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Question: The height of a mercury barometer is \(75cm\) at sea level and \(50cm\) at the top of a hill. Ratio ...

The height of a mercury barometer is 75cm75cm at sea level and 50cm50cm at the top of a hill. Ratio of the density of mercury to that of air is 104{{10}^{4}}. The height of the hill is
(A). 1.25km1.25km
(B). 2.5km2.5km
(C). 250m250m
(D). 750m750m

Explanation

Solution

The pressure in a fluid depends on the density of the fluid, acceleration due to gravity and height. The pressure measured by the barometer will be the same as the pressure of the air column. Equating both pressures and substituting corresponding values, height of the hill can be calculated. Convert units as required.
Formula used:
ΔP=(h1h2)gρHg\Delta P=({{h}_{1}}-{{h}_{2}})g{{\rho }_{Hg}}
ΔP=hgρair\Delta P=hg{{\rho }_{air}}

Complete answer:
Pressure is the force applied per unit area normally. Its SI unit is pascal (PP).
P=FAP=\dfrac{F}{A}
Here, PP is the pressure
FF is the force
AA is the area of cross section
The pressure inside a fluid is given by-
P=ρghP=\rho gh --- (1)
Here, ρ\rho is the density of the fluid
gg is acceleration due to gravity
hh is the height
Given, the height of mercury at sea level is 75cm75cm, height of mercury at hill top is 50cm50cm.
Therefore from eq (1), the difference in pressure at sea level and hilltop can be calculated as-
ΔP=(h1h2)gρHg\Delta P=({{h}_{1}}-{{h}_{2}})g{{\rho }_{Hg}}
We substitute given values in the above equation to get,
ΔP=(7550)×102×10×ρHg\Delta P=(75-50)\times {{10}^{-2}}\times 10\times {{\rho }_{Hg}}
ΔP=250×102ρHg\Rightarrow \Delta P=250\times {{10}^{-2}}{{\rho }_{Hg}} --- (2)
Pressure difference due to air column at height at sea level and hill top, from eq (1), will be
ΔP=hgρair\Delta P=hg{{\rho }_{air}}
ΔP=10hρair\Rightarrow \Delta P=10h{{\rho }_{air}} ---- (3)
Equating eq (1) and eq (2), we get,
2.5ρHg=10hρair ρHgρair=10h2.5 \begin{aligned} & 2.5{{\rho }_{Hg}}=10h{{\rho }_{air}} \\\ & \Rightarrow \dfrac{{{\rho }_{Hg}}}{{{\rho }_{air}}}=\dfrac{10h}{2.5} \\\ \end{aligned}
Given that, ρHgρair=104\dfrac{{{\rho }_{Hg}}}{{{\rho }_{air}}}={{10}^{4}}, we substitute in above equation to get,
104=10h2.5 25×103=h h=2.5km \begin{aligned} & {{10}^{4}}=\dfrac{10h}{2.5} \\\ & \Rightarrow 25\times {{10}^{3}}=h \\\ & \therefore h=2.5km \\\ \end{aligned}
The height of the hill is 2.5km2.5km
Therefore, the height of the top of the hill is 2.5km2.5km.

Hence, the correct option is (B).

Note:
The force applied is always normal to the surface area on which it is being applied. The pressure is directly proportional to the height. This means as the height increases, the pressure also increases. The height of the hill is measured from the sea level. Barometer is a device which is used to measure pressure.