Question

Physics young modulus doubt... check attachment. Please help!

Answer 1

Young's Modulus is
\[E=\frac{F \times L _{o}}{A _{o}\times \Delta l}\ ................(1)\]
where
F is the force exerted
Lo is the original length of the brass wire
Ao is the original cross-sectional area of the brass wire
Delta L is the amount by which the length changes
For the steel wire put 2Lo for the original length and Ao/4 for the original cross-sectional area.
\[Let\ \Delta L _{s}=steel\ extension\]
\[Let\ \Delta L _{b}=brass\ extension\]
\[\frac{\Delta L _{s}}{\Delta L _{b}}=\frac{F \times 2L _{o}}{2.0\times 10^{11}\times \frac{A _{o}}{4}}\times \frac{1.0\times 10^{11}\times A _{o}}{F \times L _{o}}\]
You can now do the cancellation and simplification.

Answer 2

Thanks alot, i understood everything EXCEPT why u took the area as Ao/4

Answer 3

The area of a circle is
\[\frac{\pi d ^{2}}{4}\]
If the diameter is halved the area will become one quarter of the original the reason being that the area is proportional to the square of the diameter.