Answer
(a) decrease
(b) $0.87N$
Work Step by Step
(a) We know that
$T_2=\frac{T_1}{1+\frac{m}{2M}}$
We plug in the known values to obtain:
$T_2=\frac{1.1N}{1+\frac{0.160Kg}{2(0.31Kg)}}$
$T_2=0.87N$
Thus, if the mass of the pulley is doubled then the tension $T_2$ decreases to $0.87N$
(b) We have calculated the value of $T_2=0.87N$ as in part(a).
