Toricelli’s barometer used mercury. Pascal duplicated it using French wine of density $984 \;kg m^{-3}$. Determine the height (in $m$) of the wine column for normal atmospheric pressure.
Toricelli’s barometer used mercury. Pascal duplicated it using French wine of density $984 \;kg m^{-3}$. Determine the height (in $m$) of the wine column for normal atmospheric pressure.
- A
$10.5$
- B
$8.4$
- C
$12.6$
- D
$15.8$
Similar Questions
A cylindrical furnace has height $(H)$ and diameter $(D)$ both $1 \mathrm{~m}$. It is maintained at temperature $360 \mathrm{~K}$. The air gets heated inside the furnace at constant pressure $P_a$ and its temperature becomes $T=360 \mathrm{~K}$. The hot air with density $\rho$ rises up a vertical chimney of diameter $d=0.1 \mathrm{~m}$ and height $h=9 \mathrm{~m}$ above the furnace and exits the chimney (see the figure). As a result, atmospheric air of density $\rho_a=1.2 \mathrm{~kg} \mathrm{~m}^{-3}$, pressure $P_a$ and temperature $T_a=300 \mathrm{~K}$ enters the furnace. Assume air as an ideal gas, neglect the variations in $\rho$ and $T$ inside the chimney and the furnace. Also ignore the viscous effects.
[Given: The acceleration due to gravity $g=10 \mathrm{~ms}^{-2}$ and $\pi=3.14$ ]
(image)
($1$) Considering the air flow to be streamline, the steady mass flow rate of air exiting the chimney is
. . . . .$\mathrm{gm} \mathrm{s}^{-1}$.
($2$) When the chimney is closed using a cap at the top, a pressure difference $\Delta P$ develops between the top and the bottom surfaces of the cap. If the changes in the temperature and density of the hot air, due to the stoppage of air flow, are negligible then the value of $\Delta P$ is. . . . .$\mathrm{Nm}^{-2}$.
A cylindrical furnace has height $(H)$ and diameter $(D)$ both $1 \mathrm{~m}$. It is maintained at temperature $360 \mathrm{~K}$. The air gets heated inside the furnace at constant pressure $P_a$ and its temperature becomes $T=360 \mathrm{~K}$. The hot air with density $\rho$ rises up a vertical chimney of diameter $d=0.1 \mathrm{~m}$ and height $h=9 \mathrm{~m}$ above the furnace and exits the chimney (see the figure). As a result, atmospheric air of density $\rho_a=1.2 \mathrm{~kg} \mathrm{~m}^{-3}$, pressure $P_a$ and temperature $T_a=300 \mathrm{~K}$ enters the furnace. Assume air as an ideal gas, neglect the variations in $\rho$ and $T$ inside the chimney and the furnace. Also ignore the viscous effects.
[Given: The acceleration due to gravity $g=10 \mathrm{~ms}^{-2}$ and $\pi=3.14$ ]
(image)
($1$) Considering the air flow to be streamline, the steady mass flow rate of air exiting the chimney is
. . . . .$\mathrm{gm} \mathrm{s}^{-1}$.
($2$) When the chimney is closed using a cap at the top, a pressure difference $\Delta P$ develops between the top and the bottom surfaces of the cap. If the changes in the temperature and density of the hot air, due to the stoppage of air flow, are negligible then the value of $\Delta P$ is. . . . .$\mathrm{Nm}^{-2}$.
- [IIT 2023]
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Two identical cylindrical vessels with their bases at same level, each contains a liquid of density $d$ . The height of the liquid in one vessel is $ h_1$ and that in the other vessel is $h_2$ . The area of either base is $A$ . The work done by gravity in equalizing the levels when the two vessels are connected is
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[take $\rho_{water} \left.=1000 \;\mathrm{kg} / \mathrm{m}^{3}\right]$
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[take $\rho_{water} \left.=1000 \;\mathrm{kg} / \mathrm{m}^{3}\right]$
- [NEET 2019]
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A light cylindrical vessel is kept on a horizontal surface. Area of base is A. A hole of crosssectional area $'a'$ is made just at its bottom side. The minimum coefficient of friction necessary to prevent sliding the vessel due to the impact force of the emerging liquid is $(a\,<\,<\,A)$
- [JEE MAIN 2021]