Question

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

Answer 1

of course you know , Young's modulus = stress/strain
Stress is force per unit area , strain is ratio of change in length to original length.
That's why here given , Y = 4MgL/πd²l
Where M is mass, g is Acceleration due to gravity,
L is original length d is diameter of wire and l is change in length

For finding error, use formula
∆Y/Y = ∆M/M + ∆g/g + ∆L/L + 2∆d/d + ∆l/l
so, % error in Y = % error in M + % error in g + % error in L + 2 × % error in d + % error in l

Here, 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 = 0.05
Similarly,
L = 2.890 ⇒∆L = 0.01/2 = 0.005
M = 3.00 ⇒∆M = 0.1/2 = 0.05
d = 0.082⇒∆d = 0.001/2 = 0.0005
g = 9.81 ⇒ ∆g = 0.01/2 = 0.005
l = 0.087 ⇒∆l = 0.001/2 = 0.0005

Now, Let's find error in Young's modulus
Relative 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 %

Answer 2

Answer:

The answer is in the attachment

Explanation:

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