Question

A hemisherical bowl is to be pinted from inside at the rate of rs20 per 100 m^2. the total cost of painting is rs 30.80. find 1) inner surface are of bowl 2)volume of air inside the bowl

Answer 1

$\large\sf\underbrace{\red{SOLUTION:-}}$

☞ According to question,

Rs.20 is cost for paint 100 m² area .

=> Rs.1 is costed for paint (100/20) m² area .

=> Rs.30.80 is cost for painting of

(100/20 × 30.80) m² area .

=> Rs.30.80 is cost for painting of “154 m²” area .

1) So, inner surface area of the bowl is “154 m²” .

2) $\checkmark\:\rm{\underline{\purple{\boxed{Volume\:=\:\dfrac{2}{3}\:\pi\:r^3\:}}}}$

☞ we have know that,

$\rm{\boxed{2\pi{r^2}\:=\:154\:m^2\:}}$

$\rm{\implies\:r^2\:=\:\dfrac{154}{2\times{3.14}}\:}$

$\rm{\implies\:r\:=\:\dfrac{7}{\sqrt{2}}\:m}$

☞ Now, volume of the bowl is,

$\rm{V\:=\:\dfrac{2}{3}\times{3.14}\times{(\dfrac{7}{\sqrt{2}})}^3\:}$

$\rm{V\:=\:254.08\:m^3\:}$

✪ Therefore, Volume of air inside the bowl is “254.08 m³” .