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Fo For the estimation of Young’s modulus: Y=4MgL/Pi d^2*l for the specimen of a wire, the following observations were recorded: L=2.890, M=3.00, d=0.082, g=9.81, and l=0.087. Calculate the maximum percentage error in the value of Y and mention which physical quantity causes a maximum error.
FoFor the estimation of Young’s modulus: Y=4MgL/Pi d^2*l for the specimen of a wire, the following observations were recorded: L=2.890, M=3.00, d=0.082, g=9.81, and  l=0.087. Calculate the maximum percentage error in the value of Y and mention which physical quantity causes a maximum error.

```
5 months ago Rituraj Tiwari
1789 Points
```							Young's modulus = stress/strainStress is force per unit area , strain is ratio of change in length to original length.That's why here given , Y = 4MgL/πd²lWhere M is mass, g is Acceleration due to gravity,L is original length d is diameter of wire and l is change in lengthFor finding error, use formula∆Y/Y = ∆M/M + ∆g/g + ∆L/L + 2∆d/d + ∆l/lso, % error in Y = % error in M + % error in g + % error in L + 2 × % error in d + % error in lHere, we have given M, g, L , d and l values but not given ∆M, ∆g, ∆L, ∆d and ∆l values . So, just Let's least count for all of these terms .We know, if a = 4.5 then it must be written as a = (4.5 ±0.1) hence, maximum possible error is 0.1/2 = 0.05Similarly,L = 2.890 ⇒∆L = 0.01/2 = 0.005M = 3.00 ⇒∆M = 0.1/2 = 0.05d = 0.082⇒∆d = 0.001/2 = 0.0005g = 9.81 ⇒ ∆g = 0.01/2 = 0.005l = 0.087 ⇒∆l = 0.001/2 = 0.0005Now, Let's find error in Young's modulusRelative error in Y = 0.005/2.890 + 0.05/3 + 0.005/9.81 + 2 × 0.0005/0.082 + 0.0005/0.087= 0.0368∴ maximum% error= 100 × 0.0368 = 3.68 %
```
5 months ago
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