A cylinderical vessel is filled with equal amount of weight of mercury and water.The overall height of the two layer is 29.2 cm,specific gravity of mercury is 13.6.Then the pressure of the liquid at the bottom of the vessel is:

29.2 c m

of water

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\frac{29.2}{13.6} c m

of mercury

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4 c m

of mercury

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15.6 c m

of mercury

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Solution

The correct option is D $4 c m$ of mercury
$M_{H g} = M_{w}$
$P_{H g} (A \times L_{H g}) = P_{w} (A \times L_{w})$
$∴$ $L_{w} = 13.6 L_{H g}$
$L_{w} + L_{H g} = 29.2$
$\Rightarrow L_{H g} = 2 c m L_{w} = 27.2 c m = 2 c m$ of $H g$
Total pressure $= 4 c m$ of $H g$ .

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Similar questions

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A cylinderical vessel is filled with equal amount of weight of mercury and water.The overall height of the two layer is 29.2 cm,specific gravity of mercury is 13.6.Then the pressure of the liquid at the bottom of the vessel is:

The correct option is D 4cm of mercuryMHg=MwPHg(A×LHg)=Pw(A×Lw)∴ Lw=13.6LHgLw+LHg=29.2⇒LHg=2cmLw=27.2cm=2cm of HgTotal pressure =4cm of Hg.

The correct option is D $4 c m$ of mercury
$M_{H g} = M_{w}$
$P_{H g} (A \times L_{H g}) = P_{w} (A \times L_{w})$
$∴$ $L_{w} = 13.6 L_{H g}$
$L_{w} + L_{H g} = 29.2$
$\Rightarrow L_{H g} = 2 c m L_{w} = 27.2 c m = 2 c m$ of $H g$
Total pressure $= 4 c m$ of $H g$ .