Properties of Matter JEE Main PYQs (2020–2026)

Solved Properties of Matter PYQs

3 recent questions with answers. The full set of 253 is free to practise.

1. JEE Main 2026EasyStress, Strain and Young's Modulus
   Q1

   The two wires $A$ and $B$ of equal cross-section but of different materials are joined together. The ratio of Young’s modulus of wire $A$ and wire $B$ is 20/11. When the joined wire is kept under certain tension the elongations in the wires $A$ and $B$ are equal. If the length of wire $A$ is 2.2 m , then the length of wire $B$ is
   
   $\_\_\_\_$ m.

   • A1.1
   • B2.22
   • C1.21
   • D4.44

   Show answer & solution

   Correct answer: (C)

   Young’s Modulus is defined as the ratio of Stress $(\sigma )$ to Strain $(ϵ)$ .
   
   $$Y=\frac{\text{ Stress }}{\text{ Strain }}=\frac{\sigma }{ϵ}$$
   
   Stress is force per unit area, $\sigma =\frac{\text{ Force (F) }}{\text{ Area (A) }}$
   
   Strain is change in length per unit original length,
   
   $$ϵ=\frac{\text{ Change in length }(\Delta L)}{\text{ Original length }(L)}$$
   
   So, $Y=\frac{F/A}{\Delta L/L}$
   
   $$\Rightarrow \Delta L=\frac{FL}{AY}$$
   
   Since the wires are joined in series and kept under tension, the Force (F) acting on both wires is the same.
   
   The area (A) is the same for both.
   
   It is given that the elongation in both the wires is equal, $\Delta L_A=\Delta L_B$ .
   
   Since $\Delta L_A=\Delta L_B:$
   
   $$\frac{F{L}_A}{A{Y}_A}=\frac{F{L}_B}{A{Y}_B}$$
   
   $$\Rightarrow \frac{L_A}{Y_A}=\frac{L_B}{Y_B}$$
   
   $$\Rightarrow \frac{L_A}{L_B}=\frac{Y_A}{Y_B}$$
   
   The ratio of Young’s modulus is $\frac{Y_A}{Y_B}=\frac{20}{11}$ and the length of wire $A$ is $L_A=2.2m$
   
   $$\frac{2.2}{L_B}=\frac{20}{11}$$
   
   $$\Rightarrow L_B=2.2\times \frac{11}{20}$$
   
   $$\Rightarrow L_B=\frac{24.2}{20}$$
   
   $$\Rightarrow L_B=1.21m$$
   
   Therefore, the length of wire B is 1.21 m . Hence, the correct option is (C).
2. JEE Main 2026MediumSurface Tension, Surface Energy and Capillarity
   Q2

   Eight mercury drops, each of radius , coalesce to form a bigger drop. The surface energy released in this process is $\_\_\_\_$ - ( is the surface tension of mercury).

   • A$8\pi r^{2}S$
   • B$16\pi r^{2}S$
   • C$64\pi r^{2}S$
   • D$4\pi r^{2}S$

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   Correct answer: (B)

   When 8 small drops of radius $r$ combine to form one large drop of radius , the total volume remains constant.
   
   Volume of 8 small drops $=$ Volume of 1 large drop
   
   $$8\times \left(\frac{4}{3}\pi r^3\right)=\frac{4}{3}\pi R^3$$
   
   $$\Rightarrow 8r^3=R^3$$
   
   $$\Rightarrow R=2r$$
   
   The surface energy of a single drop is $E=SA$ where $S$ is the surface tension and $A$ is the surface area.
   
   Total initial area for 8 drops $=8\times \left(4\pi r^2\right)=32\pi r^2$ .
   
   So, total initial surface energy is $U_i=32\pi r^{2}S$
   
   The final drop has radius $R=2r$ .
   
   So, final surface area $=4\pi R^2=4\pi (2r{)}^2=4\pi \left(4r^2\right)=16\pi r^2$ .
   
   So, the final surface energy is $U_f=16\pi r^{2}S$
   
   The energy released is the decrease in surface energy :
   
   $$\Delta U=U_i-U_f$$
   
   $$\Rightarrow \Delta U=32\pi r^{2}S-16\pi r^{2}S$$
   
   $$\Rightarrow \Delta U=16\pi r^{2}S$$
   
   Therefore, the surface energy released in this process is $16\pi r^{2}S$ .
   
   Hence, the correct option is (B).
3. JEE Main 2026EasyFluid Dynamics and Bernoulli's Principle
   Q3

   A water spray gun is attached to a hose of cross sectional area $30{cm}^2$ . The gun comprises of 10 perforations each of cross sectional area of $15{mm}^2$ . If the water flows in the hose with the speed of $50cm/s$ , calculate the speed at which the water flows out from each perforation. (Neglect any edge effects)

   • A$100m/s$
   • B$10m/s$
   • C$1000m/s$
   • D$15\times 10^{2}m/s$

   Show answer & solution

   Correct answer: (B)

   For an incompressible fluid, the volume of fluid entering a pipe per unit time must equal the volume of fluid leaving the pipe per unit time.
   
   Volume Flow Rate (In) = Volume Flow Rate (Out)
   
   $$A_1{v}_1=A_2{v}_2$$
   
   Where A is the area and v is the velocity of the fluid.
   
   The area of hose is $A_1=30{cm}^2=30\times 10^{-4}{m}^2$ .
   
   The speed in hose is $v_1=50cm/s=0.5m/s$ .
   
   Number of perforations is $n=10$ .
   
   The area of one perforation is $a=15{mm}^2=15\times {\left(10^{-3}m\right)}^2=15\times 10^{-6}{m}^2$
   
   So, total area of exit is, $A_2=n\times a=10\times 15\times 10^{-6}=150\times 10^{-6}{m}^2$
   
   Using the equation of continuity:
   
   $$A_1{v}_1=A_2{v}_2$$
   
   $$\Rightarrow v_2=\frac{A_1{v}_1}{A_2}$$
   
   Substituting the values :
   
   $$v_2=\frac{\left(30\times 10^{-4}\right)\times 0.5}{150\times 10^{-6}}$$
   
   $$\Rightarrow v_2=\frac{15\times 10^{-4}}{150\times 10^{-6}}$$
   
   $$\Rightarrow v_2=10m/s$$
   
   Therefore, the speed at which the water flows out from each perforation is $10m/s$ .
   
   Thus, the correct option is (B).

Topic-wise question count

Where JEE Main has asked from Properties of Matter, most frequent first.

• Stress, Strain and Young's Modulus70
• Surface Tension, Surface Energy and Capillarity53
• Viscosity and Terminal Velocity37
• Bulk Modulus, Shear Modulus and Elastic Potential Energy30
• Fluid Dynamics and Bernoulli's Principle29
• Fluid Statics and Archimedes' Principle25
• Thermal Expansion and Newton's Law of Cooling9

Year-wise Properties of Matter questions in JEE Main

Number of questions from this chapter in each year's paper(s).