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JEE Advance - Physics (2014 - Paper 1 Offline - No. 7)

During Searle's experiment, zero of the Vernier scale lies between 3.20 $$ \times $$ 10-2 m and 3.25 $$ \times $$ 10-2 m of the main scale. The 20th division of the Vernier scale exactly coincides with one of the main scale divisions. When an additional load of 2 kg is applied to the wire, the zero of the Vernier scale still lies between 3.20 $$ \times $$ 10-2 m and 3.25 $$ \times $$ 10-2 m of the main scale but now the 45th division of Vernier scale coincides with one of the main scale divisions. The length of the thin metallic wire is 2 m and its cross-sectional area is 8 $$ \times $$ 10-7 m2. The least count of the Vernier scale is 1.0 $$ \times $$ 10-5 m. The maximum percentage error in the Young's modulus of the wire is
Answer
4

Explanation

The difference between the two measurements by Vernier scale gives elongation of the wire caused by the additional load of 2 kg. In the first measurement, main scale reading is MSR = 3.20 $$\times$$ 10$$-$$2 m and Vernier scale reading is VSR = 20. The least count of Vernier scale is LC = 1 $$\times$$ 10$$-$$5 m. Thus, the first measurement by Vernier scale is

L1 = MSR + VSR $$\times$$ LC

= 3.20 $$\times$$ 10$$-$$2 + 20(1 $$\times$$ 10$$-$$5)

= 3.220 $$\times$$ 10$$-$$2 m.

In the second measurement, MSR = 3.20 $$\times$$ 10$$-$$2 m and VSR = 45. Thus, the second measurement by Vernier scale is

L2 = 3.20 $$\times$$ 10$$-$$2 + 45(1 $$\times$$ 10$$-$$5)

= 3.245 $$\times$$ 10$$-$$2 m.

The elongation of the wire due to force F = 2g is

l = L2 $$-$$ L1 = 0.025 $$\times$$ 10$$-$$2 m.

The maximum error in the measurement of l is $$\Delta$$l = LC = 1 $$\times$$ 10$$-$$5 m. Young's modulus is given by Y = $${{FL} \over {lA}}$$. The maximum percentage error in the measurement of Y is

$${{\Delta Y} \over Y} \times 100 = {{\Delta l} \over l} \times 100 = {{1 \times {{10}^{ - 5}}} \over {0.025 \times {{10}^{ - 2}}}} \times 100 = 4\% $$.

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