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

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