A ring of a thin wire of cross-sectional area | vedclass

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(A) Consider a small element of the ring of length $dl$ subtending an angle $2	heta$ at the center. The surface tension force acting on this element

Vedclass · Accepted Answer

$\frac{l\sigma}{\pi aY}$

Vedclass · Answer

(A) Consider a small element of the ring of length $dl$ subtending an angle $2	heta$ at the center. The surface tension force acting on this element is $2\sigma dl$ (since the film has two surfaces). This force is balanced by the radial component of the tension $F$ in the wire,which is $2F \sin 	heta$.For small $	heta$,$\sin 	heta \approx 	heta$. Also,$dl = r(2	heta)$,where $r$ is the radius of the ring.Equating the forces: $2F 	heta = 2\sigma (2r	heta) \Rightarrow F = 2\sigma r$.Since the circumference $l = 2\pi r$,we have $r = \frac{l}{2\pi}$.Substituting $r$: $F = 2\sigma \left(\frac{l}{2\pi}ight) = \frac{l\sigma}{\pi}$.Young's modulus $Y$ is defined as $Y = \frac{	ext{Stress}}{	ext{Strain}} = \frac{F/a}{	ext{Strain}}$.Therefore,$	ext{Strain} = \frac{F}{aY} = \frac{l\sigma / \pi}{aY} = \frac{l\sigma}{\pi aY}$.

$A$ ring of a thin wire of cross-sectional area $a$ and length $l$ is dipped into a liquid of surface tension $\sigma$ and taken out so that a film of liquid is formed in the ring. If Young's modulus of the material of the wire is $Y$,then the longitudinal strain developed in the ring will be:

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