Question

A body of weight 3 kg is suspended as shown in the figure, the tension 'T' in the horizontal string (in kg-wt) is ______.

\[\begin{bmatrix} \sin30^\circ=\cos60^\circ=0.5 \\ \\ \cos30^\circ=\sin60^\circ=\frac{\sqrt{3}}{2} \end{bmatrix}\]

विकल्प

• 2

• \[\sqrt{3}\]

• \[2\sqrt{3}\]

• 3

Answer 1

A body of weight 3 kg is suspended as shown in the figure, the tension 'T' in the horizontal string (in kg-wt) is \[\sqrt{3}\].

Explanation:

Weight of the body = 3 kg-wt

Weight is balanced vertically by vertical component of tension T

\[\therefore\quad\mathrm{T}\sin60^{\circ}=3\]

\[\mathrm{T}=\frac{3}{\sin60^\circ}=\frac{3\times2}{\sqrt{3}}=2\sqrt{3}\]

Horizontal tension T, is balanced by horizontal component of tension T.

\[\therefore\quad\mathrm{T}_1=\mathrm{Tcos}60^\circ\]

\[=2\sqrt{3}\times\frac{1}{2}\]

= \[\sqrt{3}\]