Question

In the 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 $100\text{N}$

By the FBD of pulley, and writing the equilibrium equation we get,

$F=2T$
Now FBD for block,
$T=10g=100\text{N}$
Hence the force on the puley will be, $F=200\text{N}$

Now by the FBD of the system we get, for horizontal equilibrium
$T_1\cos {60}^{\circ }=T_2\cos {30}^{\circ }$
$T_1=2T_2\frac{\sqrt{3}}{2}=\sqrt{3}{T}_2...\left(i\right)$
For vertical equilibrium

$T_1\sin {60}^{\circ }+T_2\sin {30}^{\circ }=200...\left(ii\right)$ ,

$\Rightarrow T_1\sqrt{3}+T_2=400$

Using $\left(i\right)$ and $\left(ii\right)$ solve for $T_2$
$4T_2=400$
$T_2=100\text{N}$

Hence option B is the correct answer