Question

One end of a uniform wire of length L and of weight W is attached rigidly to a point in the roof and ${{W}_{1}}$ weight is suspended from the looser end. If A is area of cross-section of the wire, the stress in the wire at a height $\dfrac{L}{4}$ from the upper end is
a)$\dfrac{{{W}_{1}}+W}{a}$
b)$\dfrac{{{W}_{1}}+\dfrac{3W}{4}}{a}$
c)$\dfrac{{{W}_{1}}+\dfrac{W}{4}}{a}$
d)$\dfrac{4{{W}_{1}}+3W}{a}$

Answer 1

Hint: Stress is equal to force per unit area i.e. $\sigma =\dfrac{F}{A}$. For the given wire, we have A as area of cross section, and tension force is in equilibrium with the weight of wire. So, we can say that F = W. Now, firstly find the tension force in the wire at a height $\dfrac{L}{4}$ from the upper end, that is equal to the tension force in the wire at a height $\dfrac{L}{4}$ from the lower end. Then, by using the relation $\sigma =\dfrac{F}{A}$, find the stress in the wire at a height $\dfrac{L}{4}$ from the upper end.

Complete step by step answer:As we know that,
For L length of wire, weight = W
So, for length $\dfrac{3L}{4}$, the weight of wire $=\dfrac{W}{L}\times \dfrac{3L}{4}=\dfrac{3W}{4}$
So, total weight at length $\dfrac{3L}{4}$ is $\dfrac{3W}{4}+{{W}_{1}}$
Also, we know that:
Stress $\sigma =\dfrac{F}{A}$ , so, we get:
\[\sigma =\dfrac{\dfrac{3W}{4}+{{W}_{1}}}{a}\]
Hence, option (b) is the correct answer.

Note:Stress is defined as “The restoring force per unit area of the material”. It is a tensor quantity. Denoted by Greek letter $\sigma $ . Measured using Pascal or $N/{{m}^{2}}$.
Where,
a) F is the restoring force measured in Newton or N.
b) A is the area of cross-section measured in${{m}^{2}}$.
c) $N/{{m}^{2}}$ is the stress measured using $N/{{m}^{2}}$ or Pa.