Water rises to a height h in a capillary tube of crosssectional area A. The height to which water will rise in a capillary tube of cross-sectional area 4 A will be

Question

Water rises to a height h in a capillary tube of crosssectional area A. The height to which water will rise in a capillary tube of cross-sectional area 4 A will be (A) h (B) h / 2 (C) h / 4 (D) 4 h

Tardigrade Admin · Accepted Answer

h=(2 S cos θ/r ρ g) Cross-sectional area increases four times, which means radius gets doubled. So, h prime=(2 S/2(r ρ g))=(h/2)

Q. Water rises to a height $h$ in a capillary tube of crosssectional area $A$ . The height to which water will rise in a capillary tube of cross-sectional area $4 A$ will be

$h$

$h /2$

$h /4$

$4 h$

$h = \frac{2 S c o s θ}{r ρ g}$
Cross-sectional area increases four times, which means radius gets doubled.
So, $h^{'} = \frac{2 S}{2 ( r ρ g )} = \frac{h}{2}$

Q. Water rises to a height h in a capillary tube of crosssectional area A. The height to which water will rise in a capillary tube of cross-sectional area 4A will be

h

h/2

h/4

4h

h=rρg2Scosθ​ Cross-sectional area increases four times, which means radius gets doubled. So, h′=2(rρg)2S​=2h​

Q. Water rises to a height $h$ in a capillary tube of crosssectional area $A$ . The height to which water will rise in a capillary tube of cross-sectional area $4 A$ will be

$h$

$h /2$

$h /4$

$4 h$

$h = \frac{2 S c o s θ}{r ρ g}$
Cross-sectional area increases four times, which means radius gets doubled.
So, $h^{'} = \frac{2 S}{2 ( r ρ g )} = \frac{h}{2}$