Question

A 20 kg load is suspended from the lower end of a wire 10 cm long and 1 mm2 in cross sectional area. The upper half of the wire is made of iron and the lower half with aluminum. The total elongation in the wire is
$({Y_{iron}} = 20 \times {10^{10}}N/{m^2},{Y_{Al}} = 7 \times {10^{10}}N/{m^2})$
(A) $1.92 \times {10^{ - 4}}m$
(B) $17.8 \times {10^{ - 3}}m$
(C) $1.78 \times {10^{ - 3}}m$
(D) $1.92 \times {10^{ - 3}}m$

Answer 1

Hint
Here, we use the formula of Young’s modulus given by, $Y = \dfrac{{F/A}}{{\Delta L/L}} = \dfrac{{F \times L}}{{A \times \Delta L}}$, where F/A is the force applied per unit area (which is called stress), and ΔL/L is the change in length per unit length (which is the longitudinal strain).

Complete step by step answer
Given that, mass of the load, m = 20 kg;
Length of the wire, $L = 10 cm = 10^{-1}$ m;
Cross sectional area of the wire, $A = 1 mm^2 = 10^{-6} m^2$.
Rearrange the formula of Young’s modulus in terms of ΔL.
$\Rightarrow Y = \dfrac{{F/A}}{{\Delta L/L}} = \dfrac{{F \times L}}{{A \times \Delta L}}\\\ \Rightarrow \Delta L = \dfrac{{F \times L}}{{A \times Y}}$
Given that, the upper half of the wire is made of iron. Let, $ΔL_i$ represent the change in length of the iron part.
The lower half of the wire is made of aluminum. Let, $ΔL_A$ represent the change in length of the aluminum part.
Therefore,
$\Rightarrow \Delta {L_i} = \dfrac{{F{L_i}}}{{A{Y_i}}}$, $\Delta {L_A} = \dfrac{{F{L_A}}}{{A{Y_A}}}$and Li = LA = 0.05 m
Total elongation in the wire is,
$\Rightarrow \Delta L = \Delta {L_i} + \Delta {L_A}$
$\Rightarrow \Delta L = \dfrac{F}{A}(\dfrac{{{L_i}}}{{{Y_i}}} + \dfrac{{{L_A}}}{{{Y_A}}})$ … (1)
Substitute force, $F = mg = 20kg \times 10m/{s^2} = 200N$.
Put the given values in equation (1).
$\Rightarrow \Delta L = \dfrac{{200N}}{{{{10}^{ - 6}}{m^2}}} \times (\dfrac{{0.05m}}{{20 \times {{10}^{10}}N/{m^2}}} + \dfrac{{0.05m}}{{7 \times {{10}^{10}}N/{m^2}}})\\\ \Rightarrow \Delta L = 200 \times {10^6} \times (25 + 71) \times {10^{ - 14}}m\\\ \Rightarrow \Delta L = 200 \times {10^{ - 8}} \times 96m\\\\\therefore \Delta L = 1.92 \times {10^{ - 4}}m$
Thus, the total elongation in the wire is$1.92 \times {10^{ - 4}}m$, which is the option (A).

Note
Young’s modulus is the measure of the ability of a material to withstand changes in length when under lengthwise tension or compression. This is also referred to as the modulus of elasticity. It is equal to the ratio of longitudinal stress to the longitudinal strain.