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(A) At the lowest point $(L)$,the velocity is $v_L = \sqrt{5gl}$. The tension is $T_L = mg + \frac{mv_L^2}{l} = mg + \frac{m(5gl)}{l} = 6mg$.At the hi

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$6\,mg, 2\,mgl$

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(A) At the lowest point $(L)$,the velocity is $v_L = \sqrt{5gl}$. The tension is $T_L = mg + \frac{mv_L^2}{l} = mg + \frac{m(5gl)}{l} = 6mg$.At the highest point $(H)$,the velocity is $v_H = \sqrt{gl}$. The tension is $T_H = \frac{mv_H^2}{l} - mg = \frac{m(gl)}{l} - mg = 0$.The difference in tension is $T_L - T_H = 6mg - 0 = 6mg$.The kinetic energy at the lowest point is $K_L = \frac{1}{2}mv_L^2 = \frac{1}{2}m(5gl) = 2.5mgl$.The kinetic energy at the highest point is $K_H = \frac{1}{2}mv_H^2 = \frac{1}{2}m(gl) = 0.5mgl$.The difference in kinetic energy is $K_L - K_H = 2.5mgl - 0.5mgl = 2mgl$.Therefore,the differences are $6mg$ and $2mgl$.

$A$ particle of mass $m$ is tied to a string of length $l$ and whirled in a vertical circle. The difference in tension and kinetic energy at the highest and lowest positions of the circular path will be:

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