During Searle’s experiment, zero of the vernier scale lies between 3.20× 10 −2 m and 3.25× 10 −2 m of the main scale. The 20th division of the vernier scale exactly coincides with one of the main scale division. When an additional load of 2kg is applied to the wire ,the Zero of the vernier scale still lies between 3.20× 10 −2 m and 3.25× 10 −2 m of the main scale, but now the 45th division of vernier scale coincides with one of the main scale division . The length of the thin metallic wire is 2m and its cross-sectional area is 8× 10 −7 m 2 . The least count of the vernier scale is 1.0× 10 −5 m . The maximum percentage error in the young’s modulus of the wire is

Question

During Searle’s experiment, zero of the vernier scale lies between 3.20× 10 −2 m  and 3.25× 10 −2 m  of the main scale. The 20th division of the vernier scale exactly coincides with one of the main scale division. When an additional load of 2kg is applied to the wire ,the Zero of the vernier scale still lies between 3.20× 10 −2 m and 3.25× 10 −2 m  of the main scale, but now the 45th division of vernier scale coincides with one of the main scale division . The length of the thin metallic wire is 2m and its cross-sectional area is 8× 10 −7 m 2  . The least count of the vernier scale is 1.0× 10 −5 m  . The maximum percentage error in the young’s modulus of the wire is

Infinity Learn · Accepted Answer

Y=FLAl⇒ΔYY=Δll Where Δl is least count of Vernier scale =10-5 ml is change in length =45th-20th×10-5=25×10-5 mΔYY×100%=Δll×100%=10-525×10-5×100%=4%

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