\[\begin {array} {1 1} (a)\;\frac{(m_1-m_2)g}{(m_1+m_2)} & \quad (b)\;\frac{(m_1-m_2)g}{(m_1+m_2+I/l^2)}\\ (c)\;\frac{(m_1-m_2+I/R^2)}{(m_1+m_2)} & \quad (d)\;\frac{(m_1-m_2+I/R^2)}{(m_1+m_2)} \end {array}\]

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$m_1g - T_1=m_1a$

$T_2-m_2g =m_2 a$

$(T_1-T_2) R= I \alpha$

$\qquad= \large\frac{Ia}{R}$

Since the string does not slip over the pulley the acceleration of the string and the tangential accelerated a point on the sum of the pulley is the same .

Solving the above three equation :

$a= \large\frac{m_1 g-m_2 g}{(m_1+m_2+ I/R^2)}$

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