Question:
The area of cross-section of a wire of length 1.1 m is 1 $mm^2$. It is leaded with 1 kg of Young's modulus of copper is $1.1\times10^i Nm^\circ$ , then the increase in length will be $(ifg = 10{ms^-2})$
The area of cross-section of a wire of length 1.1 m is 1 $mm^2$. It is leaded with 1 kg of Young's modulus of copper is $1.1\times10^i Nm^\circ$ , then the increase in length will be $(ifg = 10{ms^-2})$
Updated On: Jul 7, 2022
- 0.01 mm
- 0.075 mm
- 0.1 mm
- 0.15 mm
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The Correct Option is C
Solution and Explanation
Increase in length $l=\frac{mg^2}{AY}$
$=\frac{1\times10\times1.1}{1.1\times10^{11}\times10^{-6}}m=0.1 mm$
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