General Physics Ques 42
Question 20
- In a Searle’s experiment, the diameter of the wire as measured by a screw gauge of least count $0.001 \mathrm{~cm}$ is $0.050 \mathrm{~cm}$. The length, measured by a scale of least count $0.1 \mathrm{~cm}$, is $110.0 \mathrm{~cm}$. When a weight of $50 \mathrm{~N}$ is suspended from the wire, the extension is measured to be $0.125 \mathrm{~cm}$ by a micrometer of least count $0.001 \mathrm{~cm}$. Find the maximum error in the measurement of Young’s modulus of the material of the wire from these data.
(2004, 2M)
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Answer:
Correct Answer: 42.$1.09 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}$
Solution:
Formula:
- Young’s modulus of elasticity is given by
$$ Y=\frac{\text { stress }}{\text { strain }}=\frac{F / A}{l / L}=\frac{F L}{l A}=\frac{F L}{l \frac{\pi d^{2}}{4}} $$
Substituting the values, we get
$$ \begin{aligned} Y & =\frac{50 \times 1.1 \times 4}{\left(1.25 \times 10^{-3}\right) \times \pi \times\left(5.0 \times 10^{-4}\right)^{2}} \ & =2.24 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2} \ \text { Now, } \frac{\Delta Y}{Y} & =\frac{\Delta L}{L}+\frac{\Delta l}{l}+2 \frac{\Delta d}{d} \ & =\frac{0.1}{110}+\frac{0.001}{0.125}+2 \frac{0.001}{0.05} \ & =0.0489 \ \Delta Y & =(0.0489) Y=(0.0489) \times\left(2.24 \times 10^{11}\right) \mathrm{N} / \mathrm{m}^{2} \ & =1.09 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2} \end{aligned} $$
BETA
