Question

In an experiment of determine the Young’s modulus of wire of a length exactly 1 m, the extension in the length of the wire is measured as 0.4 mm with an uncertainty of ±0.02 mm when a load of 1 kg is applied. The diameter of the wire is measured as 0.4 mm with an uncertainty of ±0.02 mm when a load of 1 kg is applied. The diameter of the wire is measured as 0.4 mm with an uncertainty of ± 0.01 mm. The error in the measurement of Young’s modulus (ΔY) is found to be x × 10¹⁰ Nm^–2. The value of x is ___.
(Take g=10 m/s²)

Answer 1

Correct Answer: 2

We know that,
\(Y = \frac {F/A }{l/L}\)
\(A=\pi D^2\)
\(\frac {ΔY}{Y}=\frac {ΔF}{F}+2\frac {ΔD}{D}+Δ\frac {l}{e}+\frac {ΔL}{L}\)

\(=2×\frac {0.01}{0.4}+\frac {0.02}{0.4}\)

\(=\frac {0.04}{0.4}\)

\(=\frac {1}{10}\)
Now,
\(Y = \frac {Fl}{AΔl}\)

\(Y = \frac {10×1}{\pi(0.1\ mm)^2×0.4\ mm}\)

\(=1.988×10^{11}\)
\(≈2×10^{11}\)
\(\frac {Δy}{y}=\frac {1}{10}\)

\(Δy=\frac {y}{10}\)

\(Δy=2×10^{10}\)

So, the answer is \(2\).