Question

A steel ball of mass 0.1 kg falls freely from a height 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 - (Specific heat of steel = 460 J/kg/ºC, g = 10 m s^–2)

• A. 0.01⁰C
• B. 0.1⁰C
• C. 1⁰C
• D. 1.1⁰C

Answer 1

Answer: (B)

0.1⁰C

Answer 2

mg(h₁ – h₂) = msDq

Dq = $\frac{g(h_{1} - h_{2})}{5}$ = $\frac{10 \times 4.4}{460}$ = $\frac{44}{460}$ = 0.1 ⁰C