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Q.
In the arrangement as shown, tension $T_{2}$ is $\left(g=10\, m / s ^{2}\right)$
Laws of Motion
Solution:
For pulley
$F=2 T$
for block
$T=10\, g =100\, N$
$F=200\, N$
For horizontal equilibrium
$T_{1} \cos 60^{\circ}=T_{2} \cos 30^{\circ}$
$T_{1}=2 T_{2} \frac{\sqrt{3}}{2}=\sqrt{3} T_{2}$ ...(i)
For vertical equilibrium
$T_{1}=\sin 60^{\circ}+T_{2} \sin 30^{\circ}=200$ ...(ii)
Using (i) and (ii) solve for $T _{2}$
$2 T_{2}=200$
$T_{2}=100\, N$
