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A long, thin, inextensible and very flexible uniform wire is lying on the rough horizontal floor. One end of the wire is bent back and then pulled backward with constant velocity V such that, at any instant of time, the moving part of the wire always remains just above the part of the wire which is still at rest at that instant on the floor as shown in diagram. If the wire has unit length and unit mass then.

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Speed of centre of mass of the moving part at all the times will remain constant

Speed of centre of mass of the moving part at all the times will be $\frac{3}{4} V$

Minimum force needed to pull the moving part will be $\frac{V^{2}}{2}$

Minimum force needed to pull the moving part will be $\frac{3}{8} V^{2}$

answer is A, B, C.

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Detailed Solution

$X_{Cm} = x + \frac{x}{2} = \frac{3x}{2}$

Velocity of point $A = \frac{d}{dt} (2x) = V; \frac{dx}{dt} = \frac{V}{2}$

$V_{Cm} = \frac{d}{dt} (X_{Cm}) = \frac{3}{2} \frac{dx}{dt} = \frac{3}{2} (\frac{V}{2}) = \frac{3V}{4}$

Which is constant

$η = λdx ; F = \frac{dp}{dt} = \frac{(λdx) V}{dt} = λV (\frac{V}{2})$

$= \frac{{λV}^{2}}{2} ∴ λ = 1, ∴ F = \frac{V^{2}}{2}$

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