Water and mercury are filled in two cylindrical vessels up to same height. Both vessels have a hole in the wall near the bottom. The velocity of water and mercury coming out of the holes are `v_1 and v_2` respectively. Then

Options

A. `v_1=v_2`
B. `v_1=13.6v_2`
C. `v_1=v_2/13.6`
D. `v_1=sqrt(13.6) v_2`

Answer

To solve the problem, we will use Torricelli's law, which states that the velocity of fluid flowing out of an orifice under the influence of gravity is given by the formula:

\[ v = \sqrt{2gh} \]

where:
- \( v \) is the velocity of the fluid,
- \( g \) is the acceleration due to gravity,
- \( h \) is the height of the fluid column above the hole.

### Step-by-Step Solution:

1. **Identify the height of the liquid column**:
Both vessels are filled with water and mercury up to the same height \( h \).

2. **Apply Torricelli's Law**:
According to Torricelli's law, the velocity of the fluid exiting the hole is given by:
\[ v_1 = \sqrt{2gh} \quad \text{(for water)} \]
\[ v_2 = \sqrt{2gh} \quad \text{(for mercury)} \]

3. **Compare the velocities**:
Since both liquids are at the same height \( h \) and are subjected to the same gravitational acceleration \( g \), we can directly compare the two velocities:
\[ v_1 = v_2 \]

4. **Conclusion**:
Therefore, the velocities of water and mercury coming out of the holes are equal:
\[ v_1 = v_2 \]

### Final Answer:
The correct option is that \( v_1 = v_2 \).

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