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A pulley having moment of inertia 1 \mathrm{~kg}-\mathrm{m}^2 is free to rotate about its horizontal geometrical axis. The radius of the pulley is 1 \mathrm{~m}. A massless thread connects two blocks of masses 2 \mathrm{~kg} and 1 \mathrm{~kg} respectively, going over the pulley. Find the speed of the masses when the height descended by the 2 \mathrm{~kg} mass is 1 \mathrm{~m}. [Given, g=10 \mathrm{~m} / \mathrm{s}^2 ]

Option: 1

\sqrt{5} \mathrm{~m} / \mathrm{s}


Option: 2

\sqrt{\frac{20}{3}} \mathrm{~m} / \mathrm{s}


Option: 3

\sqrt{\frac{5}{3}} \mathrm{~m} / \mathrm{s}


Option: 4

\sqrt{20} \mathrm{~m} / \mathrm{s}


Answers (1)

best_answer

Using the conservation of energy.

(M-m) g h=\frac{1}{2} M v^2+\frac{1}{2} m v^2+\frac{1}{2} I \omega^2

\begin{aligned} & \omega=\frac{v}{\gamma} \\ &(2-1) \times 10 \times 1=\frac{1}{2} \times 2 \times v^2+\frac{1}{2} \times 1 \times v^2+\frac{1}{2} \times 1 \times \frac{v^2}{1^2} \\ & 10=\frac{1}{2}\left[4 v^2\right] \\ & v^2=5 \\ & v=\sqrt{5} \mathrm{~m} / \mathrm{s} \end{aligned}

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sudhir kumar

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