Question

A student performs an experiment to determine young's modulus of a wire, exactly $2m$ long, by searle's method. In a particular reading, the student measures the extension in the length of the wire to be $0.8mm$ with an uncertainty of $\pm 0.05mm$ at a load of exactly $1.0kg$ . The student also measures the diameter of the wire to be $0.4mm$ with an uncertainty of $\pm 0.01mm.$ The young's modulus obtained from the reading is $[$ Take $g=9.8m{s}^{-2}]$

• A. $(2.0\pm 0.3)\times 10^{11}N{m}^{-2}$
• B. $(2.0\pm 0.2)\times 10^{11}N{m}^{-2}$
• C. $(2.0\pm 0.1)\times 10^{11}N{m}^{-2}$
• D. $(2.0\pm 0.05)\times 10^{11}N{m}^{-2}$

Answer 1

Answer: (B)

Given,
Mass of load, $F=1.0kg$
Length of Wire, $L=2m$
Extension length of wire, $\Delta L=0.8mm$
Uncertainty in length of wire = $\Delta \left(\Delta L\right)=\pm 0.5mm$
Diameter of wire measured, $D=0.4mm$
Uncertainty in diameter of wire, $\Delta D=\pm 0.01mm$
We know, $\text{Young’s Modulus =}\frac{\text{STREES}}{\text{STRAIN}}$
$\text{Young’s Modulus = }Y\frac{\frac{F}{A}}{\frac{\Delta L}{L}}\text{=}\frac{F\times L}{A\times \Delta L}\text{=}\frac{4FL}{\pi D^2\Delta L}$
On putting appropriate values we get,
$Y=\frac{(4\times (1\times 9.8)\times 2)}{\pi \times (0.4\times 10^{-3}\times \left(0.8\times 10^{-3}\right)}$
$Y=2\times 10^{11}{\text{N m}}^{-2}$
Now, Uncertainties in Length and Diameter should be taken into consideration to evaluate the uncertainty in young's modulus measurement, $\frac{\Delta Y}{Y}=\frac{2\Delta D}{D}+\frac{\Delta (\Delta L)}{\Delta L}$
On putting the value, we get,
$\Delta Y=\left[\frac{2\times 0.01}{0.4}+\frac{0.05}{0.8}\right]2\times 10^{11}$ $=0.2\times 10^{11}$
Net $Y=(2\pm 0.2)\times {\left(10\right)}^{11}{\left(\text{N m}\right)}^{-2}$
Hence, verified,