In the system of pulleys shown what should be the value of m₁ (in gram) such that 100 g remains at rest. $\text{ (Take }g=10\mathrm{m}/{\mathrm{s}}^2\text{ ) }$

Since 100 g is at rest,

lf m₁ is going up at acceleration a, we use

$2T-m_{1}g=m_{1}a$   ……(1)

For 100 g to be at rest, acceleration of 200 g mass is 2a if m₁ is going up at a

$\begin{array}{l}2-T=0.2\left(2a\right)\\ 2-1=0.4a\\ a=\frac{10}{4}=\frac{5}{2}\mathrm{m}/{\mathrm{s}}^2\end{array}$

$\text{ From (1), we have }2\left(1\right)-10m_1=m_1\left(\frac{5}{2}\right)$

$\begin{array}{l}2=10m_1+\frac{5m_1}{2}\\ 4=25m_1\\ m_1=\frac{4}{25}=0.16\mathrm{kg}=160\mathrm{g}\end{array}$