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)

$mg=2Tsin45^{\circ }$
$\Rightarrow mg=\sqrt{2}T$
$T_1\text{ cos}\theta =Tcos45^{{}^{\circ }}$
$\Rightarrow T_1\text{cos}\theta =\frac{T}{\sqrt{2}}=\frac{\text{mg}}{2}$ ...(i)
Further, $\text{Mg}+T\text{ cos }45^{{}^{\circ }}={\text{T}}_1\text{ sin}\theta \left\{\because \text{T}=\frac{\text{mg}}{\sqrt{2}}\right\}$
$\Rightarrow T_1\text{ sin}\theta =\text{Mg}+\frac{\text{mg}}{\sqrt{2}}\frac{1}{\sqrt{2}}$ ...(ii)
$\Rightarrow \text{tan}\theta =\frac{\text{Mg}+\frac{\text{mg}}{2}}{\frac{\text{mg}}{2}}=1+\frac{2\text{M}}{\text{m}}$
{dividing Eq. (ii) by Eq. (i)}