question_answer

A body weighing 13 kg is suspended by two strings 5 m and 12 m long, their other ends being fastened to the extremities of a rod 13 m long. If the rod be so held that the body hangs immediately below the middle point. The tensions in the strings are       AIEEE  Solved  Paper-2007

A)  12 kg and 13 kg  

B)         5 kg and 5 kg

C)         5 kg and 12 kg   

D)         5 kg and 13 kg

Correct Answer: C

Solution :

If three coplanar forces are in equilibrium, then each force is proportional to the sine of the angle between the other two forces. \[\because \] \[OC=CA=CB\Rightarrow \angle AOC=\angle OAC\] and \[\angle COB=\angle OBC\] \[\therefore \] \[\sin \theta =\sin A=\frac{5}{13}\]and \[\cos \theta =\frac{12}{13}\] Now, by Lami?s theorem, \[\frac{{{T}_{1}}}{\sin ({{180}^{o}}-\theta )}=\frac{{{T}_{2}}}{\sin ({{90}^{o}}+\theta )}\] \[\Rightarrow \]\[\frac{{{T}_{1}}}{\sin \theta }=\frac{{{T}_{2}}}{\cos \theta }\] \[\Rightarrow \]\[{{T}_{1}}\cos \theta ={{T}_{2}}\sin \theta \] \[\Rightarrow \]\[{{T}_{1}}\left( \frac{12}{13} \right)={{T}_{2}}\left( \frac{5}{13} \right)\]\[\Rightarrow \]\[{{T}_{1}}=\left( \frac{5}{12} \right){{T}_{2}}\] Also, \[{{T}_{1}}\sin \theta +{{T}_{2}}\cos \theta =13\] \[\Rightarrow \] \[{{T}_{2}}\left( \frac{5}{12}.\frac{5}{13}+\frac{12}{13} \right)=13\] \[\Rightarrow \] \[{{T}_{2}}\left( \frac{169}{12.13} \right)=13\] \[\Rightarrow \] \[{{T}_{2}}=12\,kg\]and \[{{T}_{1}}=5\,kg\]