It is required to make a closed cylindrical t | vedclass

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(C) The height $(h)$ of the cylindrical tank is $1 \,m$.The base diameter is $140 \,cm$,so the base radius $(r)$ is $\frac{140}{2} \,cm = 70 \,cm = 0.

Vedclass · Accepted Answer

$7.48$

Vedclass · Answer

(C) The height $(h)$ of the cylindrical tank is $1 \,m$.The base diameter is $140 \,cm$,so the base radius $(r)$ is $\frac{140}{2} \,cm = 70 \,cm = 0.7 \,m$.The area of the sheet required is equal to the total surface area of the closed cylindrical tank,which is given by the formula $2\pi r(r + h)$.Substituting the values: $	ext{Area} = 2 	imes \frac{22}{7} 	imes 0.7 	imes (0.7 + 1) \,m^2$.$	ext{Area} = 2 	imes 22 	imes 0.1 	imes 1.7 \,m^2$.$	ext{Area} = 4.4 	imes 1.7 \,m^2 = 7.48 \,m^2$.Thus,$7.48 \,m^2$ of the metal sheet is required.

It is required to make a closed cylindrical tank of height $1 \,m$ and base diameter $140 \,cm$ from a metal sheet. How many square metres of the sheet are required for the same (in $,m^2$)? Assume $\pi = \frac{22}{7}$.

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