Radius of a capillary is 2 times 10^{-3} ,m. | vedclass

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(B) The upward force due to surface tension acting on the liquid column is given by $F = (2 \pi R) T \cos 	heta_C$. In equilibrium,the upward force d

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$5 	imes 10^{-2} \,N / m$

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(B) The upward force due to surface tension acting on the liquid column is given by $F = (2 \pi R) T \cos 	heta_C$. In equilibrium,the upward force due to surface tension balances the weight of the liquid column: $(2 \pi R) T \cos 	heta_C = W$. Assuming the contact angle $	heta_C = 0^{\circ}$ for a liquid that rises in a capillary,we have $\cos 0^{\circ} = 1$. Thus,$T = \frac{W}{2 \pi R}$. Substituting the given values: $T = \frac{6.2 	imes 10^{-4}}{2 	imes 3.14 	imes 2 	imes 10^{-3}}$. $T = \frac{6.2 	imes 10^{-4}}{12.56 	imes 10^{-3}} \approx 0.04936 \,N/m \approx 5 	imes 10^{-2} \,N/m$.

Radius of a capillary is $2 \times 10^{-3} \,m$. $A$ liquid of weight $6.2 \times 10^{-4} \,N$ may remain in the capillary. Then the surface tension of the liquid will be $:-$

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