Question

Two wires A and b are of the same materials. Their lengths are in the ratio \[1:2\] and the diameters are in the ratio $2:1$. When stretched by force ${F_A}$ and ${F_B}$ respectively they get equal increase in their lengths. Then the ratio $\dfrac{{{F_A}}}{{{F_B}}}$ should be:
A) $1:2$
B) $1:1$
C) $2:1$
D) $8:1$

Answer 1

Hint: In order to get the solution of the given question, we need to use the Hooke’s law for both the wires A and B. We can relate the values of force from Hooke’s law as the ratios of length and diameter is given. Finally, after solving the equation framed, we can conclude with the required solution of the given question.

Complete step by step solution:
In the question the ratios, of the length is given as,$\dfrac{{{l_A}}}{{{l_B}}} = \dfrac{1}{2}$ and the ratios of the diameters of the wires are given as,$\dfrac{{{D_A}}}{{{D_B}}} = \dfrac{2}{1}$.
As from the Hooke’s law we know that, $Y = \dfrac{{Fl}}{{\Delta lA}}$
Area,$A = \pi \dfrac{{{D^2}}}{4}$
Now, we need to write Hooke’s law for both the wires.
For wire A,$Y = \dfrac{{{F_A}{l_A}}}{{\Delta l\pi \dfrac{{{D^2}}}{4}}}$
$ \Rightarrow {F_A} = \dfrac{{Y\pi {D_A}^2\Delta l}}{{4{l_A}}}$
Similarly, for wire B, the force can be written as ${F_B} = \dfrac{{Y\pi {D_B}^2\Delta l}}{{4{l_B}}}$
Now, we need to take the ratio of forces on both the wires.
Therefore,$\dfrac{{{F_A}}}{{{F_B}}} = \dfrac{{Y\pi {D_A}^2\Delta l}}{{4{l_A}}} \times \dfrac{{4{l_B}}}{{Y\pi {D_B}^2\Delta l}}$
$ \Rightarrow \dfrac{{{F_A}}}{{{F_B}}} = \dfrac{{D_A^2}}{{D_B^2}} \times \dfrac{{{l_B}}}{{{l_A}}}$
$ \Rightarrow \dfrac{{{F_A}}}{{{F_B}}} = \dfrac{{{2^2}}}{{{1^2}}} \times \dfrac{2}{1}$
$\therefore \dfrac{{{F_A}}}{{{F_B}}} = \dfrac{8}{1}$
Therefore, the required ratio, $\dfrac{{{F_A}}}{{{F_B}}}$=$8:1$

Hence, option (D), i.e. $8:1$ is the correct choice for the given question.

Note: Hooke’s law is also known as the law of elasticity. According to Hooke’s law when the deformation in a body is very small, then the displacement is directly proportional to the force applied to deform the body. We can express Hooke’s law in the form of stress and strain also. The expression of Hooke’s law can also be given as, $F = - kx$ where $'k'$ is a constant. With the help of Hooke’s law we can define elasticity of a material. Hooke’s law also helps us to understand the behavior of a body when it is compressed or stretched.