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From a metallic charged body a current is drawn. The rate of increase of current at an instant is equal to the charge on the body at that instant. If the initial charge on the body is QA. the minimum time it will take for the charge to become zero is `(pi)/(2)sec`B. the minimum time it will take for the charge to become zero is 2 secC. The value of the current when the charge on the body is `Q//2` is `Q sqrt((3)/(2))`D. The value of the current when the charge on the body is `Q//2` is `(Q)/(sqrt(3))` |
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Answer» Correct Answer - A Since current is going out of the body, `I = - (dq)/(dt)` Given `(di)/(dt)=q` `q = (-d^(2)q)/(dt^(2)) rArr q + (d^(2)q)/(dt^(2)) =0` Thus is differential equation of SHM with `omega =1` `T = (2pi)/(omega )=2 pi sec` The min. time for the charge to become zero `= T//4 = (pi)/(2) sec` Let `q = Q cos omega t` When `q = (Q)/(2)` , `cos omega t=(1)/(2), sin omegat =sqrt(1-(1)/(4))=(sqrt(3))/(2)` `i = Q omega (sqrt(3))/(2) = (Q sqrt(3))/(2)`. |
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