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JEE-ADVANCED-2014
PAPER 1
Question
Durring Searle's experiment, zero of the Vernier scale lies between $3.20 \times 10^{-2} \mathrm{~m}$ and $3.25 \times 10^{-2} \mathrm{~m}$ of the main scale. The $20^{\text {th }}$ 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 $3.20 \times 10^{-2} \mathrm{~m}$ and $3.25 \times 10^{-2} \mathrm{~m}$ of the main scale but now the $45^{\text {th }}$ 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 $8 \times 10^{-7} \mathrm{~m}^2$. The least count of the Vernier scale is $1.0 \times 10^{-5} \mathrm{~m}$. The maximum percentage error in the Young's modulus of the wire is
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Question Tags
JEE Advance
Physics
Easy
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