A lead ball at `25^@`C is dropped from a height of 2 km. It is heated due to air resistance and it is assumed that all of its kinetic energy is used in increasing the temperature of ball. Find the final temperature of the ball.

Options

A. `283.73^@C`
B. `353.73^@C`
C. `463.73^@C`
D. `183.73^@C`

Answer

D

Explanation

If there is no air resistence, velocity of ball, on reaching ground is
`V=sqrt(2gh)=sqrt(2xx10xx2000)=200(m)/(s)`
Kinetic energy of ball is `k=(1)/(2)mv^2`
`=(1)/(2)xxmxx(200)^2=2xx10^4xxmJ`
If all the kinetic energy is absorbed by the ball it gets heated.
Due to this, temperature of ball is increased by `DeltaT`, we have
`k=msDeltaT`
or `2xx10^4xxm=mxx126xxDeltaT`
or `DeltaT=(2xx10^4)/(126)=158.73^@C`
`:. T_f=25+158.73=183.73^@C`