Laws of Motion

JEE Main level, single correct, medium

144 of 210

Question 144 of 210

A coin placed on a rotating table just slips when it is placed at a distance of 1cm1cm from the centre. If the angular velocity of the table is halved, it will just slip when placed at a distance of _____ from the centre:

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Answer: (B) 4cm4cm

The coin will slip when
μmg=mω2r\mu mg=m{\omega }^{2}r
⇒μg=ω2×1cm...(i)\Rightarrow \mu g={\omega }^{2}\times 1cm...(i)
Now, ω′=ω2\omega '=\dfrac{\omega }{2}
μg=(ω2)2R...(ii)\mu g={\left(\dfrac{\omega }{2}\right)}^{2}R...(ii)
Hence we can write from (i) and (ii)
ω2×1cm=ω24R⇒R=4cm{\omega }^{2}\times 1cm=\dfrac{{\omega }^{2}}{4}R \Rightarrow R=4cm

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