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A long, heavy, flexible rope with mass per unit length ' $ρ$ ' is stretched by a constant force 'F'.A sudden movement causes a circular loop of radius 'R' to form at one end of the rope. The loop rolls along the rope with speed 'v' as shown in the figure. Then

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$v = \sqrt{\frac{F}{ρ}}$ and kinetic energy carried by the loop is $2 πRF$

$v = \sqrt{\frac{F}{2 ρ}}$

$v = \sqrt{\frac{F}{ρ}}$ and kinetic energy carried by the loop is $πRF$

the kinetic energy carried by the loop is ${πRρv}^{2}$

answer is A.

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Detailed Solution

We know velocity of the loop is

$V = \sqrt{\frac{F}{ρ}}$

Kinetic Energy of the rope is

$E = \frac{1}{2} {MV}^{2} + \frac{1}{2} I_{C} ω^{2}$

$E = \frac{1}{2} {MV}^{2} (1 + \frac{k^{2}}{R^{2}}) = {MV}^{2}$

$\Rightarrow E = M \frac{F}{ρ} = F \times 2 πR$

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