Question

In the equilibrium arrangement as shown, tension $T_2$ is $(g=10{\text{m/s}}^2)$

• A. $50\text{N}$
• B. $100\text{N}$
• C. $50\sqrt{3}\text{N}$
• D. $100\sqrt{3}\text{N}$

Answer 1

The correct option is B $100\text{N}$

The FBDs of the pulley and the block of mass $10Kg$ is as shown
So, for pulley
$F=2T$
And, for block
$T=10\text{g}=100\text{N}$
$F=200\text{N}$
Again, from resolving all the available forces in horizontal and vertical direction we get


For horizontal equilibrium
$T_1\cos {60}^{\circ }=T_2\cos {30}^{\circ }$
$T_1=2T_2\frac{\sqrt{3}}{2}=\sqrt{3}{T}_2\dots \dots \left(i\right)$
For vertical equilibrium
$T_1\sin {60}^{\circ }+T_2\sin {30}^{\circ }=200\dots \dots \left(ii\right)$
Using (i) and (ii) solve for $T_2$
$2T_2=200$
$T_2=100\text{N}$