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Two exactly similar wires of steel and c...

Two exactly similar wires of steel and copper are stretched by equal forces. If the total elogation is 1cm. Find by how much is each wire elongated ? Given Y for steel ` = 20 xx 10^(11) dyn e//cm^2 and Y for copper = 12xx10^(11) dyn e//cm^2`

A

`1.12 xx 10^(-6) m`

B

`3.32 xx 10^(-6) m`

C

`2.32 xx10^(-3)m`

D

`23.2 xx10^(-4) m`

Text Solution

Verified by Experts

The correct Answer is:
C
Weight of the wire,` W = pl alpha g `
where `alpha` is the area of cross-section of the wire.
Here, length of the wire, l= 8 m = 800 cm
`rho` (density of copper) = 8.9 g/cc
` therefore W = 8.9 xx 800 xx alpha xx 980` dyne.
Now, as the elongation of the wire is due to its own weight, the deforming force at the lower end is zero, and at the upper end is the weight of the wire.
` therefore ` Average deforming force, `vecF = (0 +W)/(2)`
` = (8.9 xx 800 xx alpha xx 980)/(2) ` dyne
` therefore ` Average stress ` = (vecF)/(alpha) = (8.9 xx 800 xx 980)/(2) "dyne/cm"^2`
` = 400 xx 8.9 xx 980 "dyne/cm"^2`
If `Detla l` be the elongation of the wire , then
`Delta l = ("stress" xx l)/(Y) `, Y is the young.s modulus of copper
`= ((400 xx 8.9 xx 980 ) xx 800)/(12 xx 10^11) = 2.32 xx 10^(-3) cm`
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