If the velocity \((V)\), acceleration \((A)\) and force \((F)\) are taken as fundamental quantities instead…

If the velocity \((V)\), acceleration \((A)\) and force \((F)\) are taken as fundamental quantities instead of mass \((M)\), length \((L)\) and time \((T)\), the dimensions of Young's modulus \((Y)\) would be
  1. \(F A^{2} V^{-4}\)
  2. \(F A^{2} V^{-5}\)
  3. \(F A^{2} V^{-3}\)
  4. \(F A^{2} V^{-2}\)

Solution

Let \(Y=\left[V^{a} A^{b} F^{c}\right]\)
$\begin{aligned} [M L^{-1} T^{-2}] &= [L T^{-1}]^{n}[L T^{-2}]^{b}[M L T^{-2}]^{c} \\ M L^{-1} T^{-2} &= M^{c} L^{a+b+c} T^{-a-2 b-2 c} \end{aligned}$ \(\therefore c=1, a+b+c=-1,-a-2 b-2 c=-2\)
On solving, we get \(a=-4, b=2\) and \(c=1\)

Asked in: JEE Mains - Units and Dimensions - Chapter Test

Practice more Units and Dimensions questions on Aicharya