If instead of mass, length and time as fundamental quantities, we choo

1.	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» 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: please MARK as BRAINLIEST :)

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»

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:

please MARK as BRAINLIEST :)

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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:​

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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: