Answer
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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.
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