SECTION – 2: (Only Integer Value Correct Type)
This section contains 10 questions. Each question, when worked out will result in one integer from 0 to 9 (both inclusive).
During Searle ’ s experiment, zero of the Vernier scale lies between 3.20 x 10 -2 m and 3.25 x 10 -2 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 3.20 x10 -2 m and 3.25 x10 -2 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 8 x 10 -7 m 2 . The least count of the Vernier scale is 1.0 x 10 -5 m. The maximum percentage error in the Young ’ s modulus of the wire is
Text Solution
Verified by Experts4
(4)
since the experiment measures only change in the length of wire

From the observation
= MSR + 20 (LC)
= MSR + 45 (LC)
change in lengths = 25(LC)
and the maximum permissible error in elongation is one LC

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