Similar objects are dropped into two different liquids A and B. The displacement of the objects in A and B in 5 s is `S_1` and `S_2` respectively. If density of the liquid A is greater than the density of the liquid B, which among the following relation is correct?

Options

A. `S_1=S_2`
B. `S_1gtS_2`
C. `S_1ltS_2`
D. Cannot be determined

Explanation

To solve the problem, we need to analyze the motion of similar objects dropped into two different liquids, A and B, and understand how the density of the liquids affects their displacement over a given time period.

### Step-by-Step Solution:

1. **Understanding the Problem**: We have two liquids, A and B, with different densities. The density of liquid A is greater than that of liquid B. We need to find the relationship between the displacements \( S_1 \) (in liquid A) and \( S_2 \) (in liquid B) after 5 seconds.

2. **Using the Equation of Motion**: The displacement \( S \) of an object in a fluid can be described by the equation:
\[ S = ut + \frac{1}{2} a t^2 \]
where \( u \) is the initial velocity (which is 0 when the object is dropped), \( a \) is the acceleration, and \( t \) is the time.

3. **Simplifying the Equation**: Since the initial velocity \( u = 0 \), the equation simplifies to:
\[ S = \frac{1}{2} a t^2 \]
Given that \( t \) is the same for both liquids (5 seconds), we can express the displacements as:
\[ S_1 = \frac{1}{2} a_1 t^2 \quad \text{and} \quad S_2 = \frac{1}{2} a_2 t^2 \]

4. **Analyzing the Forces**: The net force acting on the object when submerged in a liquid is given by:
\[ F = mg - \rho V g \]
where:
- \( m \) is the mass of the object,
- \( \rho \) is the density of the liquid,
- \( V \) is the volume of the object,
- \( g \) is the acceleration due to gravity.

5. **Finding the Acceleration**: The acceleration \( a \) can be expressed as:
\[ a = \frac{F}{m} = \frac{mg - \rho V g}{m} = g - \frac{\rho V g}{m} \]
Since \( V \) and \( g \) are constants, the acceleration depends on the density of the liquid:
\[ a = g \left(1 - \frac{\rho V}{m}\right) \]

6. **Comparing Accelerations**: Since the density of liquid A (\( \rho_A \)) is greater than that of liquid B (\( \rho_B \)):
\[ a_1 < a_2 \quad \text{(because } \rho_A > \rho_B\text{)} \]

7. **Relating Displacements**: Since displacement is proportional to acceleration (for the same time interval), we have:
\[ S_1 \propto a_1 \quad \text{and} \quad S_2 \propto a_2 \]
Therefore, since \( a_1 < a_2 \), it follows that:
\[ S_1 < S_2 \]

8. **Conclusion**: The correct relation is:
\[ S_1 < S_2 \]

### Final Answer:
The displacement in liquid A is less than the displacement in liquid B, or \( S_1 < S_2 \).

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