Question

A stone of mass m is tied to one end of a wire of length L. The diameter of the wire is D and it is suspended vertically. The stone is now rotated in a horizontal plane and makes an angle$\theta $ with the vertical. If Young's modulus of the wire is Y, then the increase in the length of the wire is then
A. $\dfrac{{4mgL}}{{\pi {D^2}Y}}$
B. $\dfrac{{4mgL}}{{\pi {D^2}Y\sin \theta }}$
C. $\dfrac{{4mgL}}{{\pi {D^2}Y\cos \theta }}$
D. $\dfrac{{4mgL}}{{\pi {D^2}Ytan\theta }}$

Answer 1

Hint:-The wire is stretched due to the stone tied on it and hence must be tension there in wire. The wire is rotating and hence covering an area of circle. According to the Young’s modulus it is equal to the ratio of the stress acting on a substance to the strain produced. Using this we will find the strain in the wire and hence the increase in length will be calculated.

Complete step-by-step solution:Step 1:
We are given: A stone of mass m tied to one end of a wire of length L
The diameter of the wire is D and it is suspended vertically.
The stone is now rotated in a horizontal plane and makes an angle$\theta $ with the vertical and Young's modulus of the wire is Y.
From the above information a diagram is provided for better understanding. Let us see

Step 2:
According to the situation the wire is making angle$\theta $ with the vertical and we are rotating it to a horizontal plane.
Due to rotating it is making an ellipse or we can say a circle path is made by it to rotate.
There will be a downward force mg where, m is the mass and g is the gravity. The force tension must be applicable on the wire.
From here we can write T$\cos \theta $ = mg
Or we can write tension T =$\dfrac{{mg}}{{\cos \theta }}$ ……. (1)
First we need to find the stress in order to calculate the change in length.
Stress is defined as the force applied per unit area. Tension is a kind of force which we have found in equation 1, now we need to find the area covered by the stone.
The path covered by the stone is in the shape of circle and area of circle is $\pi {r^2}$ where, r is the radius of the circle
We are given D is the diameter so radius will be half of the diameter then, radius r = $\dfrac{D}{2}$
Putting this into the formula of area we get, area A = $\dfrac{{\pi {D^2}}}{4}$ …….. (2)
Substituting equation 1 and 2 in the definition of stress we get, $\dfrac{F}{A} = \dfrac{T}{A} = \dfrac{{4mg}}{{\pi {D^2}\cos \theta }}$
According to the Young’s modulus stress is directly proportional to strain. Y as young’s modulus is there for proportionality constant.
This implies $\dfrac{F}{A}\alpha \dfrac{{\Delta l}}{l}$ or $\dfrac{F}{A} = Y\dfrac{{\Delta l}}{l}$
Then, the strain is $\dfrac{{\Delta l}}{l} = \dfrac{{4mg}}{{Y\pi {D^2}\cos \theta }}$
This means the increase in the length of the wire is$\Delta l = \dfrac{{4mgl}}{{Y\pi {D^2}\cos \theta }}$
Hence option C is the correct solution.

Note:- The Young's Modulus of a material is a fundamental property of every material that cannot be changed. It is dependent upon temperature and pressure however. The Young's Modulus (or Elastic Modulus) is in essence the stiffness of a material. In other words, it is how easily it is bended or stretched.