If speed (V), acceleration (A) and force (F) are considered as fundamental units, the dimension of Young's…

If speed (V), acceleration (A) and force (F) are considered as fundamental units, the dimension of Young's modulus will be :
  1. $\mathrm{V}^{-2} \mathrm{~A}^{2} \mathrm{~F}^{-2}$
  2. $\mathrm{V}^{-2} \mathrm{~A}^{2} \mathrm{~F}^{2}$
  3. $\mathrm{V}^{-4} \mathrm{~A}^{-2} \mathrm{~F}$
  4. $\mathrm{V}^{-4} \mathrm{~A}^{2} \mathrm{~F}$

Solution

Let [Y] = [V] $^{a}[\mathrm{~F}]^{\mathrm{b}}[\mathrm{A}]^{\mathrm{c}}$ $\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]=\left[\mathrm{LT}^{-1}\right]^{\mathrm{a}}\left[\mathrm{MLT}^{-2}\right]^{\mathrm{b}}\left[\mathrm{LT}^{-2}\right]^{\mathrm{c}}$ $\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]=\left[\mathrm{M}^{b} \mathrm{~L}^{\mathrm{a}+b+\mathrm{c}} \mathrm{T}^{-\mathrm{a}-2 \mathrm{~b}-2 \mathrm{c}}\right]$ Comparing power both side of similar terms we get, $b=1, a+b+c=-1,-a-2 b-2 c=-2$ solving above equations we get $a=-4, b=1, c=2$ so $[\mathrm{Y}]=\left[\mathrm{V}^{-4} \mathrm{FA}^{2}\right]=\left[\mathrm{V}^{-4} \mathrm{~A}^{2} \mathrm{~F}\right]$

Asked in: JEE Main 2019 (11 Jan Shift 2)

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