A u tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are at the same levels with 10 cm of water in one arm and 12.5 cm of spirit on the other arm as shown. If 15 cm of water and spirit each are further poured into the respective arms of the tube. Difference in the level of mercury in the two arms is (Take, relative density of mercury = 13.6, relative density of spirit=0.8)

Explanation

To solve the problem, we need to analyze the U-tube with water and methylated spirit separated by mercury. We will calculate the difference in the level of mercury in the two arms after pouring additional water and spirit.

### Step-by-Step Solution:

1. **Initial Setup**:
- The left arm has 10 cm of water.
- The right arm has 12.5 cm of methylated spirit.
- The mercury levels in both arms are initially the same.

2. **Adding Water and Spirit**:
- We add 15 cm of water to the left arm and 15 cm of spirit to the right arm.
- Thus, the new heights become:
- Left arm (water): \(10 \, \text{cm} + 15 \, \text{cm} = 25 \, \text{cm}\)
- Right arm (spirit): \(12.5 \, \text{cm} + 15 \, \text{cm} = 27.5 \, \text{cm}\)

3. **Pressure Calculation**:
- The pressure at point A (left arm) can be expressed as:
\[ P_A = P_0 + \rho_w g h_w \]
where \(h_w = 25 \, \text{cm}\) (height of water), \(\rho_w\) is the density of water, and \(P_0\) is atmospheric pressure.

- The pressure at point B (right arm) can be expressed as:
\[ P_B = P_0 + \rho_s g h_s + \rho_m g h \]
where \(h_s = 27.5 \, \text{cm}\) (height of spirit), \(\rho_s\) is the density of spirit, \(\rho_m\) is the density of mercury, and \(h\) is the height of mercury above the initial level.

4. **Equating Pressures**:
- Since the mercury levels are at the same height initially, we can set \(P_A = P_B\):
\[ \rho_w g (25) = \rho_s g (27.5) + \rho_m g h \]
- Cancel \(g\) from both sides:
\[ \rho_w (25) = \rho_s (27.5) + \rho_m h \]

5. **Substituting Densities**:
- The relative density of water is 1, so \(\rho_w = 1 \, \text{g/cm}^3\).
- The relative density of spirit is given as 0.8, so \(\rho_s = 0.8 \, \text{g/cm}^3\).
- The relative density of mercury is given as 13.6, so \(\rho_m = 13.6 \, \text{g/cm}^3\).

6. **Plugging in Values**:
\[ 1 \times 25 = 0.8 \times 27.5 + 13.6 h \]
\[ 25 = 22 + 13.6 h \]
\[ 25 - 22 = 13.6 h \]
\[ 3 = 13.6 h \]
\[ h = \frac{3}{13.6} \approx 0.2206 \, \text{cm} \]

7. **Final Result**:
- The difference in the level of mercury in the two arms is approximately \(0.22 \, \text{cm}\).