During Searle’s experiment, the zero of a Vernier scale lies between 3.20×10−2m and 3.25×10−2m of the main scale divisions. 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 3.20×10−2m and 3.25×10−2m of the main scale but now the 45th division of the 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 8.0×10−7m2. The least count of the Vernier scale is 1.0×10−5m. Find the maximum percentage error in the Young modulus of the wire.

Y=FLAlHere measurement is for l only,So, ΔYY=ΔllFrom observation, l1=MS+20(LC), and l2=MS+45(LC)Change in length l2−l1=25×LC, and the maximum permissible error in measurement to elongation in one LC.ΔYY×100%=125LC(LC)×100%=4%

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During Searle’s experiment, the zero of a Vernier scale lies between $3.20 \times 10^{- 2} m$ and $3.25 \times 10^{- 2} m$ of the main scale divisions. The $20^{t h}$ 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 $3.20 \times 10^{- 2} m$ and $3.25 \times 10^{- 2} m$ of the main scale but now the $45^{t h}$ division of the 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 $8.0 \times 10^{- 7} m^{2}$ . The least count of the Vernier scale is $1.0 \times 10^{- 5} m$ . Find the maximum percentage error in the Young modulus of the wire.

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Detailed Solution

$Y = \frac{F L}{A l}$

Here measurement is for $l$ only,
So, $\frac{Δ Y}{Y} = \frac{Δ l}{l}$
From observation, $l_{1} = M S + 20 (L C),$ and $l_{2} = M S + 45 (L C)$
Change in length $l_{2} - l_{1} = 25 \times L C$ , and the maximum permissible error in measurement to elongation in one LC.