The height of a mercury column in a barometer at a place is 74 cm. If a liquid of density $5.44 g c m^{- 3}$ is used then find the height of the liquid column.

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Solution

Given,
For mercury For liquid
$H_{1} = 74 c m$ $h_{2} = ?$
$D_{1} = 13.6 c m^{- 3}$ $d_{2} = 5.44 g c m^{- 3}$
Since pressure at a place is same
$P_{1} = P_{2}$
$∴ h_{1} d_{1} g = h_{2} d_{2} g (∴ P = h d g)$
$74 \times 13.6 = h_{2} \times 5.44$
$h_{2} = 185 c m .$

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The height of a mercury column in a barometer at a place is 74 cm. If a liquid of density 5.44gcm−3 is used then find the height of the liquid column.

Given,For mercury For liquidH1=74cm h2=?D1=13.6cm−3 d2=5.44gcm−3Since pressure at a place is sameP1=P2∴h1d1g=h2d2g(∴P=hdg)74×13.6=h2×5.44h2=185cm.

The height of a mercury column in a barometer at a place is 74 cm. If a liquid of density $5.44 g c m^{- 3}$ is used then find the height of the liquid column.

Given,
For mercury For liquid
$H_{1} = 74 c m$ $h_{2} = ?$
$D_{1} = 13.6 c m^{- 3}$ $d_{2} = 5.44 g c m^{- 3}$
Since pressure at a place is same
$P_{1} = P_{2}$
$∴ h_{1} d_{1} g = h_{2} d_{2} g (∴ P = h d g)$
$74 \times 13.6 = h_{2} \times 5.44$
$h_{2} = 185 c m .$