The internal diameter of the tube of a mercury barometer is $3 m m$ . Find the correct barometric reading of the observed reading is $76.56 c m$ , Surface tension of mercury= $0.48 N / m$ , Angle of contact of mercury in glass= $140$ and density of mercury= $13600 k g / m^{3} . g = 9.8 m / s^{- 1}$

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Solution

$D = 3 m m$
Observed= $76.56 c m$
$T = 0.48 N / m$
Angle of contact= $140 °$
$△ h = \frac{2 - cos θ}{r s g} = \frac{2 \times 0.48 \times cos 40}{3 {\times 10}^{- 3} \times 10 \times 13600} = \frac{0.735 \times 2}{3 \times 136} △ h = 0.36 c m h_{C} = 76.56 + 0.36 = 76.92 c m$

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The internal diameter of the tube of a mercury barometer is 3mm. Find the correct barometric reading of the observed reading is 76.56cm, Surface tension of mercury=0.48N/m, Angle of contact of mercury in glass=140 and density of mercury=13600kg/m3.g=9.8m/s−1

D=3mmObserved=76.56cmT=0.48N/mAngle of contact=140°△h=2−cosθrsg=2×0.48×cos403×10−3×10×13600=0.735×23×136△h=0.36cmhC=76.56+0.36=76.92cm

The internal diameter of the tube of a mercury barometer is $3 m m$ . Find the correct barometric reading of the observed reading is $76.56 c m$ , Surface tension of mercury= $0.48 N / m$ , Angle of contact of mercury in glass= $140$ and density of mercury= $13600 k g / m^{3} . g = 9.8 m / s^{- 1}$

$D = 3 m m$
Observed= $76.56 c m$
$T = 0.48 N / m$
Angle of contact= $140 °$
$△ h = \frac{2 - cos θ}{r s g} = \frac{2 \times 0.48 \times cos 40}{3 {\times 10}^{- 3} \times 10 \times 13600} = \frac{0.735 \times 2}{3 \times 136} △ h = 0.36 c m h_{C} = 76.56 + 0.36 = 76.92 c m$