Question

A uniform wire (Y-given) is subjected to longitudinal tensile stress of 'S'. the overall volume change in the wire is 0.02% the fractional decrease in the radius of the wire is?

Answer 1

@ganeshie8 @samigupta8 @Rushwr

Answer 2

i am getting an answer..but its not correct..

Answer 3

@ParthKohli

Answer 4

Hi :)

Answer 5

ok..i did like this: (though its not correct! pls tell me where i went wrong..)
first i used stress/strain = Y i found (delta L/ L)
and then, \[(\frac{ \Delta r }{ r })^2 = (\Delta V / V)*(L/\Delta L)\]
then i got the req quantity from this..

Answer 6

its cylindrical..not spherical..right?

Answer 7

yes..

Answer 8

The values of Y ans S are given..

Answer 9

no..

Answer 10

this is the full Q except for the numerical values..

Answer 11

\[Y\rm \Delta l/l = S\]\[\rm \Rightarrow l' = l\left(1 + \frac{S}Y\right)\]\[\rm \Delta V = \pi r^2 l - \pi r'^2 l' \]\[\Rightarrow \rm \frac{\Delta V}{V}= 1 - \frac{r'^2 l'}{r^2 l }\]\[\Rightarrow \rm \frac{\Delta V}{V} = 1- \frac{(r - \Delta r)^2(1 + S/Y)}{r^2}\]\[\Rightarrow \rm \frac{\Delta V}V = 1 - (1 + S/Y)\left(1 - \frac{\Delta r}r\right)^2\]You're given \(\rm \Delta V/V\) and you require \(\rm -\Delta r/r\).

Cheers!

Answer 12

Yes, I've considered everything in that answer.

Answer 13

wondering if you might also go down this road....

\(V = \pi r^2 l\)

\(\delta V = 2 \pi r l \; \delta r + \pi r^2 \delta l\)


\(\dfrac{\delta V }{ V} = \dfrac{2 \pi r l \; \delta r}{\pi r^2 l} + \dfrac{\pi r^2 \delta l}{\pi r^2 l} = 2 \dfrac{\delta r }{r} + \dfrac{S}{Y} \)

all we've done here is linearise

Answer 14

thats a very nice idea! i got the answer! but i have a doubt why overall volume change means delta V/V and not just delta V?