Question

A bullet of mass $20\text{g}$ moving with a speed of $20\text{m/s}$ hits on ice block of mass $990\text{g}$ kept on a frictionless floor and gets stuck in it. How much ice will melt if $50\text{\%}$ of the lost $\text{KE}$ goes to ice?
(initial temperature of the ice block and bullet $=0{}^{\circ }\text{C}$ and the latent heat of fusion of ice is, $L=336000\text{J/kg})$

• A. $0.001\text{g}$
• B. $0.003\text{g}$
• C. $0.006\text{g}$
• D. $0.008\text{g}$

Answer 1

The correct option is C $0.006\text{g}$

The heat energy required to melt ice of $m\text{gram}$ is

$Q=(m\times L)$

$\Rightarrow Q=[m\times (336000\left)\right]\text{J}$

$\text{K.E.}$ of bullet,

$\text{KE}=\frac{1}{2}{m}_{bullet}{u}^2=\frac{1}{2}\times (20\times {10}^{-3})\times {20}^2$

$\Rightarrow \text{KE}=4\text{J}$

$50\text{\%}$ of $\text{KE}$ goes to melting of ice.

$\therefore \left(\text{KE}\right)\times \frac{1}{2}=[m\times (336000\left)\right]$

$\Rightarrow 2=m\times \left(336000\right)$

$\therefore m=\frac{2}{336000}\approx 6\times {10}^6\text{kg}=0.006\text{g}$

Hence, option $\left(\text{C}\right)$ is correct.