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

Question

A bullet of mass 10× 10-3kg moving with a speed of 20 text ms-1 hits an ice block (0?C) of 990 g kept at rest on a frictionless floor and gets embedded in it. If ice takes 50% of KE lost by the system, the amount of ice melted (in grams) approximately is: (J = 4.2 J/cal) (Latent heat of ice = 80 cal/g). (A) 6 (B) 3 (C)  6× 10-3  (D)  3× 10-3

Tardigrade Admin · Accepted Answer

Kinetic energy of bullet =(1/2)mv2 =(1/2)× 10× 10-3× 20× 20 =2J=(2/4.2) cal Heat gained by ice to melt =(2/4.2)× (50/100) or mL=(1/4.2) or m=(1/4.2× 80)g =0.002976 g =0.002976 g

6

3

$6 \times 10^{- 3}$

$3 \times 10^{- 3}$

Kinetic energy of bullet $= \frac{1}{2} m v^{2}$ $= \frac{1}{2} \times 10 \times 10^{- 3} \times 20 \times 20$ $= 2 J = \frac{2}{4.2} c a l$ Heat gained by ice to melt $= \frac{2}{4.2} \times \frac{50}{100}$ or $m L = \frac{1}{4.2}$ or $m = \frac{1}{4.2 \times 80} g$ $= 0.002976 g$ $= 0.002976 g$

6

3

6×10−3

3×10−3

Kinetic energy of bullet =21​mv2 =21​×10×10−3×20×20 =2J=4.22​cal Heat gained by ice to melt =4.22​×10050​ or mL=4.21​ or m=4.2×801​g =0.002976g =0.002976g

$6 \times 10^{- 3}$

$3 \times 10^{- 3}$

Kinetic energy of bullet $= \frac{1}{2} m v^{2}$ $= \frac{1}{2} \times 10 \times 10^{- 3} \times 20 \times 20$ $= 2 J = \frac{2}{4.2} c a l$ Heat gained by ice to melt $= \frac{2}{4.2} \times \frac{50}{100}$ or $m L = \frac{1}{4.2}$ or $m = \frac{1}{4.2 \times 80} g$ $= 0.002976 g$ $= 0.002976 g$