Properties of Matter — NEET UG practice

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Sample questions with solutions

Q1 · 2026

Water flows in a streamline motion through a horizontal pipe of circular cross-section as shown in the figure. The pressure difference of water between $P$ and $Q$ is $15 \mathrm{Nm}^{-2}$ . The area of cross-section at $P$ and $Q$ are $40 \mathrm{~cm}^2$ and $20 \mathrm{~cm}^2$ , respectively. The rate of flow of water through the pipe, in $\mathrm{cm}^3 \mathrm{~s}^{-1}$ , is:

[Take density of water $=1000 \mathrm{~kg} \mathrm{~m}^{-3}$ ]

A.
400
B.
100
C.
200
D.
300

Answer: A

Since the fluid is incompressible and flows through a pipe of varying cross-section, the continuity equation applies, meaning the volume flow rate stays the same at $P$ and $Q$ .

A_P V_P = A_Q V_Q

Given the areas, so

40 V_P = 20 V_Q \implies V_Q = 2V_P

Since the pipe is horizontal, Bernoulli's equation relates the pressure difference to the change in velocity.

P_P - P_Q = \frac{1}{2}\rho(V_Q^2 - V_P^2)

Substituting the given values,

15 = \frac{1}{2}\times 1000 \times (4V_P^2 - V_P^2)

Solving this equation gives the velocity at $P$ .

V_P = \sqrt{\frac{15}{1500}} = 0.1\ \mathrm{m/s}

Therefore

V_Q = 0.2\ \mathrm{m/s}

The rate of flow through the pipe equals $A_P V_P$ .

\text{Rate} = 40\times10^{-4}\times0.1 = 4\times10^{-4}\ \mathrm{m^3/s}

Converting to $\mathrm{cm^3/s}$ ,

4\times10^{-4}\times10^6 = 400\ \mathrm{cm^3/s}

Hence, the answer is A. 400.

Q2 · 2026

In the measurement of viscosity of liquids using terminal velocity experiment, spherical balls of same radius but having different densities are used. The variation of the terminal velocity ( $v$ ) with the ratio of density of spherical ball ( $\sigma$ ) to density of the liquid ( $\rho$ ), is best represented by:

Answer: B

Recall the expression for terminal velocity from Stokes' law, where $r$ is the radius of the ball, $h$ is the coefficient of viscosity, $\sigma$ is the density of the ball, and $\rho$ is the density of the liquid.

V_T = \frac{2r^2}{9h}(\sigma-\rho)

To compare $V_T$ with $\dfrac{\sigma}{\rho}$ , rewrite the expression by factoring out $\rho$ .

V_T = \frac{2r^2}{9h}\rho\left(\frac{\sigma}{\rho}-1\right)

This has the same form as the equation of a straight line, where $y = V_T$ , $x = \dfrac{\sigma}{\rho}$ , the slope $m = \dfrac{2r^2\rho}{9h}$ is positive, and the intercept $c = -\dfrac{2r^2\rho}{9h}$ is negative.

y = mx + c

A positive slope with a negative intercept means the graph is a straight line that crosses the x-axis (at $\frac{\sigma}{\rho}=1$ ) and increases afterward.

Hence, the answer is B.

Q3 · 2026

Match List I with List II:

	List I		List II
A.	Young's Modulus	I.	$\frac{\Delta d}{\Delta L}\left(\frac{L}{d}\right)$
B.	Compressibility	II.	$\frac{FL}{A(\Delta L)}$
C.	Bulk Modulus	III.	$-\frac{1}{\Delta P}\left(\frac{\Delta V}{V}\right)$
D.	Poisson's Ratio	IV.	$-P\left(\frac{V}{\Delta V}\right)$

Choose the correct answer from the options given below:

A.
A-IV, B-I, C-II, D-III
B.
A-III, B-II, C-I, D-IV
C.
A-I, B-IV, C-III, D-II
D.
A-II, B-III, C-IV, D-I

Answer: D

Young's Modulus is defined as the ratio of longitudinal stress to longitudinal strain.

Y = \frac{FL}{A\,\Delta L}

So A matches II.

Compressibility is the reciprocal of Bulk Modulus.

\text{Compressibility} = -\frac{1}{\Delta P}\left(\frac{\Delta V}{V}\right)

So B matches III.

Bulk Modulus is defined as negative of the pressure change divided by the fractional volume change.

K = -P\left(\frac{V}{\Delta V}\right)

So C matches IV.

Poisson's Ratio is defined as lateral strain divided by longitudinal strain.

\mu = \frac{\Delta d}{\Delta L}\left(\frac{L}{d}\right)

So D matches I.

Hence, the correct matching is A-II, B-III, C-IV, D-I, so the answer is D.

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