Question

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

Answer 1

Hello!

• Inner radius, r = $\bf\small{1 \:m}$
• Outer radius, R = $\bf\small{1.01 \:m}$

Then,

$\underline{\bf{Volume\:of\:Iron}}$ :

$\sf \dfrac{2}{3}πr^{3}(R^{3}-r^{3}) \\ \\ \implies \dfrac{2}{3} \times \dfrac{22}{7} \times [(1.01)^{3}-(1)^{3}] \implies \boxed{\red{\bf{0.006348\:m^{3}}}}$

Cheers!

Answer 2

⇒ Answer :- 0.063487 cm^3

⇒ Given :-

Inner radius = 1 m

Iron sheet is 1 cm Thick

⇒ Solution :-

Let outer radius be R = Inner radius + thickness of iron sheet

= 101 cm

Inner radius be r = 1m = 100 cm

∴ Volume of the iron = (⅔)×(22/7)× (R^3-r^3)

(2/3)×(22/7)×(101^3-100^3)

44/21×30301 cm^3

63487.80952 cm^3

0.063487 cm^3