Falling Behind in Studies? Join Our Performance Improvement Batch. Enroll For Free
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
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]
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -