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JEE Main Physics question

PhysicsSystems of Particles and Rotational MotionJEE Main

A bullet is fired from a gun. The force on the bullet is given by F = 600 − 2 × 10⁵t, where F is in newtons and t is in seconds. The force on the bullet becomes zero as soon as it leaves the barrel. What is the average impulse imparted to the bullet?

Answer: (C)

Solution

F = 0 when t = 600/(2×10⁵) = 0.003 s. Impulse = ∫₀^0.003 (600 − 2×10⁵t)

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Question ID VK-153277