Question

The young’s modulus of steel is twice that of brass, two of sample length and same area of cross-section, one of steel and one of brass one suspended from the same roof. If we want the lower ends of the wires to be at the same level, then the weight added to the steel and brass wires must be in the ratio of:
(A) $1:1$
(B) $1:3$
(C) $2:1$
(D) $4:1$

Answer 1

Hint:Young’s modulus is a mechanical property that measures the stiffness of a solid material. It defines the relationship between stress and strain in a material. It is a measurement of the ability of a material to withstand changes in length when under lengthwise tension or compression. If I equal longitudinal stress divided by strain. ${\text{E}}\;{\text{ = }}\dfrac{\sigma }{ \in },$ where E is young’s modulus, $\sigma $ is longitudinal stress and E is strain. The SI unit of young’s modulus is Pascal and it’s dimension is ${\text{M}}{{\text{L}}^{ - 1}}{{\text{T}}^{ - 2}}$

Complete step by step answer:
We have been told that the lower ends wire one same, which the elongation of the wires is same,
$\Delta L = \Delta {L_S} = \Delta {L_B}$
Where $\Delta L$ is the elongation, $\Delta {L_S}$ and $\Delta {L_B}$ is elongation of steel and brass respectively.
We also know that their cross-section areas and lengths are equal,
$
  A = {A_S} = {A_B} \\
  l = {l_S} = {l_B} \\
$
${A_s}$ = Area of cross section steel
${A_B}$ = Area of cross section steel
${l_S}$ = Length of steel wire
${l_B}$ = Length of steel wire

We know,
\[Y = \dfrac{\sigma }{ \in } = \dfrac{{\dfrac{F}{A}}}{{\dfrac{{\Delta L}}{l}}}\]
\[Where{\text{ }}F{\text{ }} = {\text{ }}force\], \[A{\text{ }} = {\text{ }}area,\] \[\Delta L = elongation{\text{ }}and{\text{ }}length{\text{ }}of{\text{ }}wire,\]
Now,
 \[\dfrac{{\dfrac{F}{A}}}{{\dfrac{{\Delta L}}{l}}} = \dfrac{F}{A} \times \dfrac{l}{{\Delta L}} = \dfrac{{w\Delta L}}{{A\Delta L}}\left[ {w = weight} \right]\]
Given, ${Y_S} = 2{Y_B}$
$ \Rightarrow \dfrac{{{W_S}\Delta L}}{{A\Delta L}} = 2\dfrac{{{W_B}\Delta L}}{{A\Delta L}}$
$ \Rightarrow \dfrac{{{W_S}}}{{{W_B}}} = \dfrac{2}{1}$
${W_S}$ = weight added in steel wire
${W_B}$ = weight added in brass wire

The answer is (C) $2:1$

Note: Stress is defined as the force acting on the unit area of a material of how much an object is deformed. Now, because of young’s modulus is the ratio of stress and strain, we can say that young’s modulus defines the stiffness of a material. This means if an object’s young’s modulus is higher we have to put greater force to deform that object, in other words, that object is very durable.