A student performs an experiment to determine the 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  ±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  ±0.01 mm. Take g=9.8 m/s2 (exact). The Young’s modulus obtained from the reading is[ Take π=3.14 ]

# A student performs an experiment to determine the 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  $±0.05\text{\hspace{0.17em}mm}$ at a load of exactly  1.0kg. The student also measures the diameter of the wire to be  0.4mm with an uncertainty of  $±0.01\text{\hspace{0.17em}mm}$. Take $g=9.8\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}$ (exact). The Young’s modulus obtained from the reading is[ Take $\text{π}=3.14$ ]

1. A

$\left(2.0±0.3\right)×{10}^{11}\text{\hspace{0.17em}N}/{\text{m}}^{\text{2}}$

2. B

$\left(2.0±0.2\right)×{10}^{11}\text{\hspace{0.17em}N}/{\text{m}}^{\text{2}}$

3. C

$\left(2.0±0.1\right)×{10}^{11}\text{\hspace{0.17em}N}/{\text{m}}^{\text{2}}$

4. D

$\left(2.0±0.05\right)×{10}^{11}\text{\hspace{0.17em}N}/{\text{m}}^{\text{2}}$

Register to Get Free Mock Test and Study Material

+91

Verify OTP Code (required)

### Solution:

Given Length of wire $l=2\text{\hspace{0.17em}m}$
Extension= $0.8\text{\hspace{0.17em}mm}$
Uncertainty =0.05 mm
Diameter of the wire   $d=0.4\text{\hspace{0.17em}mm}$
Mass  $m=1.0\text{\hspace{0.17em}kg}$
Acceleration due to gravity
Using Hooke’s Law
$Y=\frac{FL}{Al}=\frac{4FL}{\pi {d}^{2}l}.....\left(1\right)$
Also   $Y=\frac{FL}{Al}=\frac{mgL}{\text{π}{r}^{2}l}\left(\begin{array}{l}\because F=mg\\ \because A=\pi {r}^{2}\text{\hspace{0.17em}}\left(or\right)\text{\hspace{0.17em}}\pi \frac{{d}^{2}}{4}\end{array}\right)$
Where  L= length of the wire
l= elongation of the wire
d= diameter of the wire
$\begin{array}{l}Y=\frac{1.0×9.8×2}{3.14×{\left(0.2\right)}^{2}×0.8}\\ Y=194.96×{10}^{9}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1.95×{10}^{11}\text{\hspace{0.17em}N}/{\text{m}}^{\text{2}}\end{array}$
From equation (1)
$Y=\frac{FL}{Al}=\frac{4FL}{\pi {d}^{2}l}$
Taking log on both sides
$⇒\mathrm{log}Y=\mathrm{log}4FL-\mathrm{log}\text{π}{d}^{2}l$
Now partially differentiating
$\begin{array}{c}⇒\frac{\mathrm{\Delta }Y}{Y}=2\frac{\mathrm{\Delta }d}{d}+\frac{\mathrm{\Delta }l}{l}\\ =2\left(\frac{0.01}{0.4}\right)+\frac{0.05}{0.8}=\frac{9}{80}\\ ⇒\mathrm{\Delta }Y=\frac{9}{80}×Y\\ =\frac{9}{80}×1.95×{10}^{11}\\ =2.19×{10}^{11}\mathrm{N}/{\mathrm{m}}^{2}\\ Y=\left(2±0.219\right)×{10}^{11}\mathrm{N}/{\mathrm{m}}^{2}\\ \therefore Y\approx \left(2±0.2\right)×{10}^{11}\mathrm{N}/{\mathrm{m}}^{2}\end{array}$
Therefore, the correct answer is (B).

Register to Get Free Mock Test and Study Material

+91

Verify OTP Code (required)