Kinetic Theory

JEE Main level, single correct, medium

94 of 168

Question 94 of 168

A bullet of mass 5gram5gram, travelling with a speed of 210ms−1210m{s}^{-1} strikes a fixed wooden target. One half of its kinetic energy is converted into heat in the wood. The rise of temperature of the bullet if the specific heat of its material is 0.030(gram∘C)−1(1calorie=4.2×107ergs)0.030{\left(gram^{\circ}C\right)}^{-1}\left(1calorie=4.2\times {10}^{7}ergs\right) close to :

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Answer: (A) 87.5∘C87.5^{\circ}C

As per given condition Half of kinetic energy is transferred into heat energy and raises the temperature of bullet
12×12mv2=msΔT\dfrac{1}{2}\times \dfrac{1}{2}m{v}^{2}=ms\Delta T
Δt=V24s\Delta t=\dfrac{{V}^{2}}{4s}
=210×2104×4.2×0.3×1000=87.5∘C=\dfrac{210\times 210}{4\times 4.2\times 0.3\times 1000}=87.5^{\circ}C

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