Given relation

$E=\frac{B-x^2}{At}$

where:

• $[E] =$ energy $= M L^2 T^{-2}$
• $[x] = L$
• $[t] = T$

Find dimension of $B$

Since $B$ and $x^2$ are being subtracted, they must have the same dimensions.

$[B] = [x^2] = L^2$

Use the equation to find dimension of $A$

From $E=\frac{B-x^2}{At}$ dimensionally, $[E] = \frac{L^2}{[A] \cdot T}$

Therefore, $[A] = \frac{L^2}{[E] \cdot T}$

Substitute $[E] = M L^2 T^{-2}$:

$[A] = \frac{L^2}{(M L^2 T^{-2})(T)} = \frac{L^2}{M L^2 T^{-1}} = M^{-1} T^1$

Find dimension of $AB$

$[AB] = [A][B] = (M^{-1} T)(L^2) = L^2 M^{-1} T^1$

Match with options

$[AB] = L^2 M^{-1} T^1$

This corresponds to Option B.