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. $m_1=2m_2=m_3$
• B. $2m_1=m_2=2m_3$
• 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 $AC$ and $AE$ 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 $AO,O$ being the center of the loop; otherwise there would be a component tangential to the loop.
Hence
$\angle OAE=\angle OAC=\angle AOE=30^{\circ }$
The same argument applies to the segments $BD$ and $BE$.
Then by symmetry the point $E$ at which the string carries the third weight must be on the radius $HO,H$ being the highest point of the loop, and the tensions in the segments $BD$ and $BE$ 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.