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A hemispherical vessel of radius 14 cm is fully filled with sand. This sand is poured on a level ground. The heap of sand forms a cone shape of height 7cm. Calculate the area of ground occupied by the circular base of the heap of the sand.

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2464 {cm}^{2}

2364 {cm}^{2}

2264 {cm}^{2}

2164 {cm}^{2}

answer is A.

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Detailed Solution

Given,
The hemisphere vessel of Radius R = 14 cm
The hemisphere vessel's volume

\begin{matrix} V_{1} = \frac{2}{3} π \times R^{3} \\ \Rightarrow V_{1} = \frac{2}{3} \times \frac{22}{7} \times 14^{3} \end{matrix}

Now,
Consider radius of cone base =r radius.
Cone of height

h = 7 cm

Thus, the cone volume

\begin{matrix} V_{2} = \frac{1}{3} π \times r^{2} \times h \\ V_{2} = \frac{1}{3} \times \frac{22}{7} \times r^{2} \times 7 \end{matrix}

As, volume remain same after recasting. So,

V_{1} = V_{2} \frac{2}{3} \times \frac{22}{7} \times 14^{3} = \frac{1}{3} \times \frac{22}{7} \times r^{2} \times 7 2 \times 14^{3} = r^{2} \times 7 r^{2} = \frac{2 \times 14^{3}}{7} r^{2} = 784 r = \sqrt{784} r = 28 cm

Now, area of the earth that the sand heap's circular base occupies,

A = π \times r^{2}

A = \frac{22}{7} \times 28^{2}

A = 2464 {cm}^{2}

Hence, option 1 is correct.

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