Question

A lead bullet of unknown mass is fired with a speed of 180 ms-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 (Specific heat of lead $ =\text{ }0.120\text{ }J{{g}^{-1}}{{C}^{-1}} $ )

• A. $ 140{}^\circ C $
• B. $ 106{}^\circ C $
• C. $ 90{}^\circ C $
• D. $ 100{}^\circ C $

Answer 1

Answer: (C)

Since specific heat of lead is given in joule. Specific heat of lead $ =0.120\,J/g{{\,}^{o}}C=120\,J/kg $ The two-third of head produced goes into the bullet So, $ m\times s\times \Delta \theta =\frac{2}{3}\times \frac{1}{2}m{{v}^{2}} $ $ \Delta \theta =\frac{{{v}^{2}}}{3\times s} $ $ =\frac{180\times 180}{3\times 120}=90{{\,}^{o}}C $