A lead bullet (specific heat $= 0.32 c a l . g m^{o} C$ ) is completely stopped when it strikes a target with a velocity of $300 m / s$ . The heat generated is equally shared by the bullet and the target. The rise in temperature of bullet will be-

{16.7}^{o} C

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{1.67}^{o} C

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{167.4}^{o} C

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{267.4}^{o} C

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Solution

The correct option is D ${167.4}^{o} C$
Let m be the mass of the bullet. The kinetic energy :
$K . E = \frac{1}{2} m v^{2} = \frac{1}{2} \times m \times {(300)}^{2} = 45000 J$
According to the law of conservation of energy, $θ$ being the rise in
temperature
$\begin{matrix} \frac{m \times 45000}{2} = m \times 4.2 \times 10^{3} \times 0.32 \times θ θ = \frac{45000}{2 \times 4.2 \times 10^{3} \times 0.032} = {167.41}^{\circ} C \end{matrix}$

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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-

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

A lead bullet (specific heat $= 0.32 c a l . g m^{o} C$ ) is completely stopped when it strikes a target with a velocity of $300 m / s$ . The heat generated is equally shared by the bullet and the target. The rise in temperature of bullet will be-