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Q.

A bullet of mass 10×10^–3kg moving with a speed of 20ms^–1 hits an ice block (0°C) of 990g kept at rest on a frictionless floor and gets embedded in it. If ice takes 50% of K.E, the amount of ice melted (in grams) approximately is (J=4.2J/cal) (Latent heat of ice = 80cal/g)

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a

3 × 10^–3

b

6 × 10^–3

c

6

d

3

answer is D.

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Detailed Solution

$\begin{array}{l} m_{b} = 10^{- 2} kg; V = 20 m / s \\ \frac{1}{2} (\frac{1}{2} m_{b} V^{2}) = {Jm}_{iec} L_{ice} \\ \frac{1}{4} \times 10^{- 2} \times 400 = 4 .2 m_{ice} \times 80 \\ m_{ice} = \frac{1}{366} = 0 .0029 \approx 0 .003 = 3 \times 10^{- 3} gm \end{array}$

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