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

Explanation

To solve the problem step by step, we will analyze the situation before and after immersing the stone in the water.

### 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 properties of the stone
- The mass of the stone is given as 0.5 kg.
- The density of the stone is given as \(10^4 \, \text{kg/m}^3\).

### Step 3: Calculate the volume of the stone
- The volume \(V\) of the stone can be calculated using the formula:
\[ V = \frac{\text{mass}}{\text{density}} = \frac{0.5 \, \text{kg}}{10^4 \, \text{kg/m}^3} = 5 \times 10^{-5} \, \text{m}^3 \]

### Step 4: Calculate the buoyant force acting on the stone
- The buoyant force \(F_b\) acting on the stone when it is immersed in water can be calculated using Archimedes' principle:
\[ F_b = \text{density of water} \times V \times g \]
where the density of water is approximately \(10^3 \, \text{kg/m}^3\) and \(g\) (acceleration due to gravity) is approximately \(10 \, \text{m/s}^2\).

Substituting the values:
\[ F_b = 10^3 \, \text{kg/m}^3 \times 5 \times 10^{-5} \, \text{m}^3 \times 10 \, \text{m/s}^2 = 0.5 \, \text{N} \]

### Step 5: Calculate the new reading on the spring balance
- When the stone is immersed in the water, the spring balance will read the weight of the beaker with water plus the weight of the stone minus the buoyant force acting on the stone.

- The initial weight (reading) on the balance is:
\[ W_1 = 1.5 \, \text{kg} \]

- The new reading \(N\) can be expressed as:
\[ N = W_1 + \text{weight of the stone} - F_b \]

- The weight of the stone is:
\[ \text{Weight of the stone} = \text{mass} \times g = 0.5 \, \text{kg} \times 10 \, \text{m/s}^2 = 5 \, \text{N} \]

- Since we are working in kg, we need to convert the buoyant force into kg equivalent:
\[ F_b = \frac{0.5 \, \text{N}}{10 \, \text{m/s}^2} = 0.05 \, \text{kg} \]

- Now substituting the values:
\[ N = 1.5 \, \text{kg} + 0.5 \, \text{kg} - 0.05 \, \text{kg} = 1.5 + 0.5 - 0.05 = 1.95 \, \text{kg} \]

### Final Answer
The new reading on the spring balance after immersing the stone in water will be **1.95 kg**.