In figure calculate the linear acceleration of the blocks.
Mass of block `B= 8kg`
mass of disc shaped pulley `=2kg` (take `g=10m//s^(2)`)

Explanation

Let `R` on the radius of the pulley and `T_(1)` and `T_(2)` be the tension in the left and right portions of the string Unlike laws of motion where tension in striln is the same on both the sides of the pulley. In this case, the string does not slide and the pulley rotates. the tensions on both the sides of the pulley will be different. this This different tensions will provide angular acceleration to pulley.

Let `m_(1)=8kg, m_(2)=10 kg, M=2kg`
Let `a` be the acceleration of blocks.
for the blocks (linear motion)
`T_(1)-m_(1)g=m_(1)a`.......i
`m_(2)g-T_(2)=m_(2)a`...........ii
For the pulley (rotation)
Net torque `tau_(net)=Ialpha`
`T_(1)R-T_(1)R=1/2MR^(2)alpha`
The linear acceleration of the blocks is the same as the tangetial acceleration of any point on the circumference of the pulley which is `Ralpha`.
`a=Ralpha`
Dividing eqn iii by `R` and addinng to eqn i and ii we get
`m_(2)g-m_(1)g=m_(2)a+m_(1)a+M/2Ralphaimplies m_(2)g-m_(1)g`
`=(m_(2)=m_(1)+M/2)a`
`a=(m_(2)-m_(1))/(m_(2)+m_(1)+M/2g) =((10-8)g)/(10+8+2/2)=20/19m//s^(2)`