Two spheres of radii in the ratio `1 : 2` and densities in the ratio `2 : 1` and of same specific heat, are heated to same temperature and left in the same surronding. There are of cooling will be in the ratio

Options

A. `2 : 1`
B. `1 : 1`
C. `1 : 2`
D. `1 : 4`

Explanation

To solve the problem, we need to find the ratio of the rates of cooling of two spheres with given properties. Let's break it down step by step.

### Step 1: Understand the Given Ratios
- The radii of the spheres are in the ratio \( r_1 : r_2 = 1 : 2 \).
- The densities of the spheres are in the ratio \( \rho_1 : \rho_2 = 2 : 1 \).
- Both spheres have the same specific heat capacity.

### Step 2: Calculate the Volumes of the Spheres
The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
Let the radius of the first sphere be \( r_1 = r \) and the radius of the second sphere be \( r_2 = 2r \).

- Volume of the first sphere:
\[ V_1 = \frac{4}{3} \pi r^3 \]

- Volume of the second sphere:
\[ V_2 = \frac{4}{3} \pi (2r)^3 = \frac{4}{3} \pi (8r^3) = \frac{32}{3} \pi r^3 \]

### Step 3: Calculate the Mass of the Spheres
The mass \( m \) of a sphere can be calculated using the formula:
\[ m = \rho V \]
Let the density of the first sphere be \( \rho_1 = 2 \rho \) and the density of the second sphere be \( \rho_2 = \rho \).

- Mass of the first sphere:
\[ m_1 = \rho_1 V_1 = (2\rho) \left(\frac{4}{3} \pi r^3\right) = \frac{8}{3} \pi \rho r^3 \]

- Mass of the second sphere:
\[ m_2 = \rho_2 V_2 = \rho \left(\frac{32}{3} \pi r^3\right) = \frac{32}{3} \pi \rho r^3 \]

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

- Surface area of the first sphere:
\[ A_1 = 4 \pi r^2 \]

- Surface area of the second sphere:
\[ A_2 = 4 \pi (2r)^2 = 4 \pi (4r^2) = 16 \pi r^2 \]

### Step 5: Apply Newton's Law of Cooling
According to Newton's Law of Cooling, the rate of cooling \( \frac{dT}{dt} \) is proportional to the surface area and inversely proportional to the mass of the object:
\[ \frac{dT}{dt} \propto \frac{A}{m} \]

### Step 6: Calculate the Ratios of Cooling Rates
Now we can find the ratio of the rates of cooling for both spheres:
\[ \frac{\frac{dT_1}{dt}}{\frac{dT_2}{dt}} = \frac{\frac{A_1}{m_1}}{\frac{A_2}{m_2}} = \frac{A_1 \cdot m_2}{A_2 \cdot m_1} \]

Substituting the values:
\[ \frac{dT_1}{dt} : \frac{dT_2}{dt} = \frac{(4 \pi r^2) \cdot \left(\frac{32}{3} \pi \rho r^3\right)}{(16 \pi r^2) \cdot \left(\frac{8}{3} \pi \rho r^3\right)} \]

### Step 7: Simplify the Expression
\[ = \frac{(4)(32)}{(16)(8)} = \frac{128}{128} = 1 \]

### Conclusion
The ratio of the rates of cooling of the two spheres is:
\[ \frac{dT_1}{dt} : \frac{dT_2}{dt} = 1 : 1 \]