\[\begin {array} {1 1} (a)\;\frac{6g}{7} & \quad (b)\;\frac{5g}{4} \\ (c)\;\frac{3g}{2} & \quad (d)\;\frac{2g}{3} \end {array}\]

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Since the rod is massless , the tensions in the 2 strings will be the same.

Writing torque equation :

$5 g \times l/2- 2g \times l/2=I \alpha$

$\qquad= [m_1 \times (l/2) ^2 +m_2 (l/2)^2] \alpha$

$3g l/2=\bigg[\large\frac{5l^2}{4} +\frac{2l^2}{4} \bigg] \alpha$

$3g l /2=\large\frac{7l^2}{4} \alpha$

$\alpha = \large\frac{6g}{7}$

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