Question

If the time period of oscillation of a liquid drop is given by $T=k\sqrt{\frac{\rho r^{4}}{\sigma}}$, where $\rho$ refers to its density, $r$ its radius and $k$ is a dimensionless constant, then $\sigma$ has the units of

• A. surface
• B. restoring force
• C. coefficient of viscosity
• D. vapour pressure
• E. surface tension

Hint:

While the symbol $\sigma$ is often used for surface tension ($[MT^{-2}]$), in dimensional analysis problems, you must trust the math of the provided equation over common symbols. Here, the extra $r$ factor in $r^4$ shifts the result from surface tension to force.

Answer 1

The Correct Option is B

Concept:
Physics - Dimensional Analysis.
Step 1: Isolate $\sigma$ in the given equation.
Given: $T = k\sqrt{\frac{\rho r^4}{\sigma}}$
Square both sides to remove the root: $$ T^2 = k^2 \frac{\rho r^4}{\sigma} $$ Rearrange to solve for $\sigma$: $$ \sigma = \frac{k^2 \rho r^4}{T^2} $$
Step 2: Substitute dimensions of known quantities.

• $k$ is dimensionless.
• Density ($\rho$): $[ML^{-3}]$
• Radius ($r$): $[L]$
• Time ($T$): $[T]$

Step 3: Calculate dimensions of $\sigma$.
$$ [\sigma] = \frac{[ML^{-3}] \cdot [L]^4}{[T]^2} $$ $$ [\sigma] = \frac{[ML^{1}]}{[T]^2} = [MLT^{-2}] $$
Step 4: Identify the physical quantity.
The dimensions $[MLT^{-2}]$ correspond to Force.
Comparing this with the given options, $\sigma$ behaves as a force in this specific formula context.