Properties of Matter 1 Question 20
20. In Searle’s experiment, which is used to find Young’s modulus of elasticity, the diameter of experimental wire is $D=0.05 cm$ (measured by a scale of least count $0.001 cm$ ) and length is $L=110 cm$ (measured by a scale of least count $0.1 cm)$. A weight of $50 N$ causes an extension of $l=0.125 cm$ (measured by a micrometer of least count $0.001 cm$ ). Find maximum possible error in the values of Young’s modulus. Screw gauge and meter scale are free from error.
(2004, 2M)
Show Answer
Answer:
Correct Answer: 20. $1.09 \times 10^{10} \mathrm{~N} / \mathrm{m}^2$
Solution:
- Young’s modulus of elasticity is given by
$$ Y=\frac{\text { Stress }}{\text { Strain }}=\frac{F / A}{l / L}=\frac{F L}{l A}=\frac{F L}{l\left(\frac{\pi d^{2}}{4}\right)} $$
Substituting the values, we get
$$ \begin{aligned} Y & =\frac{50 \times 1.1 \times 4}{\left(1.25 \times 10^{-3}\right) \times \pi \times\left(5.0 \times 10^{-4}\right)^{2}} \\ & =2.24 \times 10^{11} N / m^{2} \end{aligned} $$
Now, $\frac{\Delta Y}{Y}=\frac{\Delta L}{L}+\frac{\Delta l}{l}+2 \frac{\Delta d}{d}$
$$ =\left(\frac{0.1}{110}\right)+\left(\frac{0.001}{0.125}\right)+2\left(\frac{0.001}{0.05}\right)=0.0489 $$
$\Delta Y=(0.0489) Y=(0.0489) \times\left(2.24 \times 10^{11}\right) N / m^{2}$
$$ =1.09 \times 10^{10} N / m^{2} $$
