A stone of relative density K is released from rest on the surface of a lake. If viscous effects are lgnored, the stone sinks in water with an acceleration of

Options

A. `g(1-K)`
B. `g(1+K)`
C. `g(1-(1)/(K))`
D. `g(1+(1)/(K))`

Explanation

To solve the problem of a stone of relative density \( K \) sinking in water with acceleration, we can follow these steps:

### Step 1: Define the Variables
- Let the density of the stone be \( \sigma \).
- Let the density of water be \( \rho \).
- The relative density \( K \) is defined as \( K = \frac{\sigma}{\rho} \).

### Step 2: Identify the Forces Acting on the Stone
When the stone is released in water, two main forces act on it:
1. The weight of the stone (\( W \)):
\[ W = Mg = \sigma V g \]
where \( V \) is the volume of the stone and \( g \) is the acceleration due to gravity.

2. The buoyant force (\( F_b \)):
\[ F_b = \text{Volume of the stone displaced} \times \text{Density of water} \times g = V \rho g \]

### Step 3: Apply Newton's Second Law
According to Newton's second law, the net force acting on the stone is equal to the mass of the stone multiplied by its acceleration (\( A \)):
\[ \sigma V A = W - F_b \]
Substituting the expressions for \( W \) and \( F_b \):
\[ \sigma V A = \sigma V g - V \rho g \]

### Step 4: Simplify the Equation
We can factor out \( V \) from both sides (assuming \( V \neq 0 \)):
\[ \sigma A = \sigma g - \rho g \]

### Step 5: Solve for Acceleration \( A \)
Rearranging the equation gives:
\[ A = g \left( \frac{\sigma - \rho}{\sigma}\right) \]
Substituting \( \sigma = K \rho \) (from the definition of relative density):
\[ A = g \left( \frac{K \rho - \rho}{K \rho}\right) = g \left( \frac{(K - 1) \rho}{K \rho}\right) \]
This simplifies to:
\[ A = g \left( \frac{K - 1}{K}\right) \]

### Step 6: Final Expression for Acceleration
Thus, the acceleration of the stone as it sinks in water is:
\[ A = g \left(1 - \frac{1}{K}\right) \]

### Conclusion
The stone sinks in water with an acceleration of:
\[ A = g \left(1 - \frac{1}{K}\right) \]

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