The question asks for the maximum volume of water that can be boiled by a heater supplying heat at a constant rate \(R\).

Water boils at \(100^{\circ}C\). For water to continuously boil (or be maintained at \(100^{\circ}C\) while still receiving heat), the rate at which heat is supplied by the heater must be equal to the rate at which heat is lost to the surroundings at the boiling temperature.

We are given:

• Heat supplied rate by the heater = \(R\)
• Boiling temperature of water = \(100^{\circ}C\)
• Heat loss rate to the atmosphere is proportional to both the temperature of the water and the volume of water. The proportionality constant is \(K\).

Let \(V\) be the volume of water in the container and \(T\) be the temperature of the water.

The heat loss rate can be expressed as:

\(\text{Heat loss rate} = K \times T \times V\)

When the water is boiling, its temperature is \(T = 100^{\circ}C\). So, the heat loss rate at the boiling point for a volume \(V\) is:

\(\text{Heat loss rate at } 100^{\circ}C = K \times 100^{\circ}C \times V\)

The maximum volume of water that can be boiled by the heater is limited by the heat supply rate \(R\). At the maximum volume, the entire heat supplied by the heater is used to compensate for the heat lost to the atmosphere at the boiling temperature (\(100^{\circ}C\)). If the volume were larger, the heat loss at \(100^{\circ}C\) would exceed \(R\), and the heater wouldn't be able to maintain the water at \(100^{\circ}C\).

So, at the maximum volume \(\left(V_{\text{max}}\right)\), the heat supplied rate equals the heat loss rate at \(100^{\circ}C\):

\(R = K \times 100 \times V_{\text{max}}\)

Now, we need to solve for \(V_{\text{max}}\):

\(V_{\text{max}} = \frac{R}{K \times 100}\)

\(V_{\text{max}} = \frac{R}{100K}\)

Comparing this result with the given options:

• Option 1: \(100/KR\)
• Option 2: \(K/100R\)
• Option 3: \(R/100K\)
• Option 4: \(KR/100\)

The calculated maximum volume matches Option 3.

Heater Heat Balance at Boiling Point

For sustained boiling at \(100^{\circ}C\), the energy supplied by the heater must equal the energy lost to the surroundings. The heater supplies energy at rate \(R\). The energy loss rate is given by \(K \times \text{Temperature} \times \text{Volume}\). At \(100^{\circ}C\), this loss is \(K \times 100 \times V\). Setting the supply equal to the loss gives \(R = 100KV\).

Maximum Volume Calculation Steps

Here are the steps to determine the maximum volume:

1. Identify the temperature at which boiling occurs (\(100^{\circ}C\)).
2. Note the rate of heat supply by the heater (\(R\)).
3. Formulate the heat loss rate based on the given proportionality (\(K \times \text{Temperature} \times \text{Volume}\)).
4. Substitute the boiling temperature (\(100^{\circ}C\)) into the heat loss formula: \(K \times 100 \times V\).
5. Equate the heat supplied rate to the heat loss rate at the maximum volume (\(V_{\text{max}}\)) to find the condition for equilibrium at \(100^{\circ}C\): \(R = K \times 100 \times V_{\text{max}}\).
6. Solve the equation for \(V_{\text{max}}\): \(V_{\text{max}} = \frac{R}{100K}\).

This formula gives the maximum volume of water that the heater can maintain at \(100^{\circ}C\) given the heat loss characteristics.