Q154 / 160 · TS EAMCET 2014 · Physics
A mass $M \mathrm{~kg}$ is suspended by a weightless string. The horizontal force required to hold the mass at $60^{\circ}$ with the vertical is
- A$M g$
- Correct answerB$M g \sqrt{3}$
- C$M g(\sqrt{3}+1)$
- D$\frac{M g}{\sqrt{3}}$
Solution
By Newton's law
$$
\begin{gathered}
F=T \sin \theta \\
M g=T \cos \theta
\end{gathered}
$$
Dividing Eq. (i) by Eq. (ii), we get
$$
\begin{aligned}
\frac{F}{M g} & =\frac{T \sin \theta}{T \cos \theta} \\
F & =M g \tan \theta \quad\left(\because \theta=60^{\circ} \text { given }\right) \\
F & =M g \tan 60^{\circ} \\
F & =\sqrt{3} M g
\end{aligned}
$$
Hence, the correct answer is option: 'B'.