Question

# A lead bullet penetrates into a solid object and melts. Assuming that 50% of its kinetic energy was used to heat it, calculate the initial speed of the bullet. The initial temperature of the bullet is 27∘C and its melting point is 327∘C. Latent heat of fusion of lead = 2.5 × 104 J kg−1 and specific heat capacity of lead = 125 J kg−1 K−1.

A

480 m/s

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B

5000 m/s

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C

100 m/s

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D

1000m/s

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Solution

## The correct option is A 480 m/s Let the mass of the bullet = m. Heat required to take the bullet from 27∘C to 327∘C. = m × (125 Jkg−1K−1)(300K) = m × (3.75 × 104 J kg−1). Heat required to melt the bullet = m × (2.5 × 104Jkg−1) If the initial speed by v, the kinetic energy is 12mv2 and hence the heat developed is 12(12mv2) = 14mv2. Thus, 14mv2 = m(3.75+2.5) × 104Jkg−1 or, v = 480 m s−1.

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