Q. A hammer of mass $1kg$ having speed of $50\ m/s,$ hit a iron nail of mass $200\ gm.$ If specific heat of iron is $0.105\ cal/gm℃$ and half the energy is converted into heat, the raise in temperature of nail is
A) $7.1℃$
B) $9.2℃$
C) $10.5℃$
D) $12.1℃$
Explanation
(a) $W = JQ \Rightarrow \frac{1}{2}\left( \frac{1}{2}Mv^{2} \right) = J(m.c.\mathrm{\Delta}\theta)$
$\Rightarrow \frac{1}{4} \times 1 \times (50)^{2} = 4.2\lbrack 200 \times 0.105 \times \mathrm{\Delta}\theta\rbrack \Rightarrow \mathrm{\Delta}\theta = 7.1℃$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.