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Q.

Two identical particles are attached at the ends of a light string which passes through a hole at the center of a table. One of the particles is made to move in a circle on the table with angular velocity ω1 and the other is made to move in a horizontal circle as a contact pendulum with angular velocity ω2. If l1 and l2 are the length of the string over and under the table, then in order to particle under the table neither moves down nor moves up, the ratio l1/l2 is

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a

ω1ω2

b

ω2ω1

c

ω12ω22

d

ω22ω12

answer is D.

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Detailed Solution

T=mω12l1    .......(i) Tsin⁡θ=mω22l2sin⁡θ⇒T=mω22l2   .......(ii)⇒ l1l2=ω22ω12
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