Question

A steel ball of mass 0.1 kg falls freely from a height of 10m and bounces to a height of 5.4 m from the ground. If the dissipated energy in this process is absorbed by the ball, the rise in its temperature is:

(Specific heat of steel = 460J/kg/${}^0$C, g=10ms${}^{-2}$)

• A. ${0.01}^{0}C$
• B. ${0.1}^{0}C$
• C. $1^{0}C$
• D. ${1.1}^{0}C$

Answer 1

The correct option is B ${0.1}^{0}C$

Change in potential energy $=$ Energy absorbed by the ball

$mg(h_1-h_2)=mc\Delta T$

$\Delta T=\frac{\left(10\right)(10-5.4)}{460}={0.1}^{o}C$