Question:medium
A string A of length 0.314 m and Young's modulus \(2 \times 10^{10}\) N/m² 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/s²)
A string A of length 0.314 m and Young's modulus \(2 \times 10^{10}\) N/m² 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/s²)
Updated On: Jun 6, 2026
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
When two strings are connected in series and subjected to a load, the tension \(F\) is uniform throughout both strings (neglecting their own weight). The total extension is the sum of the individual extensions of each string.
Step 2: Key Formula or Approach:
Young's Modulus is given by \(Y = \frac{F/A}{\Delta L / L}\).
Rearranging this, the change in length is:
\(\Delta L = \frac{F L}{A Y}\)
The total change in length is \(\Delta L_{\text{total}} = \Delta L_A + \Delta L_B\).
Step 3: Detailed Explanation:
Let's list the known properties for String A:
Length \(L_A = 0.314 \text{ m}\). Notice that \(0.314 \approx \frac{\pi}{10}\).
Young's Modulus \(Y_A = 2 \times 10^{10} \text{ N/m}^2\).
Radius \(r = 0.2 \text{ mm} = 2 \times 10^{-4} \text{ m}\).
Cross-sectional area \(A = \pi r^2 = \pi (2 \times 10^{-4})^2 = \pi \times 4 \times 10^{-8} \text{ m}^2\).
Properties for String B:
Length \(L_B = 2 L_A\).
Young's Modulus \(Y_B = 2 Y_A\).
Area \(A\) is identical since radii are the same.
Force applied by load \(F = mg = 0.8 \text{ kg} \times 10 \text{ m/s}^2 = 8 \text{ N}\).
Calculate \(\Delta L_A\):
\(\Delta L_A = \frac{F L_A}{A Y_A} = \frac{8 \times (\pi / 10)}{(\pi \times 4 \times 10^{-8}) \times (2 \times 10^{10})} = \frac{0.8 \pi}{\pi \times 8 \times 10^2} = \frac{0.8}{800} = 10^{-3} \text{ m} = 1 \text{ mm}\).
Calculate \(\Delta L_B\):
\(\Delta L_B = \frac{F L_B}{A Y_B} = \frac{F (2 L_A)}{A (2 Y_A)} = \frac{2}{2} \left( \frac{F L_A}{A Y_A} \right) = \Delta L_A = 1 \text{ mm}\).
Total extension:
\(\Delta L_{\text{total}} = \Delta L_A + \Delta L_B = 1 \text{ mm} + 1 \text{ mm} = 2 \text{ mm}\).
Step 4: Final Answer:
The net change in length of the combination is 2 mm.
When two strings are connected in series and subjected to a load, the tension \(F\) is uniform throughout both strings (neglecting their own weight). The total extension is the sum of the individual extensions of each string.
Step 2: Key Formula or Approach:
Young's Modulus is given by \(Y = \frac{F/A}{\Delta L / L}\).
Rearranging this, the change in length is:
\(\Delta L = \frac{F L}{A Y}\)
The total change in length is \(\Delta L_{\text{total}} = \Delta L_A + \Delta L_B\).
Step 3: Detailed Explanation:
Let's list the known properties for String A:
Length \(L_A = 0.314 \text{ m}\). Notice that \(0.314 \approx \frac{\pi}{10}\).
Young's Modulus \(Y_A = 2 \times 10^{10} \text{ N/m}^2\).
Radius \(r = 0.2 \text{ mm} = 2 \times 10^{-4} \text{ m}\).
Cross-sectional area \(A = \pi r^2 = \pi (2 \times 10^{-4})^2 = \pi \times 4 \times 10^{-8} \text{ m}^2\).
Properties for String B:
Length \(L_B = 2 L_A\).
Young's Modulus \(Y_B = 2 Y_A\).
Area \(A\) is identical since radii are the same.
Force applied by load \(F = mg = 0.8 \text{ kg} \times 10 \text{ m/s}^2 = 8 \text{ N}\).
Calculate \(\Delta L_A\):
\(\Delta L_A = \frac{F L_A}{A Y_A} = \frac{8 \times (\pi / 10)}{(\pi \times 4 \times 10^{-8}) \times (2 \times 10^{10})} = \frac{0.8 \pi}{\pi \times 8 \times 10^2} = \frac{0.8}{800} = 10^{-3} \text{ m} = 1 \text{ mm}\).
Calculate \(\Delta L_B\):
\(\Delta L_B = \frac{F L_B}{A Y_B} = \frac{F (2 L_A)}{A (2 Y_A)} = \frac{2}{2} \left( \frac{F L_A}{A Y_A} \right) = \Delta L_A = 1 \text{ mm}\).
Total extension:
\(\Delta L_{\text{total}} = \Delta L_A + \Delta L_B = 1 \text{ mm} + 1 \text{ mm} = 2 \text{ mm}\).
Step 4: Final Answer:
The net change in length of the combination is 2 mm.
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