Question

dimention of E×L²÷m⁵G​

Answer 1

Explanation:

Suppose,

P = E.I^2.m^-5.G^-2

Dimensions of P will be given by formula,

[P] = [EI^2M^-5G^-2]

We know,

[E] = [ML^2T^-2]

[G] = [M^-1L^3T-2]

[I] = [ML^2T^-1]

Putting values in eqn.(1),

[P] = [ML^2T^-2][ML^2T^-1]^2[M^-5][M^-1L^3T^-2]^-2

[P] = [ML^2T^-2][M^2L^4T^-2][M^-5][M^2L^-6T^4]

[P] = [M^(1+2-5+2)L(2+4-6)T(-2-2+4)]

[P] = [M^0L^0T^0]

Comparing LHS=RHS,

[P] = null.

Hence, P is dimensionless quantity.

Hope that helps you.

~~anushka Here