Two bars A and B of circular section and the same volume and made of the same material are subjected to tension. If the diameter of A is half that of B and if the force supplied to both the rods is the same and is within the elastic limit, the ratio of extension of A to that of B will be-
(A) $16$
(B) $8$
(C) $4$
(D) $2$

Hint The young’s modulus defines the elasticity or the stress required to produce a given amount of strain for a material. It is the ratio of stress applied on the body to the strain produced in it. Same volume of the bars while having different diameters would mean that the length of bars is not the same, thus the elongation will also be different.

Complete Step by step solution
Let the diameter of rod A be $d$.
Then the diameter of rod B would be $2d$
Let the force applied to both of the rods be $F$
Let the area of rod A be ${a_1}$
And let the area of rod B be ${a_2}$
Therefore,
${a_1} = \dfrac{\pi }{4}{d^2}$
${a_2} = \dfrac{\pi }{4}{(2d)^2}$
${a_2} = \dfrac{\pi }{4} \times 4{d^2} = \pi {d^2}$
Comparing ${a_1}$ and ${a_2}$, we get,
$4{a_1} = {a_2}$
Stress is defined as the force applied by the molecules of a material when it is subjected to a deformation. The stress in an object is given by the applied force per unit area.
So in an object, applied stress$\left( \sigma \right)$ is given by-
$\sigma = \dfrac{F}{a}$
In rod A, the stress is-
${\sigma _1} = \dfrac{F}{{{a_1}}}$
\[{\sigma _1} = \dfrac{F}{{\dfrac{{\pi {d^2}}}{4}}} = \dfrac{{4F}}{{\pi {d^2}}}\]
In rod B, the stress is-
${\sigma _2} = \dfrac{F}{{{a_2}}}$
${\sigma _2} = \dfrac{F}{{\pi {d^2}}}$
The strain is defined as the deformation produced in a body after a stress is applied to it. It is represented as a ratio of change in length to the previous length of the object.
Let an object of length L be elongated by amount $\Delta L$then the strain produced in it is given by-
$\varepsilon = \dfrac{{\Delta l}}{l}$
It is given that the volume of both rods is the same.
Volume V is given by-
$V = la$
Where $l$is the length of rod and $a$is the area of the rod.
Let ${l_1}{a_1}$be the volume of rod A and ${l_2}{a_2}$be the volume for rod B, then
${l_1}{a_1} = {l_2}{a_2}$
$\dfrac{{{l_1}}}{{{l_2}}} = \dfrac{{{a_2}}}{{{a_1}}}$
We know that,
$4{a_1} = {a_2}$
Therefore, we get-
$\dfrac{{{l_1}}}{{{l_2}}} = \dfrac{{4{a_1}}}{{{a_1}}}$
${l_1} = 4{l_2}$
The young’s modulus (Y) is defined as the ratio of stress to strain. It is a property of material, so it will be same for both rods-
\[Y = \dfrac{\sigma }{\varepsilon } = \dfrac{{F/a}}{{\Delta l/l}} = \dfrac{{Fl}}{{\Delta la}}\]
Equating the young’s modulus for both rods=
${Y_1} = {Y_2}$
$\dfrac{{F{l_1}}}{{\Delta {l_1}{a_1}}} = \dfrac{{F{l_2}}}{{\Delta {l_2}{a_2}}}$
$\dfrac{{\Delta {l_1}}}{{\Delta {l_2}}} = \dfrac{{{l_1}{a_2}}}{{{l_2}{a_1}}}$
On putting the values we get,
$\dfrac{{\Delta {l_1}}}{{\Delta {l_2}}} = \dfrac{{4{l_2} \times 4{a_2}}}{{{l_2} \times {a_2}}}$
$\dfrac{{\Delta {l_1}}}{{\Delta {l_2}}} = \dfrac{{16}}{1}$
Therefore the ratio of extension of rod A is $16$.

Option A is the correct answer.

Note The change in length of a rod is directly proportional to the length of the rod and inversely proportional to the change in its area, since area is given by the square of the diameter, the change in length is inversely proportional to the square of change in diameter.