A student determined Young’s Modulus of elasticity using the formula Y =  The value of g is taken to be 9.8 m/s2, without any significant error, his observation are as follows:

                   

Physical Quantity

Least count of the Equipment used for measurement

Observed value

Mass (M)

1 g

2 kg

Length of bar (L)

1 mm

1 m

 

Breadth of bar (b)

0.1 mm

4 cm

Thickness of bar (d)

0.01 mm

0.4 cm

Depression  

0.01 mm

5 mm

Then the fractional error in the measurement of Y is:

0 1 View | Posted 2 months ago
Asked by Shiksha User

  • 1 Answer

  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    2 months ago

    Y = M g L 3 4 b d 3 ?              

    For significant error in Y

    = ? M M + 3 ? L L + ? b b + 3 ? d d + ? ? ?           

    = 1 * 1 0 ? 3 2 + 3 * 1 0 ? 3 1 + 1 0 ? 2 4 + 3 * 0 . 0 1 * 1 0 ? 1 0 . 4 + 1 0 ? 2 5              

    = 0.0155

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