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One end of a horizontal thick copper wire of length $2L$ and radius $2R$ is welded to an end of another horizontal thin copper wire of length $L$ and radius $R$. When the arrangement is stretched by applying forces at two ends, the ratio of the elongation in the thin wire to that in the thick wire is:
A) $0.25$
B) $0.50$
C) $2.00$
D) $4.00$

Last updated date: 17th May 2024
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Hint: The ratio of elongation is directly proportional to length, force and it is indirectly proportional to Young’s modulus and the square of radius. By applying Hooke’s Law, we have to find the elongation and substitute it according to the formula.

Formula Used:
We will be using the formula of Hooke’s Law i.e., stress $\alpha $ strain.

Complete step by step answer:
It is given that the first wire length is $2L$ and the radius is $2R$. For the second wire, the length is $L$ and the radius $R$. To find the ratio of elongation in thin wire to thick wire. The longitudinal stress and longitudinal strain will be produced when the forces are applied on the wire and they have been deformed.
Stress: The stress is defined as the net elastic force acting per unit areas of the surface subject to deformation. In simple, stress $ = \dfrac{{Force}}{{Area}}$
Strain: The strain is defined as the change in dimension of a body subject to deforming forces. It is a scalar quantity. In wire, longitudinal strain is produced.
${\text{strain = }}\dfrac{{{\text{change in length}}}}{{{\text{original length}}}}$
Hooke’s Law states that the stress is directly proportional to the strain. Under deforming forces, the more is the stress then will be the more in the strain.
Stress $\alpha $ Strain
$\dfrac{{{\text{Stress}}}}{{{\text{Strain}}}}{\text{ = Y}}$
The constant is known as the modulus of elasticity. For a change in length of wire, the constant is called Young’s modulus of elasticity
$ Y = \dfrac{{{\text{Longitudinal stress}}}}{{{\text{Longitudinal strain}}}}$
 $ \Rightarrow Y = \dfrac{{F \times L}}{{A \times l}}$
 Where l is the elongation of wire
$ \Rightarrow l = \dfrac{{F \times L}}{{A \times Y }}$
We know that, $A = \pi {r^2}$
$ \Rightarrow l = \dfrac{{F \times L}}{{\pi {r^2} \times Y }}$
For first wire, ${l_1} = \dfrac{{{F_1} \times {L_1}}}{{\pi r_1^2 \times {Y _1}}}$
For second wire, ${l_2} = \dfrac{{{F_2} \times {L_2}}}{{\pi r_2^2 \times {Y _2}}}$
Then the ratio of elongation will be,
$ \Rightarrow \dfrac{{{l_1}}}{{{l_2}}} = \dfrac{{{F_1}}}{{{F_2}}} \times \dfrac{{{L_1}}}{{{L_2}}} \times \dfrac{{{Y _1}}}{{{Y _2}}} \times \left[ {\dfrac{{r_2^2}}{{r_1^2}}} \right]$
Both wires are made up of the same copper wire, so the force and the Young modulus is the same for the two wires.
Also, we can write it as, $\dfrac{{{L_1}}}{{{L_2}}} = \dfrac{2}{1}$
$ \Rightarrow \dfrac{{{r_1}}}{{{r_2}}} = \dfrac{2}{1}$
Substitute these values,
$ \Rightarrow \dfrac{{{l_1}}}{{{l_2}}} = \dfrac{2}{1} \times \dfrac{1}{4}$
On cancel the term we get,
$ \Rightarrow \dfrac{{{l_1}}}{{{l_2}}} = \dfrac{1}{2} = 0.50$
Therefore, the ratio of elongation in the thin wire to that of the thick wire is $0.50$

Hence the correct option (B), $0.50.$

Note: Elasticity is defined as the property of material by which it regains its original position after the deforming force has been removed. The young’s modulus which is the property of the material depends on the material.