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A stress of 1kg/mm^-2 is applied to a wire which Young’s modulus is 10^11 Nm^-2 .Find the percentage incerease in length?

A stress of 1kg/mm^-2  is applied to a wire which Young’s modulus is 10^11 Nm^-2 .Find the percentage incerease in length?

Grade:12

3 Answers

SAHIL
3778 Points
7 years ago

We have,

F/A = 1 kg/mm2 = 106 kg/m2

Y = 1011 N/m2

Now,

Y = (F/A)/(ΔL/L)

=> (ΔL/L) = (F/A)/Y = 106/1011 = 10-5

Therefore, (ΔL/L) × 100% = 10-3 %

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Bhomik Singh rana
11 Points
6 years ago
Sol^n- F/A=9.8*10^-6 Y=10^11 Y=(F/A)/(l/L)(l/L)=9.8*10^-6/10^11 =9.8*10^5 Now, = ( l/L)*100 =(9.8*10^-5)*100 =0.0098%
Unknown
26 Points
2 years ago
We know, Young's Modulus = Stress/Strain
Here, Strain = 1 kg/mm²
Converting it to SI units, we get, 1 x9.8 x 10⁶ N/m²
Note that 9.8 is being multiplied to convert Kg into Newton (1 kg = 9.81 N, here, for simplification, it is taken as 9.8 N)
 
Stress = Extension/ Original length
Y = Stress / Strain
Y{(=10¹¹ N/m²) as given} is (F/A)/(Delta L/L)
10¹¹ N/m² = (L/Delta L) x 9.8 x 10⁶ N/m²
9.8 x 10⁵ = (L/Delta L)
Percentage increase 
= Extension/original length x 100 percent
= (Delta L/L) x 100 percent
= 9.8 x 10^(-5) x 100 percent
= 9.8 x 10^(-5) x 100 percent
9.8 x 10^(-3) percent
O.OO98 percent.
 
Note that we need to multiply 9.8 N to be able to proceed with the calculation, so as the produce the same unit.
 
Thank you.

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