A lead bullet of unknown mass is fired with a speed of $180 m s^{- 1}$ into a tree in which it stops. Assuming that in this process two.third of heat produced goes into the bullet and one-third into wood. The temperature of the bullet rises by

140^{o} C

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106^{o} C

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90^{o} C

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100^{o} C

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Solution

The correct option is C $90^{o} C$
Specific heat of lead $S = 0.120 J / g^{o} C$
Thus we get $S = 120 J / k g$
The two-third of heat produced goes into the bullet.
So, $m \times S \times Δ θ = \frac{2}{3} \times \frac{1}{2} m v^{2}$
where $Δ θ$ is the rise in temperature of bullet.
$Δ θ = \frac{v^{2}}{3 \times S} = \frac{180 \times 180}{3 \times 120} = 90^{o} C$

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A lead bullet of unknown mass is fired with a speed of 180ms−1 into a tree in which it stops. Assuming that in this process two.third of heat produced goes into the bullet and one-third into wood. The temperature of the bullet rises by

The correct option is C 90oCSpecific heat of lead S=0.120J/goCThus we get S=120J/kgThe two-third of heat produced goes into the bullet.So, m×S×Δθ=23×12mv2where Δθ is the rise in temperature of bullet.Δθ=v23×S=180×1803×120=90oC

A lead bullet of unknown mass is fired with a speed of $180 m s^{- 1}$ into a tree in which it stops. Assuming that in this process two.third of heat produced goes into the bullet and one-third into wood. The temperature of the bullet rises by