# A spherical liquid drop is placed on a horizontal plane. A small disturbance causes the volume of the drop to oscillate. The time period of oscillation \[\left( T \right)\] of the liquid drop depends on radius $\left( r \right)$ of the drop, density $\left( \rho \right)$ and surface tension $\left( s \right)$ of the liquid. Which among the following will be a possible expression for $T$ (where $k$ is a dimensionless constant)?

A. $k\sqrt {\dfrac{{\rho r}}{s}} $

B. $k\sqrt {\dfrac{{{\rho ^2}r}}{s}} $

C. $k\sqrt {\dfrac{{\rho {r^3}}}{s}} $

D. $k\sqrt {\dfrac{{\rho {r^3}}}{{{s^2}}}} $

Answer

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**Hint:-**As time period of oscillation \[\left( T \right)\] of the liquid drop depends on radius $\left( r \right)$ of the drop, density $\left( \rho \right)$ and surface tension $\left( s \right)$ of the liquid

So, the time period of oscillation can be given by the expression $T = k{r^x}{\rho ^y}{s^z}$ where k, x, y, z are dimensionless constants.

Now, equate the dimensions according to the expression and find the value of x, y and z.

**Complete step-by-step solution:-**

As given in the question that the time period of oscillation \[\left( T \right)\] of the liquid drop depends on radius $\left( r \right)$ of the drop, density $\left( \rho \right)$ and surface tension $\left( s \right)$ of the liquid

So, the time period of oscillation can be given by the expression $T = k{r^x}{\rho ^y}{s^z}$ where k, x, y, z are dimensionless constants.

Now, we equate the dimensions according to the expression to find the value of x, y and z

We know that the dimension of time period of oscillation $T$ is ${T^1}$ and that of radius $r$ is ${L^1}$

As we know density $\rho = \dfrac{{Mass}}{{Volume}}$ , so the dimension of $\rho $ will be $M{L^{ - 3}}$

And the surface tension is the force per unit length, so its dimension will be $M{T^{ - 2}}$

Now, substituting these dimension in the expression for $T$ we have

${T^1} = k{\left( {{L^1}} \right)^x}{\left( {M{L^{ - 3}}} \right)^y}{\left( {M{T^{ - 2}}} \right)^z}$

On simplifying we have

${T^1} = k{M^{y + z}}{L^{x - 3y}}{T^{ - 2z}}$

Now, as LHS has on time dimension so the other dimension in RHS must be zero

For M, $y + z = 0$ or $y = - z$ …… (1)

For L, $x - 3y = 0$ …… (2)

For T, $ - 2z = 1$ or $z = - \dfrac{1}{2}$ …… (3)

Now, substituting the value of z from equation (3) in equation (1) we get

$y = \dfrac{1}{2}$

Now, substituting the value of y in equation (2) we get

$x = 3y = \dfrac{3}{2}$

Now, substituting the value of x, y and z in the expression of $T$ we have

\[T = k{r^{3/2}}{\rho ^{1/2}}{s^{ - 1/2}}\]

On simplifying we get

$T = k\sqrt {\dfrac{{\rho {r^3}}}{s}} $

**Hence, option C is correct.**

**Note:-**Dimensions of any physical quantity are those raised powers on base units to specify its unit. Dimensional formula is the expression which shows how and which of the fundamental quantities represent the dimensions of a physical quantity.

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