During Searle's experiment, zero of the Vernier scale lies between 3.20 × 10 - 2   m and 3.25…

During Searle's experiment, zero of the Vernier scale lies between 3.20×10-2 m and 3.25×10-2 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 3.20×10-2 m and 3.25×10-2 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 8×10-7 m2 . The least count of the Vernier scale is 1.0× 10-5 m . The maximum percentage error in the Young's modulus of the wire is

Solution

Y=FLlA since the experiment measures only change in the length of wire
∴∆YY×100=∆ll×100
From the observation l1=MSR+20 LC
l2=MSR+45LC
⇒ change in lengths =25LC
and the maximum permissible error in elongation is one LC
∴∆YY×100=LC25LC×100=4 .

Asked in: JEE Advanced 2014 (Paper 1)

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