SATHEE: General Physics Question 20

General Physics Question 20

Question 20

In a Searle’s experiment, the diameter of the wire as measured by a screw gauge of least count $0.001 cm$ is $0.050 cm$ . The length, measured by a scale of least count $0.1 cm$ , is $110.0 cm$ . When a weight of $50 N$ is suspended from the wire, the extension is measured to be $0.125 cm$ by a micrometer of least count $0.001 cm$ . Find the maximum error in the measurement of Young’s modulus of the material of the wire from these data.

(2004, 2M)

Show Answer

Answer:

Correct Answer: 21. $1.09 \times 10^{10} N / m^{2}$

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

$\begin{aligned} Y & = \frac{50 \times 1.1 \times 4}{(1.25 \times 10^{- 3}) \times π \times {(5.0 \times 10^{- 4})}^{2}} & = 2.24 \times 10^{11} N / m^{2} Now, \frac{Δ Y}{Y} & = \frac{Δ L}{L} + \frac{Δ l}{l} + 2 \frac{Δ d}{d} & = \frac{0.1}{110} + \frac{0.001}{0.125} + 2 \frac{0.001}{0.05} & = 0.0489 Δ Y & = (0.0489) Y = (0.0489) \times (2.24 \times 10^{11}) N / m^{2} & = 1.09 \times 10^{10} N / m^{2} \end{aligned}$

General Physics Question 20

Question 20

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

General Physics Question 20

Question 20

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu