Time period of an oscillating drop of radius r, density `rho` and surface tension T is `t=ksqrt((rhor^5)/T)`. Check the correctness of the relation

Answer

To check the correctness of the relation \( t = k \sqrt{\frac{\rho r^5}{T}} \) for the time period of an oscillating drop, we will perform dimensional analysis.

### Step-by-Step Solution:

1. **Identify the dimensions of each quantity:**
- The dimension of density (\(\rho\)) is \( [\rho] = M L^{-3} \).
- The dimension of radius (\(r\)) is \( [r] = L \).
- The dimension of surface tension (\(T\)) is \( [T] = M L^{0} T^{-2} \).
- The dimension of time (\(t\)) is \( [t] = T \).

2. **Substitute the dimensions into the expression:**
The right-hand side of the equation is:
\[ k \sqrt{\frac{\rho r^5}{T}} \]
We need to find the dimensions of \( \frac{\rho r^5}{T} \):
- The dimension of \( r^5 \) is \( [r^5] = L^5 \).
- Therefore, the dimension of \( \rho r^5 \) is:
\[ [\rho r^5] = [\rho] \cdot [r^5] = (M L^{-3}) \cdot (L^5) = M L^{2} \]
- Now, we divide this by the dimension of surface tension:
\[ \frac{\rho r^5}{T} = \frac{M L^{2}}{M L^{0} T^{-2}} = L^{2} T^{2} \]

3. **Take the square root of the expression:**
Now, we take the square root:
\[ \sqrt{\frac{\rho r^5}{T}} = \sqrt{L^{2} T^{2}} = L T \]

4. **Include the constant \(k\):**
Since \(k\) is a dimensionless constant, the dimensions of the right-hand side become:
\[ [k \sqrt{\frac{\rho r^5}{T}}] = [k] \cdot [L T] = L T \]

5. **Compare dimensions:**
The left-hand side has the dimension of time:
\[ [t] = T \]
The right-hand side has the dimension:
\[ [k \sqrt{\frac{\rho r^5}{T}}] = L T \]

6. **Conclusion:**
Since the dimensions of the left-hand side (\(T\)) do not match the dimensions of the right-hand side (\(L T\)), the relation \( t = k \sqrt{\frac{\rho r^5}{T}} \) is dimensionally incorrect.

### Final Answer:
The relation \( t = k \sqrt{\frac{\rho r^5}{T}} \) is dimensionally incorrect.