Question

Using dimensional analysis, find the value of $x$ in the relation, $Y=\frac{T^x.\cos \theta .\tau }{L^3}$ where, $Y$ is Young's Modulus, $T$ is the time period, $\tau$ is torque and $L$ is length.

• A. $0$
• B. $\frac{1}{2}$
• C. $1$
• D. $2$

Answer 1

The correct option is A $0$

The Young's modulus is given by
Young's Modulus $=\frac{\text{Stress}}{\text{Strain}}$
$\Rightarrow Y=\frac{\frac{Force\left(F\right)}{\text{Area (A)}}}{\frac{\text{change in length}\left(\Delta L\right)}{\text{Original length}\left(L_0\right)}}=\frac{\frac{F}{A}}{\frac{\Delta L}{L_0}}$
$\Rightarrow Y=\frac{F{L}_0}{A\Delta L}$
The dimensions of respective quantities are
$\left[F\right]=\left[M^1{L}^1{T}^{-2}\right]$
$\left[A\right]=\left[M^0{L}^2{T}^0\right]$
$\left[\Delta L\right]=\left[M^0{L}^1{T}^0\right]$
$\left[L_0\right]=\left[M^0{L}^1{T}^0\right]$
$\left[Y\right]=\frac{\left[M^1{L}^1{T}^{-2}\right]}{\left[M^0{L}^2{T}^0\right]}\times \frac{\left[M^0{L}^1{T}^0\right]}{\left[M^0{L}^1{T}^0\right]}$
$\left[Y\right]=\left[M^1{L}^{-1}{T}^{-2}\right]$

Also as $T$ is the time period $\left[T\right]=\left[T^1\right]$
As $L$ is the length , $\left[L\right]=\left[L^1\right]$
Here $\tau$ is the torque, $\tau =F.r\sin \theta$
$\left[\tau \right]=\left[M^1{L}^1{T}^{-2}\right]\left[L^1\right]=\left[M^1{L}^2{T}^{-2}\right]$
And here $\cos \theta$ is dimensionless
The given relation is :
$Y=\frac{T^x\cos \theta .\tau }{L^3}$
$\Rightarrow T^x=\frac{Y{L}^3}{\cos \theta .\tau }$
$\Rightarrow \left[T^x\right]=\frac{\left[M^1{L}^{-1}{T}^{-2}\right]\left[L^3\right]}{\left[M^1{L}^2{T}^{-2}\right]}=\frac{\left[M^1{L}^2{T}^{-2}\right]}{\left[M^1{L}^2{T}^{-2}\right]}$
$\Rightarrow \left[T^x\right]=\left[M^0{L}^0{T}^0\right]$
$\Rightarrow x=0$

Hence, option (a) is correct.