A lead bullet, of initial temperature 27⁰ C and speed ' V' kmph pnetrates into a solid object and melts. If 50% of the kinetic energy is used to heat it, the value of V in kmph is (for leads melting point=600K, latent heat of fusion = 2.5×10⁴JKg^–1; specific heat=125J kg^–1 K^–1)

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1200

1000

3600

1800

answer is A.

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Detailed Solution

$\begin{array}{l} T_{1} = 273 + 27 = 300 k; T_{2} = 600 k \\ ΔT = T_{2} - T_{1} = 600 - 300 = 300 k \\ L = 2 .5 \times 10^{4} J / kg; S = 125 J / kg - k \\ \frac{50}{100} (\frac{1}{2} {mv}^{2}) = mL + mSΔt \\ v^{2} = 4 [L + SΔt] = 4 [25000 + 125 \times 300] \\ v^{2} = (2 \times 250)^{2} = 500 m / s \\ ∴ v = 500 \times \frac{18}{5} = 1800 kmph \end{array}$