Spheres `P` and `Q` are uniformally constructed from the same material which is good conductor of heat and the radius of `Q` is thrice the radius of `P`. The rate of fall of temperature of `P` is `x` times that of `Q` when both are at the same surface temperture. Find the value of `'x'` :

Options

A. 27
B. 1
C. 3
D. 9

Answer

To find the value of \( x \), which represents the rate of fall of temperature of sphere \( P \) compared to sphere \( Q \), we can follow these steps:

### Step 1: Understand the relationship between the spheres
We know that spheres \( P \) and \( Q \) are made of the same material, and the radius of \( Q \) is three times that of \( P \). Let the radius of sphere \( P \) be \( r \) and the radius of sphere \( Q \) be \( R = 3r \).

### Step 2: Recall the formula for heat transfer
The rate of heat transfer (power) for a body can be given by the Stefan-Boltzmann law for black bodies, which states:
\[ P = E \cdot \sigma \cdot A \cdot T^4 \]
where:
- \( P \) is the power (rate of heat transfer),
- \( E \) is the emissivity of the material,
- \( \sigma \) is the Stefan-Boltzmann constant,
- \( A \) is the surface area,
- \( T \) is the absolute temperature.

### Step 3: Calculate the surface areas
The surface area \( A \) of a sphere is given by:
\[ A = 4\pi r^2 \]
For sphere \( P \):
\[ A_P = 4\pi r^2 \]
For sphere \( Q \):
\[ A_Q = 4\pi (3r)^2 = 36\pi r^2 \]

### Step 4: Calculate the mass of the spheres
The mass \( m \) of a sphere can be calculated as:
\[ m = \text{Volume} \times \text{Density} = \frac{4}{3}\pi r^3 \cdot \rho \]
For sphere \( P \):
\[ m_P = \frac{4}{3}\pi r^3 \cdot \rho \]
For sphere \( Q \):
\[ m_Q = \frac{4}{3}\pi (3r)^3 \cdot \rho = 36 \cdot \frac{4}{3}\pi r^3 \cdot \rho \]

### Step 5: Relate the rates of temperature change
The rate of change of temperature with respect to time can be expressed as:
\[ \frac{dT}{dt} = \frac{P}{m \cdot c} \]
where \( c \) is the specific heat capacity of the material.

For sphere \( P \):
\[ \frac{dT_P}{dt} = \frac{E \cdot \sigma \cdot A_P \cdot T^4}{m_P \cdot c} \]
For sphere \( Q \):
\[ \frac{dT_Q}{dt} = \frac{E \cdot \sigma \cdot A_Q \cdot T^4}{m_Q \cdot c} \]

### Step 6: Substitute the values
Substituting the areas and masses we calculated:
\[ \frac{dT_P}{dt} = \frac{E \cdot \sigma \cdot (4\pi r^2) \cdot T^4}{\left(\frac{4}{3}\pi r^3 \cdot \rho\right) \cdot c} \]
\[ \frac{dT_Q}{dt} = \frac{E \cdot \sigma \cdot (36\pi r^2) \cdot T^4}{\left(36 \cdot \frac{4}{3}\pi r^3 \cdot \rho\right) \cdot c} \]

### Step 7: Simplify the expressions
After simplifying both expressions, we find:
\[ \frac{dT_P}{dt} = \frac{3E \cdot \sigma \cdot T^4}{r \cdot \rho \cdot c} \]
\[ \frac{dT_Q}{dt} = \frac{E \cdot \sigma \cdot T^4}{r \cdot \rho \cdot c} \]

### Step 8: Find the ratio
Now, we can find the ratio of the rates of temperature change:
\[ \frac{dT_P}{dt} = 3 \cdot \frac{dT_Q}{dt} \]

### Conclusion
Thus, we conclude that:
\[ x = 3 \]