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PhysicsInteger+3Question 12 of 60

JEE Advanced 2014 · Paper 1, Physics question 12

During Searle's experiment, zero of the Vernier scale lies between 3.20×10−23.20\times10^{-2} m and 3.25×10−23.25\times10^{-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 3.20×10−23.20\times10^{-2} m and 3.25×10−23.25\times10^{-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 m28\times10^{-7}\ \mathrm m^2. The least count of the Vernier scale is 1.0×10−51.0\times10^{-5} m. The maximum percentage error in the Young's modulus of the wire is

Answer and worked solution

Official answer: 4

Extension. The main-scale reading does not change, and the coinciding Vernier division moves from 20 to 45:
ℓ=(45−20)×1.0×10−5=25×10−5 m.\ell=(45-20)\times1.0\times10^{-5}=25\times10^{-5}\text{ m}.

Error. Y=FLAℓY=\dfrac{FL}{A\ell}. The load, length and area are taken as exact, and the extension is uncertain by one least count:
ΔYY=Δℓℓ=1.0×10−525×10−5=4%.\frac{\Delta Y}Y=\frac{\Delta\ell}\ell=\frac{1.0\times10^{-5}}{25\times10^{-5}}=4\%.

Answer: 4

Topics:Young's modulus by Searle's methodErrors in measurementLeast count
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