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(C) Inner radius $(r) = 1 \,m$.Thickness $= 1 \,cm = 0.01 \,m$.Outer radius $(R) = 1 \,m + 0.01 \,m = 1.01 \,m$.The volume of iron used is the differe

Vedclass · Accepted Answer

$0.06348$

Vedclass · Answer

(C) Inner radius $(r) = 1 \,m$.Thickness $= 1 \,cm = 0.01 \,m$.Outer radius $(R) = 1 \,m + 0.01 \,m = 1.01 \,m$.The volume of iron used is the difference between the outer volume and the inner volume of the hemispherical shell.Volume of iron $= \frac{2}{3} \pi R^3 - \frac{2}{3} \pi r^3 = \frac{2}{3} \pi (R^3 - r^3)$.Volume $= \frac{2}{3} 	imes \frac{22}{7} 	imes ((1.01)^3 - (1)^3) \,m^3$.Volume $= \frac{44}{21} 	imes (1.030301 - 1) \,m^3$.Volume $= \frac{44}{21} 	imes 0.030301 \,m^3 \approx 0.06348 \,m^3$.

$A$ hemispherical tank is made up of an iron sheet $1 \,cm$ thick. If the inner radius is $1 \,m$,then find the volume of the iron used to make the tank. (in $,m^3$)

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