Question

A hemispherical tank is made up of an iron sheet 1 cm thick the inner radius is 1 metre then find the volume of iron used to make the tank.

Answer 1

let the inner radius be r

thickness of the iron sheet =1 cm = 0.01 m

Outer radius (R) = inner radius (r) + thickness of the iron sheet

1m +0.01m = 1.01m

Volume of the iron used to make the tank =2/3πr³ (R³-r³)

= 2/3*22/7 { (1.01)³-1³}

= 0.06348 m³ (approx)

hope it helps you.

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Answer 2

$\textbf{Answer :}$

0.06348 m³

$\textbf{Step-by-step explanation :}$

$r_{1}$ (radius of outer hemisphere) = 1 cm

$r_{2}$ (radius of inner + outer hemisphere) = 1 m + 1 cm

= 1 m + $\dfrac{1}{100}$ m

= 1 m + 0.01 m

= 1.01 m

Now...

» Volume of iron used = Volume of outer hemisphere - Volume of inner hemisphere

= $\dfrac{2}{3}$π$r_{2}^{3}$ - $\dfrac{2}{3}$π$r_{1}^3}$

Take common from both

= $\dfrac{2}{3}$ π $(r_{2}^{3}$ - $r_{1}^{3})$

= $\dfrac{2}{3}$ × $\dfrac{22}{7}$ × [(1.01)³ - (1)³]

= $\dfrac{44}{21}$ (1.030301 - 1)

= $\dfrac{44}{21}$ (0.030301)

= $\textbf{0.06348}$ $\bold{m}^{3}$