Question

A block of mass $10kg$ is suspended by three strings as shown in the figure. The tension $T_2$ is (Take $g=10m/s^2$)

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

Answer 1

The correct option is D $50\sqrt{3}N$


For horizontal equilibrium: $T_3\sin {60}^{\circ }=T_2\sin {30}^{\circ }$ $\therefore T_3=\frac{T_2}{\sqrt{3}}$
Now, for vertical equilibrium:
$\begin{array}{}{T}_2\cos {30}^{\circ }+T_3\cos {60}^{\circ }=T_1& \text{(1)}\end{array}$
And $T_1=mg=10\times 10=100N$
Substituting the value of $T_1$ and $T_3$ in equation $\left(1\right)$, we get
$\frac{\sqrt{3}{T}_2}{2}+\frac{T_2}{\sqrt{3}}\times \frac{1}{2}=100$
$\Rightarrow \frac{4T_2}{2\sqrt{3}}=100$
$\Rightarrow T_2=50\sqrt{3}N$