Linear acceleration of mass \[{m_2}\]is \[{a_2}\], then angular acceleration of \[{\alpha _1}\]is (given that there is no slipping).
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A. \[\dfrac{{{a_2}}}{R}\]
B. \[\dfrac{{\left( {{a_2} + g} \right)}}{R}\]
C. \[\dfrac{{2\left( {{a_2} + g} \right)}}{R}\]
D. None of these

Answer
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Hint: For no slipping condition or pure rotation condition the surface in contact is at instantaneous rest, i.e. the net force acting at that point is zero and hence there is no relative acceleration of the point.

Complete step by step solution:
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Image: Two rings joined by common string

Let the tension in the string connected with mass \[{m_1}\] is \[{T_1}\] and the tension in the string connecting both the masses is \[{T_2}\]. \[{\alpha _1}\] is the acceleration of the mass \[{m_1}\], \[{a_2}\] is the linear acceleration of the mass \[{m_2}\] and \[{\alpha _2}\] is the angular acceleration of the mass \[{m_2}\].

For mass \[{m_2}\] writing the Newton’s 2nd law of motion,
\[\begin{array}{c}{F_{net}} = ma\\ \Rightarrow {m_2}g - {T_2} = {m_2}{a_2} \ldots \ldots \left( i \right)\end{array}\]
The moment of inertia of the circular ring about its centre of mass is given as,
\[I = M{R^2}\]
Here, \[M\] is the mass and \[R\] is the radius of the ring.
So,
\[\begin{array}{c}{I_1} = {m_1}{R^2}\\{I_2} = {m_2}{R^2}\end{array}\]

Net torque acting about the centre of mass \[{m_2}\] is \[{T_2}R\] and the direction is clockwise. For no slipping condition,
\[{a_2} = {\alpha _2}R\]
So, the angular acceleration will be,
\[{\alpha _2} = \dfrac{{{a_2}}}{R}\]
For mass\[{m_1}\],
Using no slipping condition,
\[{c}{\alpha _1} = {\alpha _2}\]
\[\therefore {\alpha _1}= \dfrac{{{a_2}}}{R}\]
Therefore, the correct option is A.

Note: We need to assume that the string connecting both the rings are massless and inextensible. If the string is not massless and inextensible then the tension at the point of contact will be different for both the rings. If the motion is not purely rolling/slipping then we can’t use the used relation between the angular acceleration and the linear acceleration for both the rings. The linear acceleration of mass \[{m_1}\] should be zero as its centre is attached to an inextensible and immovable string.