Liquid A rises to a height of 10 cm in a cap | vedclass

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(B) The height of capillary rise is given by the formula: $h = \frac{2 S \cos 	heta}{r ho g}$,where $r$ is the radius of the tube.For liquid $A$: $

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$2 \sqrt{2}$

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(B) The height of capillary rise is given by the formula: $h = \frac{2 S \cos 	heta}{r ho g}$,where $r$ is the radius of the tube.For liquid $A$: $h_A = 10 \ cm$,$ho_A = 1 \ g/cm^3$,$	heta_A = 0^{\circ}$.$10 = \frac{2 S_A \cos(0^{\circ})}{r 	imes 1 	imes g} = \frac{2 S_A}{r g}$  --- $(1)$For liquid $B$: $h_B = -2 \ cm$ (fall),$ho_B = 10 \ g/cm^3$,$	heta_B = 135^{\circ}$.$-2 = \frac{2 S_B \cos(135^{\circ})}{r 	imes 10 	imes g} = \frac{2 S_B (-1/\sqrt{2})}{10 r g} = -\frac{S_B}{5 \sqrt{2} r g}$$2 = \frac{S_B}{5 \sqrt{2} r g}$  --- $(2)$Dividing $(2)$ by $(1)$:$\frac{2}{10} = \frac{S_B / (5 \sqrt{2} r g)}{2 S_A / (r g)} = \frac{S_B}{5 \sqrt{2} r g} 	imes \frac{r g}{2 S_A} = \frac{S_B}{10 \sqrt{2} S_A}$$\frac{1}{5} = \frac{S_B}{10 \sqrt{2} S_A} \Rightarrow \frac{S_B}{S_A} = \frac{10 \sqrt{2}}{5} = 2 \sqrt{2}$.

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