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प्रश्न
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उत्तर
Given,initial velocity, u = 70 m/s
final velocity, v = 0 m/s
sp. heat capacity of lead,s = 140 J/kgK
Let Δt be the change in temperature
K.E possessed by the bullet = `1/2` m (70)2 = 2450m joules
Heat energy = 80% of K.E = `80/100 xx 2450`m = 1960m joules
We know that, H = m s Δt
or, 1960m = m x 140 x Δt
or, Δt = `1960/40` = 14 °C
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