Let m1 and m2 be the masses suspended at the ends of a light inextensible string passing over the pulley.
m1=8kg,m2=12kg,
Let T be the tension in the string and a be the common acceleration with which m1 moves upwards and m2 moves downward = ?
The equation of motion of m1 and m2 are given by
T–m1g=m1a ...(i)
and m2g–T=m2a ...(ii)
Adding (i) and (ii), we get
(m2–m1)g=(m1+m2)a
∴a=(m2−m1)gm1+m2...(iii)
From (i) and (iii), we get
T=m1g+m1

(m2−m1)g

m1+m2

=

m1g

m1+m2

(m1+m2+m2−m1)
or T=

2m1m2

m1+m2

g...(iv)
Putting m1=8 kg and m2=12 kg and g=10 m s−2, in equations (iii), (iv), we get
a=(

12−8

8+12

)×10
=

4

20

×10=2 m s−2
and T=

2×8×12

8+12

×10
=

2×96×10

20

=96N
∴a=2m s−2,T=96 N