Physics Q34 – JEE-MAIN 2026

Home Physics Mechanical Properties of SolidsQuestion #283

A string

A

of length

0.314 m

and Young's modulus

2 \times 10^{10} {N/m}^{2}

is connected to another string

B

of length and Young's modulus both twice of those of

A

. This series combination of strings is then suspended from a rigid support and its free end is fixed to a load of mass

0.8 kg

. The net change in length of the combination is _______

mm

. (radius of both the strings is

0.2 mm

and acceleration due to gravity

= 10 {m/s}^{2}

) (Mass of both strings is to be neglected as compared to the mass of load)

Clue 1 of 3: Conceptual Principle

For strings connected in series, the total elongation is the sum of the individual elongations, and the tensile force acting on each string is the same.

🧠 Full Solution Path

Step 1: Calculate Applied Force and Cross-sectional Area✦ Active

The force applied to both strings is due to the mass of the load. The mass is $m = 0.8 kg$ and acceleration due to gravity is $g = 10 {m/s}^{2}$ .

F = m g = 0.8 kg \times 10 {m/s}^{2} = 8 N

The radius of both strings is $r = 0.2 mm = 0.2 \times 10^{- 3} m$ . The cross-sectional area $A$ for both strings is:

A = π r^{2} = π (0.2 \times 10^{- 3} m)^{2} = π (0.04 \times 10^{- 6}) m^{2} = 4 π \times 10^{- 8} m^{2}

Step 2: Calculate Elongation of Each String○ Expand

For string A: Length $L_{A} = 0.314 m$ and Young's modulus $Y_{A} = 2 \times 10^{10} {N/m}^{2}$ . Its elongation $Δ L_{A}$ is:

Δ L_{A} = \frac{F L_{A}}{A Y_{A}} = \frac{8 N \times 0.314 m}{4 π \times 10^{- 8} m^{2} \times 2 \times 10^{10} {N/m}^{2}} = \frac{8 \times 0.314}{8 π \times 10^{2}} m = \frac{0.314}{100 π} m

Using $π \approx 3.14$ :

Δ L_{A} = \frac{0.314}{100 \times 3.14} m = \frac{1}{1000} m = 1 mm

For string B: Length $L_{B} = 2 L_{A} = 2 \times 0.314 m = 0.628 m$ and Young's modulus $Y_{B} = 2 Y_{A} = 2 \times (2 \times 10^{10}) {N/m}^{2} = 4 \times 10^{10} {N/m}^{2}$ . Its elongation $Δ L_{B}$ is:

Δ L_{B} = \frac{F L_{B}}{A Y_{B}} = \frac{8 N \times 0.628 m}{4 π \times 10^{- 8} m^{2} \times 4 \times 10^{10} {N/m}^{2}} = \frac{8 \times 0.628}{16 π \times 10^{2}} m = \frac{0.628}{200 π} m

Using $π \approx 3.14$ :

Δ L_{B} = \frac{0.628}{200 \times 3.14} m = \frac{0.628}{628} m = \frac{1}{1000} m = 1 mm

Alternatively, since $L_{B} = 2 L_{A}$ and $Y_{B} = 2 Y_{A}$ , we have $Δ L_{B} = \frac{F (2 L_{A})}{A (2 Y_{A})} = \frac{F L_{A}}{A Y_{A}} = Δ L_{A} = 1 mm$ .

💡 Teacher's Secret Hint

Ensure consistent units throughout the calculation. Convert mm to m for area calculation and then back to mm for the final answer.

Step 3: Calculate Total Change in Length○ Expand

The net change in length of the combination is the sum of the individual elongations:

Δ L_{t o t a l} = Δ L_{A} + Δ L_{B} = 1 mm + 1 mm = 2 mm

💡 Teacher's Secret Hint

Remember that for series combinations, the total elongation is the sum of individual elongations, while the force is the same for each component.

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Physics Question 34 – JEE-MAIN 2026

Clue 1 of 3: Conceptual Principle

🧠 Full Solution Path