Q1:

Two bodies of masses m₁ and m₂ are communicated by a light inextensible string passing over a small smooth fixed pulley; m₁ > m₂. What is the acceleration of the system ?

ANS:D - g (m₁ - m₂)/(m₁ + m₂)

Check
A tension force induces in the string on both the masses in an upward direction.
Let tension force is T.

T = m1g-m1a.........(1)
T = m2g+m2a........(2)

Acceleration is same for both masses but it will be in opposite direction.
Thus;
a2 = -a1.

(1)=(2).
m1g-m1a=m2g+m2a,
m1g - m2g = m1a+m2a,
g(m1-m2) = a(m1+m2),

a = g(m1-m2)/(m1+m2).