Answer
Verified
407.7k+ views
Hint: 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:
$\Delta P=({{h}_{1}}-{{h}_{2}})g{{\rho }_{Hg}}$
$\Delta P=hg{{\rho }_{air}}$
Complete answer:
Pressure is the force applied per unit area normally. Its SI unit is pascal ($P$).
$P=\dfrac{F}{A}$
Here, $P$ is the pressure
$F$ is the force
$A$ is the area of cross section
The pressure inside a fluid is given by-
$P=\rho gh$ --- (1)
Here, $\rho $ is the density of the fluid
$g$ is acceleration due to gravity
$h$ is the height
Given, the height of mercury at sea level is $75cm$, height of mercury at hill top is $50cm$.
Therefore from eq (1), the difference in pressure at sea level and hilltop can be calculated as-
$\Delta P=({{h}_{1}}-{{h}_{2}})g{{\rho }_{Hg}}$
We substitute given values in the above equation to get,
$\Delta P=(75-50)\times {{10}^{-2}}\times 10\times {{\rho }_{Hg}}$
$\Rightarrow \Delta P=250\times {{10}^{-2}}{{\rho }_{Hg}}$ --- (2)
Pressure difference due to air column at height at sea level and hill top, from eq (1), will be
$\Delta P=hg{{\rho }_{air}}$
$\Rightarrow \Delta P=10h{{\rho }_{air}}$ ---- (3)
Equating eq (1) and eq (2), we get,
$\begin{align}
& 2.5{{\rho }_{Hg}}=10h{{\rho }_{air}} \\
& \Rightarrow \dfrac{{{\rho }_{Hg}}}{{{\rho }_{air}}}=\dfrac{10h}{2.5} \\
\end{align}$
Given that, $\dfrac{{{\rho }_{Hg}}}{{{\rho }_{air}}}={{10}^{4}}$, we substitute in above equation to get,
$\begin{align}
& {{10}^{4}}=\dfrac{10h}{2.5} \\
& \Rightarrow 25\times {{10}^{3}}=h \\
& \therefore h=2.5km \\
\end{align}$
The height of the hill is $2.5km$
Therefore, the height of the top of the hill is $2.5km$.
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.
Formula used:
$\Delta P=({{h}_{1}}-{{h}_{2}})g{{\rho }_{Hg}}$
$\Delta P=hg{{\rho }_{air}}$
Complete answer:
Pressure is the force applied per unit area normally. Its SI unit is pascal ($P$).
$P=\dfrac{F}{A}$
Here, $P$ is the pressure
$F$ is the force
$A$ is the area of cross section
The pressure inside a fluid is given by-
$P=\rho gh$ --- (1)
Here, $\rho $ is the density of the fluid
$g$ is acceleration due to gravity
$h$ is the height
Given, the height of mercury at sea level is $75cm$, height of mercury at hill top is $50cm$.
Therefore from eq (1), the difference in pressure at sea level and hilltop can be calculated as-
$\Delta P=({{h}_{1}}-{{h}_{2}})g{{\rho }_{Hg}}$
We substitute given values in the above equation to get,
$\Delta P=(75-50)\times {{10}^{-2}}\times 10\times {{\rho }_{Hg}}$
$\Rightarrow \Delta P=250\times {{10}^{-2}}{{\rho }_{Hg}}$ --- (2)
Pressure difference due to air column at height at sea level and hill top, from eq (1), will be
$\Delta P=hg{{\rho }_{air}}$
$\Rightarrow \Delta P=10h{{\rho }_{air}}$ ---- (3)
Equating eq (1) and eq (2), we get,
$\begin{align}
& 2.5{{\rho }_{Hg}}=10h{{\rho }_{air}} \\
& \Rightarrow \dfrac{{{\rho }_{Hg}}}{{{\rho }_{air}}}=\dfrac{10h}{2.5} \\
\end{align}$
Given that, $\dfrac{{{\rho }_{Hg}}}{{{\rho }_{air}}}={{10}^{4}}$, we substitute in above equation to get,
$\begin{align}
& {{10}^{4}}=\dfrac{10h}{2.5} \\
& \Rightarrow 25\times {{10}^{3}}=h \\
& \therefore h=2.5km \\
\end{align}$
The height of the hill is $2.5km$
Therefore, the height of the top of the hill is $2.5km$.
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Which are the Top 10 Largest Countries of the World?
Give 10 examples for herbs , shrubs , climbers , creepers
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
How do you graph the function fx 4x class 9 maths CBSE
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Change the following sentences into negative and interrogative class 10 english CBSE