A point mass m is attached to one end of a st | vedclass

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(D) The force acting on the point mass due to the string is radial,so the angular momentum remains conserved even when the string is pulled.Let the ma

Vedclass · Accepted Answer

$4:1$

Vedclass · Answer

(D) The force acting on the point mass due to the string is radial,so the angular momentum remains conserved even when the string is pulled.Let the mass of the particle be $m$,the initial radius be $r_1 = 2 \ m$,the initial linear velocity be $v_1 = 4 \ m/s$,and the initial angular velocity be $\omega_1 = v_1 / r_1 = 4 / 2 = 2 \ rad/s$.Let the final radius be $r_2 = 1 \ m$,the final linear velocity be $v_2$,and the final angular velocity be $\omega_2$.By the principle of conservation of angular momentum:$L_1 = L_2$$m r_1 v_1 = m r_2 v_2$$r_1 v_1 = r_2 v_2$Substituting the values:$2 	imes 4 = 1 	imes v_2$$v_2 = 8 \ m/s$The new angular velocity is:$\omega_2 = v_2 / r_2 = 8 / 1 = 8 \ rad/s$The ratio of the final kinetic energy $(K_2)$ to the initial kinetic energy $(K_1)$ is:$K_2 / K_1 = (\frac{1}{2} m v_2^2) / (\frac{1}{2} m v_1^2)$$K_2 / K_1 = v_2^2 / v_1^2$$K_2 / K_1 = (8)^2 / (4)^2 = 64 / 16 = 4$Thus,the ratio is $4:1$.

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