Question 61 of 210

A body weighing $13 \mathrm{~kg}$ is suspended by two strings $5 \mathrm{~m}$ and $12 \mathrm{~m}$ long, their other ends being fastened to the extremities of a rod $13 \mathrm{~m}$ long. If the rod be so held that the body hangs immediately below the middle point. The tensions in the strings are

A$12 \mathrm{~kg}$ and $13 \mathrm{~kg}$B$5 \mathrm{~kg}$ and $5 \mathrm{~kg}$C$5 \mathrm{~kg}$ and $12 \mathrm{~kg}$D$5 \mathrm{~kg}$ and $13 \mathrm{~kg}$

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Answer: (C) $5 \mathrm{~kg}$ and $12 \mathrm{~kg}$

$\mathrm{T}_2 \cos \left(\frac{\pi}{2}-\theta\right)=\mathrm{T}_1 \cos \theta \Rightarrow \mathrm{T}_1 \cos \theta=\mathrm{T}_2 \sin \theta$
$\mathrm{T}_1 \sin \theta+\mathrm{T}_2 \cos \theta=13$.
$\because \mathrm{OC}=\mathrm{CA}=\mathrm{CB}$
$\Rightarrow \angle \mathrm{AOC}=\angle \mathrm{OAC}$ and $\angle \mathrm{COB}=\angle \mathrm{OBC}$
$\therefore \sin \theta=\sin \mathrm{A}=\frac{5}{13}$ and $\cos \theta=\frac{12}{13}$
$\Rightarrow \frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{5}{12} \Rightarrow \mathrm{T}_1=\frac{5}{12} \mathrm{~T}_2$
$\mathrm{~T}_2\left(\frac{5}{12} \cdot \frac{5}{13}+\frac{12}{13}\right)=13$
$\mathrm{~T}_2\left(\frac{169}{12 \cdot 13}\right)=13$
$\mathrm{~T}_2=12 \mathrm{~kgs} . \Rightarrow \mathrm{T}_1=5 \mathrm{~kgs}$.