To find the percentage error in the measurement of the time period \( T \) of small oscillations of the surface of a liquid drop, we start with the relationship given by:
\[
T = c s^x \rho^y r^z
\]
where:
- \( s \) is the surface tension,
- \( \rho \) is the density of the liquid,
- \( r \) is the mean radius of the drop,
- \( c \) is a constant,
- \( x, y, z \) are the powers to which \( s, \rho, r \) are raised.
### Step 1: Identify the errors in measurements
The problem states the following errors:
- Error in mean radius \( r \): \( 2\% \)
- Error in surface tension \( s \): \( 1\% \)
- Error in density \( \rho \): \( 1\% \)
### Step 2: Express the percentage error in \( T \)
The percentage error in \( T \) can be expressed in terms of the individual percentage errors in \( s, \rho, \) and \( r \) as follows:
\[
\text{Percentage error in } T = \left| x \cdot \frac{\Delta s}{s} \right| + \left| y \cdot \frac{\Delta \rho}{\rho} \right| + \left| z \cdot \frac{\Delta r}{r} \right|
\]
### Step 3: Determine the values of \( x, y, z \)
From the problem, we need to find the values of \( x, y, z \) based on dimensional analysis.
Given the relationships:
- The unit of surface tension \( s \) is \( \text{M} \cdot \text{T}^{-2} \)
- The unit of density \( \rho \) is \( \text{M} \cdot \text{L}^{-3} \)
- The unit of radius \( r \) is \( \text{L} \)
We can set up equations based on the dimensions of \( T \):
1. \( x + y - 2 = 0 \) (from the dimension of \( s \))
2. \( -3y + z = 0 \) (from the dimension of \( \rho \))
3. \( z = 1 \) (from the dimension of \( r \))
Solving these equations, we find:
- From \( z = 1 \), substituting into the second equation gives \( -3y + 1 = 0 \) → \( y = \frac{1}{3} \).
- Substituting \( y = \frac{1}{3} \) into the first equation gives \( x + \frac{1}{3} - 2 = 0 \) → \( x = \frac{5}{3} \).
### Step 4: Substitute the errors into the formula
Now substituting the values of \( x, y, z \) into the percentage error formula:
\[
\text{Percentage error in } T = \left| \frac{5}{3} \cdot 1\% \right| + \left| \frac{1}{3} \cdot 1\% \right| + \left| 1 \cdot 2\% \right|
\]
Calculating each term:
- Contribution from \( s \): \( \frac{5}{3} \cdot 1\% = \frac{5}{3}\% \approx 1.67\% \)
- Contribution from \( \rho \): \( \frac{1}{3} \cdot 1\% = \frac{1}{3}\% \approx 0.33\% \)
- Contribution from \( r \): \( 1 \cdot 2\% = 2\% \)
### Step 5: Add the contributions
Now, summing these contributions gives:
\[
\text{Total percentage error in } T = 1.67\% + 0.33\% + 2\% = 4\%
\]
### Final Answer
Thus, the percentage error in the measurement of the time period \( T \) is:
\[
\boxed{4\%}
\]
