Question

A hypothetical experiment was conducted to determine Young's Formula $Y=\frac{T^x.\tau \cos \theta }{l^3}$. If $Y$ is young's modulus ($T$ is time, $\tau$ torque, $l=$ length). Find the value of $x$.
$\left(\right[Y]=M{L}^{-1}{T}^{-2},[\tau ]=M{L}^2{T}^{-2})$

• A. $1$
• B. $5$
• C. $0$
• D. $9$

Answer 1

The correct option is C $0$

It is given that, $Y=\frac{T^x.\tau \cos \theta }{l^3}$.... $\left(\text{i}\right)$
,also given that the dimensional formulae for Young's modulus and torque is $\left(\right[Y]=M{L}^{-1}{T}^{-2},[\tau ]=M{L}^2{T}^{-2})$

Dimensional formulae for length $\left[l\right]=L^1$, time $\left[T\right]=T^1$ and $\theta$ is dimensionless.

Thus equation $\left(\text{i}\right)$ becomes

$M{L}^{-1}{T}^{-2}=\frac{\left[T^1{]}^x\right[M{L}^2{T}^{-2}]}{[L{]}^3}$

$\Rightarrow M{L}^{-1}{T}^{-2}=M{L}^{-1}{T}^{-2+x}$

Comparing LHS and RHS in the above equation we have $-2+x=-2$
$\Rightarrow x=0$