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)`
Efficiency of kettle B is .

Options

A. `82.5%`
B. `72.5%`
C. `92.5%`
D. `62.5%`

Explanation

To find the efficiency of kettle B, we need to follow these steps:

### Step 1: Calculate the mass of water in kettle B
Given that the volume of water is 0.4 L and the density of water is 1000 kg/m³, we can find the mass of water (Mw) using the formula:

\[ Mw = \text{Volume} \times \text{Density} \]

Converting 0.4 L to cubic meters:
\[ 0.4 \, \text{L} = 0.4 \times 10^{-3} \, \text{m}^3 = 0.0004 \, \text{m}^3 \]

Now, calculating the mass:
\[ Mw = 0.0004 \, \text{m}^3 \times 1000 \, \text{kg/m}^3 = 0.4 \, \text{kg} = 400 \, \text{g} \]

### Step 2: Calculate the heat required to boil the water (Qw)
Using the formula for heat:
\[ Qw = Mw \times Cw \times \Delta T \]
where:
- \( Cw = 4200 \, \text{J/kg/K} \) (specific heat of water)
- \( \Delta T \) is the change in temperature (from initial temperature to boiling point, which we will denote as T).

Thus,
\[ Qw = 400 \, \text{g} \times 4200 \, \text{J/kg/K} \times T \]

### Step 3: Calculate the total heat supplied (Qs)
The total heat supplied (Qs) is the sum of the heat required to boil the water and the heat absorbed by the kettle (Qk):
\[ Qs = Qw + Qk \]
Where:
\[ Qk = Mk \times Ck \times \Delta T \]
with:
- \( Mk = 400 \, \text{g} \) (mass of kettle B)
- \( Ck = 2450 \, \text{J/kg/K} \) (specific heat of kettle B)

Thus,
\[ Qk = 400 \, \text{g} \times 2450 \, \text{J/kg/K} \times T \]

Now substituting back into the equation for Qs:
\[ Qs = (400 \times 4200 \times T) + (400 \times 2450 \times T) \]

### Step 4: Calculate the efficiency (η)
The efficiency (η) is defined as:
\[ \eta = \frac{Qw}{Qs} \]

Substituting the values of Qw and Qs:
\[ \eta = \frac{400 \times 4200 \times T}{(400 \times 4200 \times T) + (400 \times 2450 \times T)} \]

### Step 5: Simplifying the efficiency equation
The T and 400 can be canceled out:
\[ \eta = \frac{4200}{4200 + 2450} \]

Calculating the denominator:
\[ \eta = \frac{4200}{6650} \]

### Step 6: Calculate the final efficiency
Now, calculating the efficiency:
\[ \eta = 0.6303 \text{ or } 63.03\% \]

### Conclusion
The efficiency of kettle B is approximately **63.03%**.