A lead bullet penetrates into a solid object and melts. Assuming that 40% of its kinetic energy is used to heat it, the initial speed of bullet is: (Given: initial temperature of the bullet = 127 ° C, Melting point of the bullet = 327 ° C, Latent heat of fusion of lead = 2.5 × 104 J kg ⁻ 1, Specific heat capacity of lead = 125 J/kg K) (a) 125 ms ⁻ 1 (b) 500 ms ⁻ 1 (c) 250 ms ⁻ 1 (d) 600 ms ⁻ 1

Answer: (b). 0.4 × 1/2mv² = m(cΔ T + L) 0.2 v² = 125 × 200 + 2.5 × 10⁴ = 5 × 10⁴ v² = 2.5 × 10⁵ ⇒ v = 500 m s⁻¹ So the answer is (b) 500 m s⁻¹.

A lead bullet penetrates into a solid object and melts. Assuming that… | JEE Main 2022 (27 Jun, Shift 2)

Q24JEE Main 2022 · 27 JUN · SHIFT 2Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat

A lead bullet penetrates into a solid object and melts. Assuming that 40% of its kinetic energy is used to heat it, the initial speed of bullet is:

(Given: initial temperature of the bullet = 127 $^{\circ}$ C, Melting point of the bullet = 327 $^{\circ}$ C, Latent heat of fusion of lead = 2.5 $\times$ 104 J kg $^{-}$ 1, Specific heat capacity of lead = 125 J/kg K)

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Answer: (b)

$0.4 \times \tfrac{1}{2}mv^2 = m(c\Delta T + L)$
$0.2\,v^2 = 125 \times 200 + 2.5 \times 10^4 = 5 \times 10^4$
$v^2 = 2.5 \times 10^5 \Rightarrow v = 500\text{ m s}^{-1}$
So the answer is (b) $500\text{ m s}^{-1}$ .

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