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(B) The height of the water column in a capillary tube is given by $h = \frac{2S \cos 	heta}{ho rg}$.Initially,the water rises to $h_1 = 8 \ cm$ wi

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$6 \ cm, \cos^{-1} \frac{3}{4}$

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(B) The height of the water column in a capillary tube is given by $h = \frac{2S \cos 	heta}{ho rg}$.Initially,the water rises to $h_1 = 8 \ cm$ with an angle of contact $	heta_1 = 0^{\circ}$ (for water and glass).When the tube is cut at $h_2 = 6 \ cm$,the water will fill the entire tube up to the top because $6 \ cm Thus,the new height of the water column is $6 \ cm$.Since $h \propto \cos 	heta$,we have $\frac{h_1}{\cos 	heta_1} = \frac{h_2}{\cos 	heta_2}$.Substituting the values: $\frac{8}{\cos 0^{\circ}} = \frac{6}{\cos 	heta_2}$.Since $\cos 0^{\circ} = 1$,we get $\cos 	heta_2 = \frac{6}{8} = \frac{3}{4}$.Therefore,the new angle of contact is $	heta_2 = \cos^{-1} \left( \frac{3}{4} ight)$.

The lower end of a glass capillary tube is dipped in water. Water rises to a height of $8 \ cm$. The tube is then broken at a height of $6 \ cm$. The new height of water column and new angle of contact will be $:-$

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