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In an experiment for measuring the young's modulus (y), following readings were taken. Load =3 kg, length = 2.820 m , diameter = 0.041 cm and extension = 0.87. Determine percentage error in measurement of Y.

```
11 years ago

```							Dear vishrant,

I'm not sure whether this is the correct method to analyse the error,

The relation is Y = (FL) / (A.dl) where F=mg and A = Pi.D2 /4 (take g = 10m.sec2 )

to calculate Y , given F=30 N ,dL/L=0.87 and A=1.32x10-7 m2

so Y = 261.23 MPa

consider dl , this term is a change in lenth, so error is nullified as we are taking the change in length : d(dl)=0

also change in gravity dg = 0

as Y = (4mgL) / (Pi.D2.dl)

ln Y = ln 4 + ln m + ln g + ln L - 2.ln D - ln (dl)

dY/Y = dm/m  +dL/L + dL/L + 2.dD/D   ( as d(dl)=0 and dg =0 )

Taking the least counts for measuring load ( using pan balance range 5kgs,LC =1g)  = 1 gram =10-3kg

Taking the least counts for measuring length ( using meter scale range 1m,LC =0.1cm)  = 1 mm =10-3m

Taking the least counts for measuring diameter ( using vernier calipers ,LC =0.01cm)  = 0.01 mm =10-5m

dY/Y = dm/m  +dL/L + dL/L + 2.dD/D
= 10-3 / 3 +10-3 / 2.82 + 2*10-5/41*10-5
=0.0495
so (Y±dY )= (261.23 ± 12.93) MPa

or percentage error is : dY/Y x 100 = 4.95 %

-----
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Regards,
Naga Ramesh
IIT Kgp - 2005 batch
```
11 years ago
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