ML-1-T-2- is the dimensional formula ofcoefficient of frictionmodulus of elasticityforceenergy

Question

Toppr Admin · Accepted Answer

Modulus of elasticity is given by-xA0-xA0-xA0-xA0- Y-FA-L-x394-LDimension of force-xA0- F-MLT-x2212-2-Dimension of Area-xA0- A-L2-Dimension of length-xA0- L-L-Thus dimension of modulus of elasticity-xA0-xA0- Y-MLT-x2212-2-L-L2-L-ML-x2212-1T-x2212-2-

$M L^{- 1} T^{- 2}$ is the dimensional formula of
coefficient of friction
modulus of elasticity
force
energy

Modulus of elasticity is given by $Y = \frac{F}{A} . \frac{L}{Δ L}$
Dimension of force $F = [M L T^{- 2}]$
Dimension of Area $A = [L^{2}]$
Dimension of length $L = [L]$
Thus dimension of modulus of elasticity $Y = \frac{[M L T^{- 2}] [L]}{[L^{2}] [L]} = [M L^{- 1} T^{- 2}]$

ML−1T−2 is the dimensional formula ofcoefficient of frictionmodulus of elasticityforceenergy

Modulus of elasticity is given by Y=FA.LΔLDimension of force F=[MLT−2]Dimension of Area A=[L2]Dimension of length L=[L]Thus dimension of modulus of elasticity Y=[MLT−2][L][L2][L]=[ML−1T−2]

$M L^{- 1} T^{- 2}$ is the dimensional formula of
coefficient of friction
modulus of elasticity
force
energy

Modulus of elasticity is given by $Y = \frac{F}{A} . \frac{L}{Δ L}$
Dimension of force $F = [M L T^{- 2}]$
Dimension of Area $A = [L^{2}]$
Dimension of length $L = [L]$
Thus dimension of modulus of elasticity $Y = \frac{[M L T^{- 2}] [L]}{[L^{2}] [L]} = [M L^{- 1} T^{- 2}]$