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A capillary tube of inner radius 0.5 mm ...

A capillary tube of inner radius `0.5 mm` is dipped keeping it vertical in a mercury of soecific gravity 13.6, surface tension `545 dyn e//cm` and angle of contact `135^(@)` . Find the depression or elevation of liquid in the tube.

Text Solution

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Here, `r=0.5//2 = 0.25 mm = 0.0025 cm`,
Density of mercury , `rho =` specific . gravity `xx` density ofwater` = 13.6 xx 1 = 13.6 g//c c`
Surface tension , `S=545 dyn e//cm`, angle of contact `theta= 135^(@)`. The height of liquid in tube is
`h=(2S cos theta)/(r rho g) = (2 xx 545 xx cos 135^(@))/((0.025)xx13.6 xx 980) = (2xx545xx(-1//sqrt(2)))/(0.025xx13.6xx980)=-2.31 cm`.
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Knowledge Check

  • A capillary tube of inner diameter 0.5 mm is dipped in a liquid of specific gravity 13.6 and having surface tension 545 dyn/cm(angle of contact 130^(@) ) . Find the depression or elevation in the tube .

    A
    depressed 2.11 cm
    B
    elevated 2.11 cm
    C
    depressed , 3.71 cm
    D
    elevated , 3.71 cm

  • A capillary tube of radius 0.25 mm is dipped vertically in a liquid of density 800kgm^(-3) and of surface tension 3xx10^(-2)Nm^(-2) . The angle of contact of liquid -glass is given by costheta=0.3 If g=10ms^(-2) the rise of liquid in the capillary tube is.. Cm

    A
    9
    B
    0.9
    C
    `9xx10^(-3)`
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  • A capillary tube of the radius 0.5 mm is immersed in a beaker of mercury . The level inside the tube is 0.8 cm below the level in beaker and angle of contact is 120^@ . What is the surface tension of mercury , if the mass density of mercury is rho= 13.6 xx 10^3 kg m^3 and acceleration due to gravity is g = 10 m s^(-2) ?

    A
    `0.225 "N m"^(-1)`
    B
    `0.544 "N m"^(-1)`
    C
    `0.285 "N m"^(-1)`
    D
    `0.375 "N m"^(-1)`

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