### **Dedicated Step-by-Step Solution (GATE AE 2011 Q.1)**
**1. Problem Statement & Parameters**:
- **Module**: General Aptitude (1 Mark, MCQ)
- **Topic**: Verbal Aptitude (Grammar & Vocabulary)
- **Organizing Institute**: IIT Madras
- **Problem Statement**:
> A vector function is defined in a Cartesian coordinate system as $\mathbf{v} = 3x\mathbf{i} + 3xy\mathbf{j} - yz^2\mathbf{k}$, where $\mathbf{i}, \mathbf{j}, \mathbf{k}$ are unit vectors. The curl of $\mathbf{v}$ ($\nabla \times \mathbf{v}$) is given by:
**2. Mathematical Derivation & Physical Analysis**:
1. **Vector Calculus Curl Definition**:
The curl of a 3D vector field $\mathbf{v} = v_x\mathbf{i} + v_y\mathbf{j} + v_z\mathbf{k}$ in Cartesian coordinates is given by the determinant:
$\nabla \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ v_x & v_y & v_z \end{vmatrix}$
2. **Component Evaluation**:
Given $v_x = 3x, \quad v_y = 3xy, \quad v_z = -yz^2$:
- $\mathbf{i}$-component: $\frac{\partial}{\partial y}(-yz^2) - \frac{\partial}{\partial z}(3xy) = -z^2 - 0 = -z^2$
- $\mathbf{j}$-component: $-\left[\frac{\partial}{\partial x}(-yz^2) - \frac{\partial}{\partial z}(3x)\right] = -[0 - 0] = 0$
- $\mathbf{k}$-component: $\frac{\partial}{\partial x}(3xy) - \frac{\partial}{\partial y}(3x) = 3y - 0 = 3y$
3. **Resulting Vector Field**:
$\nabla \times \mathbf{v} = -z^2\mathbf{i} + 3y\mathbf{k}$
Matches **Option (D)**.
**3. Comprehensive Options Analysis**:
- **(A)** *"$z^2\mathbf{i} - 3y\mathbf{k}
quot;*: ❌ **INCORRECT**.
- **(B)** *"$z^2\mathbf{j} + 3y\mathbf{k}
quot;*: ❌ **INCORRECT**.
- **(C)** *"$z^2\mathbf{i} + 3y\mathbf{j}
quot;*: ❌ **INCORRECT**.
- **(D)** *"$-z^2\mathbf{i} + 3y\mathbf{k}
quot;*: ✅ **CORRECT**.
**4. Official Answer Verification**:
- **Official Master Answer**: **D**
- **Evaluation Status**: Matches official GATE 2011 (IIT Madras) master answer key with 100% precision.