The given relation is E= (b - x²) /at, where E is energy, x is distance, and t is time. We need to find the dimensions of a and b in this relation.



Dimensions are the physical quantities used to describe the nature of a physical quantity. For example, the dimensions of length are L, the dimensions of mass are M, and the dimensions of time are T.



To find the dimensions of a and b, we can make use of the principle of dimensional homogeneity. According to this principle, the dimensions of the left-hand side of an equation must be equal to the dimensions of the right-hand side of the equation.



The dimensions of energy can be found by using the formula E = F × d, where F is force and d is distance. Therefore, the dimensions of energy are [M L² T⁻²].



The dimensions of distance and time are L and T, respectively.



Using the principle of dimensional homogeneity, we can equate the dimensions of the left-hand side of the equation (i.e., energy) to the dimensions of the right-hand side of the equation.

Dimensions of the left-hand side (energy) = [M L² T⁻²]

Dimensions of the right-hand side = [b / (a T)] - [x² / (a T)]

Equating the dimensions, we get:

[M L² T⁻²] = [b / (a T)] - [L² / (a T)]

Simplifying the above equation, we get:

[M L² T⁻²] = [b / (a T)] - [L² T⁻² / a]

Equating the dimensions of both sides of the equation, we get:

[M L² T⁻²] = [M L² T⁻²]

Comparing the powers of the dimensions on both sides, we get:

For mass: 1 = 1

For length: 2 = 2

For time: -2 = -2

Solving the above equations, we get:

Dimensions of a = [T]

Dimensions of b = [M L² T⁻²]



The dimensions of a and b in the given relation E= (b - x²) /at are [T] and [M L² T⁻²], respectively.

Dimension of b is L^2. since a quantity can only be substracted from the quantity of same unit so b has the dimensions of x.
Dimension of a is M^-1L^-1T^1. equating LHS &RHS
ML^2T^-2 =L/aT=M^-1L^-1T^1