A uniform conducting ring of mass `m=2kg`, radius `r=2m` and resistance `8Omega` is kept on smooth, horizontal surface. A time varying magnetic field `vecB=(hati+t^(2)hatj)` Tesla is present in the region, where `t` is time in second and take vertical as `y`-axis. (Take `pi^(2)=10`). Then

Options

A. Time when ring starts toppling is 1/2 sec
B. Time when ring starts toppling is `3//4` sec
C. Heat generated through the ring till the instant when the ring start toppling is `40//3` Joule.
D. Heat generated through the ring till the instant when the ring start toppling is `80//3` Joule

Explanation

To solve the problem, we need to determine the time when the conducting ring starts to topple due to the induced current in the presence of a time-varying magnetic field. We will analyze the torques acting on the ring due to its weight and the magnetic field.

### Step-by-Step Solution:

1. **Identify the Forces and Torques:**
- The ring has a mass \( m = 2 \, \text{kg} \).
- The weight of the ring \( F_w = mg = 2 \times 10 = 20 \, \text{N} \) acts downwards.
- The torque due to the weight \( \tau_w \) is given by:
\[ \tau_w = r \times F_w = 2 \, \text{m} \times 20 \, \text{N} = 40 \, \text{N m} \]

2. **Calculate the Magnetic Flux:**
- The magnetic field is given by \( \vec{B} = \hat{i} + t^2 \hat{j} \).
- The area \( A \) of the ring is \( A = \pi r^2 = \pi (2^2) = 4\pi \, \text{m}^2 \).
- The magnetic flux \( \Phi \) through the ring is:
\[ \Phi = \int \vec{B} \cdot d\vec{A} = \int (1 \hat{i} + t^2 \hat{j}) \cdot (0 \hat{i} + 1 \hat{j}) \, dA = 4\pi t^2 \]

3. **Calculate the Induced EMF:**
- The induced EMF \( \mathcal{E} \) is given by:
\[ \mathcal{E} = -\frac{d\Phi}{dt} = -\frac{d}{dt}(4\pi t^2) = -8\pi t \]

4. **Calculate the Induced Current:**
- The resistance \( R \) of the ring is \( 8 \, \Omega \).
- The induced current \( I \) is given by:
\[ I = \frac{\mathcal{E}}{R} = \frac{-8\pi t}{8} = -\pi t \]

5. **Calculate the Force due to the Magnetic Field:**
- The force \( F_B \) on the ring due to the magnetic field is given by:
\[ F_B = I A \times \vec{B} = (-\pi t)(4\pi) \hat{j} \times (\hat{i} + t^2 \hat{j}) = -4\pi^2 t \hat{k} \]

6. **Calculate the Torque due to the Magnetic Field:**
- The torque \( \tau_B \) due to the magnetic field is:
\[ \tau_B = r \times F_B = 2 \, \hat{r} \times (-4\pi^2 t \hat{k}) = 8\pi^2 t \, \text{N m} \]

7. **Set the Torques Equal to Find the Time:**
- The ring will start to topple when the torque due to the magnetic field equals the torque due to the weight:
\[ \tau_B = \tau_w \implies 8\pi^2 t = 40 \]
- Substituting \( \pi^2 = 10 \):
\[ 80t = 40 \implies t = \frac{1}{2} \, \text{s} \]

### Final Answer:
The time when the ring starts to topple is \( t = \frac{1}{2} \, \text{s} \).