ICSE Class 9 Physics Question 16 of 19

Pressure in Fluids and Atmospheric Pressure — Question 2

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Question

Question 2

At a given place, a mercury barometer records a pressure of 0.70 m of Hg. What would be the height of water column if mercury in barometer is replaced by water ? Take density of mercury to be 13.6 x 103 kg m-3

Answer

We know that,

Pressure = h ρ g

Given,

ρ = 13.6 x 103 kg m-3

g = 9.8 m s -2

h = 0.70 m

Substituting the values in the formula above we get,

P=(0.70)×(13.6×103)×(9.8)P=93.3×103 Pa\text P = (0.70) \times (13.6 \times 10^{3}) \times (9.8) \\[0.5em] \Rightarrow \text P = 93.3 \times 10^{3} \text { Pa} \\[0.5em]

Hence, P = 93.3 x 103Pa

Now, if we use a water barometer,

P = 93.3 x 103 Pa

ρ = 1 x 103 kg m-3

g = 9.8 m s -2

Substituting the values in the formula above we get,

93.3×103=(h)×(1×103)×(9.8)h=93.3×1031×103×9.8h=9.52 m93.3 \times 10^{3} = (\text h) \times (1 \times 10^{3}) \times (9.8) \\[0.5em] \text h = \dfrac{93.3 \times 10^{3}}{1 \times 10^{3} \times 9.8} \\[0.5em] \text h = 9.52 \text { m} \\[0.5em]

Hence, height of water colum = 9.52 m