Question

A metal string is fixed between rigid supports. It is initially at negligible tension. Its young’s modulus is Y density is ρ and coefficient of linear expansion is α. It is now cooled through a temperature t transverse waves will move along it with a speed of:
$\begin{align}
  & A.\text{ }\sqrt{\dfrac{Y\alpha t}{\rho }} \\
 & B.\text{ }Y\sqrt{\dfrac{\alpha t}{\rho }} \\
 & C.\text{ }\alpha \sqrt{\dfrac{Yt}{\rho }} \\
 & D.\text{ }t\sqrt{\dfrac{\rho }{Y\alpha }} \\
\end{align}$

Answer 1

Hint: First we have to use wave velocity formula $v=\sqrt{\dfrac{T}{m}}$ in which we have to find tension (T) and mass (m) by using formula of young’s modulus, stress and strain which are acting on metal string.

Formula used:
$Y=\dfrac{\sigma }{\varepsilon }$
$\to $$v=\sqrt{\dfrac{T}{m}}$

Complete step by step solution:
$\to $We know that formula for wave velocity
$v=\sqrt{\dfrac{T}{m}}....\left( 1 \right)$
$\to $Where, v = wave velocity
$\to $T = tension
$\to $m = mass per length

$\to $ Now in order to find tension (T) we will use young’s modulus formula which is
$Y=\dfrac{\sigma }{\varepsilon }...\left( 2 \right)$
Where $\to $Y = young’s modulus
$\to $σ = stress
$\to $$\varepsilon $ = strain
$\to $Here σ is the stress which is generated due to tension force in metal string, l
$\to $Hence the formula for the stress is
$\sigma =\dfrac{force}{Area}=\dfrac{T}{A}....\left( 3 \right)$

$\to $Now here strain which is expressed as
$\begin{align}
  & \varepsilon =\dfrac{\text{change in the length}}{\text{original length}} \\
 & \varepsilon =\dfrac{\Delta l}{l} \\
\end{align}$
$\to $Here according to thermal expansion change in length will be
$\Delta l=l\alpha t$

$\to $Now strain,
$\begin{align}
  & \Rightarrow \varepsilon =\dfrac{l\alpha t}{l} \\
 & \therefore \varepsilon =\alpha t....\left( 4 \right) \\
\end{align}$

$\to $Now substitute value of equation (3) and equation (4) into equation (2) we will get
$\begin{align}
  & \Rightarrow Y=\dfrac{I}{\dfrac{A}{\alpha t}} \\
 & \therefore T=YA\alpha t.....\left( 5 \right) \\
\end{align}$

$\to $Formula for the mass per length is
$m=\dfrac{\rho .v}{l}$

$\to $Where volume = area × length
$\begin{align}
  & \Rightarrow m=\dfrac{\rho .A\times l}{l} \\
 & \therefore m=\rho A....\left( 6 \right) \\
\end{align}$

$\to $Now substitute value of equation (5) and equation (6) in equation (1)
$\begin{align}
  & \Rightarrow v=\sqrt{\dfrac{YA\alpha t}{\rho A}} \\
 & \therefore v=\sqrt{\dfrac{Y\alpha t}{\rho }} \\
\end{align}$

Hence option (A) $\sqrt{\dfrac{Y\alpha t}{\rho }}$ is correct.

Additional information:
Definition of stress:
$\to $Stress is the force acting on the unit area of a material. We can simply say that when we apply load on any material or body the body is resisting the force which applied on the body or the area of the body. This resistance offered by the body or any material is called stress.
Definition of strain:
$\to $It is defined as the amount of deformation experienced by the body in the direction of force applied, divided by initial dimensions of the body.
Definition Young's Modulus:
$\to $It is in essence the stiffness of a material. Or we can say it is a of stress and strain

Note:
When we are using expression for mass per length divide the equation by l because we are finding mass per length (l) for example don’t put $m=\rho .v$ the correct way is $m=\dfrac{\rho .v}{l}$.