InterviewSolution
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InterviewSolution
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A ring of radius R having unifromly distributed charge Q is mounted on a rod suspended by two identical strings. The tension in strings in equilibrium is `T_0.` Now a vertical magnetic field is switched on and ring is rotated at constant angular velocity `omega`. Find the maximum `omega` with which the ring can be rotated if the strings can withstand a maximum tension of `3T_0 //2.` |
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Answer» Correct Answer - B::D In equilibrium, `2T_0=mg` or `T_0=(mg)/2`…….i Magnetic moment `M=iA=(omega/(2pi)Q)(piR^2)` `tau=MBsin90^@=(omegaQR^2)/2` Let `T_1` and `T_2` be the tension in the two strings when magnetic field is switched on `(T_1ltT_2)`. For translation equilibrium of ring in vertical direction. `T_1+T_2=mg`...........i For rotationasl equilibrium `(T_1-T_2)D/2=tau=(omegaBQR^2)/2` or `T_1-T_2=(omegaBQR^2)/2`.........iii Solving eqn ii and iii, we have `T_1=(mg)/2+(omegaBQR^2)/(2D)` As `T_1gtT_2` and maximum values of `T_1` and `(3T_0)/2`, We have `(3T_0)/2=T_0+(omega_maxBQR^2)/(2D)` `:. omega_max=(DT_0)/(BQR^2)` |
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