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`).

Options

A. `410 m//sec`
B. `1230 m//sec`
C. `307.5 m//sec`
D. none of the above

Answer

A

Explanation

If mass of the bullet is `m gm`. Then total heat required for bullet to just melt down
`Q_(1) = mc Delta theta +mL = mxx0.03(327-27)+mxx6`
`=15 m cal = (15m xx 4.2) J`
Now when bullet is stopped by the obstacle, the loss in its machanical energy `=1/2 (mxx10^(-3))v^(2)J`
(As `m gm = m xx 10^(-3)kg`)
As `25%` of this energy is absorbed by obstacle,
The energy absorbed by the bullet
`Q_(2) = 75/100 xx 1/2 mv^(2) xx 10^(-3) = 3/8 mv^(2) xx 10^(-3)J J`
Now the bullet will melt if `Q_(2) ge Q_(1)`
i.e., `3/8 mv^(2) xx 10^(-3) ge 15 m xx 4.22 rArr v_(min) = 410 m//s`.