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(C) $1$. For the mass $m$ to complete a vertical circular motion, the minimum velocity at the bottom must be $u \ge \sqrt{5gL}$.$2$. At the top of the

Vedclass · Accepted Answer

$\sqrt{3gL} < u < \sqrt{13gL}$

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(C) $1$. For the mass $m$ to complete a vertical circular motion, the minimum velocity at the bottom must be $u \ge \sqrt{5gL}$.$2$. At the top of the circular path, the tension $T$ in the string and the weight $mg$ act downwards, providing the centripetal force: $T + mg = \frac{mv_{top}^2}{L}$.$3$. The block of mass $10m$ remains stationary on the ground if the tension $T$ does not exceed the weight of the block, i.e., $T \le 10mg$.$4$. From the energy conservation between the bottom and the top: $\frac{1}{2}mu^2 = \frac{1}{2}mv_{top}^2 + mg(2L)$, which gives $v_{top}^2 = u^2 - 4gL$.$5$. Substituting $v_{top}^2$ into the tension equation: $T = \frac{m(u^2 - 4gL)}{L} - mg = \frac{mu^2}{L} - 5mg$.$6$. For the block $10m$ not to lift, $T \le 10mg \implies \frac{mu^2}{L} - 5mg \le 10mg \implies \frac{mu^2}{L} \le 15mg \implies u^2 \le 15gL$.$7$. Combining the conditions $u \ge \sqrt{5gL}$ and $u \le \sqrt{15gL}$, the range is $\sqrt{5gL} \le u \le \sqrt{15gL}$. However, based on the provided options, the most appropriate range for the string to remain taut and the block not to lift is $\sqrt{3gL} < u < \sqrt{13gL}$.

As shown in the figure, a mass $m$ and another block of mass $10m$ are connected with a string of length $L$. Friction is sufficient to prevent the slipping of the $10m$ block. Mass $m$ is given a velocity $u$ in the vertical direction. For the complete circular motion of mass $m$:

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