Q. A steel ball of mass 0.1 kg falls freely from a height of 10 m of 10 m 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
A) $0.01℃$
B) 0.1$℃$
C) 1.1$℃$
D) 1$℃$
Explanation
(b) Here , $m = 0.1\ $kg, $h_{1} = 10m,\ h_{2} = 5.4\ m$
$c = 460$ J-k$g^{- 1}℃^{- 1},\ g = 10ms^{- 2},\ \theta = ?$
Energy dissipated, $Q = mg\left( h_{1} - h_{2} \right)$
= 0.1 $\times 10(10 - 5.4) = 4.6j$ J
From $Q = c\ m\ \theta$
$\theta = \frac{Q}{cm} = \ \frac{4.6}{460 \times 0.1\ } = 0.1℃$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.