Of the following quantities, which one has dimensions different from the remaining three ?

To determine which of the given quantities has dimensions different from the remaining three, we will find the dimensional formula for each of the options.

Let the fundamental dimensions be mass [M], length [L], and time [T].

1. Energy per unit volume:
Energy has the dimensions of work done, which is force multiplied by distance:
$\text{Energy}=\; \backslash text\{Force\}\; \backslash times\; \backslash text\{Distance\}\; =\; [\backslash text\{M\}\; \backslash text\{L\}\; \backslash text\{T\}^\{-2\}]\; \backslash times\; [\backslash text\{L\}]\; =\; [\backslash text\{M\}\; \backslash text\{L\}^\{2\}\; \backslash text\{T\}^\{-2\}]$
Volume has the dimensions of length cubed:
$\text{Volume}=\; [\backslash text\{L\}^\{3\}]$
Therefore, the dimensions of energy per unit volume are:
$\frac{\left[\text{M}{\text{L}}^2{\text{T}}^{-2}\right]}{\left[{\text{L}}^3\right]}=\left[\text{M}{\text{L}}^{-1}{\text{T}}^{-2}\right]$

2. Force per unit area:
Force has the dimensions:
$\text{Force}=\; [\backslash text\{M\}\; \backslash text\{L\}\; \backslash text\{T\}^\{-2\}]$
Area has the dimensions of length squared:
$\text{Area}=\; [\backslash text\{L\}^\{2\}]$
Therefore, the dimensions of force per unit area are:
$\frac{\left[\text{M}\text{L}{\text{T}}^{-2}\right]}{\left[{\text{L}}^2\right]}=\left[\text{M}{\text{L}}^{-1}{\text{T}}^{-2}\right]$

3. Product of voltage and charge per unit volume:
The product of voltage (electric potential, V) and charge (q) is equal to electrical energy:
$\text{Voltage}\times \text{Charge}=\text{Energy}=\left[\text{M}{\text{L}}^2{\text{T}}^{-2}\right]$
Thus, the product of voltage and charge per unit volume has the same dimensions as energy per unit volume:
$\frac{[\text{Voltage}\times \text{Charge}]}{\left[\text{Volume}\right]}=\frac{\left[\text{M}{\text{L}}^2{\text{T}}^{-2}\right]}{\left[{\text{L}}^3\right]}=\left[\text{M}{\text{L}}^{-1}{\text{T}}^{-2}\right]$

4. Angular momentum:
Angular momentum (L) is defined as the product of moment of inertia and angular velocity, or position vector crossed with linear momentum (r × p):
$\text{Angular momentum}=\text{Distance}\times \text{Mass}\times \text{Velocity}$
$\text{Angular momentum}=\left[\text{L}\right]\times \left[\text{M}\right]\times \left[\text{L}{\text{T}}^{-1}\right]=\left[\text{M}{\text{L}}^2{\text{T}}^{-1}\right]$

Comparing the dimensions of all four quantities:
- Energy per unit volume: $\left[\text{M}{\text{L}}^{-1}{\text{T}}^{-2}\right]$
- Force per unit area: $\left[\text{M}{\text{L}}^{-1}{\text{T}}^{-2}\right]$
- Product of voltage and charge per unit volume: $\left[\text{M}{\text{L}}^{-1}{\text{T}}^{-2}\right]$
- Angular momentum: $\left[\text{M}{\text{L}}^2{\text{T}}^{-1}\right]$

The first three quantities have the same dimensions of pressure/energy density, while Angular momentum has different dimensions.

Therefore, the correct option is Angular momentum.