Question

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

• A. $4:3$
• B. $1:4$
• C. $1:2$
• D. $1:3$

Answer 1

The correct option is C $1:2$

Outer part is the shorter part , $l=\frac{L}{3}$

Mass connected to this part $=m$

$T_1=\frac{m{v}^2}{l}=\frac{3m{v}^2}{L}$

Inner part is longer part, $l^{\prime }=\frac{2L}{3}$

Mass connected to this part$=2m$

$T_2=\frac{6m{v}^2}{2L}+T_1=\frac{6m{v}^2}{L}$

$\frac{T_1}{T_2}=\frac{1}{2}$