Question

A bullet travelling at 100 $m{s}^{-1}$ suddenly hits a concrete wall. If its K.E. is converted completely into heat, the raise in temperature is $(s=100Jk{g}^{-1}{K}^{-1})$ :

• A. 20K
• B. 40K
• C. 50K
• D. 60K

Answer 1

The correct option is C 50K

Kinetic energy of the bullet is given by:
$KE=\frac{1}{2}m{v}^2$
The kinetic energy completely changes into heat.
$Q=ms\Delta T$
$KE=Q$
$\frac{1}{2}m{v}^2=ms\Delta T$
$\Delta T=\frac{v^2}{2s}=\frac{{100}^2}{2\times 100}=50K$

Hence option C is correct.