SATHEE: Properties of Matter 1 Question 20

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 $D = 0.05 c m$ (measured by a scale of least count $0.001 c m$ ) and length is $L = 110 c m$ (measured by a scale of least count $0.1 c m)$ . A weight of $50 N$ causes an extension of $l = 0.125 c m$ (measured by a micrometer of least count $0.001 c m$ ). Find maximum possible error in the values of Young’s modulus. Screw gauge and meter scale are free from error.

(2004, 2M)

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Answer:

Correct Answer: 20. $1.09 \times 10^{10} N / m^{2}$

Solution:

Young’s modulus of elasticity is given by

$Y = \frac{Stress}{Strain} = \frac{F / A}{l / L} = \frac{F L}{l A} = \frac{F L}{l (\frac{π d^{2}}{4})}$

Substituting the values, we get

$\begin{aligned} Y & = \frac{50 \times 1.1 \times 4}{(1.25 \times 10^{- 3}) \times π \times {(5.0 \times 10^{- 4})}^{2}} \\ = 2.24 \times 10^{11} N / m^{2} \end{aligned}$

Now, $\frac{Δ Y}{Y} = \frac{Δ L}{L} + \frac{Δ l}{l} + 2 \frac{Δ d}{d}$

$= (\frac{0.1}{110}) + (\frac{0.001}{0.125}) + 2 (\frac{0.001}{0.05}) = 0.0489$

$Δ Y = (0.0489) Y = (0.0489) \times (2.24 \times 10^{11}) N / m^{2}$

$= 1.09 \times 10^{10} N / m^{2}$

Properties of Matter 1 Question 20

Answer:

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Properties of Matter 1 Question 20

Answer:

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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