A body weighing 13 ~kg is suspended by two strings 5 ~m and 12 ~m...

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 body weighing 13 ~kg is suspended by two strings 5 ~m and 12 ~m long, their other ends being fastened to the

A body weighing 13 ~kg is suspended by two strings 5 ~m and 12 ~m long, their other ends being fastened to the

Options

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 $13 \mathrm{~kg}$

Correct answer

C. $5 \mathrm{~kg}$ and $12 \mathrm{~kg}$

Step-by-step 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}

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