A lead bullet just melts when stopped by an obstacle. Assuming that 25 per cent of the heat is absorbed by the obstacle, find the velocity of the bullet if its initial temperature is `27^@C` . (Melting point of lead `= 327^@C`, specific heat of lead `=0.03 cal//g.^@C`, latent heat of fusion of lead `=6 cal//g, J = 4.2 J//cal`).

Answer

12.96

Explanation

`75%` heat is retained by bullet
`(3)/(4)[(1)/(2)mv^(2)]=msDeltatheta+mLorv=sqrt(((8sDeltatheta+8L))/(3))`
Substituting the values, we have `v=sqrt(((8xx0.03xx4.2xx300)+(8xx6xx4.2))/(3))=12.96m//s`