Question

Two masses $m$ and $M$ are attached with strings as shown in the figure. For the system to be in equilibrium, we must have

• A. $\text{tan}\theta =1+\frac{2 M}{m}$
• B. $\text{tan}\theta =1-\frac{2 m}{M}$
• C. $\text{tan}\theta =1-\frac{M}{2 m}$
• D. $\text{tan}\theta =1+\frac{m}{2 M}$

Answer 1

Answer: (A)

$mg=2Tsin45^\circ $
$\Rightarrow mg=\sqrt{2}T$
$T_{1}\text{ cos}\theta =Tcos45^{^\circ }$
$\Rightarrow T_{1}\text{cos}\theta =\frac{T}{\sqrt{2}}=\frac{\textit{mg}}{2}$ ...(i)
Further, $\textit{Mg}+T\text{ cos }45^{^\circ }=\textit{T}_{1}\text{ sin}\theta \left\{\because \textit{T} = \frac{\textit{mg}}{\sqrt{2}}\right\}$
$\Rightarrow T_{1}\text{ sin}\theta =\textit{Mg}+\frac{\textit{mg}}{\sqrt{2}}\frac{1}{\sqrt{2}}$ ...(ii)
$\Rightarrow \text{tan}\theta =\frac{\textit{Mg} + \frac{\textit{mg}}{2}}{\frac{\textit{mg}}{2}}=1+\frac{2 \textit{M}}{\textit{m}}$
{dividing Eq. (ii) by Eq. (i)}