An alpha particle enters a hollow tube of 4 m length with an initial speed of 1 km/s. It is accelerated in the tube and comes out of it with a speed of 9 km/s. The time for which it remains inside the tube is

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8×10−3s

80×10−3s

800×10−3s

8×10−4s

answer is D.

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Detailed Solution

initial velocity=u=1000m/s;final velocity=v=9000m/s; length of tube=4mv2=u2+2as⇒(9000)2−(1000)2=2×a×4⇒a=107m/s2 Now t=v−ua⇒t=9000−1000107=8×10−4sec

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