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Physics · Fluid Mechanics

JEE Main 2026 — 21 January, Morning Shift — Question 28

Water flows through a horizontal tube as shown in the figure. The difference in height between the water columns in vertical tubes is 5 cm and the area of cross-sections at AA and BB are 6 cm26 \mathrm{~cm}^{2} and 3 cm23 \mathrm{~cm}^{2} respectively. The rate of flow will be ____\_\_\_\_ cm3/s\mathrm{cm}^{3} / \mathrm{s}. (take g=10 m/s2\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2} )

Question figure
  1. Option A:

    2003\frac{200}{\sqrt{3}}

  2. Option B:

    2006200 \sqrt{6}

  3. Option C:

    2003200 \sqrt{3}

    Correct
  4. Option D:

    1003100 \sqrt{3}

Answer: C

Step-by-step solution

From continuity equation AAVA=ABVB⇒6VA=3VB⇒VB=2VAA_{A} V_{A}=A_{B} V_{B} \Rightarrow 6 V_{A}=3 V_{B} \Rightarrow V_{B}=2 V_{A} Applying Bernoullis equation between A& B\mathrm{A} \& \mathrm{~B},

PA+12ρVA2=PB+12ρVB2⇒ρg×0.05=12ρ[VB2−VA2]=12ρ(3VA2)⇒VA=2g×0.053 m/s=13 m/s=1003 cm/s\begin{aligned} & P_{A}+\frac{1}{2} \rho V_{A}^{2}=P_{B}+\frac{1}{2} \rho V_{B}^{2} \Rightarrow \quad & \rho g \times 0.05=\frac{1}{2} \rho\left[V_{B}^{2}-V_{A}^{2}\right]=\frac{1}{2} \rho\left(3 V_{A}^{2}\right) \Rightarrow \quad & V_{A}=\sqrt{\frac{2 g \times 0.05}{3}} \mathrm{~m} / \mathrm{s}=\frac{1}{\sqrt{3}} \mathrm{~m} / \mathrm{s}=\frac{100}{\sqrt{3}} \mathrm{~cm} / \mathrm{s} \end{aligned}

⇒ Volume flow rate =AAVA=6×1003 cm3/sec=A_{A} V_{A}=\frac{6 \times 100}{\sqrt{3}} \mathrm{~cm}^{3} / \mathrm{sec}

=2003 cm3/sec=200 \sqrt{3} \mathrm{~cm}^{3} / \mathrm{sec}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
21 January, Morning Shift
Subject
Physics
Chapter
Fluid Mechanics
Topic
Bernoulli's Equation and its Applications