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
615.3k+ 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
The given figure shows two endocrine glands marked class 11 biology NEET_UG

Match columnI with columnII and select the correct class 11 biology NEET

Match column I with column II and select the correct class 11 biology NEET_UG

Which floral family has left 9 right + 1 arrangement class 11 biology NEET_UG

Which is not a variety of sheep A Lohi B Beetal C Nellore class 11 biology NEET_UG

Match column I with column II and select the correct class 11 biology NEET_UG

Trending doubts
Difference Between Prokaryotic Cells and Eukaryotic Cells

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Two of the body parts which do not appear in MRI are class 11 biology CBSE

10 examples of friction in our daily life

Draw a diagram of nephron and explain its structur class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

