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(A) The maximum height $h$ to which water can rise in a capillary tube is given by $h = \frac{2T \cos 	heta}{r ho g}$.Assuming the contact angle $\

Vedclass · Accepted Answer

$70$

Vedclass · Answer

(A) The maximum height $h$ to which water can rise in a capillary tube is given by $h = \frac{2T \cos 	heta}{r ho g}$.Assuming the contact angle $	heta = 0^{\circ}$ for water and glass,the maximum height is $h_{max} = \frac{2T}{r ho g}$.Substituting the values: $h_{max} = \frac{2 	imes 0.07}{0.25 	imes 10^{-3} 	imes 1000 	imes 9.8} = \frac{0.14}{2.45} = 0.0571 \, m = 5.71 \, cm$.Since the tube projects only $2 \, cm$ above the water surface,the water will rise to the top of the tube and adjust its contact angle $	heta$ to satisfy the equilibrium condition $h' = \frac{2T \cos 	heta}{r ho g}$,where $h' = 2 \, cm = 0.02 \, m$.Thus,$\cos 	heta = \frac{h' r ho g}{2T} = \frac{0.02 	imes 0.25 	imes 10^{-3} 	imes 1000 	imes 9.8}{2 	imes 0.07} = \frac{0.049}{0.14} = 0.35$.Calculating the angle,$	heta = \cos^{-1}(0.35) \approx 69.5^{\circ} \approx 70^{\circ}$.

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