Previous year questions: Mechanical Properties of Solids

Ten questions follow from the official JEE Main 2026 papers and the NCERT Exemplar for this unit. Each one is solvable with the formulas on the first page of this chapter. The paper and shift are stated in every heading, and each answer is worked in full so that the method behind the correct option is clear.

Q1 (JEE Main 2026, 4 Apr Shift 1, 4 marks)

The pressure on a sample of water is increased by 6.3 x 10^7 N m^-2 and its volume decreases. The bulk modulus of water is 2.1 x 10^9 N m^-2. Find the percentage decrease in volume.

Answer: Bulk modulus is B = P/(fractional decrease of volume), so the fractional decrease is P/B = (6.3 x 10^7)/(2.1 x 10^9) = 3.0 x 10^-2. As a percentage this is 3 percent. The option 3 percent is correct. The reciprocal relation is worth writing first: fractional change = pressure divided by bulk modulus, with no minus sign needed when only the size of the change is asked.

Q2 (JEE Main 2026, 5 Apr Shift 1, 4 marks)

A steel wire of radius r and length L has Young's modulus Y. The radius and the length are both doubled. What happens to the Young's modulus?

Answer: Young's modulus is a property of the material, fixed by the atomic bonding, so it does not depend on the dimensions of the wire. Changing r and L changes the extension for a given load, but the ratio of stress to strain stays Y. The correct choice is that Y remains unchanged. The extension of the new wire would be one half of the old extension for the same load, because area becomes four times and length becomes two times.

Q3 (JEE Main 2026, 5 Apr Shift 2, 4 marks)

A copper wire of length 3 m is stretched by 3 mm by an external force. The volume of the wire is 600 x 10^-6 m^3 and the Young's modulus of copper is 1.1 x 10^11 N m^-2. Find the elastic potential energy stored in the wire.

Answer: Strain = delta L/L = (3 x 10^-3)/3 = 1.0 x 10^-3. Stress = Y x strain = 1.1 x 10^11 x 1.0 x 10^-3 = 1.1 x 10^8 N m^-2. Energy density = (1/2) stress x strain = 0.5 x 1.1 x 10^8 x 1.0 x 10^-3 = 5.5 x 10^4 J m^-3. Multiply by the volume: U = 5.5 x 10^4 x 600 x 10^-6 = 33 J.

Q4 (JEE Main 2026, 6 Apr Shift 1, 4 marks)

A lift of mass 1600 kg hangs from a thick iron wire of radius 4 mm. The maximum stress the wire can withstand is 4 x 10^8 N m^-2. Find the maximum acceleration the lift can have, taking g = 10 m s^-2 and pi = 3.14.

Answer: The largest tension the wire can carry is stress x area = 4 x 10^8 x pi (4 x 10^-3)^2 = 4 x 10^8 x 5.024 x 10^-5 = 2.01 x 10^4 N. The weight of the lift is 1600 x 10 = 1.6 x 10^4 N. The extra upward force available is 2.01 x 10^4 - 1.6 x 10^4 = 4.1 x 10^3 N, so the maximum acceleration is 4.1 x 10^3/1600 = 2.56 m s^-2. The option 2.56 is correct.

Q5 (JEE Main 2026, 6 Apr Shift 1, 4 marks)

Two wires A and B of the same cross-sectional area but different materials are joined end to end. The ratio of their Young's moduli is Y_A/Y_B = 20/11. Under a certain tension the elongations are equal. If the length of wire A is 2.2 m, find the length of wire B.

Answer: Equal tension and equal area give equal stress in both wires. Equal elongation with the same original length would mean equal strain, but the lengths are not equal. Write the extensions as F L_A/(A Y_A) and F L_B/(A Y_B) and set them equal. Then L_B = L_A Y_B/Y_A = 2.2 x 11/20 = 1.21 m. A shorter wire of the softer material can stretch by the same amount as a longer wire of the stiffer one.

Q6 (JEE Main 2026, 4 Apr Shift 2, 4 marks)

A metal string A hangs from a rigid support with a block of mass M at its end. A second block of mass 2M hangs from the first block by string B. The cross-sectional areas of A and B are equal, the lengths are in the ratio L_A/L_B = 2, and the Young's moduli are in the ratio Y_A/Y_B = 0.5. Find the ratio of the elongations of A and B.

Answer: String A carries both blocks, so its tension is 3Mg; string B carries only the lower block, so its tension is 2Mg. Extension is F L/(A Y), hence the ratio is (3Mg L_A/(A Y_A)) / (2Mg L_B/(A Y_B)) = (3/2)(L_A/L_B)(Y_B/Y_A) = 1.5 x 2 x 2 = 6. The answer is 6, so the upper, longer string stretches six times as much as the lower one.

Q7 (NCERT Exemplar, 1 mark)

The modulus of rigidity of an ideal liquid is (a) infinity, (b) zero, (c) unity, (d) some finite small non-zero value. Choose the correct option and justify it.

Answer: The modulus of rigidity is the shear modulus, the ratio of shear stress to shear strain. An ideal liquid cannot support a shear stress at rest, so it shows no elastic shear strain and flows instead. The ratio is therefore zero, and option (b) is correct. This is also why the shear modulus of gases is taken as zero.

Q8 (NCERT Exemplar, 2 marks)

A wire of length L and radius r is clamped at one end and pulled at the other end by a force f, so its length increases by l. A second wire of the same material has length 2L and radius 2r and is pulled by a force 2f. Find its increase in length.

Answer: The extension is f L/(A Y), with A proportional to r^2. For the first wire, l = f L/(pi r^2 Y). For the second wire, the force is 2f, the length is 2L and the area is 4A, so the extension is (2f)(2L)/((4A)Y) = f L/(A Y) = l. The longer and thicker wire, loaded twice as hard, stretches by exactly the same amount because the area grows faster than the load and length together.

Q9 (NCERT Exemplar, 3 marks)

A steel rod of length 1 m and cross-sectional area 1 cm^2 is heated from 0 degrees C to 200 degrees C while both ends are held firmly, so it cannot expand. Take Y = 2.0 x 10^11 N m^-2 and the coefficient of linear expansion alpha = 1.0 x 10^-5 per degree C. Find the tension produced in the rod.

Answer: If the rod were free it would try to expand by alpha x delta T x L, and blocking that expansion is equivalent to compressing it back by the same amount. The thermal strain is alpha x delta T = 1.0 x 10^-5 x 200 = 2.0 x 10^-3. The stress is Y times that strain, 2.0 x 10^11 x 2.0 x 10^-3 = 4.0 x 10^8 N m^-2. The area is 1 cm^2 = 1.0 x 10^-4 m^2, so the force is 4.0 x 10^8 x 1.0 x 10^-4 = 4.0 x 10^4 N, which is about 40 kN.

Q10 (NCERT Exemplar, 2 marks)

To what depth must a rubber ball be taken in sea water so that its volume decreases by 0.1 percent? The bulk modulus of rubber is 9.8 x 10^8 N m^-2 and the density of sea water is 10^3 kg m^-3. Take g = 9.8 m s^-2.

Answer: The fractional change of volume is 0.1/100 = 1.0 x 10^-3. The pressure needed is B x (delta V/V) = 9.8 x 10^8 x 1.0 x 10^-3 = 9.8 x 10^5 Pa. The pressure at depth h is rho g h, so h = P/(rho g) = 9.8 x 10^5/(10^3 x 9.8) = 100 m. A rubber ball taken 100 m under the sea has its volume reduced by about a tenth of a percent, which shows that rubber is far more compressible than steel.