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A particle tied to a string describes a vertical circular motion of radius \(r\) continually. If it has a velocity \(\sqrt{3 \mathrm{gr}}\) at the highest point, then the ratio of the respective tensions in the string holding it at the highest and lowest points is
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The correct answer is:
\(1: 4\)
Tension at the highest point
\(\mathrm{T}_{\text {top }}=\frac{\mathrm{mv}^2}{\mathrm{r}}-\mathrm{mg}=2 \mathrm{mg}\left(\therefore \mathrm{v}_{\text {top }}=\sqrt{3 \mathrm{gr}}\right)\)
Tension at the lowest point
\(\begin{aligned}
& \mathrm{T}_{\text {bottom }}=2 \mathrm{mg}+6 \mathrm{mg}=8 \mathrm{mg} \\
& \therefore \frac{\mathrm{T}_{\text {top }}}{\mathrm{T}_{\text {bottom }}}=\frac{2 \mathrm{mg}}{8 \mathrm{mg}}=\frac{1}{4} .
\end{aligned}\)
\(\mathrm{T}_{\text {top }}=\frac{\mathrm{mv}^2}{\mathrm{r}}-\mathrm{mg}=2 \mathrm{mg}\left(\therefore \mathrm{v}_{\text {top }}=\sqrt{3 \mathrm{gr}}\right)\)
Tension at the lowest point
\(\begin{aligned}
& \mathrm{T}_{\text {bottom }}=2 \mathrm{mg}+6 \mathrm{mg}=8 \mathrm{mg} \\
& \therefore \frac{\mathrm{T}_{\text {top }}}{\mathrm{T}_{\text {bottom }}}=\frac{2 \mathrm{mg}}{8 \mathrm{mg}}=\frac{1}{4} .
\end{aligned}\)
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