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# To one end of a light inextensible string oflength L is attached a particleof mass m. The particle is on a smooth horizintal table. The string passes through a hole in the table and to its other end is attached a small particle of equal mass m. The system is set in motion with the first particle describing a circle on the table with constant angular velocity omega1 and the second particle moving in a hrizintla circle as a conical pendulum with constant angular velocity omrga2.(a) then the length of the poritond of string on either side of the hole are in the ratio omega2^2:omega1^2(b) omega1 and omega2 satisfy the relation1/omega1^2 +1/omega2^2 > L/g(c) tension in the two parts of the string are equal(d) omega1 and omega2 satisfy the relation1/omega1^2 +1/omega2^2 < L/g

509 Points
10 years ago

let the length of thread over the table is L1 & below the table is L2 ...

L1+L2 = L        ...........1

tension in upper part as wel as lower part will be same because the thread is massless  , let it beT ....

mass m is circulating with angular velocity W1 so

T = mW12L1         ........2                  (L1 is the radius of circle)

(tension in the thread provides necessary centripital force )

now ,

another mass is moving in conical pendulam , let it is making an angle @ with verticle then

Tcos@ = mg         ..........3

Tsin@ = mW22 R            ........4               (R is the radius of circular path)

R = L2sin@

putting value of R in eq4 we get

T = mW22L2            ........5

equating 2 & 5

L1/L2 = W22/W12                                 ans  (a)

1/W12 + 1/W22 = mL1/T + mL2/T                                  (from eq 2 & 5 )

=m(L1+L2)/T

=mL/T                                                  (L1+L2=L)

=Lcos@/g                                               ( substituting T from eq 3)

cos@ = max =1 so

1/W12+1/W22 < L/g                          ans (d )

thus option a,c,d are correct