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 divisions. When an additional load of 2 kg is applied to the wire, the zero of the Vernier scale still lies between 3.20 × 10–2 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 divisions. The length of the thin metallic wire is 2 m and its cross-sectional area is 8 × 10–7 m2. 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

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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 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^–2 m and 3.25 × 10^–2 m of the main scale but now the 45^th 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 × 10^–7 m². 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

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$Y = \frac{FL}{AΔL}$

$\frac{dY}{Y} = \frac{d (ΔL)}{ΔL}$

x₁ = 3.20 × 10^–2 + 20 × 10^–5 = 3.220

x₂ = 3.20 × 10^–2 + 4.5 × 10^–5 = 3.245 × 10^–2 m

$∴ ∆$ L = 0.025 × 10^–2 m

$∴ \frac{dY}{Y} = \frac{10^{- 5}}{0 .025 \times 10^{- 2}} \times 100 = 4$