Question

Two massless rings slide on a smooth circular loop of the wire whose axis lies in a horizontal plane. $A $ smooth massless inextensible string passes through the rings, which carries masses $m_{1}$ and $m_{2}$ at the two ends and mass $m_{3}$ between the rings. If there is equilibrium when the line connecting each ring with centre subtends an angle $30^{\circ}$ with vertical as shown in figure. Then the ratio of masses are

• A. $m_{1}=2 m_{2}=m_{3}$
• B. $2 m_{1}=m_{2}=2 m_{3}$
• C. $m_{1}=m_{2}=m_{3}$
• D. None of these

Answer 1

Answer: (C)

As no friction is involved, the tensions in the segments $A C$ and $A E$ of the string must be the same.
Let its magnitude be $T$. For the ring A to be at rest on the smooth loop, the resultant force on it must be along $A O, O$ being the center of the loop; otherwise there would be a component tangential to the loop.
Hence
$\angle O A E=\angle O A C=\angle A O E=30^{\circ}$
The same argument applies to the segments $B D$ and $B E$.
Then by symmetry the point $E$ at which the string carries the third weight must be on the radius $H O, H$ being the highest point of the loop, and the tensions in the segments $B D$ and $B E$ are also $T$.
Now consider the point $E$. Each of the three forces acting on it, which are in equilibrium, are at an angle of $120^{\circ}$ to the adjacent one.
As two of the forces have magnitude $T$, the third force must also have magnitude $T$.
Therefore, the three weights carried by the string are equal.