1000 small drops of water each of radius 'r' are joined together to form a single drop of water During the process the change in energy has raised the temerature . If T is the surface tension of the water , d is density , the rise in temperature is

Options

A. `(100T)/(Jr)`
B. `(2.7T)/(Jrd)`
C. `(10T)/(Jr)`
D. `T/(Jrd)s`

Explanation

To solve the problem of finding the rise in temperature when 1000 small drops of water, each of radius 'r', are combined to form a single larger drop, we can follow these steps:

### Step 1: Calculate the Volume of the Small Drops
The volume \( V \) of a single small drop of water (sphere) is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
For 1000 small drops, the total volume \( V_{small} \) is:
\[ V_{small} = 1000 \times \frac{4}{3} \pi r^3 = \frac{4000}{3} \pi r^3 \]

### Step 2: Calculate the Volume of the Larger Drop
Since there is no loss of volume when the small drops combine into one larger drop, the volume \( V_{large} \) of the larger drop must equal the total volume of the small drops:
\[ V_{large} = \frac{4}{3} \pi R^3 \]
Setting the volumes equal gives:
\[ \frac{4000}{3} \pi r^3 = \frac{4}{3} \pi R^3 \]
Cancelling \( \frac{4}{3} \pi \) from both sides:
\[ 1000 r^3 = R^3 \]
Taking the cube root:
\[ R = 10 r \]

### Step 3: Calculate the Surface Energy of the Small Drops
The surface area \( A \) of a single small drop is:
\[ A = 4 \pi r^2 \]
Thus, the total surface area of 1000 small drops is:
\[ A_{small} = 1000 \times 4 \pi r^2 = 4000 \pi r^2 \]
The surface energy \( E_{small} \) of the small drops is given by:
\[ E_{small} = T \times A_{small} = T \times 4000 \pi r^2 \]

### Step 4: Calculate the Surface Energy of the Larger Drop
The surface area of the larger drop is:
\[ A_{large} = 4 \pi R^2 = 4 \pi (10r)^2 = 400 \pi r^2 \]
The surface energy \( E_{large} \) of the larger drop is:
\[ E_{large} = T \times A_{large} = T \times 400 \pi r^2 \]

### Step 5: Calculate the Change in Surface Energy
The change in surface energy \( \Delta E \) when the small drops combine into one large drop is:
\[ \Delta E = E_{large} - E_{small} = (T \times 400 \pi r^2) - (T \times 4000 \pi r^2) \]
\[ \Delta E = T \times 400 \pi r^2 - T \times 4000 \pi r^2 = T \times (400 - 4000) \pi r^2 = -T \times 3600 \pi r^2 \]

### Step 6: Relate Change in Energy to Temperature Rise
The change in surface energy is converted into heat, which raises the temperature of the water. The heat \( Q \) absorbed by the water is given by:
\[ Q = m \cdot c \cdot \Delta T \]
Where:
- \( m \) is the mass of the water,
- \( c \) is the specific heat capacity,
- \( \Delta T \) is the change in temperature.

The mass \( m \) of the water can be calculated using density \( d \) and volume:
\[ m = d \times V_{large} = d \times \frac{4}{3} \pi R^3 = d \times \frac{4}{3} \pi (10r)^3 = d \times \frac{4000}{3} \pi r^3 \]

### Step 7: Substitute and Solve for Temperature Rise
Substituting \( Q \) into the equation:
\[ -T \times 3600 \pi r^2 = \left(d \times \frac{4000}{3} \pi r^3\right) \cdot c \cdot \Delta T \]
Cancelling \( \pi \) from both sides:
\[ -T \times 3600 r^2 = \left(d \times \frac{4000}{3} r^3\right) \cdot c \cdot \Delta T \]
Rearranging for \( \Delta T \):
\[ \Delta T = \frac{-T \times 3600 r^2}{d \times \frac{4000}{3} r^3 \cdot c} \]
\[ \Delta T = \frac{-T \times 3600 \cdot 3}{d \cdot 4000 \cdot r \cdot c} \]
\[ \Delta T = \frac{-10800 T}{4000 d c r} \]
\[ \Delta T = \frac{-27 T}{10 d c r} \]

### Final Result
The rise in temperature \( \Delta T \) is:
\[ \Delta T = \frac{27 T}{10 d c r} \]