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 need to calculate the new reading on the spring balance after immersing the stone in the water.

### Step 1: Understand the initial conditions
The initial reading on the spring balance is given as 1.5 kg. This reading represents the combined weight of the beaker and the water inside it.

### Step 2: Determine the properties of the stone
The stone has a mass of 0.5 kg and a density of \(10^4 \, \text{kg/m}^3\). We need to calculate the volume of the stone to find the buoyant force acting on it when it is immersed in water.

### Step 3: Calculate the volume of the stone
The volume \(V\) of the stone can be calculated using the formula:
\[ V = \frac{m}{\rho} \]
where \(m\) is the mass of the stone and \(\rho\) is its density.

Substituting the values:
\[ V = \frac{0.5 \, \text{kg}}{10^4 \, \text{kg/m}^3} = 5 \times 10^{-5} \, \text{m}^3 \]

### Step 4: Calculate the buoyant force (upthrust)
The buoyant force \(F_b\) acting on the stone is equal to the weight of the water displaced by the stone. This can be calculated using the formula:
\[ F_b = V \cdot \rho_{\text{water}} \cdot g \]
where \(\rho_{\text{water}} = 10^3 \, \text{kg/m}^3\) (density of water) and \(g \approx 9.81 \, \text{m/s}^2\).

Substituting the values:
\[ F_b = 5 \times 10^{-5} \, \text{m}^3 \cdot 10^3 \, \text{kg/m}^3 \cdot 9.81 \, \text{m/s}^2 \]
\[ F_b = 5 \times 10^{-5} \cdot 10^3 \cdot 9.81 = 0.4905 \, \text{N} \]

### Step 5: Convert buoyant force to kg
To find the equivalent mass that this force represents, we can use the relation:
\[ \text{mass} = \frac{F_b}{g} \]
Substituting the values:
\[ \text{mass} = \frac{0.4905 \, \text{N}}{9.81 \, \text{m/s}^2} \approx 0.05 \, \text{kg} \]

### Step 6: Calculate the new reading on the spring balance
The new reading on the spring balance will be the initial reading plus the additional weight due to the buoyant force:
\[ \text{New Reading} = 1.5 \, \text{kg} + 0.05 \, \text{kg} = 1.55 \, \text{kg} \]

### Conclusion
The new reading on the spring balance after immersing the stone in water will be approximately **1.55 kg**.

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