SATHEE: General Physics Question 12

General Physics Question 12

Question 12

A student performs an experiment to determine the Young’s modulus of a wire, exactly $2 m$ long, by Searle’s method. In a particular reading, the student measures the extension in the length of the wire to be $0.8 mm$ with an uncertainty of $\pm 0.05 mm$ at a load of exactly $1.0 kg$ . The student also measures the diameter of the wire to be $0.4 mm$ with an uncertainty of $\pm 0.01 mm$ . Take $g = 9.8 m / s^{2}$ (exact). The Young’s modulus obtained from the reading is close to

(2007, 3M) (a) $(2.0 \pm 0.3) 10^{11} N / m^{2}$ (b) $(2.0 \pm 0.2) \times 10^{11} N / m^{2}$ (c) $(2.0 \pm 0.1) \times 10^{11} N / m^{2}$ (d) $(2.0 \pm 0.05) \times 10^{11} N / m^{2}$

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Answer:

Correct Answer: 12. (b)

Solution:

$\begin{aligned} Y & = \frac{F L}{A l} = \frac{4 F L}{π d^{2} l} = \frac{(4) (1.0 \times 9.8) (2)}{π {(0.4 \times 10^{- 3})}^{2} (0.8 \times 10^{- 3})} & = 2.0 \times 10^{11} N / m^{2} \end{aligned}$

Further $\frac{Δ Y}{Y} = 2 \frac{Δ d}{d} + \frac{Δ l}{l}$

$∴ Δ Y = 2 \frac{Δ d}{d} + \frac{Δ l}{l} Y$

$= 2 \times \frac{0.01}{0.4} + \frac{0.05}{0.8} \times 2.0 \times 10^{11}$

$= 0.225 \times 10^{11} N / m^{2}$

$= 0.2 \times 10^{11} N / m^{2}$

(By rounding off)

or $(Y + Δ Y) = (2 \pm 0.2) \times 10^{11} N / m^{2}$

General Physics Question 12

Question 12

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General Physics Question 12

Question 12

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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