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
457.2k+ 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.
Recently Updated Pages
Master Class 11 Accountancy: Engaging Questions & Answers for Success
Glucose when reduced with HI and red Phosphorus gives class 11 chemistry CBSE
The highest possible oxidation states of Uranium and class 11 chemistry CBSE
Find the value of x if the mode of the following data class 11 maths CBSE
Which of the following can be used in the Friedel Crafts class 11 chemistry CBSE
A sphere of mass 40 kg is attracted by a second sphere class 11 physics CBSE
Trending doubts
Define least count of vernier callipers How do you class 11 physics CBSE
The combining capacity of an element is known as i class 11 chemistry CBSE
Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE
Find the image of the point 38 about the line x+3y class 11 maths CBSE
Can anyone list 10 advantages and disadvantages of friction
Distinguish between Mitosis and Meiosis class 11 biology CBSE