Question

A hemispherical bowl made of wood has inner diameter of $10.5$cm. Find the cost of painting it on the inside at the rate of Rs.$12$ per $100c{m}^2$.

Answer 1

We know that the surface area of hemisphere with radius $r$ is $A=2\pi r^2$.

It is given that the diameter of the hemispherical bowl is $10.5$ cm,

therefore, radius is $\frac{10.5}{2}=5.25$ cm, thus,

$A=2\pi r^2=2\times \frac{22}{7}\times (5.25{)}^2$

$=2\times \frac{22}{7}\times 27.26$

$=\frac{1212.64}{7}$

$=173.23$cm${}^2$

Therefore, the surface area of hemispherical bowl is $173.23$ cm${}^2$.

Now, the cost of painting inside the hemisphere is $12$ per $100$cm${}^2$,

therefore, cost of painting $173.23$ cm${}^2$ is:

$\frac{173.23\times 12}{100}=20.79$

Hence, cost of painting is Rs.$20.79$.