Question

A steel wire is suspended vertically from a rigid support. When loaded with a weight in the air, it is extended by ${l_a}$ and when the weight is immersed completely in water, the extension is reduced to ${l_w}$. Then the relative density of the material of the weight is:
A. $\dfrac{{{l_a}}}{{{l_w}}}$
B. $\dfrac{{{l_a}}}{{{l_a} - {l_w}}}$
C. $\dfrac{{{l_w}}}{{({l_a} - {l_w})}}$
D. $\dfrac{{{l_w}}}{{{l_a}}}$

Answer 1

Elasticity is the property of a material by the virtue of which it opposes the change in its shape size or length and regains its initial state as the applied force is removed. According to Hooke's law, if the strain is small, the stress produced in the body is proportional to the strain. Modulus of elasticity is defined for the material as the ratio of stress and strain.

Complete step by step answer:
It is given that the extended length in air is ${l_a}$ and the reduced length in water is ${l_w}$. We have to find the relative density of the weight.The ratio of longitudinal stress to longitudinal strain is called young's modulus of elasticity. Young modulus is given by the following formula.
$Y = \dfrac{{Fl}}{{A\Delta l}}$
Here, $F$ is the force, $l$ is the initial length of the wire, $A$ is the area of cross-section, and $\Delta l$ is the change in the length of wire.

So, we can write the young's modulus for both air and water.
$Y = \dfrac{{Fl}}{{A{l_a}}} = \dfrac{{mgl}}{{A{l_a}}}$…………….(1)
$\Rightarrow Y = \dfrac{{{F^1}l}}{{A{l_w}}} \\\ \Rightarrow Y = \dfrac{{\left( {mg - \rho Vg} \right)}}{{A{l_w}}}$ ……………..(2)
Now, we know that young's modulus is constant for the material, i.e., equation (1) is equal to equation (2).
$\dfrac{{mgl}}{{A{l_a}}} = \dfrac{{\left( {mg - \rho Vg} \right)}}{{A{l_w}}}$
Let us further simplify this.
$\dfrac{{ml}}{{{l_a}}} = \dfrac{{\left( {m - \rho V} \right)}}{{{l_w}}} \\\ \Rightarrow \rho = \dfrac{m}{V} \times \left( {1 - \dfrac{{{l_w}}}{{{l_a}}}} \right)$
We know, density is mass divided by volume.
$\rho = {\rho ^1} \times \left( {1 - \dfrac{{{l_w}}}{{{l_a}}}} \right) \\\ \therefore \dfrac{\rho }{{{\rho ^1}}} = \dfrac{{{l_a}}}{{{l_a} - {l_w}}}$

Hence, the correct option is B.

Note: When an object is hanging the force acting on it is its weight $mg$.In the water, buoyancy force acts on an object which is equal to the weight of the object minus the upward thrust by the water.