A certain block weights `22` N in air . It weights `17` N when immersed in water. When immersed in another liquid it weights `18` N . Density of other liquid is `800 "kg m"^(-3)` .
Calculate the relative density of the other liquid .

Options

A. `0.8`
B. `0.67`
C. `0.2`
D. `1`

Answer

To calculate the relative density of the other liquid, we can follow these steps:

### Step 1: Understand the Problem
We are given the weight of a block in air, in water, and in another liquid. We need to find the relative density of the other liquid.

### Step 2: Identify Given Values
- Weight of the block in air (W_air) = 22 N
- Weight of the block in water (W_water) = 17 N
- Weight of the block in another liquid (W_liquid) = 18 N
- Density of the other liquid (ρ_liquid) = 800 kg/m³

### Step 3: Calculate Upthrust in Water
The upthrust (buoyant force) when the block is immersed in water can be calculated as:
\[ \text{Upthrust in water} = W_{\text{air}} - W_{\text{water}} \]
\[ \text{Upthrust in water} = 22 \, \text{N} - 17 \, \text{N} = 5 \, \text{N} \]

### Step 4: Calculate Upthrust in the Other Liquid
Similarly, the upthrust when the block is immersed in the other liquid is:
\[ \text{Upthrust in liquid} = W_{\text{air}} - W_{\text{liquid}} \]
\[ \text{Upthrust in liquid} = 22 \, \text{N} - 18 \, \text{N} = 4 \, \text{N} \]

### Step 5: Relate Upthrust to Volume and Density
The upthrust is equal to the weight of the liquid displaced, which can be expressed as:
\[ \text{Upthrust} = V \cdot \rho \cdot g \]
where \( V \) is the volume of the block, \( \rho \) is the density of the liquid, and \( g \) is the acceleration due to gravity.

For water:
\[ 5 \, \text{N} = V \cdot \rho_{\text{water}} \cdot g \]
For the other liquid:
\[ 4 \, \text{N} = V \cdot \rho_{\text{liquid}} \cdot g \]

### Step 6: Set Up the Ratio
Now we can set up the ratio of the densities using the upthrust values:
\[ \frac{\rho_{\text{liquid}}}{\rho_{\text{water}}} = \frac{4 \, \text{N}}{5 \, \text{N}} \]

### Step 7: Calculate Relative Density
Since the density of water (ρ_water) is approximately 1000 kg/m³, we can express the relative density (RD) as:
\[ RD = \frac{\rho_{\text{liquid}}}{\rho_{\text{water}}} = \frac{4}{5} = 0.8 \]

### Step 8: Conclusion
Thus, the relative density of the other liquid is 0.8.

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