JEE Advanced Fluid Flow, Reynold’s Number & Bernoulli’s Principle PYQ

Q 1 :

An open glass tube is immersed in mercury in such a way that a length of 8 cm extends above the mercury level. The open end of the tube is then closed and sealed and the tube is raised vertically up by additional 46 cm. What will be the length of the air column above mercury in the tube now?

(Atmospheric pressure = 76 cm of Hg) [2012]

16 cm
22 cm
38 cm
6 cm

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[IMAGE 392]

$Length of the air column above mercury in the tube is,$

$P_{f} + x = P_{0} \Rightarrow P_{f} = (76 - x)$

$As T = cons. \Rightarrow P V = cons.$

$\Rightarrow P_{i} V_{i} = P_{f} V_{f}$ $\Rightarrow (8 \times A) \times 76 = (76 - x) \times A \times (54 - x)$

$∴$ $x = 38$

$Thus, length of air column = 54 - 38 = 16 cm$

Q 2 :

Water is filled in a container upto height 3 m. A small hole of area $' a'$ is punched in the wall of the container at a height 52.5 cm from the bottom. The cross sectional area of the container is $A$ . If $a / A = 0.1$ then $v^{2}$ is (where $v$ is the velocity of water coming out of the hole) [2005]

50
51
48
51.5

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Q 3 :

A large open tank has two holes in the wall. One is a square hole of side $L$ at a depth $y$ from the top and the other is a circular hole of radius $R$ at a depth $4 y$ from the top. When the tank is completely filled with water, the quantities of water flowing out per second from both holes are the same.

Then, $R$ is equal to [2000]

$\frac{L}{\sqrt{2 π}}$
$2 π L$
$L$
$\frac{L}{2 π}$

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$As we know, velocity of efflux V = \sqrt{2 g h}$

$From equation of continuity A_{1} V_{1} = A_{2} V_{2}$

$\sqrt{(2 g y)} \times L^{2} = \sqrt{(2 g \times 4 y)} π R^{2}$

$\Rightarrow L^{2} = 2 π R^{2}$

$∴$ $R = \frac{L}{\sqrt{2 π}}$

Q 4 :

Two large, identical water tanks, 1 and 2, kept on the top of a building of height $H$ , are filled with water up to height $h$ in each tank. Both the tanks contain an identical hole of small radius on their sides, close to their bottom. A pipe of the same internal radius as that of the hole is connected to tank 2, and the pipe ends at the ground level. When the water flows from the tanks 1 and 2 through the holes, the times taken to empty the tanks are $t_{1}$ and $t_{2}$ , respectively. If $H = (\frac{16}{9}) h,$ then the ratio $\frac{t_{1}}{t_{2}}$ is ______. [2024]

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[IMAGE 393]

$From principle of continuity a_{1} v_{1} = a_{2} v_{2}$

$a \sqrt{2 g h} = - A \frac{d h}{d t}$

$d t = - \frac{A}{a} \frac{d h}{\sqrt{2 g h}}$

$t = \int d t = - \frac{2 A}{a \sqrt{2 g}} (\sqrt{h_{f}} - \sqrt{h_{i}})$

$t = \frac{2 A}{a \sqrt{2 g}} (\sqrt{h_{i}} - \sqrt{h_{f}})$

$For tank 1: h_{i} = h, h_{f} = 0$

$∴$ $t_{1} = \frac{2 A}{a \sqrt{2 g}} (\sqrt{h})$

$For tank 2: h_{f} = \frac{16 h}{9}, h_{i} = h + H = \frac{25 h}{9}$

$t_{2} = \frac{2 A \sqrt{h}}{a \sqrt{2 g}} (\frac{5}{3} - \frac{4}{3}) = \frac{2 A \sqrt{h}}{a \sqrt{2 g}} \times \frac{1}{3}$

$∴$ $\frac{T_{1}}{T_{2}} = 3$

Q 5 :

A cylindrical vessel of height 500 mm has an orifice (small hole) at its bottom. The orifice is initially closed and water is filled in it up to height $H$ . Now the top is completely sealed with a cap and the orifice at the bottom is opened. Some water comes out from the orifice and the water level in the vessel becomes steady with height of water column being 200 mm. Find the fall in height (in mm) of water level due to opening of the orifice.

[Take atmospheric pressure = $1.0 \times 10^{5} {N/m}^{2}$ , density of water = $1000 {kg/m}^{3}$ and $g = 10 {m/s}^{2}$ . Neglect any effect of surface tension.] [2009]

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[IMAGE 394]

$Initially, pressure of air column above water$

$P_{1} = 10^{5} {N m}^{- 2}$ $and$ $V_{1} = (500 - H) A,$

$where A is the area of cross-section of the vessel.$

$Finally, the volume of air column above water$ $V_{2} = (500 - 200) A = 300 A .$

$If P_{2} is the pressure of air then$ $P_{2} + ρ g h = P_{1}$

$∴$ $P_{2} + 10^{3} \times 10 \times \frac{200}{1000} = 10^{5}$

$∴$ $P_{2} = 9.8 \times 10^{4} {N/m}^{2}$

$Assuming the temperature remains constant, according to Boyle's law$

$P_{1} V_{1} = P_{2} V_{2}$

$∴$ $10^{5} \times (500 - H) A = (9.8 \times 10^{4}) \times 300 A$ $\Rightarrow H = 206 mm$

$∴$ $Fall in height of water level due to the opening of orifice$ $= 206 - 200 = 6 mm$

Q 6 :

A spray gun is shown in the figure where a piston pushes air out of a nozzle. A thin tube of uniform cross section is connected to the nozzle. The other end of the tube is in a small liquid container. As the piston pushes air through the nozzle, the liquid from the container rises into the nozzle and is sprayed out. For the spray gun shown, the radii of the piston and the nozzle are 20 mm and 1 mm respectively. The upper end of the container is open to the atmosphere.

[IMAGE 395]

Q. If the piston is pushed at a speed of $5 {mm s}^{- 1}$ , the air comes out of the nozzle with a speed of [2014]

$0.1 {m s}^{- 1}$
$1 {m s}^{- 1}$
$2 {m s}^{- 1}$
$8 {m s}^{- 1}$

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$Here, piston is pushed at a speed, v_{1} = 5 m/s$

$Let air comes out of nozzle with a speed v_{2}$

$From principle of continuity,$

$a_{1} v_{1} = a_{2} v_{2}$

$\Rightarrow π r_{1}^{2} v_{1} = π r_{2}^{2} v_{2} \Rightarrow r_{1}^{2} v_{1} = r_{2}^{2} v_{2}$

$\Rightarrow {(20)}^{2} \times 5 = {(1)}^{2} \times v_{2}$

$∴$ $v_{2} = 2000 {mm s}^{- 1} = 2 {m s}^{- 1}$

Q 7 :

A spray gun is shown in the figure where a piston pushes air out of a nozzle. A thin tube of uniform cross section is connected to the nozzle. The other end of the tube is in a small liquid container. As the piston pushes air through the nozzle, the liquid from the container rises into the nozzle and is sprayed out. For the spray gun shown, the radii of the piston and the nozzle are 20 mm and 1 mm respectively. The upper end of the container is open to the atmosphere.

[IMAGE 396]

Q. If the density of air is $ρ_{a}$ , and that of the liquid $ρ_{l}$ , then for a given piston speed the rate (volume per unit time) at which the liquid is sprayed will be proportional to [2014]

$\sqrt{\frac{ρ_{a}}{ρ_{l}}}$
$\sqrt{ρ_{a} ρ_{l}}$
$\sqrt{\frac{ρ_{l}}{ρ_{a}}}$
$ρ_{l}$

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[IMAGE 397]

$P_{X} - P_{Y} = \frac{1}{2} ρ_{a} v_{a}^{2}$

$P_{Z} - P_{Y} = \frac{1}{2} ρ_{l} v_{l}^{2}$

$But P_{Z} = P_{X}$

$∴$ $\frac{1}{2} ρ_{l} v_{l}^{2} = \frac{1}{2} ρ_{a} v_{a}^{2}$ $\Rightarrow v_{l} = \sqrt{\frac{ρ_{a}}{ρ_{l}}} \times v_{a}$

$∴$ $Volume flow rate \propto \sqrt{\frac{ρ_{a}}{ρ_{l}}}$

Q 8 :

STATEMENT-1: The stream of water flowing at high speed from a garden hose pipe tends to spread like a fountain when held vertically up, but tends to narrow down when held vertically down.

STATEMENT-2: In any steady flow of an incompressible fluid, the volume flow rate of the fluid remains constant. [2008]

Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
Statement-1 is True, Statement-2 is False
Statement-1 is False, Statement-2 is True

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The volume flow rate (Q) of an incompressible fluid in steady flow remains constant.

From equation of continuity, $a v$ = constant

$∴$ $Q = a \times v = constant$ $or, a \propto \frac{1}{v}$

where $a$ = area of cross-section and $v$ = velocity

If $v$ decreases, $a$ increases and vice-versa.

When stream of water moves up, its speed $(v)$ decreases and therefore $' a'$ increases i.e. the water spreads out as a fountain. When stream of water from hose pipe moves down, its speed increases and therefore area of cross-section decreases.

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Q 2 : Water is filled in a container upto height 3 m. A small hole of area 'a' is punched in the wall of the container at a height 52.5 cm from the bottom. The cross sectional area of the container is A. If a/A=0.1 then v2 is (where v is the velocity of water coming out of the hole) [2005]

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