An alpha particle enters a hollow tube of 4 m length with an initial speed of 1kms. It is accelerated in the tube and comes out of it with a speed of 9kms. The time for which it remains inside the tube is
A
8×10−3s
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B
80×10−3s
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C
800×10−3s
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D
8×10−4s
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Solution
The correct option is D8×10−4s v2=u2+2as⇒(9000)2−(1000)2=2×a×4 ⇒a=107ms2 Now t=v−ua ⇒t=9000−1000107=8×10−4sec