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

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 $2$.