Water drops dripping from a faucet oscillate (bulge in and out) as they are falling. The time for one oscillation, T (measured in seconds), depends only on the surface tension (measured in kg/s2) and the mass M of the drop. Which of the following could be the correct formulan for the period?

To determine the correct formula for the period \( T \) of oscillation for a water drop, we will use **dimensional analysis**. The period \( T \) depends on two variables:

1. **Surface tension (\( \sigma \))**: measured in \( \text{kg/s}^2 \)
2. **Mass of the drop (\( M \))**: measured in \( \text{kg} \)

### Step 1: Analyze the dimensions of the variables

- The dimensions of **\( T \)** (the period) are time \([T] = \text{s}\).
- The dimensions of **surface tension \( \sigma \)** are:
  \[
  [\sigma] = \frac{\text{mass}}{\text{time}^2} = \frac{\text{kg}}{\text{s}^2}
  \]
- The dimensions of **mass \( M \)** are:
  \[
  [M] = \text{kg}
  \]

### Step 2: Form a relationship between \( T \), \( M \), and \( \sigma \)

We assume that \( T \) is proportional to a product of powers of \( M \) and \( \sigma \), i.e.,
\[
T = k M^a \sigma^b
\]
where \( k \) is a dimensionless constant, and \( a \) and \( b \) are the powers to be determined using dimensional analysis.

### Step 3: Apply dimensional consistency

Substitute the dimensions of \( T \), \( M \), and \( \sigma \) into the equation:

\[
[T] = [M]^a [\sigma]^b
\]

This becomes:

\[
\text{s} = (\text{kg})^a \left(\frac{\text{kg}}{\text{s}^2}\right)^b
\]

Simplify the right-hand side:

\[
\text{s} = \text{kg}^a \cdot \text{kg}^b \cdot \text{s}^{-2b}
\]

Combine terms with the same units:

\[
\text{s} = \text{kg}^{a+b} \cdot \text{s}^{-2b}
\]

### Step 4: Equate powers of dimensions

To satisfy dimensional consistency, we equate the powers of mass (kg) and time (s) on both sides:

- For mass (kg): \( a + b = 0 \)
- For time (s): \( -2b = 1 \)

From \( -2b = 1 \), we find \( b = -\frac{1}{2} \).

Substitute \( b = -\frac{1}{2} \) into \( a + b = 0 \):

\[
a - \frac{1}{2} = 0 \implies a = \frac{1}{2}
\]

### Step 5: Write the final formula

Thus, the period \( T \) is:

\[
T = k M^{\frac{1}{2}} \sigma^{-\frac{1}{2}}
\]

This can be written as:

\[
T = k \frac{\sqrt{M}}{\sqrt{\sigma}}
\]

### Conclusion

The period of oscillation \( T \) is proportional to \( \frac{\sqrt{M}}{\sqrt{\sigma}} \).

---

Would you like further details or have any questions about this solution?

### Related Questions:
1. How does surface tension affect the formation of water droplets?
2. Can the mass of a droplet change during free fall?
3. What is the physical significance of dimensional analysis in physics?
4. How would the period change if the mass of the drop doubled?
5. What are other examples of oscillations that depend on mass and tension?

**Tip**: Dimensional analysis is a powerful tool to derive relations between physical quantities without solving complex equations.

Water drops dripping from a faucet oscillate (bulge in and out) as they are falling. The time for one oscillation, T (measured in seconds), depends only on the surface tension (measured in kg/s2) and the mass M of the drop. Which of the following could be the correct formula for the period?

Explore the formula for the oscillation period of a water drop, derived using dimensional analysis. Learn how surface tension and mass influence the oscillation, with key concepts in oscillations and dimensional consistency. Suitable for high school and college-level physics students.

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