Thermal Properties of Matter

JEE Main level, single correct, medium

63 of 126

Question 63 of 126

A lead bullet penetrates into a solid object and melts. Assuming that 4040% of its kinetic energy is used to heat it, the initial speed of bullet is
(Given, initial temperature of the bullet =127∘C=127^{\circ}C,
Melting point of the bullet =327∘C=327^{\circ}C,
Latent heat of fusion of lead =2.5×104Jkg−1=2.5\times {10}^{4}J{kg}^{-1},
Specific heat capacity of lead =125JkgK−1=125Jkg{K}^{-1})

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Answer: (B) 500ms−1500{ms}^{-1}

Loss of kinetic energy of the lead bullet==Heat generated to raise its temperature from 127∘C127^{\circ}C to 327∘C327^{\circ}C.
4040% loss of the kinetic energy of the lead bullet==Heat gained by the lead bullet to raise its temperature.
0.4(12mv2)=m×[SPb×∆T+m×LPb]0.4\left(\dfrac{1}{2}m{v}^{2}\right)=m\times \left[{S}_{Pb}\times ∆T+m\times {L}_{Pb}\right]
0.4×12mv2=mSPb(T2−t1)+mLPb0.4\times \dfrac{1}{2}m{v}^{2}=m{S}_{Pb}\left({T}_{2}-{t}_{1}\right)+m{L}_{Pb}
0.2mv2=m×[125×(327−127)+2.5×104]0.2m{v}^{2}=m\times \left[125\times \left(327-127\right)+2.5\times {10}^{4}\right]
v2=5×[125×200+2.5×104]{v}^{2}=5\times \left[125\times 200+2.5\times {10}^{4}\right]
v2=5×[25000+25000]{v}^{2}=5\times \left[25000+25000\right]
v=500ms−1v=500m{s}^{-1}

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