JEE Main 202621 JanuaryMorning Shift

Exam: JEE Main
Year: 2026
Date: 21 January
Shift: Morning Shift
Subject: Physics
Chapter: Mechanical Properties of Fluids

JEE Main 2026 · 21 January · Morning Shift · Physics · Mechanical Properties of Fluids

Water flows through a horizontal tube as shown in the figure. The...

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

A

and

B

are

6 \mathrm{~cm}^{2}

and

3 \mathrm{~cm}^{2}

respectively. The rate of flow will be

\_\_\_\_

\mathrm{cm}^{3} / \mathrm{s}

. (take

g=10 \mathrm{~m} / \mathrm{s}^{2}

)

Water flows through a horizontal tube as shown in the figure. The difference in height between the water columns in

Options

A $\frac{200}{\sqrt{3}}$
B $200 \sqrt{6}$
C $100 \sqrt{3}$
D $200 \sqrt{3}$

Correct answer

D. $200 \sqrt{3}$

Step-by-step solution

Given:

h = 5

cm,

A_1 = 6

cm²,

A_2 = 3

cm²,

g = 1000

cm/s²
By continuity:

A_1v_1 = A_2v_2 \Rightarrow v_2 = 2v_1

By Bernoulli's equation:

P_1 - P_2 = \frac{1}{2}\rho(v_2^2 - v_1^2)

Since

P_1 - P_2 = \rho gh

\rho gh = \frac{1}{2}\rho(4v_1^2 - v_1^2) = \frac{3\rho v_1^2}{2}

v_1 = \sqrt{\frac{2gh}{3}} = \sqrt{\frac{2 \times 1000 \times 5}{3}} = \frac{100}{\sqrt{3}}

cm/s
Rate of flow

Q = A_1v_1 = 6 \times \frac{100}{\sqrt{3}} = 200\sqrt{3}

cm³/s

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