Q.

A bullet is fired horizontally down an open cylinder of cross-sectional area 102m2 . When the gun is fired, bullet is at rest and the volume between the end of cylinder and bullet is 5×104m3, and pressure of gas in this volume is 8P0 where P0 is atmospheric pressure. Gaseous mixture in the cylinder has γ=1.5. The bullet moves down quickly, so that no heat is transferred to the gas. Friction between bullet and the barrel is negligible and no gas leaks around bullet. P0=105N/m2

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a

If bullet leaves with maximum kinetic energy, then minimum length of barrel required
is 20 cm

b

If bullet leaves with maximum kinetic energy, then minimum length of barrel required
is 10 cm

c

The maximum kinetic energy with which bullet can leave is 250 J

d

The maximum kinetic energy with which bullet can leave is 150 J

answer is B, C.

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Detailed Solution

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Since no heat transferred  process is adiabatic.

Watm +Wgas =ΔKEbullet Wgas =PiViPfVfγ1=8P0×5×104P0Vf0.5 Also 8P0×5×104γ=P0Vfγ Vf=82/3×5×104m3=20×104m3Wgas =8P0×5×104P0×20×1040.5 =P0×104×20(21)0.5=400J
Given 102×x=5×104x=5×102m
Also 102L=20×104L=20×102m
Watm=P0VfVi=P015×104=150J

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