A beaker containing water is placed on the platform of a spring balance. The balance reads `1.5 kg.` A stone of mass `0.5 kg` and density `10^(4) kg//m^(3)` is immersed in water without touching the walls of the beaker. What will be the balance reading now?

Options

A. 2 kg
B. 2.5 kg
C. 1 kg
D. 3 kg

Answer

To solve the problem step by step, we will analyze the situation involving the beaker, the water, and the stone.

### Step 1: Understand the initial conditions
The initial reading of the spring balance is given as 1.5 kg. This reading represents the weight of the beaker filled with water.

### Step 2: Determine the weight of the stone
The stone has a mass of 0.5 kg. The weight of the stone can be calculated using the formula:
\[ \text{Weight} = \text{mass} \times g \]
where \( g \) (acceleration due to gravity) is approximately \( 9.81 \, \text{m/s}^2 \). However, since we are only concerned with the mass for the balance reading, we can simply use the mass of the stone, which is 0.5 kg.

### Step 3: Understand the buoyancy force
When the stone is immersed in water, it experiences a buoyant force equal to the weight of the water displaced by the stone. The buoyant force acts upwards and is equal to the weight of the stone when it is fully submerged.

### Step 4: Calculate the new reading on the balance
When the stone is immersed in the water, it does not touch the walls of the beaker, which means it does not exert any additional forces on the walls. The total weight that the spring balance will read now includes:
1. The initial weight of the beaker and water (1.5 kg).
2. The weight of the stone (0.5 kg).

Thus, the new reading on the balance will be:
\[ \text{New reading} = \text{Initial reading} + \text{Weight of the stone} \]
\[ \text{New reading} = 1.5 \, \text{kg} + 0.5 \, \text{kg} = 2.0 \, \text{kg} \]

### Conclusion
The new reading on the spring balance after immersing the stone in water will be **2.0 kg**.

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