Two wires are made of the same material and have the same volume. However wire $1$ has cross sectional area A and wire $2$ has cross-sectional area $3A$. If the length of wire $1$ increases by $\Delta x$ on applying force $F$, how much force is needed to stretch wire $2$ by the same amount?
- 4 F
- 6 F
- 9 F
- F
The Correct Option is C
Solution and Explanation
For wire $1, Y=\frac{F / A}{\Delta x / 1} \ldots$ (i)
For wire $2, Y =\frac{ F ^{\prime} / 3 A }{\Delta x /(1 / 3)} \ldots$ (ii)
From (i) and (ii),
$\frac{F}{A} \times \frac{1}{\Delta x}=\frac{F^{\prime}}{3 A} \times \frac{1}{3 \Delta x} $
$\Rightarrow F^{\prime}=9 F$.
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Concepts Used:
Stress and Strain
Stress and Strain are the terms in physics, which are used to explain deformation of solids.
What is Stress?
Force applied per unit area is known as stress.
- σ=F/A
- σ is stress applied
- F is force applied
- A is that the area of force applied
- Stress is measured by unit N/m2
What is Strain?
As a result of stress, change of shape is observed in the body. The change or deformity consequential to the stress acting on the body is called strain. Strain can be defined as the amount or measure of deformity that takes place due to the force applied on the object.
Strain is denoted with (ε). It has no units.
Longitudinal Strain = Δ L/L
Relation Between Stress and Strain
The English scientist Robert Hooke, while studying spring and elasticity, noticed that many materials displayed an identical property when the stress-strain relationship was studied. There exists a linear region where the force required to stretch the material was proportional to the extension of the material; this is called Hooke’s law. Mathematically, the law is presented as:
F = -k.x
Where, F = the force
x = the extension length
k = spring constant in N/m