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Question

$2 \mathrm{T} \cos \frac{	heta}{2}=\mathrm{Mg}$ for equillibrium

$\mathrm{T}=\frac{\mathrm{Mg}}{2 \cos \frac{	heta}{2}}$

for $T$ to be maximum

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$2 \mathrm{T} \cos \frac{	heta}{2}=\mathrm{Mg}$ for equillibrium

$\mathrm{T}=\frac{\mathrm{Mg}}{2 \cos \frac{	heta}{2}}$

for $T$ to be maximum cos $\frac{	heta}{2}$ must be minimum so

$\frac{	heta}{2}$ must be maximum

A boy is hanging from branch of a tree as shown. Angle between his arm, for tension to be maximum in arms must be ............ $^o$

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