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An iron rocket fragment initially at -100^oC enters the earth's atmosphere almost horizontally and quickly fuses completely in atmospheric friction. Specific heat of iron is 0.11 kcal/kg ^oC its melting point is 1535^oC and the latent heat of fusion is 3 kcals/kg. The minimum velocity with which the fragment must have entered the atmosphere is

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0.45 km/s

1.32 km/s

2.32 km/s

zero

answer is B.

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Detailed Solution

The entire kinetic energy of the fragment is changed to heat. Expressing mass m in kilograms everywhere, we have

$\begin{array}{l} \frac{1}{2} {mv}^{2} = [m (3) + m (0 .11) (1535^{\circ} + 100^{\circ})] 4184 \\ v^{2} = (8368) [3 + 180] ≃ 1 .76 \times 10^{6} \\ \Rightarrow v = (\sqrt{1 .76}) \times 10^{3} = 1 .32 km / s \end{array}$

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