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

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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}$

Similar Questions

How much force is required to produce an increase of $0.2\%$ in the length of a brass wire of diameter $0.6\, mm$ (Young’s modulus for brass = $0.9 \times {10^{11}}N/{m^2}$)