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)

$m g=2 T \sin 45^{\circ}$
$\Rightarrow m g=\sqrt{2} T$
$T_{1} \cos \theta=T \cos 45^{\circ}$
$\Rightarrow T_{1} \cos \theta=\frac{T}{\sqrt{2}}=\frac{m g}{2} \ldots$ (i)
Further, $M g+T \sin 45^{\circ}=T_{1} \sin \theta\left\{\because T=\frac{m g}{\sqrt{2}}\right\}$
$\Rightarrow T_{1} \sin \theta=M g+\frac{m g}{\sqrt{2}} \frac{1}{\sqrt{2}} \ldots$ (ii)
$\Rightarrow \tan \theta=\frac{M_{g}+\frac{m g}{2}}{\frac{m g}{2}}=1+\frac{2 M}{m}$
[dividing Eq. (ii) by Eq. (i) ]