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A particle of mass $m$ is tied to string of length $l$ and whirled in vertical circle. The difference of tension and kinetic energy at highest and lowest position of circular path will be
$2\,mg$, $5\, mgl$
$5\,mg$, $3\, mgl$
$6\,mg$, $2\, mgl$
$3\,mg$, $5\, mgl$
Solution
$\mathrm{T}_{\mathrm{L}}=\mathrm{mg}+\frac{\mathrm{m}(\sqrt{5 \mathrm{g} 1})^{2}}{\ell}$
$\mathrm{T}_{\mathrm{L}}=6 \mathrm{mg}$
$\mathrm{T}_{\mathrm{H}}+\mathrm{mg}=\frac{\mathrm{m}(\sqrt{\mathrm{g} 1})^{2}}{\ell}$
$\mathrm{T}_{\mathrm{H}}=\mathrm{mg}-\mathrm{mg}=0$
$\mathrm{T}_{\mathrm{L}}-\mathrm{T}_{\mathrm{H}}=6 \mathrm{mg}$
$\mathrm{K}_{\mathrm{L}}=\frac{1}{2} \mathrm{m}(\sqrt{5 \mathrm{g} 1})^{2}$
$\mathrm{K}_{\mathrm{H}}=\frac{1}{2} \mathrm{m}(\sqrt{\mathrm{g} 1})^{2}$
$\mathrm{K}_{\mathrm{L}}-\mathrm{K}_{\mathrm{H}} \Rightarrow \frac{\mathrm{m}}{2}(5 \mathrm{g} 1-\mathrm{g} 1)$
$\Rightarrow 2 \mathrm{mg} 1$
