KVPY 2019 SA Exam Previous Papers

Question : 22 of 80

Marks: +1, -0

A ball is moving uniformly in a circular path of radius 1m with a time period of 1.5 s. If the ball is suddenly stopped at t = 8.3 s, the magnitude of the displacement of the ball with respect to its position at t = 0 s is closest to :

Validate

Solution:

∆ t = 8.3 - 7.5 = 0.8

sec.

T={2π}/ω={2×3.14}/ω=1.5

ω={2×3.14}/1.5

θ=ω∆ t

={2×3.14}/ 1.5×0.8≈3.14≈π

rad
So, displacement = 2R = 2m

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