Question

A ball is dropped on a floor from a height of $2$m. After the collision it rises upto a height of $1$m. Assuming that $20\%$ of mechanical energy is lost in the form of thermal energy. If the specific heat capacity of the ball is $800$ $J/K$ then the rise in temperature of the ball during collision is (Take $g=10m/s^2$).

• A. $2.5\times {10}^{-3}{}^{o}C$
• B. $7.5\times {10}^{-3}{}^{o}C$
• C. $1.5\times {10}^{-3}{}^{o}C$
• D. $5\times {10}^{-3}{}^{o}C$

Answer 1

The correct option is A $2.5\times {10}^{-3}{}^{o}C$

Loss in mechanical energy$=$ change in PE
$=mg(h_2-h_1)$
Also $20\%$ loss in ME$=$heat gained by ball
$\frac{20}{100}\times mg(h_2-h_1)=ms\Delta T$
$\frac{1}{5}\times 10(2-1)=800\times \Delta T$
$\Delta T=\frac{10}{5\times 800}=\frac{1}{400}$
$\Delta T=2.5\times {10}^{-3}{}^{o}C$.