Question

A rod of length 4 m is suspended from two wires
as shown in figure. One of them is made of steel
and other of copper of cross-sectional area 10^-2m^2
and 4 x 10^-2 m^2. Find distance x if equal stress is
experienced by both the wires (Ysteel = 2 10^11 N/m^2
and Y Copper = 1.1 * 10^11 N/m^2
Steel






niche bs W hai​

Answer 1

Answer:

We know that the young's modulus is having a formulae of y=stress/strain so we will get that strain = stress/ y.

So, the stress of the steel wire = stress of the copper wire which will be given as σs/ys = σc/yc which will be σs/σc = ys/yc which will be 2*10^11/1.11*10^11 so the σs/σc = 1.801. Hence, σs=1.801σc.

Again, it is known that the σ=T/A where T is the tension and A is the area. So, we get that Ts=σs*As or Tc=σc*Ac, Therefore, if we divide both of them we will get that Ts/Tc =σs*As / σc*Ac or 1.801*As/Ac which is 10^-2/4*10^-2 * 1.801 that on solving we will get that 0.45. So, Ts= 0.45Tc. So, if the tension in the steel wire is 0.45 times of the copper wire tension. Accordingly the value of distance will change and will not remain at the center.

Answer 2

Answer: 3.2m

Explanation:HELLO MATE,

HERE YOUR ANSWER IS

Given,

Ysteel=2x10¹¹ N/m²

Ycopper=1.1x 10¹¹ N/m²

Cross sectional area of steel [As]= 10^-2 m²

Cross sectional area of copper[Ac]=4x10^-2 m²

Since it is said that equal stress is experienced by both wires. So,,

Stress of steel=stress of copper

And we know that stess = Tension/Area

so,,

Ts/As=Tc/Ac

Ts/10^-2= Tc/ 4x10^-2

Tc=4Ts

Now here length of the rod is 4m

So by applying the concept of centre of mass, we get that,

Ts x [X]= Tc x [4-X]

Ts x [X]= 4 Ts x [4-X] [Tc= 4Ts]

X= 16-4X

5X=16

X=3.2 m

Hope it helps.....

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