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4-1.Newton's Laws of Motion
hard
$A$ flexible chain of weight $W$ hangs between two fixed points $A$ & $B$ which are at he same horizontal level. The inclination of the chain with the horizontal at both the points of support is $\theta$ . What is the tension of the chain at the mid point?
A
$\frac{W}{2}. cosec \theta$
B
$\frac{W}{2} . tan \theta$
C
$\frac{W}{2} cot \theta$
D
none
Solution
$\mathrm{T} \sin \theta=\frac{\mathrm{W}}{2}$
$\therefore \quad \mathrm{T}=\frac{\mathrm{W}}{2 \sin \theta}$
Also, $\mathrm{T} \cos \theta=\mathrm{T}_{0}$
$\therefore \quad \mathrm{T}_{0}=\mathrm{T} \cos \theta$
$\mathrm{T}_{0}=\frac{\mathrm{W}}{2 \sin \theta} \cos \theta$
$=\frac{W}{2} \cot \theta$
Standard 11
Physics
