JEE Advanced 2021 — Paper 2 — Question 1

$\Sigma \mathrm{F}_{\mathrm{x}}=0$ $\mathrm{R}_{1}=\mathrm{T} \cos 45^{\circ}$ $\begin{gathered} \mathrm{R}_{1}=\frac{\mathrm{T}}{\sqrt{2}} \end{gathered}$ $\Sigma \mathrm{F}_{\mathrm{y}}=0$ $\mathrm{R}_{2}+\mathrm{T} \sin 45^{\circ}=\mathrm{W}+\alpha \mathrm{W}$ $\begin{gathered} \mathrm{R}_{2}+\frac{\mathrm{T}}{\sqrt{2}}=\mathrm{W}(1+\alpha) \end{gathered}$

$\Sigma \tau_{0}=0$ $W \frac{\mathrm{~L}}{2}+\alpha \mathrm{WL}=\frac{\mathrm{T}}{\sqrt{2}} \mathrm{~L}$ $\begin{gathered} \mathrm{T}=\sqrt{2} \mathrm{~W}\left[\alpha+\frac{1}{2}\right] \end{gathered}$

From (ii) and (iii) $\mathrm{R}_{2}+\mathrm{W}\left[\alpha+\frac{1}{2}\right]=\mathrm{W}(1+\alpha)$

$\mathrm{R}_{2}=\frac{\mathrm{W}}{2}$ Hence, option (A) is correct. From (i) and (iii) $\mathrm{R}_{1}=\mathrm{W}\left[\alpha+\frac{1}{2}\right]$

$\alpha=0.5, \mathrm{R}_{1}=\mathrm{W}$ Hence, option (B) is correct. From equation (iii) if $\alpha=0.5$ $\mathrm{T}=\sqrt{2} \mathrm{~W}$

$\mathrm{T}_{\text {max }}=2 \sqrt{2} \mathrm{~W}$ For rope to break $\mathrm{T}>2 \sqrt{2} \mathrm{~W}$

$\sqrt{2} \mathrm{~W}\left[\alpha+\frac{1}{2}\right]>2 \sqrt{2} \mathrm{~W}$ $\alpha>\frac{3}{2}$ Hence, option (D) is correct.