General Physics Ques 38
Question 17
- During Searle’s experiment, zero of the vernier scale lies between $3.20 \times 10^{-2} \mathrm{~m}$ and $3.25 \times 10^{-2} \mathrm{~m}$ of the main scale. The 20th division of the vernier scale exactly coincides with one of the main scale divisions. When an additional load of $2 \mathrm{~kg}$ is applied to the wire, the zero of the vernier scale still lies between $3.20 \times 10^{-2} \mathrm{~m}$ and $3.25 \times 10^{-2} \mathrm{~m}$ of the main scale but now the 45 th division of vernier scale coincides with one of the main scale divisions. The length of the thin metallic wire is $2 \mathrm{~m}$ and its cross-sectional area is $8 \times 10^{-7} \mathrm{~m}^{2}$. The least count of the vernier scale is $1.0 \times 10^{-5} \mathrm{~m}$. The maximum percentage error in the Young’s modulus of the wire is
(2014 Adv.)
Analytical & Descriptive Questions
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Answer:
Correct Answer: 38.(4)
Solution:
Formula:
- $Y=\frac{F / A}{\frac{\Delta l}{l}}, \Delta l=25 \times 10^{-5} \mathrm{~m}$
$$ \begin{gathered} Y=\frac{F}{A} \cdot \frac{l}{\Delta l} \ \frac{\Delta Y}{Y} \times 100=\frac{10^{-5}}{25 \times 10^{-5}} \times 100=4 % \end{gathered} $$
BETA
