.$5$ per $100{cm}^2$

$\therefore ,$Cost of painting the solid$=678.5\times \frac{5}{100}=$Rs.$33.90$

Hence, the approximate cost of painting the solid so formed is Rs.$33.90$