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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.

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