Let us consider a system of units in which mass and angular momentum are dimensionless. If length has dimensions of $L$, which of the following statement(s) is/are correct?

Step 1: Identify the given dimensions.

In the given system of units, mass and angular momentum are dimensionless, and length has the dimension $L$.

$$[M] = 1$$ $$[J] = 1$$ $$[\text{length}] = L$$

Step 2: Determine the dimensions of velocity and time.

The angular momentum $J$ of a particle of mass $m$ moving with velocity $v$ in a circular path of radius $r$ is given by:

$$J = mvr$$

Taking dimensions on both sides:

$$[J] = [M][v][r]$$

Substitute the known dimensions:

$$1 = 1 \cdot [v] \cdot L$$ $$[v] = L^{-1}$$

Using the relation for time $[t] = \frac{[r]}{[v]}$:

$$[t] = \frac{L}{L^{-1}}$$ $$[t] = L^{2}$$

Step 3: Determine the dimension of linear momentum.

Linear momentum is defined as $p = mv$.

$$[p] = [M][v]$$ $$[p] = 1 \cdot L^{-1} = L^{-1}$$

This matches option (2).

Step 4: Determine the dimension of force.

Force can be expressed using the centripetal force formula $F = \frac{mv^2}{r}$.

$$[F] = \frac{[M][v]^2}{[r]}$$ $$[F] = \frac{1 \cdot (L^{-1})^2}{L}$$ $$[F] = L^{-3}$$

This does not match option (1).

Step 5: Determine the dimension of energy.

Energy has the same dimensions as kinetic energy, $E = \frac{1}{2}mv^2$.

$$[E] = [M][v]^2$$ $$[E] = 1 \cdot (L^{-1})^2 = L^{-2}$$

This matches option (3).

Step 6: Determine the dimension of power.

Power is defined as energy per unit time, $P = \frac{E}{t}$.

$$[P] = \frac{[E]}{[t]}$$ $$[P] = \frac{L^{-2}}{L^2} = L^{-4}$$

This does not match option (4).

Answer: The correct statements are (2) (linear momentum has dimension $L^{-1}$) and (3) (energy has dimension $L^{-2}$).