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Question

A lead bullet (specific heat =0.32cal.gmoC) is completely stopped when it strikes a target with a velocity of 300m/s. The heat generated is equally shared by the bullet and the target. The rise in temperature of bullet will be-

A
16.7oC
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B
1.67oC
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C
167.4oC
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D
267.4oC
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Solution

The correct option is D 167.4oC
Let m be the mass of the bullet. The kinetic energy :
K.E=12mv2=12×m×(300)2=45000J
According to the law of conservation of energy, θ being the rise in
temperature
m×450002=m×4.2×103×0.32×θθ=450002×4.2×103×0.032=167.41C

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