A tall cylinder with a cross-sectional area 11.0 cm^2 is partially filled with mercury; the surface of the mercury is 4.00 cm above the bottom of the cylinder. Water is slowly poured in on top of the mercury, and the two fluids don't mix.
What volume of water must be added to double the gauge pressure at the bottom of the cylinder?
What volume of water must be added to double the gauge pressure at the bottom of the cylinder?
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Pressure at the bottom = h dg.
Thus pressure depends only upon the height.
The height of mercury is 4cm.
To double the pressure another 4cm of mercury has to be added or equivalent water has to be added.
4cm of mercury = 4*13.54cm of water = 54.16 cm of water.
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Volume of water = height* area = 54.16cm*11.0 cm^2 = 595.76[cm]^3
Thus pressure depends only upon the height.
The height of mercury is 4cm.
To double the pressure another 4cm of mercury has to be added or equivalent water has to be added.
4cm of mercury = 4*13.54cm of water = 54.16 cm of water.
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Volume of water = height* area = 54.16cm*11.0 cm^2 = 595.76[cm]^3
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Density of Mercury = 13.53 gram/cm^3
Mercury is 13.53 times as heavy as water
The volume of water must 13.53 times as much as the mercury
4.00 cm* 13.53 = 54.12 cm
The cylinder must be filled with water till the water level is 58.12 cm above the bottom of the cylinder
Mercury is 13.53 times as heavy as water
The volume of water must 13.53 times as much as the mercury
4.00 cm* 13.53 = 54.12 cm
The cylinder must be filled with water till the water level is 58.12 cm above the bottom of the cylinder