Mechanical Properties of Solids & Liquids — Previous Year Questions (JEE Main 2026)
Solved JEE Main 2026 previous year questions on Mechanical Properties of Solids and Liquids — Young's modulus, bulk modulus, elasticity, surface tension, viscosity, Bernoulli and buoyancy — with step-by-step solutions.
Solved JEE Main 2026 questions from the Mechanical Properties of Solids and Liquids chapter, covering elastic moduli, surface tension, viscosity, fluid flow and buoyancy, each with a concise step-by-step solution.
Solutions are AI-generated and pending review.
Solution
Bulk modulus relates pressure change to fractional volume change:
$$B = \frac{\Delta P}{-\dfrac{\Delta V}{V}} \implies \left|\frac{\Delta V}{V}\right| = \frac{\Delta P}{B}$$Substitute the values:
$$\left|\frac{\Delta V}{V}\right| = \frac{6.3 \times 10^7}{2.1 \times 10^9} = 0.03$$As a percentage:
$$0.03 \times 100 = 3\%$$Answer: B (3 %)
Solution
Both strings carry the same tension $F = mg = 0.8 \times 10 = 8\ \text{N}$ and have the same radius, so the same area:
$$A = \pi r^2 = \pi (0.2 \times 10^{-3})^2 = 1.257 \times 10^{-7}\ \text{m}^2$$Elongation of $A$:
$$\Delta L_A = \frac{F L_A}{A Y_A} = \frac{8 \times 0.314}{(1.257 \times 10^{-7})(2 \times 10^{10})} \approx 1.0\ \text{mm}$$For string $B$, both length and Young’s modulus are doubled, so $L_B/Y_B = L_A/Y_A$ and:
$$\Delta L_B = \frac{F L_B}{A Y_B} = \frac{F (2L_A)}{A (2Y_A)} = \Delta L_A \approx 1.0\ \text{mm}$$Total change:
$$\Delta L = \Delta L_A + \Delta L_B \approx 2\ \text{mm}$$Answer: B (2)
Solution
A soap bubble has two surfaces, so the work equals surface tension times the increase in total area:
$$W = T \cdot \Delta A = T \cdot 2 \cdot 4\pi (r_2^2 - r_1^2) = 8\pi T (r_2^2 - r_1^2)$$With $r_1 = 0.01\ \text{m}$, $r_2 = 0.02\ \text{m}$:
$$r_2^2 - r_1^2 = (4 - 1)\times 10^{-4} = 3 \times 10^{-4}\ \text{m}^2$$$$W = 8 \times \frac{22}{7} \times 3.5 \times 10^{-2} \times 3 \times 10^{-4}$$$$W = 8 \times 11 \times 10^{-2} \times 3 \times 10^{-4} = 264 \times 10^{-6}\ \text{J}$$Answer: 264
Solution
Maximum tension the wire can bear:
$$T_{\max} = \sigma_{\max} \cdot A = \sigma_{\max} \cdot \pi r^2$$$$T_{\max} = 4 \times 10^8 \times 3.14 \times (4 \times 10^{-3})^2 = 4 \times 10^8 \times 5.024 \times 10^{-5} = 20096\ \text{N}$$For an accelerating lift, $T = m(g + a)$:
$$a_{\max} = \frac{T_{\max}}{m} - g = \frac{20096}{1600} - 10 = 12.56 - 10 = 2.56\ \text{m/s}^2$$Answer: A (2.56)
Solution
The wires are in series, so they carry the same tension $F$ and have the same area $A$. Elongation:
$$\Delta L = \frac{F L}{A Y}$$Equal elongations ($\Delta L_A = \Delta L_B$) give:
$$\frac{L_A}{Y_A} = \frac{L_B}{Y_B} \implies L_B = L_A \cdot \frac{Y_B}{Y_A}$$Given $\dfrac{Y_A}{Y_B} = \dfrac{20}{11}$, so $\dfrac{Y_B}{Y_A} = \dfrac{11}{20}$:
$$L_B = 2.2 \times \frac{11}{20} = 1.21\ \text{m}$$Answer: C (1.21)
Solution
From the definition of bulk modulus $B = -V\dfrac{dP}{dV}$ (taken constant), integrating gives the excess pressure needed at volume $V$:
$$P = -B\ln\!\left(\frac{V}{V_0}\right)$$Work done on the gas during compression from $V_0$ to $V_f = V_0/3$:
$$W = \int_{V_f}^{V_0} P\, dV = -B\int_{V_f}^{V_0}\ln\!\left(\frac{V}{V_0}\right) dV$$Evaluating the integral gives:
$$W = B V_0\left(\frac{2}{3} - \frac{1}{3}\ln 3\right)$$With $B = 3 \times 10^5$ and $V_0 = 3 \times 10^{-3}\ \text{m}^3$ (so $BV_0 = 900$):
$$W = 900\left(0.6667 - 0.3662\right) = 900 \times 0.3005 \approx 270\ \text{J}$$Answer: 270
Solution
Initial drop radius $R = 1\ \text{mm} = 10^{-3}\ \text{m}$. Volume is conserved, so for $n = 512$ droplets of radius $r$:
$$\frac{4}{3}\pi R^3 = n \cdot \frac{4}{3}\pi r^3 \implies r = \frac{R}{n^{1/3}} = \frac{R}{8}$$Change in surface energy = surface tension times increase in total area:
$$\Delta E = T\left(n \cdot 4\pi r^2 - 4\pi R^2\right) = 4\pi R^2 T\left(n^{1/3} - 1\right)$$$$\Delta E = 4\pi (10^{-3})^2 (0.08)(8 - 1) = 4\pi \times 10^{-6} \times 0.08 \times 7$$$$\Delta E \approx 7.04 \times 10^{-6}\ \text{J} \implies \alpha \approx 7$$Answer: B (7)
Solution
At a cross-section a distance $\dfrac{l}{3}$ from the top, the tension equals the total weight hanging below that section: the load $W$ plus the weight of the wire below.
Length of wire below the section is $l - \dfrac{l}{3} = \dfrac{2l}{3}$, so its weight is $\dfrac{2}{3}w$.
$$\text{Tension} = W + \frac{2}{3}w$$Stress:
$$\sigma = \frac{W + \frac{2}{3}w}{A} = \frac{W}{A} + \frac{2}{3}\frac{w}{A}$$Comparing with $\dfrac{W}{A} + \dfrac{2}{\gamma}\dfrac{w}{A}$ gives $\gamma = 3$.
Answer: 3
Solution
Adding the coin increases the extra submerged depth by $\Delta h = 3.87\ \text{cm}$. The additional buoyant force supports the coin’s weight, so the mass of displaced water in the extra submerged slab equals the coin’s mass:
$$m_{\text{coin}} = \rho_{\text{water}} \cdot A \cdot \Delta h$$where $A = 10 \times 10 = 100\ \text{cm}^2$:
$$m_{\text{coin}} = 1 \times 100 \times 3.87 = 387\ \text{g}$$(The wood density is not needed for the coin’s mass — it only fixes the initial submersion.)
Answer: 387
Solution
Tension in each string equals the total weight hanging below it:
- String $A$ (top) carries both blocks: $F_A = (M + 2M)g = 3Mg$
- String $B$ carries only the lower block: $F_B = 2Mg$
Elongation $\Delta L = \dfrac{F L}{A Y}$ with equal cross-sectional areas:
$$\frac{\Delta L_A}{\Delta L_B} = \frac{F_A}{F_B}\cdot\frac{L_A}{L_B}\cdot\frac{Y_B}{Y_A}$$Given $\dfrac{L_A}{L_B} = 2$ and $\dfrac{Y_A}{Y_B} = 0.5 \Rightarrow \dfrac{Y_B}{Y_A} = 2$:
$$\frac{\Delta L_A}{\Delta L_B} = \frac{3M}{2M}\times 2 \times 2 = \frac{3}{2}\times 4 = 6$$Answer: D (6)
Solution
By the equation of continuity, the volume flow rate is conserved:
$$A_{\text{hose}}\, v_{\text{hose}} = n\, a_{\text{hole}}\, v_{\text{out}}$$Convert to SI: $A_{\text{hose}} = 30\ \text{cm}^2 = 30 \times 10^{-4}\ \text{m}^2$, $a_{\text{hole}} = 15\ \text{mm}^2 = 15 \times 10^{-6}\ \text{m}^2$, $v_{\text{hose}} = 0.5\ \text{m/s}$, $n = 10$.
$$v_{\text{out}} = \frac{A_{\text{hose}}\, v_{\text{hose}}}{n\, a_{\text{hole}}} = \frac{30 \times 10^{-4} \times 0.5}{10 \times 15 \times 10^{-6}} = \frac{1.5 \times 10^{-3}}{1.5 \times 10^{-4}} = 10\ \text{m/s}$$Answer: B (10 m/s)
Solution
The condition $T_3 + T_1 = 2T_2$ rearranges to $T_3 - T_2 = T_2 - T_1 = \Delta T$, so both temperature steps are equal.
Let $L_0$ be the length at $T_1$. First expansion:
$$\Delta L_1 = L_0\,\alpha\,\Delta T$$For the second step, the strip already has length $L_1 = L_0(1 + \alpha\Delta T)$ at $T_2$:
$$\Delta L_2 = L_1\,\alpha\,\Delta T = L_0(1 + \alpha\Delta T)\,\alpha\,\Delta T$$$$\Delta L_2 = \Delta L_1\,(1 + \alpha\Delta T)$$Answer: D ($\Delta L_1[1 + \alpha\Delta T]$)
Solution
Radii: $r_1 = 1\ \text{cm} = 0.01\ \text{m}$, $r_2 = 3\ \text{cm} = 0.03\ \text{m}$.
A soap bubble has two surfaces:
$$W = 8\pi T (r_2^2 - r_1^2)$$$$r_2^2 - r_1^2 = (9 - 1)\times 10^{-4} = 8 \times 10^{-4}\ \text{m}^2$$$$W = 8\pi \times 0.03 \times 8 \times 10^{-4} = 1.92\pi \times 10^{-4}\ \text{J}$$Comparing with $\alpha\pi \times 10^{-4}$ gives $\alpha = 1.92$.
Answer: C (1.92)
Solution
For a bubble rising at terminal velocity, the (upward) buoyant force balances viscous drag; taking the bubble’s density as negligible, Stokes’ law gives:
$$v = \frac{2}{9}\frac{r^2 (\rho_{\text{liq}} - \rho_{\text{bubble}})\,g}{\eta} \approx \frac{2}{9}\frac{r^2 \rho_{\text{liq}}\,g}{\eta}$$Solve for $\eta$ with $r = 1\ \text{mm} = 10^{-3}\ \text{m}$, $v = 0.5\ \text{cm/s} = 5 \times 10^{-3}\ \text{m/s}$:
$$\eta = \frac{2}{9}\frac{r^2 \rho\,g}{v} = \frac{2}{9}\cdot\frac{(10^{-3})^2 (2000)(10)}{5 \times 10^{-3}}$$$$\eta = \frac{2}{9}\cdot\frac{2 \times 10^{-2}}{5 \times 10^{-3}} = \frac{2}{9}\times 4 = 0.888\ \text{Pa·s}$$Convert to Poise ($1\ \text{Pa·s} = 10\ \text{Poise}$):
$$\eta = 0.888 \times 10 = 8.88 \approx 8.8\ \text{Poise}$$Answer: B (8.8)
Solution
Vessel height from $V = \pi r^2 H$, with $V = 528\ \text{dm}^3 = 0.528\ \text{m}^3$, $r = 0.40\ \text{m}$ (using $\pi = 22/7$):
$$H = \frac{V}{\pi r^2} = \frac{0.528}{\frac{22}{7}(0.16)} = 1.05\ \text{m}$$Efflux speed (Torricelli), with depth of hole below water surface $h = 0.70\ \text{m}$:
$$v = \sqrt{2gh} = \sqrt{2 \times 10 \times 0.70} = \sqrt{14}\ \text{m/s}$$Fall height: the hole is $H - 0.70$ above the vessel base, and the vessel sits on a block of height $H$, so the hole is at:
$$y = (H - 0.70) + H = 2(1.05) - 0.70 = 1.40\ \text{m above the ground}$$Time of flight: $t = \sqrt{\dfrac{2y}{g}} = \sqrt{\dfrac{2 \times 1.40}{10}} = \sqrt{0.28}\ \text{s}$
Range:
$$R = v\,t = \sqrt{14}\times\sqrt{0.28} = \sqrt{3.92} = 1.98\ \text{m} \approx 140\sqrt{2}\ \text{cm}$$Answer: B ($140\sqrt{2}$)
Solution
Young’s modulus is a material property — it depends only on the material, not on the geometry (dimensions) of the sample.
$$Y = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{\Delta L / L}$$Although stress and strain individually depend on $A$ and $L$, their ratio $Y$ is fixed for a given material at a given temperature. Doubling the radius and length changes the wire’s stiffness but not its Young’s modulus.
$$Y \text{ remains unchanged}$$Answer: C (remains unchanged)
Solution
For shear, the modulus of rigidity relates shear stress to shear strain:
$$\eta = \frac{F/A}{x/L} \implies x = \frac{F L}{A \eta}$$The tangential force acts on the top face of area $A = L^2 = (0.05)^2 = 2.5 \times 10^{-3}\ \text{m}^2$, and $L = 0.05\ \text{m}$:
$$x = \frac{10 \times 0.05}{(2.5 \times 10^{-3})(10^5)} = \frac{0.5}{250} = 2 \times 10^{-3}\ \text{m}$$$$x = 2\ \text{mm}$$Answer: 2
Solution
Volume conservation fixes the big drop’s radius $R$:
$$8 \cdot \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \implies R = 8^{1/3} r = 2r$$Surface energy released = surface tension times decrease in total area:
$$\Delta E = S\left(8 \cdot 4\pi r^2 - 4\pi R^2\right) = 4\pi S\left(8r^2 - (2r)^2\right)$$$$\Delta E = 4\pi S\left(8r^2 - 4r^2\right) = 4\pi S \cdot 4r^2 = 16\pi r^2 S$$Answer: B ($16\pi r^2 S$)
Solution
Elastic potential energy per unit volume $= \dfrac{1}{2}\,(\text{stress})(\text{strain}) = \dfrac{1}{2}Y(\text{strain})^2$, so:
$$U = \frac{1}{2}Y\left(\frac{\Delta L}{L}\right)^2 V$$Strain:
$$\frac{\Delta L}{L} = \frac{3 \times 10^{-3}}{3} = 10^{-3}$$$$U = \frac{1}{2}(1.1 \times 10^{11})(10^{-3})^2 (600 \times 10^{-6})$$$$U = \frac{1}{2}(1.1 \times 10^{11})(10^{-6})(6 \times 10^{-4}) = \frac{1}{2}\times 66 = 33\ \text{J}$$Answer: 33
Solution
Continuity gives the speed at $B$. Convert areas: $A_A = 1.0\ \text{cm}^2 = 10^{-4}\ \text{m}^2$, $A_B = 20\ \text{mm}^2 = 20 \times 10^{-6}\ \text{m}^2$; $v_A = 0.10\ \text{m/s}$.
$$v_B = \frac{A_A v_A}{A_B} = \frac{10^{-4}\times 0.10}{20 \times 10^{-6}} = 0.5\ \text{m/s}$$Bernoulli (same height, so no potential term):
$$P_A - P_B = \frac{1}{2}\rho\left(v_B^2 - v_A^2\right)$$$$P_A - P_B = \frac{1}{2}(600)\left(0.5^2 - 0.1^2\right) = 300\,(0.25 - 0.01) = 300 \times 0.24 = 72\ \text{Pa}$$Answer: D (72)
Solution
Terminal velocity from Stokes’ law (drag balancing weight) scales as the square of the radius:
$$v_t = \frac{2}{9}\frac{r^2(\rho - \sigma)g}{\eta} \implies v_t \propto r^2$$Volume conservation for $n = 64$ identical droplets:
$$r = \frac{R}{n^{1/3}} = \frac{R}{64^{1/3}} = \frac{R}{4}$$Therefore:
$$\frac{v_1}{v_2} = \left(\frac{R}{r}\right)^2 = \left(\frac{R}{R/4}\right)^2 = 4^2 = 16$$Answer: D (16)