Question

The minimum velocity of capillary waves on the surface of the water is $\left( {Surface\,tension\,of\,water = 7.2 \times {{10}^{ - 2}}N/m} \right)$
$\left( A \right){\text{ 0}}{\text{.23m/s}}$
$\left( B \right){\text{ 0}}{\text{.46m/s}}$
$\left( C \right){\text{ 0}}{\text{.69m/s}}$
$\left( D \right){\text{ 0}}{\text{.92m/s}}$

Answer 1

Since we know that the capillary wave is a wave that travels beside the phase border of a liquid. So by using the formula of minimum velocity which is given by ${v_{\min }} = \sqrt {2{{\left( {\dfrac{{{T_g}}}{\rho }} \right)}^{1/2}}}$ . And on substituting the values, we will be able to get the solution.

Formula used
The minimum velocity of capillary waves,
${v_{\min }} = \sqrt {2{{\left( {\dfrac{{{T_g}}}{\rho }} \right)}^{1/2}}}$
${v_{\min }}$ , will be the minimum velocity of capillary wave
${T_g}$ , will be the surface tension of water
$\rho$ , will be the density.

Complete Step By Step Answer:
So we have the question in which we have to find the minimum velocity of the water and for this, the surface tension of water is given and we know that the density of water is given by ${10^3}kg/{m^3}$ . So from the formula, we have the equation as,
$\Rightarrow {v_{\min }} = \sqrt {2{{\left( {\dfrac{{{T_g}}}{\rho }} \right)}^{1/2}}}$
So on substituting the values, we get
$\Rightarrow {v_{\min }} = \sqrt {2{{\left( {\dfrac{{7.2 \times {{10}^{ - 2}} \times 9.8}}{{{{10}^3}}}} \right)}^{1/2}}}$
On solving it will get the above expression as
$\Rightarrow {v_{\min }} = 1.414{\left( {\dfrac{{7.2 \times {{10}^{ - 2}} \times 9.8}}{{{{10}^3}}}} \right)^{1/4}}$
And again solving it, we will get
$\Rightarrow {v_{\min }} = 0.23m/s$
Hence, the minimum velocity of capillary waves on the surface of the water is $0.23m/s$ .

Note:
Capillary waves are produced on the surface of fluid which is in a gravitational field. Usually, it is formed in water bodies, like lakes. It is produced by the interplay between gravitation and surface tension and hydrodynamics of the fluid. It is well-known by their wavelength but this is somewhat arbitrary.