Fluid Mechanics | Surface Tension And Viscosity | JEE Main previous year questions with Solutions

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Topic Question Set

Q 11 :

The surface of water in a water tank of cross-sectional area $750 {cm}^{2}$ on the top of a house is $h$ m above the tap level. The speed of water coming out through the tap of cross-sectional area $500 {mm}^{2}$ is $30 cm/s$ . At that instant, $\frac{d h}{d t}$ is $x \times 10^{- 3} m/s$ . The value of $x$ will be _______. [2023]

0%

(2)

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

$750 \times 10^{- 4} v_{1} = 500 \times 10^{- 6} \times 0.3$

$v_{1} = \frac{500 \times 3 \times 10^{- 3}}{750} m/s = 2 \times 10^{- 3} m/s$

$\Rightarrow \frac{d h}{d t} = - 2 \times 10^{- 3} m/s$

Q 12 :

Figure below shows a liquid being pushed out of the tube by a piston having area of cross section $2.0 {cm}^{2}$ . The area of cross section at the outlet is $10 {mm}^{2}$ . If the piston is pushed at a speed of ${4 cm s}^{- 1}$ , the speed of outgoing fluid is _________ ${cm s}^{- 1}$ . [2023]

0%

(80)

$By equation of continuity$

$A_{1} V_{1} = A_{2} V_{2}$

$\Rightarrow V_{2} = \frac{2 \times 4}{10 \times 10^{- 2}} = 80 cm/s$

Q 13 :

Glycerine of density $1.25 \times 10^{3} {kg m}^{- 3}$ is flowing through the conical section of a pipe. The area of cross-section of the pipe at its ends is $10 {cm}^{2}$ and $5 {cm}^{2}$ and the pressure drop across its length is ${3 N m}^{- 2}$ . The rate of flow of glycerine through the pipe is $x \times 10^{- 5} m^{3} s^{- 1}$ . The value of $x$ is _________. [2023]

0%

(4)

$Δ P = P_{1} - P_{2} = {3 N/m}^{2} (given)$

${By continuity eq}^{n}, A_{1} v_{1} = A_{2} v_{2}$

$∴$ $v_{1} = \frac{A_{2}}{A_{1}} v_{2}$ ...(i)

${By Bernoulli's eq}^{n},$

$P_{1} + \frac{1}{2} ρ v_{1}^{2} = P_{2} + \frac{1}{2} ρ v_{2}^{2}$

$P_{1} - P_{2} = \frac{1}{2} ρ (v_{2}^{2} - v_{1}^{2})$

$\Rightarrow Δ P = \frac{1}{2} ρ [1 - {(\frac{A_{2}}{A_{1}})}^{2}] v_{2}^{2}$

$3 = \frac{1}{2} \times 1.25 \times 10^{3} [1 - {(\frac{5}{10})}^{2}] v_{2}^{2}$

$∴$ $v_{2} = 8 \times 10^{- 2} m/s$

$So discharge rate = A_{2} v_{2} = 5 \times 10^{- 4} \times 8 \times 10^{- 2} = 4 \times 10^{- 5} m^{3} / s$

$Correct answer is x = 4$

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