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**Hint:**Here we need a relation between the moment of inertia and angular force i.e. torque. It is given by $\tau = I\alpha $; Where \[\tau \]= torque; $I$= Moment of Inertia, $\alpha $= Angular Acceleration. The moment of inertia for the above diagram would be $I = \left( {\dfrac{{M{L^2}}}{{12}} + M{{\left( {\dfrac{L}{4}} \right)}^2}} \right)$. Also, the relation between linear force and torque is \[\tau = r \times F\]. Where \[\tau \]= torque; r = distance; F = force. Equate the two relations together and you will get the acceleration.

**Complete step by step solution:**

Find out the acceleration:

$\tau = I\alpha $;

Also,

\[\tau = r \times F\];

Here the torque will apply a downward angular force:

\[\tau = rF\sin 90\] ….(After the rope is cut the angle would be $90^\circ $)

Here F = mg; $r = h = \dfrac{L}{4}$;

\[\tau = mgh\]; ….(Here a = g)

Put the given values in the above equation:

\[\tau = mg\dfrac{L}{4}\];

Put the above relation in the equation $\tau = I\alpha $;

\[I\alpha = mg\dfrac{L}{4}\];

Here $I = \dfrac{{M{L^2}}}{{12}} + \dfrac{{M{L^2}}}{{16}}$ ;

Put the value of $I$in the equation: $\tau = I\alpha $;

\[I\alpha = mg\dfrac{L}{4}\];

\[\left( {\dfrac{{M{L^2}}}{{12}} + \dfrac{{M{L^2}}}{{16}}} \right)\alpha = mg\dfrac{L}{4}\];

Take the common out:

\[\left( {\dfrac{L}{3} + \dfrac{L}{4}} \right)\dfrac{{ML}}{4}\alpha = mg\dfrac{L}{4}\];

Here: M = m. Cancel out the common factor:

\[\left( {\dfrac{L}{3} + \dfrac{L}{4}} \right)\alpha = g\];

Take LCM:

\[\left( {\dfrac{{4L + 3L}}{{12}}} \right)\alpha = g\];

Do the calculation:

\[\left( {\dfrac{{7L}}{{12}}} \right)\alpha = g\];

Take the value along with L on the RHS:

\[\alpha = \left( {\dfrac{{12g}}{{7L}}} \right)\];

Now, $\alpha = \alpha r$; $r = \dfrac{L}{2} + \dfrac{L}{4} = \dfrac{{3L}}{4}$ ;

So,

\[\alpha = \left( {\dfrac{{12g}}{{7L}}} \right) \times \dfrac{{3L}}{4}\];

\[\alpha = \left( {\dfrac{{12g}}{{7L}}} \right) \times \dfrac{{3L}}{4}\]

Solve the above equation we get:

\[\alpha = \left( {\dfrac{{9g}}{7}} \right) \downarrow \]; …(torque is acting downwards therefore angular acceleration is down)

**Hence, Option (B) is correct.**

A uniform beam of length L and mass m is supported as shown. If the cable suddenly breaks, then acceleration on the end B is \[\left( {\dfrac{{9g}}{7}} \right) \downarrow \].

**Note:**In this question we have been given two options i.e. either we can find out the acceleration or the reaction force. Acceleration has already been taken out in the solution so, for the reaction force the net force on the pin would be linear force F – The reaction force R which would be equal to the mass times acceleration at the center of mass i.e. pin. The acceleration at centre of mass would be the acceleration due to gravity g minus (-).

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