A container of cross-section area `'S'` and height `'h'` is filled with mercury up to the brim. Then the container is sealed airtight and a hole of small cross section area `'S//n'` (where `'n'` is a positive constant) is puched in its bottom. Find out the time interval upto which the mercury will come out from the bottom hole.
[Take the atmospheric pressure to be equal to `h_(0)` height of mercury column: `h gt h_(0)`]

Explanation

To solve the problem, we need to find the time interval during which mercury will flow out of the hole at the bottom of the sealed container. We will use principles from fluid mechanics, particularly Bernoulli's theorem and the relationship between the velocities of the fluid.

### Step-by-Step Solution:

1. **Understanding the System**:
- We have a container with cross-sectional area \( S \) and height \( h \) filled with mercury.
- A small hole with cross-sectional area \( \frac{S}{n} \) is punched at the bottom.
- The container is sealed, so the pressure at the top is constant.

2. **Velocity Relationship**:
- Let \( V_1 \) be the velocity at which the mercury level decreases (the velocity of mercury inside the container).
- Let \( V_2 \) be the velocity of mercury flowing out of the hole.
- By the principle of conservation of mass, we have:
\[ S \cdot V_1 = \frac{S}{n} \cdot V_2 \implies V_2 = n \cdot V_1 \]

3. **Applying Bernoulli's Theorem**:
- We apply Bernoulli's theorem at two points:
- Point 1: At height \( y \) inside the container.
- Point 2: At the hole at the bottom.
- The equation becomes:
\[ P_1 + \frac{1}{2} \rho V_1^2 + \rho g y = P_2 + \frac{1}{2} \rho V_2^2 + \rho g \cdot 0 \]
- Since the container is sealed, the pressure at the top \( P_1 \) is 0. The atmospheric pressure at the hole \( P_2 \) is given as \( \rho g h_0 \):
\[ 0 + \frac{1}{2} \rho V_1^2 + \rho g y = \rho g h_0 + \frac{1}{2} \rho V_2^2 \]

4. **Simplifying the Equation**:
- Rearranging the equation gives:
\[ \frac{1}{2} \rho V_1^2 + \rho g y = \rho g h_0 + \frac{1}{2} \rho V_2^2 \]
- Canceling \( \rho \) from all terms, we have:
\[ \frac{1}{2} V_1^2 + g y = g h_0 + \frac{1}{2} V_2^2 \]

5. **Substituting for \( V_2 \)**:
- Substitute \( V_2 = n V_1 \):
\[ \frac{1}{2} V_1^2 + g y = g h_0 + \frac{1}{2} (n V_1)^2 \]
- This simplifies to:
\[ \frac{1}{2} V_1^2 + g y = g h_0 + \frac{1}{2} n^2 V_1^2 \]
- Rearranging gives:
\[ g y - g h_0 = \frac{1}{2} (n^2 - 1) V_1^2 \]

6. **Finding \( V_1 \)**:
- Solving for \( V_1^2 \):
\[ V_1^2 = \frac{2g (y - h_0)}{n^2 - 1} \]
- Thus,
\[ V_1 = \sqrt{\frac{2g (y - h_0)}{n^2 - 1}} \]

7. **Relating \( V_1 \) to Height Change**:
- Since \( V_1 = -\frac{dy}{dt} \):
\[ -\frac{dy}{dt} = \sqrt{\frac{2g (y - h_0)}{n^2 - 1}} \]
- Rearranging gives:
\[ dt = -\frac{dy}{\sqrt{\frac{2g (y - h_0)}{n^2 - 1}}} \]

8. **Integrating**:
- Integrate from \( y = h \) to \( y = h_0 \):
\[ t = -\sqrt{\frac{n^2 - 1}{2g}} \int_{h}^{h_0} \frac{dy}{\sqrt{y - h_0}} \]
- The integral evaluates to:
\[ t = -\sqrt{\frac{n^2 - 1}{2g}} \left[ 2\sqrt{y - h_0} \right]_{h}^{h_0} \]
- This gives:
\[ t = 2\sqrt{\frac{n^2 - 1}{2g}} \left( \sqrt{h - h_0} \right) \]

9. **Final Result**:
- Therefore, the time interval \( t \) during which the mercury will come out from the bottom hole is:
\[ t = 2 \sqrt{\frac{n^2 - 1}{2g}} \sqrt{h - h_0} \]