Physics - NLM Question with Solution | TestHub

As there is a load at P , so tension in AP and PB will be different. Let these be ${\mathrm{T}}_1$ and ${\mathrm{T}}_2$ respectively. For vertical equilibrium of P ,${\mathrm{T}}_2\cos 60^{\circ }=\mathrm{Mg}$ i.e., ${\mathrm{T}}_2=2\mathrm{Mg}$ ...(i) And for horizontal equilibrium of P , ${\mathrm{T}}_1={\mathrm{T}}_2\sin 60^{\circ }={\mathrm{T}}_2(\sqrt{3}/2)$ ...(i) Substituting the value of ${\mathrm{T}}_2$ from Eq. (i), ${\mathrm{T}}_1=(2\mathrm{Mg})\times (\sqrt{3}/2)=\sqrt{3}\mathrm{Mg}$