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Q.

Velocity of a particle of mass 2 kg varies with time 't' according to the equation v→=(2ti^−4j^)m/s m/s. Here 'r' is in seconds. The impulse imparted to the particle in the time interval 0 ≤ t ≤ 2s is:

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a

(4i^)kg−m/s

b

(8i^−4j^)kg−m/s

c

(8i^+4j^)kg−m/s

d

(8i^)kg−m/s

answer is D.

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Detailed Solution

Impulse = Change in momentum=2[(2×2i^−4j^)−(2×0i^−4j^)]=(8i^)kg−m/s
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