Question

In the system of pulleys shown what should be value of $m_1$ such that 100 gm remains at rest : $(Takeg=10{m/s}^2)$

• A. $180gm$
• B. $160gm$
• C. $100gm$
• D. $200gm$

Answer 1

The correct option is D $200gm$

$\begin{array}{c}\text{From diagram,}\\ \text{in}{P}_2\text{pulley}100gm\\ \text{will lifted up by}200gm\text{.}\\ \text{So, resultant acceleration}\end{array}$

$\begin{array}{c}\text{of pulley}{p}_2\text{should be at the downwards}\\ \text{to remain}100gm\text{. block at rest.}\\ 200a=200g-T\\ 100a=T-100g\\ 300a=100g\\ a=g/cm/s^2\end{array}$

$\begin{array}{cc}\therefore T& =\frac{1009}{3}+100g\\ 0qT& =\frac{9009}{3}\end{array}$

-for pulley $P$. $m_{1}a=2T-m_{1}q$ on $m_1\frac{g}{3}=\frac{800}{3}g-m_{1}g$ on $\frac{m_1}{3}+m_1=\frac{800}{3}$ in $q{m}_1=800$ on $m_1=200$

Ans. Option D