Copper of fixed bolume 'V' is drawn into wire of length 'l'. When the wire is subjected to a constant force 'F', the extension produced in the wire is'\Delta l'. Which of the following graphs is a straight line?

  • Option 1)

    \Delta l\ \text{versus}\ 1/l

  • Option 2)

    \Delta l\ \text{versus}\ l^{2}

  • Option 3)

    \Delta l\ \text{versus}\ 1/l^{2}

  • Option 4)

    \Delta l\ \text{versus}\ l

 

Answers (1)

As we learnt in

Young Modulus -

Ratio of normal stress to longitudnal strain

it denoted by Y

Y= frac{Normal : stress}{longitudnal: strain}

- wherein

Y=frac{F/A}{Delta l/L}

F -  applied force

A -  Area

Delta l -  Change in lenght

l - original length

 

 

 

 

 

V=Al= constant

F=Y-\left ( \frac{V}{l} \right ).\frac{\Delta l}{l}=\left ( \frac{YV}{l^{2}} \right )\Delta l

\Delta l=\left (\frac{F}{YV} \right ).l^{2}

\Delta l \ \alpha \ l^{2}

graph between \Delta l versus l^{2} will be a straight line.


Option 1)

\Delta l\ \text{versus}\ 1/l

this option is incorrect

Option 2)

\Delta l\ \text{versus}\ l^{2}

this option is correct

Option 3)

\Delta l\ \text{versus}\ 1/l^{2}

this option is incorrect

Option 4)

\Delta l\ \text{versus}\ l

this option is incorrect

Preparation Products

JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Buy Now
Knockout JEE Main April 2021 (Subscription)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
Buy Now
Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Knockout BITSAT 2021

It is an exhaustive preparation module made exclusively for cracking BITSAT..

₹ 4999/- ₹ 2999/-
Buy Now
Exams
Articles
Questions