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(B) First, calculate the Total Surface Area $(TSA)$ of the closed cylindrical tank:$TSA = 2\pi r(r + h)$Given $r = 2.1 	ext{ m}$ and $h = 4.5 	ext{

Vedclass · Accepted Answer

$95.04$

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(B) First, calculate the Total Surface Area $(TSA)$ of the closed cylindrical tank:$TSA = 2\pi r(r + h)$Given $r = 2.1 	ext{ m}$ and $h = 4.5 	ext{ m}$,$TSA = 2 	imes \frac{22}{7} 	imes 2.1 	imes (2.1 + 4.5) = 2 	imes \frac{22}{7} 	imes 2.1 	imes 6.6 = 87.12 	ext{ m}^2$.Let $A$ be the actual steel used. Since $1/12$ of the steel was wasted, the steel used for the tank is $A - \frac{1}{12}A = \frac{11}{12}A$.Equating this to the surface area of the tank:$\frac{11}{12}A = 87.12$$A = \frac{87.12 	imes 12}{11} = 7.92 	imes 12 = 95.04 	ext{ m}^2$.Thus, the actual steel used is $95.04 	ext{ m}^2$.

Find how much steel was actually used to make a closed cylindrical tank (radius $2.1 \text{ m}$, height $4.5 \text{ m}$), if $1/12$ of the steel actually used was wasted in making the tank. [Assume $\pi = 22/7$] (in $\text{ m}^2$)

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