A uniform rope of length l =12 m is suspended between two fixed nails A and B that are in the same horizontal level. The rope makes an angle $θ = 37^{0}$ with the horizontal at the nails. The radius of curvature of the rope at the lowest point is.

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4 m

2 m

8 m

6 m

answer is D.

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Detailed Solution

$F_{0} \cos θ = F \to (1)$

$F_{0} \sin θ = \frac{λ \lg}{2} \to (2)$

Solving 1, 2 , we get
$F = \frac{λ \lg}{2} \cot θ \to (3)$
Let the radius of curvature of the rope at the lowest point be r
$F d θ = λ r d θ g F = λ g r r = \frac{F}{λ g} = \frac{l}{2} \cot θ r = \frac{12}{2} \times \frac{4}{3} = 8 m$

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