A bullet is fired horizontally down an open end of cylinder (barrel) of cross-sectional area $10^{- 2} m^{2} .$ When the gun is fired, bullet is at rest and the volume between the closed end of cylinder and bullet is $5 \times 10^{- 4} m^{3}$ and pressure of gas in this volume is $8 P_{0}$ where $P_{0}$ 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. $(P_{0} = 10^{5} N / m^{2})$

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

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

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

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

answer is B, C.

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Detailed Solution

Since no heat transferred $\Rightarrow$ process is adiabatic.

$∴ W_{a t m} + W_{g a s} = Δ K E_{b u l l e t}$

$W_{g a s} = \frac{P_{i} V_{i} - P_{f} V_{f}}{γ - 1} = \frac{8 P_{0} \times 5 \times 10^{- 4} - P_{0} V_{f}}{0.5} A l s o, 8 P_{0} {(5 \times 10^{- 4})}^{γ} - P_{0} V_{f}^{γ} \Rightarrow V_{f} = 8^{2 / 3} \times 5 \times 10^{- 4} m^{3} = 20 \times 10^{- 4} m^{3}$

$∴ W_{g a s} = \frac{8 P_{0} \times 5 \times 10^{- 4} - P_{0} \times 20 \times 10^{- 4}}{0.5} = \frac{P_{0} \times 10^{- 4} \times 20 (2 - 1)}{0.5} = 400 J$

$G i v e n 10^{- 2} \times x = 5 \times 10^{- 4} \Rightarrow x = 5 \times 10^{- 2} m A l s o 10^{- 2} L = 20 \times 10^{- 4} \Rightarrow L = 20 \times 10^{- 2} m$