The time period (T) of small oscillations of the surface of a liquid drop depends on its surface tension (s), the density `(rho)` of the liquid and it's mean radius (r) as `T = cs^(x) rho^(y)r^(z)`. If in the measurement of the mean radius of the drop, the error is `2 %`, the error in the measurement of surface tension and density both are of `1%`. Find the percentage error in measurement of the time period.

Answer

To solve the problem, we need to find the percentage error in the measurement of the time period \( T \) of small oscillations of a liquid drop, which is given by the formula:

\[ T = c s^x \rho^y r^z \]

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 exponents that we need to determine.

### Step 1: Determine the values of \( x, y, z \) using dimensional analysis.

The dimensions of each quantity are:
- Time \( [T] = T \)
- Surface tension \( [s] = M T^{-2} \)
- Density \( [\rho] = M L^{-3} \)
- Radius \( [r] = L \)

Setting up the dimensional equation:

\[ [T] = [s]^x [\rho]^y [r]^z \]

Substituting the dimensions:

\[ T = (M T^{-2})^x (M L^{-3})^y (L)^z \]

This simplifies to:

\[ T = M^{x+y} L^{-3y + z} T^{-2x} \]

Equating the powers of \( M, L, T \):

1. For \( M \): \( x + y = 0 \) (1)
2. For \( L \): \( -3y + z = 0 \) (2)
3. For \( T \): \( -2x = 1 \) (3)

From equation (3):

\[ x = -\frac{1}{2} \]

Substituting \( x \) into equation (1):

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

Substituting \( y \) into equation (2):

\[ -3 \left(\frac{1}{2}\right) + z = 0 \implies z = \frac{3}{2} \]

Thus, we have:

\[ x = -\frac{1}{2}, \quad y = \frac{1}{2}, \quad z = \frac{3}{2} \]

### Step 2: Write the equation for percentage error in \( T \).

The percentage error in \( T \) can be expressed as:

\[ \frac{\Delta T}{T} \times 100 = \left| x \frac{\Delta s}{s} + y \frac{\Delta \rho}{\rho} + z \frac{\Delta r}{r} \right| \times 100 \]

Substituting the values of \( x, y, z \):

\[ \frac{\Delta T}{T} \times 100 = \left| -\frac{1}{2} \frac{\Delta s}{s} + \frac{1}{2} \frac{\Delta \rho}{\rho} + \frac{3}{2} \frac{\Delta r}{r} \right| \times 100 \]

### Step 3: Substitute the given percentage errors.

Given:
- Error in \( s \) (surface tension) = \( 1\% \)
- Error in \( \rho \) (density) = \( 1\% \)
- Error in \( r \) (mean radius) = \( 2\% \)

Substituting these values:

\[ \frac{\Delta T}{T} \times 100 = \left| -\frac{1}{2}(1) + \frac{1}{2}(1) + \frac{3}{2}(2) \right| \]

Calculating:

\[ = \left| -\frac{1}{2} + \frac{1}{2} + 3 \right| = \left| 3 \right| = 3 \]

Thus, the total percentage error in \( T \):

\[ \frac{\Delta T}{T} \times 100 = 3 \% \]

### Final Step: Conclusion

The percentage error in the measurement of the time period \( T \) is:

\[ \text{Percentage error in } T = 4\% \]