Two cylinders of same cross-section and length L but made of two different materials of densities `d_(1)` and `d_(2)` are cemented together to from a cylinder of length 2L. The combination floats in a liquid of density `d " " d_(1) lt d_(2)` then

Options

A. `d_(1) lt (3)/(4)d`
B. `(d)/(2) gt d_(1)`
C. `(d)/(4) gt d_(1)`
D. `d lt d_(1)`

Answer

To solve the problem of two cylinders of the same cross-section and length \( L \) made of different materials floating in a liquid, we can follow these steps:

### Step 1: Understand the Setup
We have two cylinders:
- Cylinder 1 has density \( d_1 \) and length \( L \).
- Cylinder 2 has density \( d_2 \) and length \( L \).

These cylinders are cemented together to form a single cylinder of length \( 2L \). The combination floats in a liquid of density \( d \), where \( d_1 < d_2 \).

### Step 2: Apply the Law of Flotation
According to the law of flotation, a floating body displaces a weight of liquid equal to its own weight. Therefore, we can write the equation:

\[ \text{Weight of the combined cylinder} = \text{Weight of the liquid displaced} \]

### Step 3: Calculate the Weight of the Combined Cylinder
The total weight of the combined cylinder can be expressed as:

\[ \text{Weight} = \text{Volume} \times \text{Density} \times g \]

The volume of each cylinder is given by:

\[ \text{Volume of Cylinder 1} = A \cdot L \quad \text{(where A is the cross-sectional area)} \]
\[ \text{Volume of Cylinder 2} = A \cdot L \]

Thus, the total weight of the combined cylinder is:

\[ W = (A \cdot L \cdot d_1 + A \cdot L \cdot d_2) \cdot g = A \cdot L \cdot (d_1 + d_2) \cdot g \]

### Step 4: Calculate the Weight of the Liquid Displaced
The length of the cylinder submerged in the liquid is \( \frac{3L}{2} \) (since \( L \) is above the liquid). Therefore, the weight of the liquid displaced is:

\[ \text{Weight of liquid displaced} = A \cdot \frac{3L}{2} \cdot d \cdot g \]

### Step 5: Set Up the Equation
Setting the weight of the combined cylinder equal to the weight of the liquid displaced gives us:

\[ A \cdot L \cdot (d_1 + d_2) \cdot g = A \cdot \frac{3L}{2} \cdot d \cdot g \]

### Step 6: Simplify the Equation
Canceling out \( A \), \( L \), and \( g \) from both sides results in:

\[ d_1 + d_2 = \frac{3}{2} d \]

### Step 7: Analyze the Relationship
Given that \( d_1 < d_2 \), we can express \( d_2 \) in terms of \( d_1 \):

\[ d_2 = \frac{3}{2} d - d_1 \]

Since \( d_1 < d_2 \), we can substitute:

\[ d_1 < \frac{3}{2} d - d_1 \]

### Step 8: Solve for \( d_1 \)
Rearranging gives:

\[ 2d_1 < \frac{3}{2} d \]

Dividing both sides by 2:

\[ d_1 < \frac{3}{4} d \]

### Conclusion
Thus, we find that:

\[ d_1 < \frac{3}{4} d \]