Properties Of Solids And Liquids Question 63
Question: A lead bullet of 10 g travelling at 300 m/s strikes against a block of wood and comes to rest. Assuming 50% of heat is absorbed by the bullet, the increase in its temperature is (Specific heat of lead = 150J/kg, K) [EAMCET 2001]
Options:
A) 100°C
B) 125°C
C) 150°C
D) 200°C
Show Answer
Answer:
Correct Answer: C
Solution:
Since specific heat of lead is given in Joules,
hence use $ W=Q $ instead of $ W=JQ $ .
therefore $ \frac{1}{2}\times ( \frac{1}{2}mv^{2} )=m.c.\Delta \theta $
therefore $ \Delta \theta =\frac{v^{2}}{4c}=\frac{{{(300)}^{2}}}{4\times 150}=150{}^\circ C $ .
