A sphere of ice at `0^(@)C` having initial radius `R` is placed in an environment having ambient temperature `gt 0^(@)C`. The ice melts uniformly, such that shape remains spherical. After a time 't' the radius of the sphere has reduced to r. which graph best depicts r(t)

Options

A.
B.
C.
D.

Explanation

To solve the problem of how the radius \( r(t) \) of a melting ice sphere changes over time, we can follow these steps:

### Step 1: Understand the Problem
We have a sphere of ice at \( 0^\circ C \) with an initial radius \( R \). It is placed in an environment with a temperature greater than \( 0^\circ C \). As the ice melts uniformly, it retains its spherical shape. We need to find out how the radius \( r \) changes with time \( t \).

### Step 2: Apply Heat Transfer Principles
The melting of the ice sphere can be analyzed using Newton's law of cooling, which states that the rate of heat transfer is proportional to the temperature difference between the object and its surroundings. The heat loss from the sphere can be expressed as:
\[ \frac{dQ}{dt} = -\sigma A (T^4 - T_0^4) \]
where:
- \( \sigma \) is the Stefan-Boltzmann constant,
- \( A \) is the surface area of the sphere,
- \( T \) is the temperature of the sphere,
- \( T_0 \) is the ambient temperature.

### Step 3: Calculate the Surface Area
The surface area \( A \) of a sphere with radius \( r \) is given by:
\[ A = 4\pi r^2 \]

### Step 4: Relate Heat Loss to Mass Loss
The heat loss can also be related to the mass loss of the ice, which can be expressed as:
\[ \frac{dQ}{dt} = \frac{dm}{dt} \cdot L \]
where \( L \) is the latent heat of fusion of ice, and \( m \) is the mass of the ice.

### Step 5: Express Mass in Terms of Radius
The mass \( m \) of the ice can be expressed as:
\[ m = \rho V = \rho \left(\frac{4}{3}\pi r^3\right) \]
where \( \rho \) is the density of ice.

### Step 6: Differentiate the Mass with Respect to Time
Differentiating the mass with respect to time gives:
\[ \frac{dm}{dt} = \rho \cdot 4\pi r^2 \frac{dr}{dt} \]

### Step 7: Set Up the Equation
Equating the two expressions for heat loss, we have:
\[ -\sigma (4\pi r^2)(T^4 - T_0^4) = \rho \cdot 4\pi r^2 \frac{dr}{dt} \cdot L \]

### Step 8: Simplify the Equation
We can cancel \( 4\pi r^2 \) from both sides (assuming \( r \neq 0 \)):
\[ -\sigma (T^4 - T_0^4) = \rho L \frac{dr}{dt} \]

### Step 9: Rearrange and Integrate
Rearranging gives:
\[ \frac{dr}{dt} = -\frac{\sigma (T^4 - T_0^4)}{\rho L} \]
This indicates that the radius decreases over time. Integrating this equation will yield the relationship between \( r \) and \( t \).

### Step 10: Determine the Graph
The relationship derived indicates that \( r(t) \) is a linear function with a negative slope. This means that as time increases, the radius decreases linearly. The graph of \( r(t) \) will be a straight line that starts at \( R \) and decreases over time.

### Conclusion
The graph that best depicts \( r(t) \) is a straight line with a negative slope, indicating that the radius decreases uniformly over time.