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 resistances of coil of kettles A and B are `R_(A) and R_(B)`, respectively, then we can say

Options

A. `R_AgtR`
B. `R_A = R_B`
C. `R_AltR_B`
D. cannot be ascertained by above data.

Explanation

To solve the problem, we need to analyze the specifications of the two kettles and their efficiencies.

### Step 1: Understand the specifications of the kettles
- **Kettle A:**
- Specific heat capacity (c_A) = 1680 J/kg·K
- Mass (m_A) = 200 g = 0.2 kg
- Time to boil (t_A) = 6 min = 360 s

- **Kettle B:**
- Specific heat capacity (c_B) = 2450 J/kg·K
- Mass (m_B) = 400 g = 0.4 kg
- Time to boil (t_B) = 8 min = 480 s

### Step 2: Calculate the mass of water in each kettle
Given that both kettles have the same volume of 0.4 L and the density of water is 1000 kg/m³:
- Mass of water (m_water) = Volume × Density = 0.4 L × 1000 kg/m³ = 0.4 kg

### Step 3: Calculate the energy required to heat the water
Using the formula for heat energy:
\[ Q = mc\Delta T \]
Assuming the temperature change (ΔT) is the same for both kettles, we can denote it as ΔT.

For Kettle A:
\[ Q_A = m_{water} \cdot c_{water} \cdot \Delta T = 0.4 \cdot 4200 \cdot \Delta T \]

For Kettle B:
\[ Q_B = m_{water} \cdot c_{water} \cdot \Delta T = 0.4 \cdot 4200 \cdot \Delta T \]

### Step 4: Calculate the total energy supplied to each kettle
The total energy supplied to each kettle can be expressed as:
\[ Q_{supplied} = V^2/R \cdot t \]
Where V is the potential difference, R is the resistance, and t is the time.

For Kettle A:
\[ Q_{supplied, A} = \frac{V^2}{R_A} \cdot t_A = \frac{V^2}{R_A} \cdot 360 \]

For Kettle B:
\[ Q_{supplied, B} = \frac{V^2}{R_B} \cdot t_B = \frac{V^2}{R_B} \cdot 480 \]

### Step 5: Set up the efficiency equations
The efficiency of each kettle can be defined as:
\[ \text{Efficiency} = \frac{Q_{water}}{Q_{supplied}} \]

For Kettle A:
\[ \text{Efficiency}_A = \frac{0.4 \cdot 4200 \cdot \Delta T}{\frac{V^2}{R_A} \cdot 360} \]

For Kettle B:
\[ \text{Efficiency}_B = \frac{0.4 \cdot 4200 \cdot \Delta T}{\frac{V^2}{R_B} \cdot 480} \]

### Step 6: Compare efficiencies
Given that the efficiency of Kettle A is higher than Kettle B:
\[ \frac{0.4 \cdot 4200 \cdot \Delta T}{\frac{V^2}{R_A} \cdot 360} > \frac{0.4 \cdot 4200 \cdot \Delta T}{\frac{V^2}{R_B} \cdot 480} \]

Cancelling out common terms:
\[ \frac{1}{R_A} \cdot 360 > \frac{1}{R_B} \cdot 480 \]

Rearranging gives:
\[ R_B > R_A \cdot \frac{480}{360} \]
\[ R_B > R_A \cdot \frac{4}{3} \]

### Step 7: Conclusion
Since Kettle A has a higher efficiency and is using the same energy over a shorter time, it implies that its resistance must be less than that of Kettle B. Thus:
\[ R_A < R_B \]

### Final Answer
The resistance of kettle A is less than the resistance of kettle B:
\[ R_A < R_B \]