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Two wires are made of the same material and have the same volume. However wire 1 has cross-sectional area A and wire 2 has cross-sectional area 3A. If the length of wire 1 increases by \Delta x on applying force  F , how much force is needed to stretch wire 2 by the same amount?

  • Option 1)

    F

  • Option 2)

    4F

  • Option 3)

    6F

  • Option 4)

    9F

 

Answers (2)

best_answer

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

 

 

 

 For the same material, Young's modulus is the same and it is given that the volume is the same and the Area of cross section for the wire 1 is A and that of 2 is 3A

v = v1 = v2

y=A\times l_{1}=3A\times l_{2}            or    l_{2}=\frac{l_{1}}{3}

y=\frac{F/A}{\Delta l/l}

\therefore\ \; F_{1}=\frac{yA\Delta l_{1}}{l_{1}}\ and\ \;F_{2}= \frac{y3A\Delta l_{2}}{l_{2}}

    \Delta l_{1}=\Delta l_{2}=\Delta x

\therefore\ \;F_{2}=\frac{y3A\Delta x}{l_{1}/3}=9\left(\frac{yA\Delta x}{l_{1}} \right )=9F_{1}=9F

Because F1 = F2

Correct option is 4.

 


Option 1)

F

This is an incorrect option.

Option 2)

4F

This is an incorrect option.

Option 3)

6F

This is an incorrect option.

Option 4)

9F

This is the correct option.

Posted by

perimeter

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Dhanush

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