A cylindrical vessel is filled with equal amounts of weight of mercury in water. The overall height of the two layers 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

cm of HG

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\frac{29.2}{13.6}

cm of HG

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4

cm of HG

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15.6

cm of HG

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Solution

The correct option is C $4$ cm of HG
$M_{H g} = M_{w}$

$ρ_{H g} (A \times L_{H g}) = ρ_{w} (A \times L_{w})$

$L_{w} = 13.6 L_{H g}$

$L_{w} + L_{H g} = 29.2, L_{H g} = 2 c m$

$L_{w} = 27.2 c m$

Total $= 4 c m o f H G$

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A cylindrical vessel is filled with equal amounts of weight of mercury in water. The overall height of the two layers 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 C 4 cm of HGMHg=Mw ρHg(A×LHg)=ρw(A×Lw) Lw=13.6LHg Lw+LHg=29.2,LHg=2cm Lw=27.2cm Total =4 cm of HG

A cylindrical vessel is filled with equal amounts of weight of mercury in water. The overall height of the two layers 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 C $4$ cm of HG
$M_{H g} = M_{w}$

$ρ_{H g} (A \times L_{H g}) = ρ_{w} (A \times L_{w})$

$L_{w} = 13.6 L_{H g}$

$L_{w} + L_{H g} = 29.2, L_{H g} = 2 c m$

$L_{w} = 27.2 c m$

Total $= 4 c m o f H G$