A lead ball at 30^ C is dropped from a height of 6.2 km . The ball...

A lead ball at

30^{\circ} \mathrm{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. If the specific heat of lead

=126 \mathrm{Jkg}^{-10} \mathrm{C}^{-1}

and melting point of lead

=130 \quad{ }^{\circ} \mathrm{C}

and suppose that any mechanical energy lost is used to heat the ball, then the latent heat of fusion of lead is

Options

A $2.4 \times 10^4 \mathrm{Jkg}^{-1}$
B $3.6 \times 10^4 \mathrm{Jkg}^{-1}$
C $7.6 \times 10^2 \mathrm{Jkg}^{-1}$
D $4.2 \times 10^3 \mathrm{Jkg}^{-1}$

Correct answer

A. $2.4 \times 10^4 \mathrm{Jkg}^{-1}$

Step-by-step solution

The gravitational potential energy of the ball

=m g h

\begin{aligned} & =m \times 10 \times 6.2 \times 10^3 \\ & =m \times 62 \times 10^4 \mathrm{~J}\end{aligned}

Now energy required to take ball from $30^{\circ} \mathrm{C} to 330^{\circ} \mathrm{C}$ is $m \times 126 \times 300=m \times 37800$ Energy required to melt the ball $=m L$ where, $L=$ latent heat

\begin{aligned} \Rightarrow m \times 6.2 \times 10^4 \times L & =m \times 37800+m L \\ L & =2.4 \times 10^4 \mathrm{Jkg}^{-1}\end{aligned}$$

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