1.Units, Dimensions and Measurement
medium

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

$FA ^{2} V ^{-2}$

(JEE MAIN-2020)

Solution

$Y = F ^{ x } A ^{ y } V ^{z}$

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

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

comparing power of $ML$ and $T$

$x=1 \ldots(1)$

$x+2 y+z=-1$$…(2)$

$-2 x-z=-2$$…(3)$

after solving

$x=1$

$y=-1$

$z=0$

$Y=F A^{-1} V^{0}$

Standard 11
Physics

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