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Q.

During Searle’s experiment, initially, zero of vernier scale lies between 3.20×102m and  3.25×102m  of the main scale. The 20th division of the vernier scale coincides with one of the main scale divisions. When an additional load of 2kg is applied to the wire, the zero of the vernier scale still lies between  3.20×102m  and  3.25×102m  of the main scale but now, 45th division of vernier scale coincides with one of the main scale divisions. Least count of vernier scale is  1.0×105m. Maximum percentage error in the measurement of deformation of the wire is : 

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answer is 8.

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Detailed Solution

e=l2l1dee×100=2×105m25×105m=8%

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During Searle’s experiment, initially, zero of vernier scale lies between 3.20×10−2m and  3.25×10−2m  of the main scale. The 20th division of the vernier scale coincides with one of the main scale divisions. When an additional load of 2kg is applied to the wire, the zero of the vernier scale still lies between  3.20×10−2m  and  3.25×10−2m  of the main scale but now, 45th division of vernier scale coincides with one of the main scale divisions. Least count of vernier scale is  1.0×10−5m. Maximum percentage error in the measurement of deformation of the wire is :