Given here,
Length, L=1m and extension in length, ΔL=0.4×10−3m
Mass of load, m=1kg and diameter of wire, d=0.4×10−3m
Young's modulus is given by
Y=AΔLFL=(4πd2)0.4×10−3(mg)×(1)
⇒Y=π(0.4×10−3)2×0.4×10−310×4
⇒Y=22×64×10−3×10−940×7
⇒Y=0.199×10−12Nm−2
Error in measurement of Young's modulus,
YΔY=FΔF+LΔL+AΔA+(ΔL)Δ(ΔL)
=0.40.02+2dΔd=40.2+2×0.40.01
=20.1+20.1=0.1
=0.199×1011≈2×1010
Hence, the value of x=2.
In an experiment of determine the Young's modulus of wire of a length exactly 1m, the extension in the length of the wire is measured as 0.4 mm with…
JEE Main 2022 — Physics Mechanics
2022integermedium
In an experiment of determine the Young's modulus of wire of a length exactly 1m, the extension in the length of the wire is measured as 0.4mm with an uncertainty of ±0.02mm when a load of 1kg is applied. The diameter of the wire is measured as 0.4mm with an uncertainty of ±0.01mm. The error in the measurement of Young's modulus (ΔY) is found to be x×1010Nm−2. The value of x is _____ .
Official previous-year question
Held on 26 Jul 2022 · Verified 1 Jul 2026.
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