Question

plz help me
I will mark u as brainlist if you and this q correctly​

Answer 1

Answer:

$\huge{\dag}{\underline{\boxed{\sf{\pink{Answer:- }}}}}$

Let r dm be the radius of the base and h dm be the height of the cylindrical tank.

$Then, h=6r (given) \\ Total \: surface \: area =2πr(r+h)=2πr(r+6r)=14π {r}^{2} \\ cost \: of \: painting =Rs.(14π {r}^{2} )× \frac{60}{100} =Rs \: \frac{42}{5} {r}^{2} \\ It \: is \: given \: that \: the \: cost \: of \: painting \: is \: Rs. 237.60 \\ ∴ \frac{42}{5} π {r}^{2} =237.60 \\ ⇒ \frac{42}{5} × \frac{22}{7} × {r}^{2} =237.60 \\ ⇒ {r}^{2} =237.60× \frac{5}{42} × \frac{7}{22} =9 \\ ⇒r=3dm \\ ∴h=6r=18dm \\ Hence, \: volume \: of \: the \: cylinder =π {r}^{2} h= \\ (π×3×3×18)d {m}^{3} =( \frac{22}{7} ×9×18)dm3 \\ =509.14dm3$