Question

a hemespherical tank is made up of an iron sheet 1cm thick. If the inner radius is 1m, then find the volume of the iron used to make the tank​

Answer 1

SOLUTION:-

Given:

•A hemispherical tank is made up of an iron sheet 1cm thick.

•The inner radius is 1m.

To find:

The volume of the iron used to make the tank.

Explanation:

•Inner radius(r)= 1m.

•Thickness of iron sheet= 1cm

In metre;

1m= 100

1cm = 1/100m

Thickness of iron sheet= 0.01m.

Outer Radius(R):

Inner radius (r) + thickness of Iron

Outer Radius=1m + 0.01m

Outer Radius= 1.01m.

Now,

Using the formula of Hemisphere of volume of the Iron used to make the tank;

$\frac{2}{3} \pi {r}^{3} ( {r}^{3} - {r}^{3} ) \: Cubic \: units \\ \\ Volume = \frac{2}{3} \times \frac{22}{7} \times ((1.01) {}^{3} - 1 {}^{3} ) \\ \\ Volume = \frac{44}{21} \times (1.030301 - 1) \\ \\ Volume = \frac{44}{21} \times 0.030301 \\ \\ Volume = \frac{1. 333244}{21} {m}^{3} \\ \\ Volume = 0.063487 {m}^{3}$

Thus,

The volume required of the tank is 0.063487m³

Answer 2

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Inner Radius of the tank, (r ) = $1m$

Outer Radius (R ) = $1.01$m

Volume of the iron used in the tank = $(\frac{2}{3})π(R^3– r^3)$

= $(\frac{2}{3})\times (\frac {22}{7})\times (1.01^3– 1^3)$

= $0.06348$

So, volume of the iron used in the hemispherical tank is $0.06348m^3$ .