Solve this following
Question:

If speed $\mathrm{V}$, area $\mathrm{A}$ and force $\mathrm{F}$ are chosen as fundamental units, then the dimension of Young’s modulus will be :

 

  1. $\mathrm{FA}^{-1} \mathrm{~V}^{0}$

  2. $\mathrm{FA}^{2} \mathrm{~V}^{-1}$

  3. $\mathrm{FA}^{2} \mathrm{~V}^{-3}$

  4. $\mathrm{FA}^{2} \mathrm{~V}^{-2}$


Correct Option: 1

Solution:

$\mathrm{Y}=\mathrm{Fx} \mathrm{A}^{\mathrm{y}} \mathrm{V} \mathrm{z}$

$\mathrm{M}^{1} \mathrm{~L}^{-1} \mathrm{~T}^{-2}=\left[\mathrm{MLT}^{-2}\right]^{x}\left[\mathrm{~L}^{2}\right]^{y}\left[\mathrm{LT}^{-1}\right]^{z}$

$\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-2}=[\mathrm{M}]^{x}[\mathrm{~L}]^{x+2 y+z}[\mathrm{~T}]^{-2 x-z}$

comparing power of $\mathrm{ML}$ and $\mathrm{T}$

$x=1 \ldots(1)$

$x+2 y+z=-1$  ………………(2)

$-2 x-z=-2$ …………(3)

after solving

$x=1$

$y=-1$

$z=0$

$\mathrm{Y}=\mathrm{FA}^{-1} \mathrm{~V}^{0}$

 

Administrator

Leave a comment

Please enter comment.
Please enter your name.