Question

In a Searle's experiment, the diameter of the wire as measured by a screw gauge with least count 0.001cm is 0.050 cm.
The length, measured by a scale of least count 0.1cm, is 110.0 cm.
When a weight of 50 N is suspended from the wire, the extension is measured to be 0.125 cm by a micrometer of least count 0.001 cm.
Find the maximum error in the measurement of Young's modulus of the material of the wire from these data.

y = (W/A) × (L/X)

Where W is the weight suspended from the wire, A is the cross section area, L is the length & X is the extension in the wire.

[IIT JEE 2004]​

Answer 1

d = 0.050vcm;

Δd = 0.001 cm

l = 110 cm

Δl = 0.1 cm

W = 50 N

δl = 0.125 cm

Δ(δl) = 0.001 cm

=> Y = Fl / π(d/2)^2δl

= 50(110)/3.14(0.050/2 * 10^-2)^2 (0.125 * 10^-2)

=2.24 * 10^13 Nm^-2

=> Now, ΔY/Y = Δl/l + 2* Δd/d + Δδl /δl = 0.1/110 + 2* 0.001/0.050 + 0.001/0.125 = 0.0489

ΔY/Y = 0.0489

ΔY = 0.0489 *  2.24 * 10^13 Nm^-2

= 0.1095 * 10^13 Nm^-2

=> Therefore,

Y = (2.24 ± 0.1095 )* 10^13 Nm^-2

Thus, the maximum error in the measurement of Young’s modulus of the material of wire from these data is (2.24 ± 0.1095 )* 10^13 Nm^-2.

Answer 2

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