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Standard XII

Physics

Surface Tension

A glass rod o...

Question

A glass rod of radius $r_{1}$ is inserted symmetrically into a vertical capillary tube of radius $r_{2}$ such that their lower ends are at the same level. The arrangement is now dipped in water. The height to which water will rise into the tube will be ( $σ =$ surface tension of water, $ρ =$ density of water)

\frac{2 σ}{(r_{2} - r_{1}) ρ g}

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\frac{σ}{(r_{2} - r_{1}) ρ g}

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\frac{2 σ}{(r_{2} + r_{1}) ρ g}

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\frac{2 σ}{({r_{2}}^{2} + {r_{1}}^{2}) ρ g}

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Solution

The correct option is B $\frac{2 σ}{(r_{2} - r_{1}) ρ g}$
Total upward force due to surface tension $= σ [2 π r_{1} + 2 π r_{2}]$
This supports the weight of the liquid column of height h
Weight of the liquid column $= h [π r_{2}^{2} - π r_{1}^{2}] ρ g$
or
$h (r_{2} - r_{1}) ρ g = 2 σ$
or
$h = \frac{2 σ}{(r_{2} - r_{1}) ρ g}$

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A glass rod of radius r1 is inserted symmetrically into a vertical capillary tube of radius r2 such that their lower ends are at the same level. The arrangement is now dipped in water. The height to which water will rise into the tube will be (σ= surface tension of water, ρ= density of water)

The correct option is B 2σ(r2−r1)ρgTotal upward force due to surface tension =σ[2πr1+2πr2]This supports the weight of the liquid column of height hWeight of the liquid column =h[πr22−πr21]ρgorh(r2−r1)ρg=2σorh=2σ(r2−r1)ρg