Question

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. 3
• B. 2
• C. 1.9
• D. 1

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
When strings are connected in series, the tension (force \(F\)) in both strings is the same. The total extension is the sum of the individual extensions of string A and string B (\(\Delta L_{total} = \Delta L_A + \Delta L_B\)).

Step 2: Key Formula or Approach:
1. Extension \(\Delta L = \frac{F L}{A Y}\) 2. Area of cross-section \(A = \pi r^2\). 3. Force \(F = mg\).

Step 3: Detailed Explanation:
1. Given for String A: \(L_A = 0.314\) m, \(Y_A = 2 \times 10^{10}\) N/m², \(r = 0.2 \times 10^{-3}\) m. - \(F = 0.8 \times 10 = 8\) N. - Area \(A = \pi (2 \times 10^{-4})^2 = 3.14 \times 4 \times 10^{-8} = 12.56 \times 10^{-8}\) m². - \(\Delta L_A = \frac{8 \times 0.314}{(3.14 \times 4 \times 10^{-8}) \times 2 \times 10^{10}} = \frac{8 \times 0.314}{0.314 \times 8} \times 10^{-3} = 1 \text{ mm}\). 2. Given for String B: \(L_B = 2 L_A\), \(Y_B = 2 Y_A\). - \(\Delta L_B = \frac{F (2 L_A)}{A (2 Y_A)} = \frac{F L_A}{A Y_A} = \Delta L_A = 1 \text{ mm}\). 3. Total Change: - \(\Delta L_{total} = 1 \text{ mm} + 1 \text{ mm} = 2 \text{ mm}\).

Step 4: Final Answer:
The net change in length is 2 mm.