Question

If the ratio of lengths, radii and Young’s moduli of steel and brass wires are \[a,{\text{ }}b\] and \[c\] respectively, their respective loads are in the ratio\[3:2\] , then the corresponding ratio of increase in their lengths would be
A. \[\dfrac{{2{a^2}c}}{b}\]
B. $\dfrac{{3a}}{{2{b^2}c}}$
C. $\dfrac{{3ac}}{{{b^2}}}$
D. $\dfrac{{3c}}{{2a{b^2}}}$

Answer 1

Hint:-In this question, we can find the Young’s modulus for the steel wire and brass wire separately. We can change the equations in the terms of change in lengths and then find the ratio by substituting the given values.

Complete step-by-step solution:
According to the question, we have the
Ratio of lengths of steel and brass wires=$\dfrac{{{L_s}}}{{{L_b}}} = a$
Ratio of radii of the steel and brass wires= $\dfrac{{{r_s}}}{{{r_b}}} = b$
Or the ratio of cross-section areas of steel and brass wires=\[\dfrac{{{A_s}}}{{{A_b}}} = \dfrac{{{r_s}^2}}{{{r_b}^2}} = {b^2}\]
Ratio of Young’s modulus of steel and brass wires=$\dfrac{{{Y_s}}}{{{Y_b}}} = c$
Ratio of respective loads on the steel and brass wires=$\dfrac{{{F_s}}}{{{F_b}}} = \dfrac{3}{2}$
We know that the if \[F\] is the load on the wire and \[L\] is the length of the wire and \[A\] is the cross-section area of the wire then Young’s Modulus \[Y\] is given as
$Y = \dfrac{{FL}}{{A\Delta L}}$
(Where $\Delta L$ is the change in the length of the wire applying the load on the wire.)
Or we can write-
$\Delta L = \dfrac{{FL}}{{AY}}$
Now, let the change in length of the lengths in the steel and brass wires are $\Delta {L_s}$and $\Delta {L_b}$.
Then for the steel wire, we have the Young’s modulus-
\[\Delta {L_s} = \dfrac{{{F_s}{L_s}}}{{{A_s}{Y_s}}}\]
And for the brass wire, we have the Young’s modulus-
\[\Delta {L_b} = \dfrac{{{F_b}{L_b}}}{{{A_b}{Y_b}}}\]
So, the ratio of $\Delta {L_s}$and $\Delta {L_b}$will be given as-
\[
  \dfrac{{\Delta {L_s}}}{{\Delta {L_b}}} = \dfrac{{{F_s}{L_s}{A_b}{Y_b}}}{{{F_b}{L_b}{A_s}{Y_s}}} \\
   \Rightarrow \dfrac{{\Delta {L_s}}}{{\Delta {L_b}}} = = \left( {\dfrac{{{F_s}}}{{{F_b}}}} \right)\left( {\dfrac{{{L_s}}}{{{L_b}}}} \right)\left( {\dfrac{{{A_b}}}{{{A_s}}}} \right)\left( {\dfrac{{{Y_b}}}{{{Y_s}}}} \right) \\
\]
Putting the values, we get-
\[
   \Rightarrow \dfrac{{\Delta {L_s}}}{{\Delta {L_b}}} = \left( {\dfrac{3}{2}} \right)\left( a \right)\left( {\dfrac{1}{{{b^2}}}} \right)\left( {\dfrac{1}{c}} \right) \\
   \Rightarrow \dfrac{{\Delta {L_s}}}{{\Delta {L_b}}} = \dfrac{{3a}}{{2{b^2}c}} \\
\]
Hence, option B is correct.

Note:- In this question, we have to keep in mind that the ratio between the radii of steel wire and brass wire is given but in the formula of Young’s modulus, we have area. So, we have to find the ratio between the areas of the wires.