The height of a mercury barometer is \[75cm\] at sea level and \[50cm\] at the top of a hill a . Ratio of density of mercury to that of air is \[{10^4}\]. The height of the hill is-
A. \[1.25Km\]
B. \[2.5Km\]
C. \[250m\]
D. \[750m\]
Answer
Verified
453.9k+ views
Hint:A mercury barometer is a device that is used to measure the atmospheric pressure at a given location. As, ratio of density of mercury to that of the air, \[\dfrac{{{\rho _{Hg}}}}{{{\rho _{Air}}}}\] is given=\[{10^4}\]. We know the equation for the change in pressure. By substituting all the given values, we can easily find the value of the height, h.
Formula used:
\[\Delta p = \left( {{h_1} - {h_2}} \right) \times {\rho _{Hg}} \times g\]
Here \[\Delta p\] is the change in pressure,${h_1}$ and ${h_2}$ are the heights of
barometer, g is gravity and \[{\rho _{Hg}}\] is the density of mercury.
Complete step by step answer:
As we know that the pressure difference between the sea level and the top of hill is-
\[\Delta p = \left( {{h_1} - {h_2}} \right) \times {\rho _{Hg}} \times g\] ---- (1)
${h_1}$ and ${h_2}$ are the heights of mercury barometer- given- \[75cm\] and \[50cm\] respectively. Now substitute all the values in the equation (1), we get-
\[\Delta p = \left( {{h_1} - {h_2}} \right) \times {\rho _{Hg}} \times g\]
\[\Rightarrow\Delta p = \left( {75 - 50} \right) \times {10^{ - 2}} \times {\rho _{Hg}} \times g\] --- (2)
Pressure difference due to h metre of air-\[\Delta p = h \times {\rho _{Air}} \times g\]-- (3)
Equate equation (2) and (3), we get-
\[h \times {\rho _{Air}} \times g\]=\[\left( {75 - 50} \right) \times {10^{ - 2}} \times {\rho _{Hg}} \times g\]
For finding the height of the hill, h we can take all terms on the right hand side, we get-
\[\dfrac{{{\rho _{Hg}}}}{{{\rho _{Air}}}} \times 25 \times {10^{ - 2}}\]
Now we know the ratio of density of mercury to the air is already given in this question,
\[\therefore h = {10^4} \times 25 \times {10^{ - 2}}\]
So, the height of the hill comes out to be \[2500m\]or \[2.5Km\] .
Hence, option B is correct.
Note:A mercury barometer is a device that is used to measure the atmospheric pressure at a given location. The barometer consists of a vertical glass tube which is closed at one end. Additionally, The air around us has weight, and it presses against everything it touches. That pressure is known as atmospheric pressure.
Formula used:
\[\Delta p = \left( {{h_1} - {h_2}} \right) \times {\rho _{Hg}} \times g\]
Here \[\Delta p\] is the change in pressure,${h_1}$ and ${h_2}$ are the heights of
barometer, g is gravity and \[{\rho _{Hg}}\] is the density of mercury.
Complete step by step answer:
As we know that the pressure difference between the sea level and the top of hill is-
\[\Delta p = \left( {{h_1} - {h_2}} \right) \times {\rho _{Hg}} \times g\] ---- (1)
${h_1}$ and ${h_2}$ are the heights of mercury barometer- given- \[75cm\] and \[50cm\] respectively. Now substitute all the values in the equation (1), we get-
\[\Delta p = \left( {{h_1} - {h_2}} \right) \times {\rho _{Hg}} \times g\]
\[\Rightarrow\Delta p = \left( {75 - 50} \right) \times {10^{ - 2}} \times {\rho _{Hg}} \times g\] --- (2)
Pressure difference due to h metre of air-\[\Delta p = h \times {\rho _{Air}} \times g\]-- (3)
Equate equation (2) and (3), we get-
\[h \times {\rho _{Air}} \times g\]=\[\left( {75 - 50} \right) \times {10^{ - 2}} \times {\rho _{Hg}} \times g\]
For finding the height of the hill, h we can take all terms on the right hand side, we get-
\[\dfrac{{{\rho _{Hg}}}}{{{\rho _{Air}}}} \times 25 \times {10^{ - 2}}\]
Now we know the ratio of density of mercury to the air is already given in this question,
\[\therefore h = {10^4} \times 25 \times {10^{ - 2}}\]
So, the height of the hill comes out to be \[2500m\]or \[2.5Km\] .
Hence, option B is correct.
Note:A mercury barometer is a device that is used to measure the atmospheric pressure at a given location. The barometer consists of a vertical glass tube which is closed at one end. Additionally, The air around us has weight, and it presses against everything it touches. That pressure is known as atmospheric pressure.
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