A block of mass `5` kg and volume `0.05"m"^(3)` floats in a liquid of density `140 "kg m"^(-3)` .
The fraction of block inside liquid is

Options

A. `(7)/(5)`
B. `(2)/(7)`
C. `(5)/(7)`
D. `(7)/(2)`

Explanation

To find the fraction of the block that is submerged in the liquid, we can follow these steps:

### Step 1: Understand the Problem
We have a block of mass \( m = 5 \, \text{kg} \) and volume \( V = 0.05 \, \text{m}^3 \) floating in a liquid with a density \( \rho_L = 140 \, \text{kg/m}^3 \). We need to find the fraction of the block's volume that is submerged in the liquid.

### Step 2: Use Archimedes' Principle
According to Archimedes' principle, the weight of the liquid displaced by the submerged part of the block is equal to the weight of the block itself when it is floating.

### Step 3: Set Up the Equation
Let \( V_0 \) be the volume of the block that is submerged in the liquid. The weight of the block is given by:
\[ \text{Weight of the block} = m \cdot g \]
where \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).

The weight of the liquid displaced by the submerged volume \( V_0 \) is:
\[ \text{Weight of displaced liquid} = V_0 \cdot \rho_L \cdot g \]

Setting these two weights equal gives us:
\[ m \cdot g = V_0 \cdot \rho_L \cdot g \]

### Step 4: Cancel \( g \) from Both Sides
Since \( g \) is present on both sides, we can cancel it out:
\[ m = V_0 \cdot \rho_L \]

### Step 5: Solve for \( V_0 \)
Rearranging the equation to solve for \( V_0 \):
\[ V_0 = \frac{m}{\rho_L} \]

### Step 6: Substitute the Values
Now, substituting the known values:
\[ V_0 = \frac{5 \, \text{kg}}{140 \, \text{kg/m}^3} = \frac{5}{140} \, \text{m}^3 = \frac{1}{28} \, \text{m}^3 \]

### Step 7: Find the Fraction of the Block Submerged
The fraction of the block submerged \( F \) is given by:
\[ F = \frac{V_0}{V} \]
Substituting the values we have:
\[ F = \frac{\frac{1}{28}}{0.05} \]

### Step 8: Calculate the Fraction
To simplify \( F \):
\[ F = \frac{1}{28} \div 0.05 = \frac{1}{28} \times \frac{1}{0.05} = \frac{1}{28} \times 20 = \frac{20}{28} = \frac{5}{7} \]

### Final Answer
The fraction of the block that is submerged in the liquid is:
\[ \frac{5}{7} \]

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