In the arrangement shown in fig. ${\mathrm{m}}_1=1 \mathrm{kg}$, ${\mathrm{m}}_2=2 \mathrm{kg}$. Pulleys are massless and string are light. For what value of $\mathrm{M}$ the mass ${\mathrm{m}}_1$ moves with constant velocity (neglect friction)

Mass $m_1$ moves with constant velocity if tension in the lower string ${\mathrm{T}}_1={\mathrm{m}}_1 \mathrm{g}=\left(1\right)\left(10\right)=10 \mathrm{N}$....(i) 
$\therefore$ Tension in the upper string is
${\mathrm{T}}_2=2{ \mathrm{T}}_1=20 \mathrm{N}$ …………….(ii)
Acceleration of block $M$ is therefore,
$\mathrm{a}=\frac{{\mathrm{T}}_2}{\mathrm{M}}=\frac{20}{\mathrm{M}}$ …………….(iii)
This is also the acceleration of pulley $2$ .

Absolute acceleration of mass ${\mathrm{m}}_1$ is zero. Thus, acceleration of ${\mathrm{m}}_1$ relative to pulley $2$ is a upwards or acceleration of ${\mathrm{m}}_2$ with respect to pulley $2$ is a downwards. Drawing free body diagram of ${\mathrm{m}}_2$ with respect to pulley $2$ . Equation of motion gives
$20-\frac{40}{M}-10=2\mathrm{a}=\frac{40}{M}$ solving this we get $\mathrm{M}=8 \mathrm{k} \mathrm{g}$.