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A body of mass ' $\mathrm{m}$ ' attached at the end of a string is just completing the loop in a vertical circle. The apparent weight of the body at the lowest point in its path is ( $\mathrm{g}=$ gravitational acceleration)
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The correct answer is:
$6 \mathrm{mg}$
The tension at the lowest point is $\mathrm{T}=\frac{\mathrm{mv}^2}{\mathrm{r}}+\mathrm{mg}$ To complete the vertical circle, the minimum velocity should be $\mathrm{v}=\sqrt{5 \mathrm{gr}}$
$$
\begin{aligned}
\therefore \quad \mathrm{T} & =\frac{\mathrm{m}(\sqrt{5 \mathrm{gr}})^2}{\mathrm{r}}+\mathrm{mg} \\
\mathrm{R} & =5 \mathrm{mg}+\mathrm{mg} \\
\therefore \quad \mathrm{T} & =6 \mathrm{mg}
\end{aligned}
$$
$$
\begin{aligned}
\therefore \quad \mathrm{T} & =\frac{\mathrm{m}(\sqrt{5 \mathrm{gr}})^2}{\mathrm{r}}+\mathrm{mg} \\
\mathrm{R} & =5 \mathrm{mg}+\mathrm{mg} \\
\therefore \quad \mathrm{T} & =6 \mathrm{mg}
\end{aligned}
$$
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