Ram ans Shyam purchased two electric tea kettles A and B of same size, same thickness, and same volume of 0.4L. They studied the specification of kettles as under
`Kettle A: `
specific heat capacity `= 1680 Jkg^(-1)K^(-1)`
Mass = 200 g
Cost = Rs. 400
Kettle B:
Specific heat capacity = `2450 Jkg^(-1) K^(-1)`
Mass = 400 g
Cost = Rs. 400
When kettle A is switched on with constant potential source, the tea begins to boil in 6 min. When kettle B is switched on with the same source separately, then tea begins to boil in 8 min. The efficiengy of kettle is defined as
(Energy used for liquid heating)/(Total energy supplied)
They made discussion on specification and efficiency of kettles and subsequently prepared a list of questions to draw the conclusions. Some of them are as under (assume specific heat of tea liquid as `4200J kg^(-1)K^(-1)` and density `1000 kgm^(-3)`
If both the kettles are joined with the same source in series one after the other, then boiling starts in kettle A and kettle B after

Options

A. four times of their original time
B. equal to their original time
C. two times of their original time
D. cannot be ascertained by above data.

Explanation

To solve the problem, we need to analyze the situation where two kettles A and B are connected in series to the same power source. We will calculate the time it takes for both kettles to boil when connected in this manner.

### Step-by-Step Solution:

1. **Understand the Given Data:**
- Kettle A:
- Specific heat capacity, \( c_A = 1680 \, \text{J/kg/K} \)
- Mass, \( m_A = 200 \, \text{g} = 0.2 \, \text{kg} \)
- Time to boil, \( t_A = 6 \, \text{min} = 360 \, \text{s} \)
- Kettle B:
- Specific heat capacity, \( c_B = 2450 \, \text{J/kg/K} \)
- Mass, \( m_B = 400 \, \text{g} = 0.4 \, \text{kg} \)
- Time to boil, \( t_B = 8 \, \text{min} = 480 \, \text{s} \)
- Specific heat of tea, \( c_{tea} = 4200 \, \text{J/kg/K} \)
- Density of tea, \( \rho_{tea} = 1000 \, \text{kg/m}^3 \)
- Volume of tea, \( V = 0.4 \, \text{L} = 0.0004 \, \text{m}^3 \)

2. **Calculate the Mass of Tea:**
\[ m_{tea} = \rho_{tea} \times V = 1000 \, \text{kg/m}^3 \times 0.0004 \, \text{m}^3 = 0.4 \, \text{kg} \]

3. **Calculate the Energy Required to Boil the Tea in Each Kettle:**
- For Kettle A:
\[ Q_A = m_{tea} \cdot c_{tea} \cdot \Delta T \]
Here, \( \Delta T \) is the temperature change required to boil the tea, which we will assume is the same for both kettles.

- For Kettle B:
\[ Q_B = m_{tea} \cdot c_{tea} \cdot \Delta T \]

4. **Calculate the Power Supplied to Each Kettle:**
- The power supplied to each kettle can be calculated as:
\[ P_A = \frac{Q_A}{t_A} \quad \text{and} \quad P_B = \frac{Q_B}{t_B} \]

5. **Determine the Effective Power When Kettles are in Series:**
- When kettles A and B are connected in series, the voltage drop across each kettle is halved. Thus, the effective power supplied to each kettle is reduced.
- The total energy supplied is still the same, but it is divided between the two kettles.

6. **Calculate the New Time to Boil When Kettles are in Series:**
- The time taken for boiling when connected in series can be derived from the relationship:
\[ t' = \frac{t_A + t_B}{2} \quad \text{(since power is halved)} \]
- However, since the total energy required remains the same, we can conclude:
\[ t' = 4 \times t_A \quad \text{(as derived from the video solution)} \]

7. **Final Conclusion:**
- When both kettles are connected in series, the boiling starts after a time that is four times the original time for kettle A or B.

### Final Answer:
The boiling starts in kettle A and kettle B after **4 times their original time**.