If speed $(\mathrm{V})$, acceleration $(\mathrm{A})$ and force $(\mathrm{F})$ are considered as fundamental units, the dimension of Young’s modulus will be :

Question:

If speed $(\mathrm{V})$, acceleration $(\mathrm{A})$ and force $(\mathrm{F})$ 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. $\mathrm{V}^{-4} \mathrm{~A}^{2} \mathrm{~F}$

JEE Main Previous Year Single Correct Question of JEE Main from Chemistry Laws of Motion chapter.

JEE Main Previous Year 2019


Correct Option: 4

Solution:

$$

\begin{aligned}

&\text { Let }[\mathrm{Y}]=[\mathrm{V}]^{\mathrm{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}^{\mathrm{b}} \mathrm{L}^{\mathrm{a}+\mathrm{b}+\mathrm{c}} \mathrm{T}^{-\mathrm{a}-2 \mathrm{~b}-2 \mathrm{c}}\right]}

\end{aligned}

$$

Comparing power both side of similar terms we get,

$\mathrm{b}=1, \mathrm{a}+\mathrm{b}+\mathrm{c}=-1,-\mathrm{a}-2 \mathrm{~b}-2 \mathrm{c}=-2$

solving above equations we get:

$$

\begin{aligned}

&\mathrm{a}=-4, \mathrm{~b}=1, \mathrm{c}=2 \\

&\text { so }[\mathrm{Y}]=\left[\mathrm{V}^{-4} \mathrm{FA}^{2}\right]=\left[\mathrm{V}^{-4} \mathrm{~A}^{2} \mathrm{~F}\right]

\end{aligned}

$$

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