Two masses m and M are attached to the string | vedclass

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(B) For mass $m$,the vertical equilibrium gives: $2 T \sin 45^{\circ} = m g$,which simplifies to $2 T (\frac{1}{\sqrt{2}}) = m g$,so $T = \frac{m g}{\

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$	an 	heta=1+\frac{2 M}{m}$

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(B) For mass $m$,the vertical equilibrium gives: $2 T \sin 45^{\circ} = m g$,which simplifies to $2 T (\frac{1}{\sqrt{2}}) = m g$,so $T = \frac{m g}{\sqrt{2}}$.For mass $M$,the forces acting on the junction are the tension $T$ (acting downwards at $45^{\circ}$),the tension $T'$ (acting upwards at angle $	heta$),and the weight $M g$ (downwards).Resolving forces horizontally: $T' \cos 	heta = T \cos 45^{\circ} = \frac{T}{\sqrt{2}}$.Resolving forces vertically: $T' \sin 	heta = T \sin 45^{\circ} + M g = \frac{T}{\sqrt{2}} + M g$.Dividing the vertical equation by the horizontal equation: $	an 	heta = \frac{\frac{T}{\sqrt{2}} + M g}{\frac{T}{\sqrt{2}}} = 1 + \frac{M g}{\frac{T}{\sqrt{2}}}$.Substituting $T = \frac{m g}{\sqrt{2}}$: $	an 	heta = 1 + \frac{M g}{\frac{m g}{\sqrt{2} \cdot \sqrt{2}}} = 1 + \frac{M g}{\frac{m g}{2}} = 1 + \frac{2 M}{m}$.

Two masses $m$ and $M$ are attached to the strings as shown in the figure. If the system is in equilibrium,then

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