Verified Solution • 2025–26 Session moderate

QJEEM22Ph56a9: JEE Main 2022 Physics - Class 11 JEEM22Ph56a9 Solution: In an experiment to determine the Young's modulus of wire of a…

In an experiment to determine the Young's modulus of wire of a length exactly 1 m1 \mathrm{~m} , the extension in the length of the wire is measured as 0.4 mm0.4 \mathrm{~mm} with an uncertainty of ± 0.02 mm\pm\, 0.02 \mathrm{~mm} when a load of 1 kg1 \mathrm{~kg} is applied. The diameter of the wire is measured as 0.4 mm0.4 \mathrm{~mm} with an uncertainty of ± 0.01 mm\pm \,0.01 \mathrm{~mm} . The error in the measurement of Young's modulus (ΔY)(\Delta \mathrm{Y}) is found to be x×1010 Nm−2x \times 10^{10}\, \mathrm{Nm}^{-2} . The value of xx is _________________. (\left(\right. take g=10 ms−2\mathrm{g}=10 \mathrm{~ms}^{-2} )

Step-by-Step Verified Solution

F/Al/L=Y, A=πD2{{F/A} \over {l/L}} = Y,\,A = \pi {D^2}

ΔYY=ΔFF+2ΔDD+Δle+ΔLL{{\Delta Y} \over Y} = {{\Delta F} \over F} + {{2\Delta D} \over D} + {{\Delta l} \over e} + {{\Delta L} \over L}

=2×0.010.4+0.020.4= 2 \times {{0.01} \over {0.4}} + {{0.02} \over {0.4}}

=0.040.4=110= {{0.04} \over {0.4}} = {1 \over {10}}

Y=FlAΔlY = {{Fl} \over {A\Delta l}}

=10×1π(0.1 mm)2×0.4 mm= {{10 \times 1} \over {\pi {{(0.1\,mm)}^2} \times 0.4\,mm}}

=1.988×1011= 1.988 \times {10^{11}}

≈2×1011\approx 2 \times {10^{11}}

Δyy=110{{\Delta y} \over y} = {1 \over {10}}

Δy=y10=2×1010\Delta y = {y \over {10}} = 2 \times {10^{10}}

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