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(C) We know that power $P$ is given by $P = Fv = m \left( \frac{dv}{dt} \right) v$.Given $m = 2\, kg$ and $P = \frac{3t^2}{2}$.Substituting the values

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(C) We know that power $P$ is given by $P = Fv = m \left( \frac{dv}{dt} ight) v$.Given $m = 2\, kg$ and $P = \frac{3t^2}{2}$.Substituting the values: $2 \left( \frac{dv}{dt} ight) v = \frac{3t^2}{2}$.Rearranging the terms: $v \, dv = \frac{3}{4} t^2 \, dt$.Integrating both sides from $t = 0$ to $t = 2$ and $v = 0$ to $v = v_f$:$\int_{0}^{v_f} v \, dv = \int_{0}^{2} \frac{3}{4} t^2 \, dt$.$\left[ \frac{v^2}{2} ight]_{0}^{v_f} = \frac{3}{4} \left[ \frac{t^3}{3} ight]_{0}^{2}$.$\frac{v_f^2}{2} = \frac{3}{4} 	imes \frac{8}{3} = 2$.$v_f^2 = 4$,which gives $v_f = 2\, m/s$.

Power supplied to a particle of mass $2\, kg$ varies with time as $P = \frac{3t^2}{2} \, W$. Here $t$ is in $seconds$. If the velocity of the particle at $t = 0$ is $v = 0$,the velocity of the particle at time $t = 2\, s$ will be ........... $m/s$.

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