Question

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

• A. $FA ^{-1} V ^{0}$
• B. $FA ^{2} V ^{-1}$
• C. $FA ^{2} V ^{-3}$
• D. $FA ^{2} V ^{-2}$

Answer 1

The Correct Option is A

$Y=F^{x} A^{y} V^{z}$
$M ^{1} L ^{-1} T ^{-2}=\left[ M L T ^{-2}\right]^{x}\left[ L ^{2}\right]^{ y }\left[ L T ^{-1}\right]^{z}$
$M ^{1} L ^{1} T ^{-2}=[ M ]^{ x }[ L ]^{ x +2 y + z }[ T ]^{-2 x - z }$
comparing power of ML and T
$x=1 \ldots(1)$
$x+2 y+z=-1 \ldots . .(2)$
$-2 x-z=-2 \ldots(3)$
after solving
$x=1$
$y=-1$
$z=0$
$Y = FA ^{-1} V ^{0}$