Question 6A 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.

Question

Accepted Answer

Given,Inner radius of the tank (r) = 1 mThickness of iron = 1 cm = 1100\dfrac{1}{100}1001​ m = 0.01 mOuter radius of the tank (R) = Inner radius + Thickness = 1 m + 0.01 m = 1.01 mVolume of iron used (V) = Volume of tank with outer radius - Volume of tank with inner radiusV=23πR3−23πr3=23π(R3−r3)=23×227×[(1.01)3−(1)3]=23×227×[1.030301−1]=23×227×0.030301=4421×0.030301=1.33324421=0.06348 m3.V = \dfrac{2}{3}πR^3 - \dfrac{2}{3}πr^3 \[1em] = \dfrac{2}{3}π(R^3 - r^3) \[1em] = \dfrac{2}{3} \times \dfrac{22}{7} \times [(1.01)^3 - (1)^3] \[1em] = \dfrac{2}{3} \times \dfrac{22}{7} \times [1.030301 - 1 ] \[1em] = \dfrac{2}{3} \times \dfrac{22}{7} \times 0.030301 \[1em] = \dfrac{44}{21} \times 0.030301 \[1em] = \dfrac{1.333244}{21} \[1em] = \text{0.06348 m}^3.V=32​πR3−32​πr3=32​π(R3−r3)=32​×722​×[(1.01)3−(1)3]=32​×722​×[1.030301−1]=32​×722​×0.030301=2144​×0.030301=211.333244​=0.06348 m3.Hence, volume of iron used to make the tank = 0.06348 m3.

Surface Areas and Volumes — Question 8

Question 6