JEE Advanced 2014 Paper 1 Physics Q12

Physics · Solids & Fluids · Foundations & Measurement

Problem

During Searle's experiment, zero of the Vernier scale lies between 3.20×10−2 m3.20\times10^{-2}\,\mathrm{m} and 3.25×10−2 m3.25\times10^{-2}\,\mathrm{m} of the main scale. The 20th20^{\mathrm{th}} division of the Vernier scale exactly coincides with one of the main scale divisions. When an additional load of 2 kg2\,\mathrm{kg} is applied to the wire, the zero of the Vernier scale still lies between 3.20×10−2 m3.20\times10^{-2}\,\mathrm{m} and 3.25×10−2 m3.25\times10^{-2}\,\mathrm{m} of the main scale but now the 45th45^{\mathrm{th}} division of Vernier scale coincides with one of the main scale divisions. The length of the thin metallic wire is 2 m2\,\mathrm{m} and its cross-sectional area is 8×10−7 m28\times10^{-7}\,\mathrm{m^2}. The least count of the Vernier scale is 1.0×10−5 m1.0\times10^{-5}\,\mathrm{m}. The maximum percentage error in the Young's modulus of the wire is

Browse the problem library