If a barometer is kept on the earth's surface then calculate the height of the barometer column $h$ .
Atmospheric Pressure ( $P_{a}$ ) is $10^{5}$ .
The density of mercury ( $ρ_{m}$ ) is $13600 k g m^{- 3}$ .
Gravitational acceleration ( $g$ ) near earth's surface is $10 m s^{- 2}$ .

64 c m

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74 c m

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84 c m

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94 c m

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Solution

The correct option is B $74 c m$
Given:
Acceleration due to gravity, $g = 10 m s^{- 2}$
Density of mercury, $ρ_{m} = 13600 k g m^{- 3}$
Atmospheric Pressure, $P_{a} = 10^{5} P a$

The weight of the liquid in the column balances the atmospheric pressure in the barometer.
$ρ_{m} g h = P_{a}$
$13600 \times h \times 10 = 10^{5}$
$h = 0.735 m \approx 74 c m$

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If a barometer is kept on the earth's surface then calculate the height of the barometer column h. Atmospheric Pressure (Pa) is 105. The density of mercury (ρm) is 13600 kg m−3. Gravitational acceleration (g) near earth's surface is 10 m s−2.

The correct option is B 74 cmGiven: Acceleration due to gravity, g=10 m s−2 Density of mercury, ρm=13600 kg m−3 Atmospheric Pressure, Pa=105 Pa The weight of the liquid in the column balances the atmospheric pressure in the barometer. ρmgh=Pa 13600×h×10=105 h=0.735 m≈74 cm

If a barometer is kept on the earth's surface then calculate the height of the barometer column $h$ .
Atmospheric Pressure ( $P_{a}$ ) is $10^{5}$ .
The density of mercury ( $ρ_{m}$ ) is $13600 k g m^{- 3}$ .
Gravitational acceleration ( $g$ ) near earth's surface is $10 m s^{- 2}$ .