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(A) Given,length of the pendulum $\ell = 10 \ m$ and acceleration due to gravity $g = 10 \ ms^{-2}$.Initially,the bob is at a horizontal position,so i

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$6 \sqrt{5} \ ms^{-1}$

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(A) Given,length of the pendulum $\ell = 10 \ m$ and acceleration due to gravity $g = 10 \ ms^{-2}$.Initially,the bob is at a horizontal position,so its initial potential energy with respect to the lowest point is $U_i = mg\ell$.The bob is released from rest,so its initial kinetic energy is $K_i = 0$.Total initial energy $E_i = mg\ell$.As the bob moves to the lowest point,it dissipates $10 \%$ of its initial energy against air resistance.Energy lost = $0.10 	imes mg\ell$.Remaining energy at the lowest point = $E_f = E_i - 0.10 	imes E_i = 0.90 	imes mg\ell$.At the lowest point,the potential energy is $0$,so the entire remaining energy is kinetic energy $K_f = \frac{1}{2}mv^2$.Equating the energies: $\frac{1}{2}mv^2 = 0.90 	imes mg\ell$.$v^2 = 2 	imes 0.90 	imes g 	imes \ell = 1.8 	imes 10 	imes 10 = 180$.$v = \sqrt{180} = \sqrt{36 	imes 5} = 6\sqrt{5} \ ms^{-1}$.

The bob of a pendulum was released from a horizontal position. The length of the pendulum is $10 \ m$. If it dissipates $10 \%$ of its initial energy against air resistance,the speed with which the bob arrives at the lowest point is : [Use $g = 10 \ ms^{-2}$]

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