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It is given in the question that the pulley rotates without friction and the rope connecting the masses $ {m_1} $ and $ {m_2} $ does not slip.

It is also mentioned that $ {m_1} > {m_2} $ . This means that the rope will move towards $ {m_1} $ and it will go down while $ {m_2} $ will accelerate upwards.

As they both are connected via rope, the acceleration of both blocks would be the same, but in opposite directions.

Therefore drawing the free body diagrams of both masses assuming these conditions,

For block $ {m_1} $ ,

Balancing the vertical forces and acceleration,

$ {m_1}a = {m_1}g - {T_1} $ …(a)

For block $ {m_2} $ ,

Balancing the vertical forces and acceleration,

$ {m_2}a = {T_2} - {m_2}g $ …(b)

Adding equations (a) and (b), we obtain-

$ {m_1}a + {m_2}a = {m_1}g - {T_1} + {T_2} - {m_2}g $

On rearranging this,

$ \left( {{m_1} + {m_2}} \right)a = \left( {{m_1} - {m_2}} \right)g - \left( {{T_1} - {T_2}} \right) $ … $ (1) $

Now, on drawing the free body diagram of the pulley,

We see an angular acceleration of the pulley, which is caused by the turning of the rope around it, therefore the linear component of this acceleration must be equal to the torque produced by the rope.

$ \tau = \left( {{T_1} - {T_2}} \right)r $ ...(c)

(The difference in the tensile forces is used as the torque to produce rotation in the pulley.)

Also,

$ \tau = I\alpha $

where $ \alpha $ is the angular acceleration of the pulley while $ I $ is the moment of inertia of the pulley.

For a circular disk, the Moment of inertia about its central axis of rotation is given by,

$ I = \dfrac{{m{r^2}}}{2} $

Keeping the mass of the pulley equal to $ M $ and its radius equal to $ r $ ,

The torque is,

$ \tau = \dfrac{{M{r^2}\alpha }}{2} $ …(d)

On equating (c) and (d) we get,

$ \tau = ({T_1} - {T_2})r = \dfrac{{M{r^2}\alpha }}{2} $

$ \Rightarrow {T_1} - {T_2} = \dfrac{{Mr\alpha }}{2} $

Writing in terms of linear acceleration using the relation,

$ a = r\alpha $

we have,

$ {T_1} - {T_2} = \dfrac{{Ma}}{2} $

$ a = \dfrac{{2({T_1} - {T_2})}}{M} $ … $ (2) $

Substituting the value of $ a $ in equation $ (1) $ , we get-

$ \left( {{m_1} + {m_2}} \right)\left( {\dfrac{{2({T_1} - {T_2})}}{M}} \right) = \left( {{m_1} - {m_2}} \right)g - \left( {{T_1} - {T_2}} \right) $

Dividing both sides with $ {T_1} - {T_2} $ ,

$ \left( {{m_1} + {m_2}} \right)\left( {\dfrac{2}{M}} \right) = \dfrac{{\left( {{m_1} - {m_2}} \right)g}}{{{T_1} - {T_2}}} - 1 $

$ \Rightarrow \dfrac{{2\left( {{m_1} + {m_2}} \right)}}{M} + 1 = \dfrac{{\left( {{m_1} - {m_2}} \right)g}}{{{T_1} - {T_2}}} $

Upon cross multiplying we get,

$ {T_1} - {T_2} = \dfrac{{\left( {{m_1} - {m_2}} \right)g}}{{\left( {\dfrac{{2({m_1} + {m_2})}}{M} + 1} \right)}} $

$ \Rightarrow {T_1} - {T_2} = \dfrac{{M\left( {{m_1} - {m_2}} \right)g}}{{2({m_1} + {m_2}) + M}} $

$ M $ is present in both, the numerator and the denominator of the relation.

As the value of $ M $ is increased, the difference in the values of $ {T_1} $ and $ {T_2} $ diminishes.

For a fixed value of $ {m_1} $ and $ {m_2} $ , since $ {m_1} > {m_2} $ the numerator always remains a positive quantity. With the increase in the value of $ M $ , the overall value of the term increases but the rate of this increment slows down.

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