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

A bullet of mass 10×10−3 kg and temperature 0∘C, moving with a sp

1.	A bullet of mass 10×10−3 kg and temperature 0∘C, moving with a speed of 20 ms−1 hits an ice block (0∘C) of mass 990 g kept at rest on a frictionless floor and gets embedded in it. If ice takes 50% of K.E lost by the system as the heat energy, the amount of ice melted (in grams) approximately is :(Latent heat of ice =80 cal/g)
Answer» A bullet of mass $10 \times 10^{- 3} kg$ and temperature $0^{\circ} C$ , moving with a speed of $20 m s^{- 1}$ hits an ice block $(0^{\circ} C)$ of mass $990 g$ kept at rest on a frictionless floor and gets embedded in it. If ice takes $50 %$ of K.E lost by the system as the heat energy, the amount of ice melted (in $grams$ ) approximately is : (Latent heat of ice $= 80 cal/g$ )

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