A lead ball at $30 ° C$ is dropped from a height of 6.2 km. The ball is heated due to the air resistance and it completely melts just before reaching the ground. The molten substance falls slowly on the ground. The latent heat of fusion of lead is (Specific heat capacity of lead $= 126 J/kg°C$ and melting point of lead $= 330 °C$ . Assume that any mechanical energy lost is used to heat the ball. Use $g = 10 {m/s}^{2}$ .)

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3.78 \times 10^{5} J/kg

3.78 \times 10^{4} J/kg

2.42 \times 10^{4} J/kg

2.4 \times 10^{5} J/kg

answer is C.

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Detailed Solution

t_{1} = 30 ° C,  h = 6.2 {km;L}_{f} = ?; S_{L} = 126 J

$t_{2} = 330 ° C;g = 10 {m/s}^{2}$

Potential energy $= mS Δ t + mL$

$m \times 10 \times 6 .2 \times 10^{3} = m (126 \times 300 + L)$

$62 \times 10^{3} = 378 \times 10^{2} + L$

$\Rightarrow (620 - 378) 10^{2} = L$

$\Rightarrow {242×10}^{2} = L$

$\Rightarrow 2.42 \times 10^{4} = L$

$∴ L = 2.42 \times 10^{4} J/kg$

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