Question:
A wire of length \( 2 \) m and area \( 1 \times 10^{-6} \, m^2 \) stretches by \( 1 \) mm under force \( 200 \) N. Young's modulus is:
A wire of length \( 2 \) m and area \( 1 \times 10^{-6} \, m^2 \) stretches by \( 1 \) mm under force \( 200 \) N. Young's modulus is:
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Always double-check your units! In elasticity problems, extensions are often given in mm or cm. Converting them to meters (\( 10^{-3} \) or \( 10^{-2} \)) immediately prevents power-of-ten errors in your final answer.
Updated On: Jun 3, 2026
- \( 2 \times 10^{11} \, Pa \)
- \( 4 \times 10^{11} \, Pa \)
- \( 1 \times 10^{11} \, Pa \)
- \( 5 \times 10^{10} \, Pa \)
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The Correct Option is B
Solution and Explanation
Concept:
Young's modulus (\( Y \)) is a measure of the stiffness of a solid material. It is defined as the ratio of longitudinal stress to longitudinal strain within the elastic limit of the material:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L / L} = \frac{F \cdot L}{A \cdot \Delta L} \]
where \( F \) is the applied force, \( A \) is the cross-sectional area, \( L \) is the original length, and \( \Delta L \) is the change in length.
Step 1: Extracting given values and converting to SI units.
Original length (\( L \)) = \( 2 \) m
Cross-sectional area (\( A \)) = \( 1 \times 10^{-6} \, m^2 \)
Force (\( F \)) = \( 200 \) N
Extension (\( \Delta L \)) = \( 1 \) mm = \( 1 \times 10^{-3} \) m
Step 2: Substituting the values into the Young's modulus formula.
\[ Y = \frac{200 \times 2}{(1 \times 10^{-6}) \times (1 \times 10^{-3})} \]
Step 3: Simplifying the calculation.
\[ Y = \frac{400}{1 \times 10^{-9}} \] \[ Y = 400 \times 10^9 \] \[ Y = 4 \times 10^{11} \, Pa \]
Step 1: Extracting given values and converting to SI units.
Original length (\( L \)) = \( 2 \) m
Cross-sectional area (\( A \)) = \( 1 \times 10^{-6} \, m^2 \)
Force (\( F \)) = \( 200 \) N
Extension (\( \Delta L \)) = \( 1 \) mm = \( 1 \times 10^{-3} \) m
Step 2: Substituting the values into the Young's modulus formula.
\[ Y = \frac{200 \times 2}{(1 \times 10^{-6}) \times (1 \times 10^{-3})} \]
Step 3: Simplifying the calculation.
\[ Y = \frac{400}{1 \times 10^{-9}} \] \[ Y = 400 \times 10^9 \] \[ Y = 4 \times 10^{11} \, Pa \]
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