Physics Elasticity Question 124
Question: A wire extends by 1 mm when a force is applied. Double the force is applied to another wire of same material and length but half the radius of cross-section. The elongation of the wire in mm will be
[EAMCET 1986]
Options:
8
4
2
1
Show Answer
Answer:
Correct Answer: A
Solution:
$ l=\frac{FL}{\pi r^{3}} $
Therefore $ l\propto \frac{F}{r^{2}} $ (Y and L are constant) $ \frac{l _{2}}{l _{1}}=\frac{F _{2}}{F _{1}}\times {{\left( \frac{r _{1}}{r _{2}} \right)}^{2}}=2\times {{(2)}^{2}}=8 $ \ $ l _{2}=8l _{1}=8\times 1=8mm $
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