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 $\pm \, 0.05 \, mm$ at a load of exactly $1.0kg$ . The student also measures the diameter of the wire to be $0.4mm$ with an uncertainty of $\pm \, 0.01 \, mm.$ The young's modulus obtained from the reading is $\left[\right.$ Take $g=9.8 \, ms^{- 2}\left]\right.$

• 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 = $\Delta \left(\Delta L\right)=\pm0.5mm$
Diameter of wire measured, $D=0.4mm$
Uncertainty in diameter of wire, $\Delta D=\pm0.01mm$
We know, $\text{Young's Modulus =}\frac{\text{STREES}}{\text{STRAIN}}$
$\text{Young's Modulus = }Y\frac{\frac{F}{A}}{\frac{\Delta L}{L}}\text{=}\frac{F \times L}{A \times \Delta L}\text{=}\frac{4 F L}{\pi D^{2} \Delta L}$
On putting appropriate values we get,
$Y=\frac{(4 \times(1 \times 9.8) \times 2)}{\pi \times\left(0.4 \times 10^{-3} \times\left(0.8 \times 10^{-3}\right)\right.}$
$Y=2\times 10^{11}\text{N m}^{- 2}$
Now, Uncertainties in Length and Diameter should be taken into consideration to evaluate the uncertainty in young's modulus measurement, $\frac{\Delta Y}{Y}=\frac{2 \Delta D}{D}+\frac{\Delta \left(\right. \Delta L \left.\right)}{\Delta L}$
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 $Y=\left(\right.2\pm0.2\left.\right)\times \left(10\right)^{11}\left(\text{N m}\right)^{- 2}$
Hence, verified,