SATHEE: General Physics Question 17

General Physics Question 17

Question 17

During Searle’s experiment, zero of the vernier scale lies between $3.20 \times 10^{- 2} m$ and $3.25 \times 10^{- 2} m$ 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 $2 kg$ is applied to the wire, the zero of the vernier scale still lies between $3.20 \times 10^{- 2} m$ and $3.25 \times 10^{- 2} m$ 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 $2 m$ and its cross-sectional area is $8 \times 10^{- 7} m^{2}$ . The least count of the vernier scale is $1.0 \times 10^{- 5} m$ . 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: 18. (4)

Solution:

$Y = \frac{F / A}{\frac{Δ l}{l}}, Δ l = 25 \times 10^{- 50} m$

$\begin{matrix} Y = \frac{F}{A} \cdot \frac{l}{Δ l} \frac{Δ Y}{Y} \times 100 = \frac{10^{- 5}}{25 \times 10^{- 5}} \times 100 = 4 \end{matrix}$

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