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Properties of Solids and Liquids: Physics | JEE Main

The figure shows a liquid of given density flowing steadily in horizontal tube of varying cross-section. Cross sectional areas at

\text{A}

is

1.5\text{ cm}^2

, and

\text{B}

is

25\text{ mm}^2

, if the speed of liquid at

\text{B}

is

60\text{ cm/s}

then

(\text{P}_\text{A} - \text{P}_\text{B})

is : (Given

\text{P}_\text{A}

and

\text{P}_\text{B}

are liquid pressures at

\text{A}

and

\text{B}

points. Density

\rho = 1000\text{ kg m}^{-3}

\text{A}

and

\text{B}

are on the axis of tube

Horizontal tube with a wider section on the left labeled A narrowing to a section on the right labeled B.

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

A

175\text{ Pa}

B

27\text{ Pa}

C

135\text{ Pa}

D

36\text{ Pa}

Step-by-step solutionView

Correct answer

Using the continuity equation to find the speed at point A ( $v_A = 0.1\text{ m/s}$ ) from $v_B = 0.6\text{ m/s}$ , Bernoulli's equation gives $P_A - P_B = \frac{1}{2}\rho (v_B^2 - v_A^2) = 175\text{ Pa}$ .

Option analysis

Why each option works or fails

A ·

175\text{ Pa}

Correct application of fluid dynamics principles. First find $v_A$ via $A_A v_A = A_B v_B$ , ensuring unit consistency ( $1.5\text{ cm}^2 = 150\text{ mm}^2$ , giving $v_A = 0.1\text{ m/s}$ ), then evaluate $P_A - P_B = \frac{1}{2}(1000)(0.6^2 - 0.1^2) = 175\text{ Pa}$ .

B ·

27\text{ Pa}

The student fails to convert units of area correctly, confusing $1.5\text{ cm}^2$ as $15\text{ mm}^2$ or miscalculating the velocity ratio. Remember that $1\text{ cm}^2 = 100\text{ mm}^2$ , so $1.5\text{ cm}^2 = 150\text{ mm}^2$ , leading to an area ratio of $6:1$ and $v_A = 10\text{ cm/s}$ .

C ·

135\text{ Pa}

The student calculates dynamic pressure using an incorrect squared velocity term or an arithmetic error when evaluating $\frac{1}{2}\rho (v_B^2 - v_A^2)$ . Compute $v_B^2 - v_A^2 = 0.6^2 - 0.1^2 = 0.36 - 0.01 = 0.35\text{ m}^2/\text{s}^2$ , which multiplied by $500\text{ kg/m}^3$ yields exactly $175\text{ Pa}$ .

D ·

36\text{ Pa}

The student forgets the factor of $1/2$ in the dynamic pressure term, or computes $\rho v_B^2 / 10$ due to mismatched units. Bernoulli's equation for horizontal flow is $P_A - P_B = \frac{1}{2}\rho(v_B^2 - v_A^2)$ . Always include the $\frac{1}{2}$ prefactor in kinetic pressure energy.

Reviewed route

Solution

StepWorking

01given

Cross-sectional area at $A$ : $A_A = 1.5\text{ cm}^2 = 1.5 \times 10^{-4}\text{ m}^2 = 150\text{ mm}^2$ . Cross-sectional area at $B$ : $A_B = 25\text{ mm}^2 = 0.25\text{ cm}^2 = 25 \times 10^{-6}\text{ m}^2$ . Speed at $B$ : $v_B = 60\text{ cm/s} = 0.6\text{ m/s}$ . Density of liquid: $\rho = 1000\text{ kg/m}^3$ . Tube is horizontal, so height $h_A = h_B$ .

02find

Calculate the pressure difference $(P_A - P_B)$ in pascals (Pa).

03strategise

First, find the velocity at $A$ ( $v_A$ ) using the equation of continuity: $A_A v_A = A_B v_B$ . Then, apply Bernoulli's principle for a horizontal streamline: $P_A + \frac{1}{2}\rho v_A^2 = P_B + \frac{1}{2}\rho v_B^2 \implies P_A - P_B = \frac{1}{2}\rho (v_B^2 - v_A^2)$ .

04execute

Using $A_A = 150\text{ mm}^2$ and $A_B = 25\text{ mm}^2$ , we get $v_A = \frac{A_B}{A_A} v_B = \frac{25}{150} \times 60 = 10\text{ cm/s} = 0.1\text{ m/s}$ .

05execute

Now compute $P_A - P_B = \frac{1}{2} \times 1000 \times ((0.6)^2 - (0.1)^2) = 500 \times (0.36 - 0.01) = 500 \times 0.35 = 175\text{ Pa}$ .

✓verify

Since cross-section $A_A > A_B$ , velocity $v_A < v_B$ , which requires $P_A > P_B$ by Bernoulli's principle. The positive value $175\text{ Pa}$ is physically consistent.

Hints that build this answer step by step

What is the fluid velocity at point A, $v_A$ , in SI units ( $ext{m/s}$ )?
$v_A = 0.1\text{ m/s}$
Which formula correctly relates the pressure difference $(P_A - P_B)$ to the velocities for this horizontal tube?
$P_A - P_B = \frac{1}{2}\rho (v_B^2 - v_A^2)$
Substituting $\rho = 1000\text{ kg/m}^3$ , $v_B = 0.6\text{ m/s}$ , and $v_A = 0.1\text{ m/s}$ , what is the value of $(P_A - P_B)$ ?
$175\text{ Pa}$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Physics
Concepts assessed: Physics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 8 September 2026

Quick checks

Students also ask

Can I keep the areas in mm² when finding the velocity ratio?

Yes, because the continuity equation involves a ratio of areas ( $A_B/A_A$ ), so any matching unit of area works.

Answer

Using the continuity equation to find the speed at point A ( $v_A = 0.1\text{ m/s}$ ) from $v_B = 0.6\text{ m/s}$ , Bernoulli's equation gives $P_A - P_B = \frac{1}{2}\rho (v_B^2 - v_A^2) = 175\text{ Pa}$ .

Why each option works or fails

A · correct: $175\text{ Pa}$ - Correct application of fluid dynamics principles. First find $v_A$ via $A_A v_A = A_B v_B$ , ensuring unit consistency ( $1.5\text{ cm}^2 = 150\text{ mm}^2$ , giving $v_A = 0.1\text{ m/s}$ ), then evaluate $P_A - P_B = \frac{1}{2}(1000)(0.6^2 - 0.1^2) = 175\text{ Pa}$ .
B: $27\text{ Pa}$ - The student fails to convert units of area correctly, confusing $1.5\text{ cm}^2$ as $15\text{ mm}^2$ or miscalculating the velocity ratio. Remember that $1\text{ cm}^2 = 100\text{ mm}^2$ , so $1.5\text{ cm}^2 = 150\text{ mm}^2$ , leading to an area ratio of $6:1$ and $v_A = 10\text{ cm/s}$ .
C: $135\text{ Pa}$ - The student calculates dynamic pressure using an incorrect squared velocity term or an arithmetic error when evaluating $\frac{1}{2}\rho (v_B^2 - v_A^2)$ . Compute $v_B^2 - v_A^2 = 0.6^2 - 0.1^2 = 0.36 - 0.01 = 0.35\text{ m}^2/\text{s}^2$ , which multiplied by $500\text{ kg/m}^3$ yields exactly $175\text{ Pa}$ .
D: $36\text{ Pa}$ - The student forgets the factor of $1/2$ in the dynamic pressure term, or computes $\rho v_B^2 / 10$ due to mismatched units. Bernoulli's equation for horizontal flow is $P_A - P_B = \frac{1}{2}\rho(v_B^2 - v_A^2)$ . Always include the $\frac{1}{2}$ prefactor in kinetic pressure energy.

Step-by-step solution

given: Cross-sectional area at $A$ : $A_A = 1.5\text{ cm}^2 = 1.5 \times 10^{-4}\text{ m}^2 = 150\text{ mm}^2$ . Cross-sectional area at $B$ : $A_B = 25\text{ mm}^2 = 0.25\text{ cm}^2 = 25 \times 10^{-6}\text{ m}^2$ . Speed at $B$ : $v_B = 60\text{ cm/s} = 0.6\text{ m/s}$ . Density of liquid: $\rho = 1000\text{ kg/m}^3$ . Tube is horizontal, so height $h_A = h_B$ .
find: Calculate the pressure difference $(P_A - P_B)$ in pascals (Pa).
strategise: First, find the velocity at $A$ ( $v_A$ ) using the equation of continuity: $A_A v_A = A_B v_B$ . Then, apply Bernoulli's principle for a horizontal streamline: $P_A + \frac{1}{2}\rho v_A^2 = P_B + \frac{1}{2}\rho v_B^2 \implies P_A - P_B = \frac{1}{2}\rho (v_B^2 - v_A^2)$ .
execute: Using $A_A = 150\text{ mm}^2$ and $A_B = 25\text{ mm}^2$ , we get $v_A = \frac{A_B}{A_A} v_B = \frac{25}{150} \times 60 = 10\text{ cm/s} = 0.1\text{ m/s}$ .
execute: Now compute $P_A - P_B = \frac{1}{2} \times 1000 \times ((0.6)^2 - (0.1)^2) = 500 \times (0.36 - 0.01) = 500 \times 0.35 = 175\text{ Pa}$ .
verify: Since cross-section $A_A > A_B$ , velocity $v_A < v_B$ , which requires $P_A > P_B$ by Bernoulli's principle. The positive value $175\text{ Pa}$ is physically consistent.

Properties of Solids and Liquids: Physics | JEE Main

What feels right?

Using the continuity equation to find the speed at point A ( $v_A = 0.1\text{ m/s}$ ) from $v_B = 0.6\text{ m/s}$ , Bernoulli's equation gives $P_A - P_B = \frac{1}{2}\rho (v_B^2 - v_A^2) = 175\text{ Pa}$ .

Why each option works or fails

Solution

01given

02find

03strategise

04execute

05execute

✓verify

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