A metal plate is lying at the bottom of a tank full of a transparent liquid. Height of tank is `100` cm but the plate appears to be at `45` cm above bottom. The rerfractive index of liquid is

Options

A. `1.00`
B. `1.53`
C. `1.81`
D. `1.32`

Answer

To find the refractive index of the liquid in which a metal plate is submerged, we can use the relationship between the real depth (D) and the apparent depth (D') of the object.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Height of the tank (D) = 100 cm (this is the real depth)
- Apparent depth (D') = 45 cm above the bottom of the tank.
- Therefore, the distance from the top of the liquid to the plate is:
\[ D' = 100 \, \text{cm} - 45 \, \text{cm} = 55 \, \text{cm} \]

2. **Understand the Relationship:**
The relationship between real depth (D) and apparent depth (D') is given by the formula:
\[ D' = \frac{\mu_2}{\mu_1} \cdot D \]
where:
- \( \mu_1 \) = refractive index of the liquid (unknown)
- \( \mu_2 \) = refractive index of air = 1 (since air is the medium above the liquid)

3. **Substituting Known Values:**
Since \( \mu_2 = 1 \), we can rewrite the equation:
\[ D' = \frac{1}{\mu_1} \cdot D \]
Rearranging gives:
\[ \mu_1 = \frac{D}{D'} \]

4. **Plugging in the Values:**
Now substitute the values of D and D':
\[ \mu_1 = \frac{100 \, \text{cm}}{55 \, \text{cm}} \]

5. **Calculating the Refractive Index:**
Performing the division:
\[ \mu_1 = \frac{100}{55} \approx 1.8181 \]

6. **Final Result:**
The refractive index of the liquid is approximately:
\[ \mu_1 \approx 1.81 \]

### Conclusion:
The refractive index of the liquid is approximately **1.81**.