Question

At $30^{\circ} C$, a lead bullet of $50\, g$, is fired vertically upwards with a speed of $840 \,m / s$. The specific heat of lead is $0.02\, cal / g ^{\circ} C$. On returning to the starting level, it strikes a cake of ice at $0^{\circ} C$. The amount of ice melted is (Assume all the energy is spent in melting only \& latent heat of fusion of ice is $80\, cal / g$ )

• A. 62.7 g
• B. 55 g
• C. 52.675 g
• D. 52.875 g

Answer 1

Answer: (D)

$m_{b} s_{b}(30-0)+\frac{1}{2} m v^{2}=m_{i c e} L$
$\Rightarrow 50 \times 0.02 \times(30-0)( cal )+\frac{1}{2} \times 50 \times 10^{-3} \times 840 \times 840( J )=m_{i c e} 80( cal )$
$\Rightarrow 30( cal )+\frac{1}{2} \times \frac{50 \times 10^{-3} \times 840 \times 840( cal )}{4.2}=m_{i c e} 80( cal )$
$\Rightarrow m_{i c e}=52.875\, g$