A lead bullet of 10 g travelling at 300 m$s^{- 1}$ strikes against a block of wood comes to rest. Assuming 50% of heat is absorbed by the bullet, the increase in is temperature is ( Specific heat of lead = 150 Jkg$K^{- 1})$

Question

Dr. Abubacker · Accepted Answer

(c) Here, $m = 10$ g =$10^{- 2}$ kg
$v = 300\ ms^{- 1},	heta = ?C, = 150$ J-k$g^{- 1}K^{- 1}$
$Q = \frac{50}{100}\left( \frac{1}{2}mv^{2} ight) = \frac{1}{4} 	imes 10^{- 2}(300)^{2} = 225\ $J
From $Q = cm\ 	heta$
$	heta\  = \frac{Q}{cm} = \frac{225}{150 	imes 10^{- 2}} = 150℃$

Q. A lead bullet of 10 g travelling at 300 m$s^{- 1}$ strikes against a block of wood comes to rest. Assuming 50% of heat is absorbed by the bullet, the increase in is temperature is ( Specific heat of lead = 150 Jkg$K^{- 1})$

Explanation

Reviewed by Dr. Abubacker

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