Physics - Error Question with Solution | TestHub

During Searle's experiment, zero of the Vernier scale lies between$3.20\times {10}^{-2}\mathrm{ }\mathrm{m}$and$3.25\times {10}^{-2}\mathrm{ }\mathrm{m}$of the main scale. The${20}^{\mathrm{t}\mathrm{h}}$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$3.20\times {10}^{-2\mathrm{ }}\mathrm{m}$and$3.25\times {10}^{-2}\mathrm{ }\mathrm{m}$of the main scale but now the${45}^{\mathrm{t}\mathrm{h}}$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$8\times {10}^{-7}\mathrm{ }{\mathrm{m}}^2$. The least count of the Vernier scale is$1.0\times \mathrm{ }{10}^{-5}\mathrm{ }\mathrm{m}$. The maximum percentage error in the Young's modulus of the wire is