Consider a circular ring of radius 1.4 ,cm ly | vedclass

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(B) When a thin ring is lifted from a liquid surface,the liquid adheres to both the inner and outer circumferences of the ring. Thus,the total length

Vedclass · Accepted Answer

$0.125$

Vedclass · Answer

(B) When a thin ring is lifted from a liquid surface,the liquid adheres to both the inner and outer circumferences of the ring. Thus,the total length of the contact line is $L = 2 \pi r + 2 \pi r = 4 \pi r$.The upward force $F$ required to lift the ring is the sum of the weight of the ring $W$ and the force due to surface tension $F_s$.$F = W + F_s = W + T 	imes L = W + T 	imes (4 \pi r)$.Given that the force required is $0.022 \,N$ greater than the weight,we have $F - W = 0.022 \,N$.Therefore,$T 	imes 4 \pi r = 0.022$.Given $r = 1.4 \,cm = 1.4 	imes 10^{-2} \,m$.$T = \frac{0.022}{4 \pi 	imes 1.4 	imes 10^{-2}} = \frac{0.022}{4 	imes 3.14159 	imes 0.014} \approx \frac{0.022}{0.1759} \approx 0.125 \,N/m$.Using $\pi \approx \frac{22}{7}$:$T = \frac{0.022}{4 	imes (22/7) 	imes 0.014} = \frac{0.022 	imes 7}{4 	imes 22 	imes 0.014} = \frac{0.154}{1.232} = 0.125 \,N/m$.

Consider a circular ring of radius $1.4 \,cm$ lying on the surface of a liquid. If a vertical force of $0.022 \,N$ greater than the weight of the ring is required to lift this ring from the liquid surface,then the surface tension of the liquid is (in $\,Nm^{-1}$)

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