There is a horizontal film of soap solution. | vedclass

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(D) Consider a small element of the thread of length $\Delta l$ subtending an angle $	heta$ at the center. The length is given by $\Delta l = R	heta

Vedclass · Accepted Answer

$2RT$

Vedclass · Answer

(D) Consider a small element of the thread of length $\Delta l$ subtending an angle $	heta$ at the center. The length is given by $\Delta l = R	heta$.The soap film has two surfaces,so the force due to surface tension acting radially outward on the element is $F_s = 2 	imes T 	imes \Delta l = 2TR	heta$.The tension $F$ in the thread acts at both ends of the element. Resolving the tension forces,the radial component is $2F \sin(	heta/2)$.For equilibrium,the radial component of the tension must balance the force due to surface tension:$2F \sin(	heta/2) = 2TR	heta$.For small angles,$\sin(	heta/2) \approx 	heta/2$. Substituting this into the equation:$2F(	heta/2) = 2TR	heta$$F	heta = 2TR	heta$$F = 2TR$.

There is a horizontal film of soap solution. On it,a thread is placed in the form of a loop. The film is pierced inside the loop,and the thread becomes a circular loop of radius $R$. If the surface tension of the soap solution is $T$,then what will be the tension in the thread?

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