An object is floating in water with fraction of its volume submerged in water filled in beaker which is at rest with respect to ground. Now this beaker is accelerated with constant acceleration downwards `(a lt g)` then

Options

A. fraction of volume submerged will not change.
B. Buoyant force on the block will remain same
C. pressure at any point in the water will be more than earlier
D. Fraction of volume submerged will change.

Explanation

To solve the problem, we need to analyze the situation where an object is floating in water in a beaker that is being accelerated downwards with a constant acceleration \( a \) (where \( a < g \)).

### Step-by-Step Solution:

1. **Understanding the Forces Acting on the Object:**
- When the beaker is at rest, the object floating in the water experiences two main forces: the gravitational force acting downwards (weight of the object) and the buoyant force acting upwards.
- The weight of the object is given by \( W = mg \), where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity.
- The buoyant force \( F_b \) is given by Archimedes' principle: \( F_b = \rho V_s g \), where \( \rho \) is the density of the fluid, \( V_s \) is the volume of the fluid displaced (which is equal to the volume of the submerged part of the object), and \( g \) is the acceleration due to gravity.

2. **Condition for Floating:**
- For the object to float, the buoyant force must equal the weight of the object:
\[ F_b = W \implies \rho V_s g = mg \]
- From this, we can derive that the fraction of the volume submerged \( \frac{V_s}{V} = \frac{m}{\rho V} \), where \( V \) is the total volume of the object.

3. **Effect of Downward Acceleration:**
- When the beaker is accelerated downwards with acceleration \( a \), the effective gravitational acceleration acting on the object becomes \( g' = g - a \).
- The new buoyant force can be expressed as:
\[ F_b' = \rho V_s g' \]
- The weight of the object remains the same, \( W = mg \).

4. **New Condition for Floating:**
- For the object to continue floating under the new conditions, we need to equate the new buoyant force to the weight:
\[ \rho V_s (g - a) = mg \]
- Rearranging gives:
\[ \rho V_s g - \rho V_s a = mg \]
- Since \( mg = \rho V_s g \), we can substitute this into the equation:
\[ \rho V_s g - \rho V_s a = \rho V_s g \]
- This implies that the term involving \( a \) does not affect the balance of forces, and therefore, the fraction of the volume submerged remains unchanged.

5. **Conclusion:**
- The fraction of the volume submerged does not change when the beaker is accelerated downwards with a constant acceleration \( a < g \). The buoyant force remains equal to the weight of the object, ensuring it continues to float.