Let $\overrightarrow{\mathrm{E}}(\mathrm{x},\mathrm{y},\mathrm{z})=2{\mathrm{x}}^2\hat{\mathrm{i}}+5\mathrm{y}\hat{\mathrm{j}}+3\mathrm{z}\hat{\mathrm{k}}$ the value of ∭v $(\overrightarrow{\mathrm{\nabla }}.\overrightarrow{E})dV$, where v is the volume enclosed by the unit cube defined by 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, and 0 ≤ z ≤ 1 is

The correct option is 10.

To find the value of the volume integral ${\iiint }_V(\overrightarrow{\nabla }\cdot \overrightarrow{E})dV$ over the unit cube $V$, we first need to compute the divergence of the electric field vector $\overrightarrow{E}(x,y,z)$.

The given vector field is:
$\overrightarrow{E}=2x^2\hat{i}+5y\hat{j}+3z\hat{k}$

The divergence of a vector field $\overrightarrow{E}=E_x\hat{i}+E_y\hat{j}+E_z\hat{k}$ is defined as:
$\overrightarrow{\nabla }\cdot \overrightarrow{E}=\frac{\partial E_x}{\partial x}+\frac{\partial E_y}{\partial y}+\frac{\partial E_z}{\partial z}$

Substituting the components of our vector field, we obtain:
$E_x=2x^2\Rightarrow \frac{\partial E_x}{\partial x}=\frac{\partial }{\partial x}\left(2x^2\right)=4x$
$E_y=5y\Rightarrow \frac{\partial E_y}{\partial y}=\frac{\partial }{\partial y}\left(5y\right)=5$
$E_z=3z\Rightarrow \frac{\partial E_z}{\partial z}=\frac{\partial }{\partial z}\left(3z\right)=3$

Summing these partial derivatives gives the divergence:
$\overrightarrow{\nabla }\cdot \overrightarrow{E}=4x+5+3=4x+8$

Now, we integrate this divergence over the volume $V$ of the unit cube defined by the boundaries $0\le x\le 1$, $0\le y\le 1$, and $0\le z\le 1$:
${\iiint }_V(\overrightarrow{\nabla }\cdot \overrightarrow{E})dV={\int }_0^1{\int }_0^1{\int }_0^1(4x+8)dzdydx$

Since the integrand depends only on $x$, we can separate the integrals for $y$ and $z$:
${\iiint }_V(\overrightarrow{\nabla }\cdot \overrightarrow{E})dV=\left[{\int }_0^1\right(4x+8\left)dx\right]\cdot \left[{\int }_0^{1}dy\right]\cdot \left[{\int }_0^{1}dz\right]$

Evaluating the integrals for $y$ and $z$:
${\int }_0^{1}dy={\left[y\right]}_0^1=1$
${\int }_0^{1}dz={\left[z\right]}_0^1=1$

Now we evaluate the integral with respect to $x$:
${\int }_0^1(4x+8)dx={[2x^2+8x]}_0^1=(2{\left(1\right)}^2+8(1\left)\right)-\left(0\right)=2+8=10$

Multiplying these results together, we get:
${\iiint }_V(\overrightarrow{\nabla }\cdot \overrightarrow{E})dV=10\cdot 1\cdot 1=10$

Thus, the value of the triple integral is 10.