question_answer

Two particles of equal mass are connected to a rope AB of negligible mass such that one is at end A and other dividing the length of rope in the ratio \[1:2\]from B. The rope is rotated about end B in a horizontal plane. Ratio of tensions in the smaller part to the other is (ignore effect of gravity)                   

A) \[4:3~\]

B) \[1:4\]

C) \[1:2\]

D) \[1:3\]

Correct Answer: A

Solution :

[a]    \[{{T}_{1}}=m{{\omega }^{2}}(3\,\ell )\] and\[{{T}_{2}}-{{T}_{1}}=m{{\omega }^{2}}\ell \] or \[{{T}_{2}}={{T}_{1}}+m{{\omega }^{2}}\ell =m{{\omega }^{2}}(3\ell )+m{{\omega }^{2}}\ell \]             \[=4m{{\omega }^{2}}\ell \] \[\therefore \frac{{{T}_{2}}}{{{T}_{1}}}=\frac{4}{3}\]