The dimensional formula for spring constant is ____.
A. \[{M^1}{L^1}{T^{ - 2}}\]
B. \[{M^1}{L^0}{T^{ - 2}}\]
C. \[{M^0}{L^1}{T^{ - 2}}\]
D. \[{M^1}{L^2}{T^{ - 1}}\]

Hint:A dimensional formula is an equation that expresses the relationship between fundamental and derived units in terms of dimensions (equation). In mechanics, the three basic dimensions of length, mass, and time are denoted by the letters L, M, and T, respectively.

Complete step by step answer:
According to Hooke’s rule, the force used to compact or expand a spring is directly proportional to the extent it is extended. The formula of spring constant is given as:
$F = - k. x$
\[k = \dfrac{{ - F}}{x}\]
Where, $F$ is the spring's restoring force, aimed at restoring equilibrium. The spring constant is $k$. The spring's displacement from its equilibrium state is $x$.

In other words, if the spring displacement is unity, the spring constant is the force applied.If a force $F$ is applied to the spring, it will stretch it to the point that the equilibrium position is displaced by $x$.
$\text{Dimension of F} = [ML{T^{ - 2}}] \\
\Rightarrow \text{Dimension of x = L} \\
\Rightarrow \text{Dimension of k} = \dfrac{F}{x} = \dfrac{{[ML{T^{ - 2}}]}}{{[L]}} \\
\therefore \text{Dimension of k} = [M{L^0}{T^{ - 2}}]$
It is expressed in Newton per meter (N/m).

Hence, option B is correct.

Note:The spring constant, k, is measured in Newtons per metre (N/m), and x is the spring's displacement from its equilibrium position. The spring constant, k, is a measure of the spring's stiffness. Spring constants are higher in stiffer (more difficult to stretch) springs.