Chapter 4 - Force and Motion - Learning Path Questions and Exercises - Exercises - Page 139: 80

$\frac{1}{3}$

Let $T_{1}$ be the tension between $m_{1}$ and $m_{3}$. Let $T_{2}$ be the tension between $m_{3}$ and $m_{2}$. So, $T_{1}=m_{1}g=0.25\times9.8=2.45N$ $T_{2}=m_{2}g=0.5\times9.8=4.9N$ Since, $T_{2}>T_{1}$ force of friction on $m_{3}$ acts towards left. So, $T_{1}+f=T_{2}$ $T_{1}+static\, friction \,coefficient\times m_{3}g=T_{2}$ Static friction coefficient $=\frac{T_{2}-T_{1}}{m_{3}g}=\frac{4.9-2.45}{0.75\times9.8}=\frac{1}{3}$