A capillary tube of radius 0-25 mm is dipped vertically in a liquid of density 800 kg m-3 and of surface tension 3- 10-2 Nm-2-The angle of contact of liquid-glass is given by cos- - 0-3- If g-10ms-2- the rise of liquid in the capillary tube is Cm-

The correct option is A 0.9 From the formula h=2Tcosθrgρ we get #160; h=2× 3× 10^ 2× (0.3).25× 10^ 3× 10× 800 =.009 #160; m =0.9 ...

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Standard VIII

Chemistry

Surface Tension

A capillary t...

Question

A capillary tube of radius 0.25 mm is dipped vertically in a liquid of density $800 k g m^{- 3}$ and of surface tension $3 \times 10^{- 2} N m^{- 2}$ .The angle of contact of liquid-glass is given by $c o s θ = 0.3$ . If $g = 10 m s^{- 2}$ , the rise of liquid in the capillary tube is______ $(C m)$ :

9

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0.9

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9 x 10^{- 3}

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0.09

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Solution

The correct option is A $0.9$
From the formula $h = \frac{2 T c o s θ}{r g ρ}$ we get
$h = \frac{2 \times 3 \times 10^{- 2} \times (0.3)}{.25 \times 10^{- 3} \times 10 \times 800}$
$= .009 m$
$= 0.9 c m$

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A capillary tube of radius 0.25 mm is dipped vertically in a liquid of density 800kgm−3 and of surface tension 3×10−2Nm−2.The angle of contact of liquid-glass is given by cosθ=0.3. If g=10ms−2, the rise of liquid in the capillary tube is______ (Cm):

The correct option is A 0.9From the formula h=2Tcosθrgρwe get h=2×3×10−2×(0.3).25×10−3×10×800=.009m=0.9cm

The correct option is A $0.9$
From the formula $h = \frac{2 T c o s θ}{r g ρ}$ we get
$h = \frac{2 \times 3 \times 10^{- 2} \times (0.3)}{.25 \times 10^{- 3} \times 10 \times 800}$
$= .009 m$
$= 0.9 c m$