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

Answer 1

Answer: (A)

$mg=2T\sin 45^{\circ }$
$\Rightarrow mg=\sqrt{2}T$
$T_1\cos \theta =T\cos 45^{\circ }$
$\Rightarrow T_1\cos \theta =\frac{T}{\sqrt{2}}=\frac{mg}{2}\dots$ (i)
Further, $Mg+T\sin 45^{\circ }=T_1\sin \theta \left\{\because T=\frac{mg}{\sqrt{2}}\right\}$
$\Rightarrow T_1\sin \theta =Mg+\frac{mg}{\sqrt{2}}\frac{1}{\sqrt{2}}\dots$ (ii)
$\Rightarrow \tan \theta =\frac{M_g+\frac{mg}{2}}{\frac{mg}{2}}=1+\frac{2M}{m}$
[dividing Eq. (ii) by Eq. (i) ]