Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • An experiment is conducted to determine Young’s modulus (Y ) of a material using the concept of elasticity. A wire of length L = 2 m and diameter d = 0.5 mm is subjected to a load F = 500 N,

An experiment is conducted to determine Young’s modulus (Y ) of a material using the concept of elasticity. A wire of length L = 2 m and diameter d = 0.5 mm is subjected to a load F = 500 N, resulting in an extension e = 1.5 mm. Calculate Young’s modulus of the material.

Option: 1

4.4 * 1012 N/m2


Option: 2

3.4 * 1012 N/m2


Option: 3

3.4 * 1011 N/m2


Option: 4

4.4 * 1011 N/m2


Answers (1)

best_answer

Given values:

Length of the wire (L) = 2 m
Diameter of the wire (d) = 0.5 mm
Load applied (F) = 500 N
Extension (e) = 1.5 mm

Step 1: Calculate the cross-sectional area (A) of the wire using its diameter:

                    A = \frac{\pi d^{2}}{4}

Step 2: Calculate the stress (\sigma) in the wire:

                    \sigma = \frac{F}{A}

Step 3: Calculate the strain (\epsilon) in the wire:

                    \epsilon = \frac{e}{L}

Step 4: Calculate Young's modulus (Y) using the relationship between stress and strain:

                    Y = \frac{\sigma}{\epsilon}

Step 5: Substitute the values and calculate Y:

                    Y = \frac{\frac {F}{A}}{\frac{e}{L}}

Step 6: Calculate the areas (A) in m2 by converting d to meters:

                            A = \frac{\pi (0.5 * 10^{-3})^{2}}{4} = 1.9635 * 10^{-7} m^{2}

Step 7: Substitute the values into the formula for Y:

                    Y = \frac{500 N}{1.9635 * 10^{-7} m^{2} } * \frac{2 m}{1.5 * 10^{-3} m}

Step 8: Calculate Y:

                    Y = 339.5 * 10^{10} N/m^{2}

Young's modulus of the material is 3.4 * 1012 N/m2.

Posted by

Rishabh

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2024 session 2 paper 2 answer key objection window closes today; fees
anam-khan

JEE Main 2024 Topper: Farmer’s son Gajare Nilkrishna says ‘keep questioning until you understand’
anam-khan

NIT Puducherry Cut-off 2024: Computer Science and Engineering had highest closing rank last year

Latest Question