If ratio of diameters, lengths and Young's modulus of steel and copper wires shown in figure are

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8.Mechanical Properties of Solids

hard

If the ratio of diameters, lengths and Young's modulus of steel and copper wires shown in the figure are $p, q$ and $s$ respectively, then the corresponding ratio of increase in their lengths would be

$\frac{{5q}}{{7{p^2}s}}$

$\;\frac{{7q}}{{5{p^2}s}}$

$\;\frac{{2q}}{{5sp}}$

$\;\frac{{7q}}{{5sp}}$

Solution

$As\,Y = \frac{{FL}}{{A\Delta L}} = \frac{{4FL}}{{\pi {D^2}\Delta L}}$

$\Delta L = \frac{{4FL}}{{\pi {D^2}Y}}$

$\therefore \frac{{\Delta {L_s}}}{{\Delta {L_c}}} = \frac{{{F_S}}}{{{F_C}}}\frac{{{L_S}}}{{{L_C}}}\frac{{D_C^2}}{{D_S^2}}\frac{{{Y_C}}}{{{Y_S}}}$

Where subscripts $'S'$ and $'C'$ refer to copper and steel respectively.

$Here,\,{F_S} = \left( {5m + 2m} \right)g = 7mg$

${F_C} = 5mg$

$\frac{{{L_S}}}{{{L_C}}} = q,\frac{{{D_S}}}{{{D_C}}} = p,\frac{{{Y_S}}}{{{Y_C}}} = s$

$\therefore \frac{{\Delta {L_S}}}{{\Delta {L_C}}} = \left( {\frac{{7mg}}{{5mg}}} \right)\left( q \right){\left( {\frac{1}{p}} \right)^2}\left( {\frac{1}{s}} \right) = \frac{{7q}}{{5{p^2}s}}$

Standard 11

Physics

Figure shows graph between stress and strain for a uniform wire at two different femperatures. Then

medium

Two wires each of radius $0.2\,cm$ and negligible mass, one made of steel and other made of brass are loaded as shown in the figure. The elongation of the steel wire is $………\times 10^{-6}\,m$. [Young's modulus for steel $=2 \times 10^{11}\,Nm ^{-2}$ and $g =10\,ms ^{-2}$ ]

hard

(JEE MAIN-2023)

A force $F$ is applied on the wire of radius $r$ and length $L$ and change in the length of wire is $l.$ If the same force $F$ is applied on the wire of the same material and radius $2r$ and length $2L,$ Then the change in length of the other wire is

easy

A copper wire of length $2.2 \;m$ and a steel wire of length $1.6\; m ,$ both of diameter $3.0 \;mm ,$ are connected end to end. When stretched by a load, the net elongation is found to be $0.70 \;mm$. Obtain the load applied in $N$.