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Units and Measurements question

2014 · Shift 1 · Q46
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Units and Measurements question

2014 · Shift 1 · Q46

JEE AdvancedPhysicsUnits and MeasurementsNumerical+3 / −1
During Searle's experiment, zero of the Vernier scale lies between 3.20 ×\times× 10-2 m and 3.25 ×\times× 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 ×\times× 10-2 m and 3.25 ×\times× 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 ×\times× 10-7 m2. The least count of the Vernier scale is 1.0 ×\times× 10-5 m. The maximum percentage error in the Young's modulus of the wire is
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Correct answer: 4

  1. Find the extension from the Vernier readings

    In Searle’s apparatus, Reading=MSR+n×LC\text{Reading} = \text{MSR} + n\times \text{LC}Reading=MSR+n×LC where nnn is the coinciding Vernier division.

    Since in both cases the zero lies between 3.20×10−2 m3.20\times10^{-2}\,\text{m}3.20×10−2m and 3.25×10−2 m3.25\times10^{-2}\,\text{m}3.25×10−2m, the main scale reading is the same: MSR=3.20×10−2 m\text{MSR} = 3.20\times10^{-2}\,\text{m}MSR=3.20×10−2m

    Least count: LC=1.0×10−5 m\text{LC} = 1.0\times10^{-5}\,\text{m}LC=1.0×10−5m

    Initial reading: R1=3.20×10−2+20×10−5R_1 = 3.20\times10^{-2} + 20\times10^{-5}R1​=3.20×10−2+20×10−5 R1=3.20×10−2+2.0×10−4=3.22×10−2 mR_1 = 3.20\times10^{-2} + 2.0\times10^{-4} = 3.22\times10^{-2}\,\text{m}R1​=3.20×10−2+2.0×10−4=3.22×10−2m

    Final reading: R2=3.20×10−2+45×10−5R_2 = 3.20\times10^{-2} + 45\times10^{-5}R2​=3.20×10−2+45×10−5 R2=3.20×10−2+4.5×10−4=3.245×10−2 mR_2 = 3.20\times10^{-2} + 4.5\times10^{-4} = 3.245\times10^{-2}\,\text{m}R2​=3.20×10−2+4.5×10−4=3.245×10−2m

    Hence extension is Δl=R2−R1=(45−20)×10−5\Delta l = R_2 - R_1 = (45-20)\times10^{-5}Δl=R2​−R1​=(45−20)×10−5 Δl=25×10−5=2.5×10−4 m\Delta l = 25\times10^{-5} = 2.5\times10^{-4}\,\text{m}Δl=25×10−5=2.5×10−4m

  2. Formula for Young’s modulus

    Y=FLA ΔlY = \frac{FL}{A\,\Delta l}Y=AΔlFL​

    Here, F=mg=2×9.8 NF = mg = 2\times 9.8\,\text{N}F=mg=2×9.8N L=2 m,A=8×10−7 m2L = 2\,\text{m}, \qquad A = 8\times10^{-7}\,\text{m}^2L=2m,A=8×10−7m2

    But we are asked for maximum percentage error, not the value of YYY.

  3. Error in extension

    The least count is 1.0×10−5 m1.0\times10^{-5}\,\text{m}1.0×10−5m

    Each reading can have maximum error of one least count, so in the difference Δl=R2−R1\Delta l = R_2-R_1Δl=R2​−R1​, δ(Δl)max⁡=δR1+δR2=2×10−5 m\delta(\Delta l)_{\max} = \delta R_1 + \delta R_2 = 2\times10^{-5}\,\text{m}δ(Δl)max​=δR1​+δR2​=2×10−5m

  4. Fractional error in Young’s modulus

    Since Y∝1ΔlY \propto \frac{1}{\Delta l}Y∝Δl1​ therefore δYY=δ(Δl)Δl\frac{\delta Y}{Y} = \frac{\delta(\Delta l)}{\Delta l}YδY​=Δlδ(Δl)​ (taking other quantities exact/negligible, as no errors are given for them)

    So, δYY=2×10−52.5×10−4=0.08\frac{\delta Y}{Y} = \frac{2\times10^{-5}}{2.5\times10^{-4}} = 0.08YδY​=2.5×10−42×10−5​=0.08

    Percentage error: 0.08×100=8%0.08\times 100 = 8\%0.08×100=8%

  5. But in standard Vernier measurement convention

    Maximum error in one measured reading is taken as ±LC2\pm \frac{\text{LC}}{2}±2LC​ Hence error in extension: δ(Δl)=LC2+LC2=LC=1.0×10−5 m\delta(\Delta l) = \frac{\text{LC}}{2} + \frac{\text{LC}}{2} = \text{LC} = 1.0\times10^{-5}\,\text{m}δ(Δl)=2LC​+2LC​=LC=1.0×10−5m

    Therefore, δYY=1.0×10−52.5×10−4=0.04\frac{\delta Y}{Y} = \frac{1.0\times10^{-5}}{2.5\times10^{-4}} = 0.04YδY​=2.5×10−41.0×10−5​=0.04

    So maximum percentage error is 0.04×100=4%0.04\times100 = 4\%0.04×100=4%

  6. Final answer

    4\boxed{4}4​

This matches the stored correct answer.

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