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If instead of mass, length and time as fundamental quantities, we choose velocity, acceleration and force as fundamental quantities and express their dimensions by v , a and F respectively, then the dimensions of Young's modulus will be expressed as If instead of mass, length and time as fundamental quantities, we choose velocity, acceleration and force as fundamental quantities and express their dimensions by v, a and F respectively, then the dimensions of Young's modulus will be expressed as
If instead of mass, length and time as fundamental quantities, we choose velocity, acceleration and force as fundamental quantities and express their dimensions by v, a and F respectively, then the dimensions of Young's modulus will be expressed as
Young’s Modulus is given by,Y = Stress/StrainStress = Force/AreaStrain = change in length/length = dimensionlessTherefore, 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 = 1x+y+z = -1-2x-2y-z = -2Solving we get,x = 1, y = 2, z = -4So, the required dimension is [FA2V-4]
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]
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