Write the dimension a/b in the relation E = (b - x²) / at , where E is energy, x is distance and t is time. Please show the method to solve this question.

here, equation is like this. E = B - (X2 / AT)

here E is made OF difference of two terms B and (X2/AT)

THEREFORE

Dimension of B is the same as E i.e.[ M1 L2 T-2] Energy = Force x Distance = Mass x Acceleration x Distance

Also, dimension of (X2/AT) is the same as E i.e. [ M1 L2 T-2]

Now, X=distance = [ m0 L1 t0 ]

& T=time = [m0 lo T1]

therefore dimension of A is found as below.

Dim.of (X2/AT) = {Dim.of X2} / {Dim. of A } x {Dim. of T}

so {M1 L2 T-2} = {M0 L2 T0} / { Ma Lb Tc } x { M0 L0 T1}

so {M1 L2 T-2} = {M0 L2 T-1} / {Ma Lb Tc}

so {M1 L2 T-2} = { M0-a L2-b T-1-c }

COMPARING

0-a = 1 so a = -1

2-b = 2 so b = 0

-1-c =-2 so c = 1

Therefore Dimensions of A is [ Ma Lb Tc ] = [ M-1 L0 T1 ] i.e. unit is s/kg.

We knew that Dimension of B is same as Energy i.e.[ M1 L2 T-2]

i.e. unit is kg-m2/s2 i.e. kg-m/s2 * m i.e.N-m i.e. Joule

THUS

Dimensions of A*B is

[ M-1 L0 T1] * [M1 L2 T-2] = [M0 L2 T-1] i.e. unit is m2/s which is the unit of Kinematic Viscosity.

hope u got it