Question

If the ratio of lengths, radii and Young’s modulus of steel and brass wire shown in the figure are a, b and c respectively, the ratio between the increase in lengths of brass and steel wires would be:

& A.\,\dfrac{{{b}^{2}}}{2c} \\\ & B.\,\dfrac{bc}{2{{a}^{2}}} \\\ & C.\,\dfrac{b{{a}^{2}}}{2c} \\\ & D.\,\dfrac{2{{b}^{2}}c}{a} \\\ \end{aligned}$$

Answer 1

Young’s modulus of a material is the ratio of the product of the force and length by the product of the area and the change in the length. The tension in the wire will be equal to the weight of the wire. The tension in the brass wire equals twice the weight of the wire, as the tension in the brass wire equals the tension due to the brass wire and the steel wire.

Formula Used
$Y=\dfrac{Fl}{A\Delta l}$

Complete step-by-step solution:
From the given information, we have the data as follows.
Young’s modulus of a material is the ratio of the product of the force and length by the product of the area and the change in the length.
$Y=\dfrac{Fl}{A\Delta l}$
Where Y is Young’s modulus, F is the force or the tension, l is the length, A is the area and $\Delta l$is the change in the length.
Now, we need to represent the above equation in terms of the change in the length, so, we have, $\Delta l=\dfrac{Fl}{AY}$
We will consider the different expressions for the steel and the brass materials.
The change in the length of the brass wire is given as follows.
$\Delta {{l}_{b}}=\dfrac{{{F}_{b}}{{l}_{b}}}{{{A}_{b}}{{Y}_{b}}}$…… (1)
The change in the length of the steel wire is given as follows.
$\Delta {{l}_{s}}=\dfrac{{{F}_{s}}{{l}_{s}}}{{{A}_{s}}{{Y}_{s}}}$…… (2)
Divide the equations (1) and (2).

& \dfrac{\Delta {{l}_{b}}}{\Delta {{l}_{s}}}=\dfrac{\dfrac{{{F}_{b}}{{l}_{b}}}{{{A}_{b}}{{Y}_{b}}}}{\dfrac{{{F}_{s}}{{l}_{s}}}{{{A}_{s}}{{Y}_{s}}}} \\\ & \therefore \dfrac{\Delta {{l}_{b}}}{\Delta {{l}_{s}}}=\dfrac{{{T}_{b}}}{{{T}_{s}}}\times \dfrac{{{l}_{b}}}{{{l}_{s}}}\times \dfrac{{{A}_{s}}}{{{A}_{b}}}\times \dfrac{{{Y}_{s}}}{{{Y}_{b}}} \\\ \end{aligned}$$ Consider the block diagrams of the steel and brass wires and the tensions in the wires. The ratio of the tensions in the strings is given as follows. $$\dfrac{{{T}_{b}}}{{{T}_{s}}}=\dfrac{2mg}{mg}=2$$ Substitute the values in the above equations. $$\begin{aligned} & \dfrac{\Delta {{l}_{b}}}{\Delta {{l}_{s}}}=2\times \dfrac{1}{a}\times {{b}^{2}}\times c \\\ & \therefore \dfrac{\Delta {{l}_{b}}}{\Delta {{l}_{s}}}=\dfrac{2{{b}^{2}}c}{a} \\\ \end{aligned}$$ $$\therefore $$The ratio of the change in the lengths of the brass and steel wires is$$\dfrac{2{{b}^{2}}c}{a}$$, thus, option (D) is correct. **Note:** The tension in the brass wire equals twice the weight of the wire, as the tension in the brass wire equals the tension due to the brass wire and the steel wire. The area of the wires should be considered in terms of the radius, as the ratio of radii is given.