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A marble block of mass 2 kg lying on ice when given a velocity of 6 m/s is stopped by friction in 10 sec. Then the coefficient of friction is
Options
(a) 0.01
(b) 0.02
(c) 0.03
(d) 0.06
Correct Answer:
0.06
Explanation:
From equation of motion, v = u – at
⇒ o = u – μgt
⇒ μ = u/gt
= 6/(10×10)
= 0.06
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Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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