Two masses ${\mathrm{m}}_1=5\mathrm{kg} \mathrm{and} {\mathrm{m}}_2=4.8\mathrm{kg}$ tied to a string are hanging over a light frictionless pulley. The acceleration of the masses when left free to move is $\left[\mathrm{g}=9.8{\mathrm{ms}}^{-2}\right]$

As ${\mathrm{m}}_1>{\mathrm{m}}_2$

$\therefore {\mathrm{m}}_1$ will go down m2 and will go up

Mass'${\mathrm{m}}_1$': $:{\mathrm{W}}_1-\mathrm{T}={\mathrm{m}}_1\mathrm{a}$

$\therefore {\mathrm{m}}_1\mathrm{g}-\mathrm{T}={\mathrm{m}}_1\mathrm{a}\to \left(1\right)$

Mass '${\mathrm{m}}_2$ ':${\mathrm{W}}_2-\mathrm{T}={\mathrm{m}}_2\mathrm{a}$$\therefore {\mathrm{m}}_2\mathrm{g}-\mathrm{T}={\mathrm{m}}_2\mathrm{a}\to \left(2\right)$

Writing equation (1) and equation (2) together we get

${\mathrm{m}}_1\mathrm{g}-\mathrm{T}={\mathrm{m}}_1\mathrm{a}\to \left(1\right)$

$\mathrm{T}-{\mathrm{m}}_2\mathrm{g}={\mathrm{m}}_2\mathrm{a}\to \left(2\right)$

${\mathrm{m}}_1\mathrm{g}-{\mathrm{m}}_2\mathrm{g}={\mathrm{m}}_1\mathrm{a}+{\mathrm{m}}_2\mathrm{a}$

$\therefore \mathrm{g}({\mathrm{m}}_1-{\mathrm{m}}_2)=\mathrm{a}({\mathrm{m}}_1+{\mathrm{m}}_2)$

$\therefore a=\frac{g\left(m_1-m_2\right)}{(m_1+m_2)}\to \left(3\right)$

Substituting the values of $\text{'}{\mathrm{m}}_1 \text{'},\text{'}{\mathrm{m}}_2\text{'}$and 'g' in equation (3) we get

$\therefore$Acceleration of the masses $=0.2 {\mathrm{ms}}^{-2}$