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 $5 \mathrm{~kg}$
• Correct answer
  C

  $5 \mathrm{~kg}$ and $12 \mathrm{~kg}$
• D

  $5 \mathrm{~kg}$ and $13 \mathrm{~kg}$

Solution

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

Hence, the correct answer is option: 'C'.