A lead bullet of mass 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 `= 150 J//kg^(@)C`)

Options

A. `100^(@)C`
B. `125^(@)C`
C. `150^(@)C`
D. `200^(@)C`

Answer

C

Explanation

Heat released = Heat absorbed
`Q = (1)/(2)K.E`
`implies msDeltat=(1)/(2)xx(1)/(2)xxmxxv^(2)`
`implies (150)Deltat=(1)/(4)xx(300)^(2)`
`therefore Deltat=150^(@)C`