Two wires each of radius 0.2 cm and negligible mass, one made of steel and other made of brass are loaded as shown in the figure. The elongation of the steel wire is _____ x 10-6 m .
[Young’s modulus for steel = 2 x 1011 Nm-2 and g = 10ms-2]
[Young’s modulus for steel = 2 x 1011 Nm-2 and g = 10ms-2]
Show Hint
Remember the formula for elongation under tensile stress: \(∆L = \frac{FL}{AY}\) consistent units throughout your calculations
Correct Answer: 20
Solution and Explanation
Step 1: Forces Acting on the System
The tension in the wire is given by:
\[ T_2 = T_1 + 20 = 20 + 11.4 = 31.4 \, \text{N} \]
Step 2: Formula for Elongation (\( \Delta L \))
The elongation in the steel wire is given by:
\[ \Delta L = \frac{T_2 L}{A Y} \]
- \( T_2 = 31.4 \, \text{N} \) (tension)
- \( L = 1.6 \, \text{m} \) (length of the wire)
- \( A = \pi (0.2 \times 10^{-2})^2 \, \text{m}^2 \) (cross-sectional area)
- \( Y = 2 \times 10^{11} \, \text{Pa} \) (Young’s modulus of steel)
Step 3: Substitute the Values
\[ \Delta L = \frac{31.4 \cdot 1.6}{\pi (0.2 \times 10^{-2})^2 \cdot 2 \times 10^{11}} \]
Step 4: Simplify the Expression
Calculate step-by-step:
\[ \Delta L = \frac{50.24}{\pi \cdot 4 \times 10^{-8} \cdot 2 \times 10^{11}} \]
\[ \Delta L = 2 \times 10^{-5} \, \text{m} \]
Step 5: Convert to Micrometers
\[ \Delta L = 20 \times 10^{-6} \, \text{m} = 20 \, \mu\text{m} \]
Final Answer:
The elongation in the steel wire is:
\( \Delta L = 20 \, \mu\text{m}. \)
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