Question

A load of mass $M kg$ is suspended from a steel wire of length $2m$ and radius $1.0mm$ in Searle's apparatus experiment. The increase in length produced in the wire is $4.0mm$. Now the load is fully immersed in a liquid of relative density 2. The relative density of the material of load is 8. The new value of the increase in the length of the steel wire is:
(A) 4 mm
(B) 3 mm
(C) 5 mm
(D) Zero

Answer 1

Hint
When the load is fully immersed in a liquid, the suspended steel wire will elongate to a certain amount. To find the increase in length, we have to compare the two cases using the formula for the Young’s modulus.
Formula used : In the solution we will be using the following formula,
$\dfrac{F}{A} = Y\left( {\dfrac{{\Delta L}}{L}} \right)$
Where ${\text{Y }}$is the Young’s modulus,
$F$ is the force and $A$ is the area of cross-section.
$L$ is the length of the wire and $\Delta L$ is the increase in length.

Complete step by step answer
In the question it is given that ΔL = 4.0 mm,
From the formula for the Young’s modulus we have,
$\dfrac{F}{A} = Y\left( {\dfrac{{\Delta L}}{L}} \right)$
Here the force on the wire will be the weight of the mass given as $Mg$
When we substitute this value in the above formula,
$\dfrac{{Mg}}{A} = Y\left( {\dfrac{{\Delta L}}{L}} \right)$
Now the mass can be written in the form of the density and volume as, $M = 8\rho v$
$\dfrac{{\left( {8\rho v} \right)g}}{A} = Y\left( {\dfrac{{\Delta L}}{L}} \right)$
And when in water the mass will be, $M = 8\rho v - 2\rho v$
$\dfrac{{\left( {8\rho vg - 2\rho vg} \right)}}{A} = {\text{Y}}\left( {\dfrac{{\Delta L'}}{L}} \right)$
Here the increase in length is $\Delta L'$
$\dfrac{{\left( {6\rho vg} \right)}}{A} = Y\left( {\dfrac{{\Delta L'}}{L}} \right)$
On dividing the equations, we get,
$\dfrac{{\dfrac{{\left( {8\rho v} \right)g}}{A}}}{{\dfrac{{\left( {6\rho vg} \right)}}{A}}} = \dfrac{{Y\left( {\dfrac{{\Delta L}}{L}} \right)}}{{Y\left( {\dfrac{{\Delta L'}}{L}} \right)}}$
On cancelling all the common terms we get,
$\dfrac{8}{6} = \dfrac{{\Delta L}}{{\Delta L'}}$
That is we have,
$\dfrac{4}{3} = \dfrac{{\Delta L}}{{\Delta L'}}$
So we can find the increase in length in the water
$\Delta L' = \left( {\dfrac{3}{4}} \right)\Delta L$
On substituting the value $\Delta L = 4mm$
$\Delta L' = \left( {\dfrac{3}{4}} \right) \times 4$
So we get,
$\Delta L' = 3mm$
Thus, the new value of the increase in the length of the steel wire is 3 mm.
Hence, the correct answer is option (B).

Note
The Young’s modulus is also called the modulus of elasticity. It measures the tensile stiffness of a solid material. It gives the relationship between the tensile stress and the axial strain. It has a unit of Newton per meter square.