JEE Advanced 2014 Physics question
During Searle's experiment, zero of the Vernier scale lies between m and 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 m and 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 . The least count of the Vernier scale is m. The maximum percentage error in the Young's modulus of the wire is
Solution
Readings: initial m; final m. Extension m. ; the
See the full step-by-step solution
Create a free account to read the complete working, and practise questions like this in a timed test.
Create a free account