JEE Challenger
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PhyMedMain 2026 • 4th Apr - Shift 1

Calculate Net Extension of Two Suspended Strings in Series

A string AA of length 0.314 m0.314 \text{ m} and Young's modulus 2×1010 N/m22 \times 10^{10} \text{ N/m}^2 is connected to another string BB of length and Young's modulus both twice of those of AA. 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 kg0.8 \text{ kg}. The net change in length of the combination is ________ mm\text{mm}. (radius of both the strings is 0.2 mm0.2 \text{ mm} and acceleration due to gravity =10 m/s2=10 \text{ m/s}^2) (Mass of both strings is to be neglected as compared to the mass of load)

Options

A

33

B

22

Correct
C

1.91.9

D

11

Topics & Concepts

Mechanical Properties of SolidsElasticity

Step-by-Step Solution

To find the net extension of the combination of the two strings connected in series, we calculate the individual extension of each string under the applied load.

1. Identify the Given Parameters

For String AA:

  • Length, LA=0.314 mL_A = 0.314\text{ m}
  • Young's modulus, YA=2×1010 N/m2Y_A = 2 \times 10^{10}\text{ N/m}^2
  • Radius, r=0.2 mm=2×10−4 mr = 0.2\text{ mm} = 2 \times 10^{-4}\text{ m}

For String BB:

  • Length, LB=2LAL_B = 2 L_A
  • Young's modulus, YB=2YAY_B = 2 Y_A
  • Radius, r=0.2 mm=2×10−4 mr = 0.2\text{ mm} = 2 \times 10^{-4}\text{ m}

System Parameters:

  • Load mass, m=0.8 kgm = 0.8\text{ kg}
  • Acceleration due to gravity, g=10 m/s2g = 10\text{ m/s}^2

2. Determine the Tension and Cross-Sectional Area

Since the mass of the strings is neglected and they are connected in series, the tension FF throughout both strings is equal to the weight of the load: F=mg=0.8×10=8 NF = m g = 0.8 \times 10 = 8\text{ N}

The cross-sectional area AsA_s for both strings (having the same radius) is: As=πr2=π×(2×10−4 m)2=4π×10−8 m2A_s = \pi r^2 = \pi \times \left(2 \times 10^{-4}\text{ m}\right)^2 = 4\pi \times 10^{-8}\text{ m}^2

3. Calculate the Extension of String AA

The extension ΔLA\Delta L_A of string AA is given by Hooke's Law: ΔLA=FLAAsYA\Delta L_A = \frac{F L_A}{A_s Y_A}

Substituting the given values and using π≈3.14\pi \approx 3.14 (so LA=0.314 m=0.1π mL_A = 0.314\text{ m} = 0.1\pi\text{ m}): ΔLA=8×0.314(4π×10−8)×(2×1010)\Delta L_A = \frac{8 \times 0.314}{(4\pi \times 10^{-8}) \times (2 \times 10^{10})} ΔLA=8×0.3148π×102=0.3143.14×100=10−3 m=1 mm\Delta L_A = \frac{8 \times 0.314}{8\pi \times 10^2} = \frac{0.314}{3.14 \times 100} = 10^{-3}\text{ m} = 1\text{ mm}

4. Calculate the Extension of String BB

The extension ΔLB\Delta L_B of string BB is: ΔLB=FLBAsYB\Delta L_B = \frac{F L_B}{A_s Y_B}

Given that LB=2LAL_B = 2 L_A and YB=2YAY_B = 2 Y_A: ΔLB=F(2LA)As(2YA)=FLAAsYA=ΔLA=1 mm\Delta L_B = \frac{F (2 L_A)}{A_s (2 Y_A)} = \frac{F L_A}{A_s Y_A} = \Delta L_A = 1\text{ mm}

5. Calculate Net Extension

The net change in length of the combination is the sum of the individual extensions: ΔLnet=ΔLA+ΔLB=1 mm+1 mm=2 mm\Delta L_{\text{net}} = \Delta L_A + \Delta L_B = 1\text{ mm} + 1\text{ mm} = 2\text{ mm}

Thus, the correct option is B (22).

Calculate Net Extension of Two Suspended Strings in Series | Physics PYQ Solution - JEE Challenger