Question

If instead of mass, length and time as fundamental quantities, we choose velocity, acceleration and force as fundamental quantities express their dimensions v,a  and f respectively, then the dimensions of young's modulus will be expressed as:​

Answer 1

Answer:

Young’s Modulus is given by,

Y = Stress/Strain

Stress = Force/Area

Strain = change in length/length = dimensionless

Therefore, dimension of Y is [ML-1T-2]

Let,

[FxAyVz]  = [ML-1T-2]

=> [MLT-2]x [LT-2]y [LT-1]z = [ML-1T-2]

=> [Mx Lx+y+z T-2x-2y-z] = [ML-1T-2]

Thus,

x = 1

x+y+z = -1

-2x-2y-z = -2

Solving we get,

x = 1, y = 2, z = -4

So, the required dimension is [FA2V-4]

Explanation:

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