Q63 / 75 · JEE Main 2025 (23 Jan Shift 1) · Physics

A gun fires a lead bullet of temperature 300 K into a wooden block. The bullet having melting temperature of 600 K penetrates into the block and melts down. If the total heat required for the process is 625 J , then the mass of the bullet is $\qquad$ grams. (Latent heat of fusion of lead $=2.5 \times 10^4 \mathrm{JKg}^{-1}$ and specific heat capacity of lead $=125 \mathrm{JKg}^{-1}$ $\left.\mathrm{K}^{-1}\right)$

  • Correct answer
    A
    $10$
  • B
    $20$
  • C
    $5$
  • D
    $15$

Solution

$\begin{aligned} & 625=\mathrm{ms} \Delta \mathrm{T}+\mathrm{mL} \\ & 625=\mathrm{m}\left[125 \times 300+2.5 \times 10^4\right] \\ & 625=\mathrm{m}[37500+25000] \\ & 625=\mathrm{m}[62500] \\ & \mathrm{m}=\frac{1}{100} \mathrm{~kg} \\ & \mathrm{M}=10 \text { grams }\end{aligned}$

Hence, the correct answer is option: 'A'.

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