A simple pendulum with a bob of mass m = 1 k | vedclass

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(D) The effective acceleration $g_{eff}$ is given by the vector sum of gravity and the electric force per unit mass.$g_{eff} = \sqrt{g^2 + (qE/m)^2} =

Vedclass · Accepted Answer

None of these

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(D) The effective acceleration $g_{eff}$ is given by the vector sum of gravity and the electric force per unit mass.$g_{eff} = \sqrt{g^2 + (qE/m)^2} = \sqrt{10^2 + (5 	imes 10^{-6} 	imes 2 	imes 10^6 / 1)^2} = \sqrt{100 + 10^2} = \sqrt{200} = 10\sqrt{2} \ m/s^2$.The condition for the particle to just complete the circle (or leave the path at the top) is $v_C^2 = g_{eff} 	imes l = 10\sqrt{2} 	imes 1 = 10\sqrt{2}$.Using the work-energy theorem between point $A$ and $C$:$\frac{1}{2}mu^2 = \frac{1}{2}mv_C^2 + m g_{eff} (2l)$$u^2 = v_C^2 + 4 g_{eff} l = 10\sqrt{2} + 4(10\sqrt{2})(1) = 50\sqrt{2} \ m^2/s^2$.$u = \sqrt{50\sqrt{2}} \ m/s$. Since this is not among the options, the correct answer is $D$.

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