Two spheres of volume 250 cc each but of relative densities 0.8 and 1.2 are connected by a string and the combination is immersed in a liquid. Find the tension T in the string. `(g=10m//s^(2))`

Options

A. `5.0 N`
B. `0.5 N`
C. `1.0 N`
D. `2.0 N`

Answer

To find the tension \( T \) in the string connecting the two spheres, we can follow these steps:

### Step 1: Understand the Problem
We have two spheres with volumes of 250 cc each, having relative densities of 0.8 and 1.2. We need to find the tension in the string when they are immersed in a liquid.

### Step 2: Convert Volume to Cubic Meters
The volume of each sphere is given as 250 cc. We convert this to cubic meters:
\[ V = 250 \, \text{cc} = 250 \times 10^{-6} \, \text{m}^3 \]

### Step 3: Calculate the Densities of the Spheres
Using the relative densities, we can find the actual densities of the spheres:
- Density of Sphere 1 (\( \rho_1 \)):
\[ \rho_1 = 0.8 \times \rho_w = 0.8 \times 1000 \, \text{kg/m}^3 = 800 \, \text{kg/m}^3 \]
- Density of Sphere 2 (\( \rho_2 \)):
\[ \rho_2 = 1.2 \times \rho_w = 1.2 \times 1000 \, \text{kg/m}^3 = 1200 \, \text{kg/m}^3 \]

### Step 4: Calculate the Weight of Each Sphere
The weight of each sphere can be calculated using the formula \( W = mg \):
- Weight of Sphere 1 (\( W_1 \)):
\[ W_1 = m_1 g = \rho_1 V g = 800 \times 250 \times 10^{-6} \times 10 = 2 \, \text{N} \]
- Weight of Sphere 2 (\( W_2 \)):
\[ W_2 = m_2 g = \rho_2 V g = 1200 \times 250 \times 10^{-6} \times 10 = 3 \, \text{N} \]

### Step 5: Calculate the Buoyant Force on Each Sphere
The buoyant force (\( F_B \)) acting on each sphere is equal to the weight of the liquid displaced, which is the same for both since they have the same volume:
\[ F_B = \rho_{liquid} V g \]
Assuming the liquid is water (\( \rho_{liquid} = 1000 \, \text{kg/m}^3 \)):
\[ F_B = 1000 \times 250 \times 10^{-6} \times 10 = 2.5 \, \text{N} \]

### Step 6: Apply Newton's Second Law to Each Sphere
For Sphere 1:
\[ F_B - T - W_1 = 0 \implies 2.5 - T - 2 = 0 \implies T = 0.5 \, \text{N} \]

For Sphere 2:
\[ T + F_B - W_2 = 0 \implies T + 2.5 - 3 = 0 \implies T = 0.5 \, \text{N} \]

### Step 7: Conclusion
The tension \( T \) in the string connecting the two spheres is:
\[ T = 0.5 \, \text{N} \]

### Final Answer
The tension in the string is \( \boxed{0.5 \, \text{N}} \).