A uniform metallic wire is elongated by 0.04 m when subjected to a linear force F. The elongation, if its length and diameter is doubled and subjected to the same force will be ________ cm.
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Correct Answer: 2
Solution and Explanation
To solve the problem, we need to understand how elongation changes with the length and diameter of the wire while keeping the force constant. The elongation \( e \) of a wire is given by:
\( e = \frac{FL}{AY} \)
where \( F \) is the force, \( L \) is the length, \( A \) is the cross-sectional area, and \( Y \) is Young's modulus. Initially, the wire elongates by 0.04 m.
When the length \( L \) is doubled to \( 2L \) and diameter \( d \) is also doubled, the new length becomes \( 2L \), and the new area \( A \) becomes \( \pi(\frac{2d}{2})^2 = 4A \).
Substitute these into the formula:
\( e' = \frac{F(2L)}{4A \cdot Y} = \frac{2FL}{4AY} = \frac{1}{2}\frac{FL}{AY} \).
Since \( \frac{FL}{AY} = 0.04 \) m, the new elongation \( e' = \frac{1}{2} \times 0.04 = 0.02 \) m.
Convert \( e' \) from meters to centimeters:
\( 0.02 \, \text{m} = 2 \, \text{cm} \).
The elongation when the length and diameter are doubled is 2 cm, which lies within the given range (2, 2).
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