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Question

Physics Question on physical world

If speed VV, area AA and force FF are chosen as fundamental units, then the dimension of Young's modulus will be :

A

FA1V0FA ^{-1} V ^{0}

B

FA2V1FA ^{2} V ^{-1}

C

FA2V3FA ^{2} V ^{-3}

D

FA2V2FA ^{2} V ^{-2}

Answer

FA1V0FA ^{-1} V ^{0}

Explanation

Solution

Y=FxAyVzY=F^{x} A^{y} V^{z}
M1L1T2=[MLT2]x[L2]y[LT1]zM ^{1} L ^{-1} T ^{-2}=\left[ M L T ^{-2}\right]^{x}\left[ L ^{2}\right]^{ y }\left[ L T ^{-1}\right]^{z}
M1L1T2=[M]x[L]x+2y+z[T]2xzM ^{1} L ^{1} T ^{-2}=[ M ]^{ x }[ L ]^{ x +2 y + z }[ T ]^{-2 x - z }
comparing power of ML and T
x=1(1)x=1 \ldots(1)
x+2y+z=1..(2)x+2 y+z=-1 \ldots . .(2)
2xz=2(3)-2 x-z=-2 \ldots(3)
after solving
x=1x=1
y=1y=-1
z=0z=0
Y=FA1V0Y = FA ^{-1} V ^{0}