A uniformly tapering conical wire is made from a material of Young’s modulus Y and has a normal, unextended length L.  The radii, at the upper and lower ends of this conical wire, have values R and 3R, respectively.  The upper end of the wire is fixed to a rigid support and a mass M is suspended from its lower end.  The equilibrium extended length, of this wire, would equal :

  • Option 1)

    L\left ( 1+\frac{2}{9} \, \frac{Mg}{\pi YR^{2}}\right )

  • Option 2)

    L\left ( 1+\frac{1}{3} \, \frac{Mg}{\pi YR^{2}}\right )

  • Option 3)

    L\left ( 1+\frac{1}{9} \, \frac{Mg}{\pi YR^{2}}\right )

  • Option 4)

    L\left ( 1+\frac{2}{3} \, \frac{Mg}{\pi YR^{2}}\right )

 

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

 

 

 

 

\frac{r-R}{x}=\frac{3R-R}{L}

r=R\left(1+\frac{2x}{L} \right )

Y=\frac{mg}{\pi R^{2}\frac{dL}{dx}}\ \; \Rightarrow\ \; dL=\frac{mg}{\pi R^{2}}\frac{dx}{\left(1+\frac{2x}{L} \right )^{2}}

\Delta L=\frac{mg}{Y\pi R^{2}}\int_{0}^{L}\frac{dx}{\left(1+\frac{2x}{L} \right )^{2}}\ \; \Rightarrow\ \; \frac{mgL}{\left(1+\frac{2x}{L} \right )^{2}}

Now, L'=L+\Delta L=L+\frac{mgL}{\left(1+\frac{2x}{L} \right )^{2}}

L'=L\left(1+\frac{1}{3}\frac{mg}{\pi R^{2}Y} \right )

Correct option is 2.


Option 1)

L\left ( 1+\frac{2}{9} \, \frac{Mg}{\pi YR^{2}}\right )

This is an incorrect option.

Option 2)

L\left ( 1+\frac{1}{3} \, \frac{Mg}{\pi YR^{2}}\right )

This is the correct option.

Option 3)

L\left ( 1+\frac{1}{9} \, \frac{Mg}{\pi YR^{2}}\right )

This is an incorrect option.

Option 4)

L\left ( 1+\frac{2}{3} \, \frac{Mg}{\pi YR^{2}}\right )

This is an incorrect option.

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