A U-tube is such that the diameter of one lim | vedclass

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(C) The capillary rise in a tube of radius $r$ is given by $h = \frac{2T \cos 	heta}{ho g r}$.For water,the contact angle $	heta \approx 0^\circ$,

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$8 	imes 10^{-3}\,m$

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(C) The capillary rise in a tube of radius $r$ is given by $h = \frac{2T \cos 	heta}{ho g r}$.For water,the contact angle $	heta \approx 0^\circ$,so $\cos 	heta = 1$.The difference in levels in a $U$-tube with limbs of radii $r_1$ and $r_2$ is $\Delta h = h_1 - h_2 = \frac{2T}{ho g} \left( \frac{1}{r_1} - \frac{1}{r_2} ight)$.Given: $T = 0.07\,N/m$,$\Delta h = 3.6\,cm = 3.6 	imes 10^{-2}\,m$,$ho = 1000\,kg/m^3$,$g = 9.8\,m/s^2$,and $d_1 = 0.4\,mm = 0.4 	imes 10^{-3}\,m$ (so $r_1 = 0.2 	imes 10^{-3}\,m$).Substituting the values: $3.6 	imes 10^{-2} = \frac{2 	imes 0.07}{1000 	imes 9.8} \left( \frac{1}{0.2 	imes 10^{-3}} - \frac{1}{r_2} ight)$.$3.6 	imes 10^{-2} = \frac{0.14}{9800} \left( 5000 - \frac{1}{r_2} ight)$.$3.6 	imes 10^{-2} = 1.428 	imes 10^{-5} 	imes (5000 - \frac{1}{r_2})$.Solving for $r_2$,we find $r_2 = 4 	imes 10^{-3}\,m$.Since $d = 2r_2$,$d = 8 	imes 10^{-3}\,m$.

$A$ $U$-tube is such that the diameter of one limb is $0.4\,mm$ and that of the other is $d\,mm$. If the surface tension of water contained in the tube is $0.07\,N/m$ and the difference in the levels of liquid in the limbs is $3.6\,cm$,then the value of $d$ is:

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