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-s

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
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 $6 {cm}^{2}$ and $3 {cm}^{2}$ respectively. The rate of flow will be ______ ${cm}^{3} / s$ . (take g = 10 m/s²) [2026]

1 $\frac{200}{\sqrt{3}}$

2 $100 \sqrt{3}$

3 $200 \sqrt{3}$

4 $200 \sqrt{6}$

Ans.
(3)

From continuity equation

$A_{A} V_{A} = A_{B} V_{B} \Rightarrow 6 V_{A} = 3 V_{B} \Rightarrow V_{B} = 2 V_{A}$

Applying Bernoulli's equation between A & B,

$P_{A} + \frac{1}{2} ρ V_{A}^{2} = P_{B} + \frac{1}{2} ρ V_{B}^{2}$

$\Rightarrow ρ g \times 0.05 = \frac{1}{2} ρ [V_{B}^{2} - V_{A}^{2}] = \frac{1}{2} ρ (3 V_{A}^{2})$

$\Rightarrow V_{A} = \sqrt{\frac{2 g \times 0.05}{3}} m/s = \frac{1}{\sqrt{3}} m/s = \frac{100}{\sqrt{3}} cm/s$

$\Rightarrow Volume flow rate = A_{A} V_{A} = \frac{6 \times 100}{\sqrt{3}} {cm}^{3} / sec$

$= 200 \sqrt{3} {cm}^{3} / sec$

Want Us to Email you About Special Offers & Updates?