The Young's modulus of steel is twice tha | vedclass

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Question

Let $'L'$ and $A$ be lenght and area of cross section of each wire. In order to have the lower ends of the wires to be at the same level $(i.e

Vedclass · Accepted Answer

$2:1$

Vedclass · Answer

Let $'L'$ and $A$ be lenght and area of cross section of each wire. In order to have the lower ends of the wires to be at the same level $(i.e.,\,same\,elongation\,is\,produced\,in\,both\,wiers)$ let weights $W_s$ and $W_b$ are added to steel and brass wires respectively. Then, By definition of $Young's$ modulus, the elongation produced in the steel wire is 

 $\Delta {L_s} = \frac{{{W_s}L}}{{{Y_s}A}}$                          $\left( {asY = \frac{{W/A}}{{\Delta L/L}}} ight)$

and that in the brass wire is   $\Delta {L_b} = \frac{{{W_b}L}}{{{Y_b}A}}$

But $\Delta {L_s} = \Delta {L_b}$                           $(given)$

$	herefore \frac{{{W_s}L}}{{{Y_s}A}} = \frac{{{W_b}L}}{{{Y_b}A}}\,\,or\,\,\frac{{{W_s}}}{{{W_b}}} = \frac{{{Y_s}}}{{{Y_b}}}$

$As\,\frac{{{Y_s}}}{{{Y_b}}} = 2$                            $(given)$

$	herefore \frac{{{W_s}}}{{{W_b}}} = \frac{2}{1}$

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Similar Questions

Read the following two statements below carefully and state, with reasons, if it is true or false. $(a)$ The Young’s modulus of rubber is greater than that of steel; $(b)$ The stretching of a coil is determined by its shear modulus.

The length of an iron wire is $L$ and area of cross-section is $A$. The increase in length is $l$ on applying the force $F$ on its two ends. Which of the statement is correct

A meter scale of mass $m$ , Young modulus $Y$ and cross section area $A$ is hanged vertically from ceiling at zero mark. Then separation between $30\ cm$ and $70\ cm$ mark will be :-( $\frac{{mg}}{{AY}}$ is dimensionless)

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$(a)$ The Young’s modulus of rubber is greater than that of steel;

$(b)$ The stretching of a coil is determined by its shear modulus.