Properties Of Solids And Liquids Question 45
Question: A bullet moving with a uniform velocity v, stops suddenly after hitting the target and the whole mass melts be m, specific heat S, initial temperature 25°C, melting point 475°C and the latent heat L. Then v is given by [NCERT 1972]
Options:
A) $ mL=mS(475-25)+\frac{1}{2}\cdot \frac{mv^{2}}{J} $
B) $ mS(475-25)+mL=\frac{mv^{2}}{2J} $
C) $ mS(475-25)+mL=\frac{mv^{2}}{J} $
D) $ mS(475-25)-mL=\frac{mv^{2}}{2J} $
Show Answer
Answer:
Correct Answer: B
Solution:
Firstly the temperature of bullet rises up to melting point, then it melts.
Hence according to $ W=JH $ .
therefore $ \frac{1}{2}mv^{2}=J.[m.c.\Delta \theta +mL]=J[mS(475-25)+mL] $
therefore $ mS(475-25)+mL=\frac{mv^{2}}{2J} $
BETA
