Question

A hypothetical experiment conducted to determine Young's formula $Y=\frac{\cos \theta\, T^{x} \cdot \tau}{l^{3}} .$ If
$Y=$ Young's modulus,
$T=$ time period,
$\tau=$ torque and
$l=$ length, then find the value of $x$

• A. zero
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (A)

We have, $Y=\frac{\cos \theta \,T^{x} \cdot \tau}{l^{3}}\,\,\,$ ...(i)
We know, $[Y]=\left[M L^{-1}\, T^{-2}\right],[T]=\left[T^{1}\right]$
$[\tau]=\left[M L^{2}\, T^{-2}\right]$
$\theta=$ dimensionless
$[l]=[L]$
Thus, Eq. (i) becomes
$\left[M L^{-1} \,T^{-2}\right]=\frac{\left[T^{1}\right]^{x}\,\left[M L^{2}\, T^{-2}\right]}{\left[L^{3}\right]}$
or $\left[M L^{-1}\,T^{-2}\right]=\left[M L^{-1}\, T^{-2+x}\right]$
Comparing the powers, we have
$-2+x=-2$
$\therefore x=0$