Question:

Time period of a particle is \( T = k\sqrt{\frac{\rho r^3}{\sigma}} \), where \( k \) is a dimensionless constant, \( \rho \) is density, and \( r \) is radius. The dimension of \( \sigma \) is same as

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This formula $T \propto \sqrt{\rho r^3/\sigma}$ specifically relates to the time period of oscillation of a liquid drop, where restoring forces are provided by surface tension $\sigma$. Knowing standard physical formulas can serve as a quick sanity check.
Updated On: Apr 21, 2026
  • surface tension
  • restoring force
  • coefficient of viscosity
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The Correct Option is A

Solution and Explanation

Step 1: Understanding the Concept:
The principle of dimensional homogeneity states that an equation is dimensionally correct if the dimensions of all terms on both sides are the same. We can use this to find the unknown dimensions of a physical quantity in a given formula.

Step 2: Key Formula or Approach:

Write the dimensional formula for each known variable in the equation $T = k\sqrt{\frac{\rho r^3}{\sigma}}$.
Isolate the unknown quantity $\sigma$ and solve for its dimensions. Compare the result with the dimensions of the given options.

Step 3: Detailed Explanation:

The dimensions of the given quantities are:
- Time period ($T$): $[T]$
- Constant ($k$): Dimensionless, so $[1]$
- Density ($\rho = \text{mass/volume}$): $[ML^{-3}]$
- Radius ($r$): $[L]$, so $[r^3] = [L^3]$
Substitute these into the given formula:
\[ [T] = \sqrt{\frac{[ML^{-3}] \cdot [L^3]}{[\sigma]}} \]
\[ [T] = \sqrt{\frac{[M]}{[\sigma]}} \]
Square both sides to remove the square root:
\[ [T]^2 = \frac{[M]}{[\sigma]} \]
Rearrange to solve for the dimensions of $\sigma$:
\[ [\sigma] = \frac{[M]}{[T]^2} = [MT^{-2}] \]
Now, let's find the dimensions of the given options:
(A) Surface tension ($S = \text{Force/Length}$): $[S] = \frac{[MLT^{-2}]}{[L]} = [MT^{-2}]$. This matches.
(B) Restoring force ($F$): $[F] = [MLT^{-2}]$. Does not match.
(C) Coefficient of viscosity ($\eta$): $[\eta] = [ML^{-1}T^{-1}]$. Does not match.

Step 4: Final Answer:

The dimension of $\sigma$ is the same as surface tension.
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