Question

The cost of painting the total outside surface of a closed cylindrical oil tank at 60
paise per sq. dm is 237.60. The height of the tank is 6 times the radius of the base of the tank. Find its volume correct to two decimal places.

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Answer 1

Step-by-step explanation:

509.14 - volume of the cylinder

Answer 2

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

$\sf Then, \;h = 6r \qquad\qquad\qquad\bigg\lgroup\bf Given \bigg\rgroup$

⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━

We know that,

$\star\;{\boxed{\sf{\purple{TSA_{\;(cylinder)} = 2 \pi rh(r + h)}}}}$

$:\implies\sf 2 \pi r(r + 6r)$

$:\implies\sf \pink{14 \pi r^2}$

Cost of painting,

$:\implies\sf 14 \pi r^2 \times \dfrac{60}{100}$

$:\implies\sf \pink{Rs.\; \dfrac{42}{5} \pi r^2}$

☯ Given that, The cost of painting is ₹ 237.60

$:\implies\sf \dfrac{42}{5} \pi r^2 = 237.60$

$:\implies\sf \dfrac{ \cancel{42}}{5} \times \dfrac{22}{ \cancel{7}} \times r^2 = 237.60$

$:\implies\sf \dfrac{6}{5} \times 22 \times r^2 = 237.60$

$:\implies\sf r^2 = 237.60 \times \dfrac{5}{6} \times \dfrac{1}{22}$

$:\implies\sf r^2 = 9$

$:\implies\sf \sqrt{r^2} = \sqrt{9}$

$:\implies\sf \green{r = 3}$

Therefore,

• h = 6r = 18 dm

⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━

☯ Now, Finding Volume of cylinder,

$\star\;{\boxed{\sf{\purple{Volume_{\;(cylinder)} = \pi r^2 h}}}}$

$:\implies\sf \dfrac{22}{7} \times 3 \times 3 \times 18$

$:\implies\sf \dfrac{22}{7} \times 9 \times 18$

$:\implies{\boxed{\frak{\pink{509.14\;dm^3}}}}\;\bigstar$

$\therefore\;{\underline{\sf{Hence,\;Volume\;of\; cylinder\;is\; \bf{509.14\;dm^3}.}}}$