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  • Please help! A compressive force, F is applied at the two ends of a long thin steel rod. It is heated, simultaneously, such that its temperature increases by ∆T. The net change in its length is zero. Let l be the length of the rod, A its area of cros

A compressive force, F is applied at the two ends of a long thin steel rod.  It is heated, simultaneously, such that its temperature increases by T.  The net change in its length is zero.  Let l be the length of the rod, A its area of cross-section, Y its Young’s modulus, and α its coefficient of linear expansion.  Then, F is equal to :

  • Option 1)

    l^{2}Y\alpha \Delta T

     

  • Option 2)

    lA\:\: Y\alpha \: \: \Delta T

  • Option 3)

     

    A\: Y\alpha \: \Delta T

  • Option 4)

    \frac{AY}{\alpha \Delta T}

     

     

 

Answers (1)

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

 

 

 

 Here l=l_{0}(l+\alpha\Delta T)

l=l_{0}+l_{0}\alpha\Delta T 

l-l_{0}=l_{0}\alpha \Delta T

\Rightarrow\ \; \Delta l=l_{0}\alpha \Delta T

\Rightarrow\ \; \frac{\Delta l}{l_{0}}=\alpha \Delta T

    Y=\frac{F}{A} \frac{l_{0}}{\Delta l}

\Rightarrow\ \; F=\frac{Y_{A}\Delta l}{l_{0}}

    F=Y_{A} \alpha.\Delta T

Correct option is 3.


Option 1)

l^{2}Y\alpha \Delta T

 

This is an incorrect option.

Option 2)

lA\:\: Y\alpha \: \: \Delta T

This is an incorrect option.

Option 3)

 

A\: Y\alpha \: \Delta T

This is the correct option.

Option 4)

\frac{AY}{\alpha \Delta T}

 

 

This is an incorrect option.

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perimeter

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