91Ó°ÊÓ

If the surface tension force of fluid is less than the force acting on it due to a body, then the body breaks the surface of the fluid.

The radius of the base of an insect’s leg is, \(r = 3.0 \times {10^{ - 5}}\;{\rm{m}}\).

The mass of the insect is, \(m = 0.016\;{\rm{g}} \times \frac{{1\;{\rm{kg}}}}{{1000\;{\rm{g}}}} = 0.016 \times {10^{ - 3}}\;{\rm{kg}}\).

Using the Table 10-4, the surface tension of water at \(20^\circ {\rm{C}}\) is given by,

\(\gamma = 0.072\;{\rm{N/m}}\)

The expression for the total force acting on the six legs of the insect is given by,

\(\begin{array}{c}\gamma = \frac{{{F_t}}}{{6L}}\\\gamma = \frac{{{F_t}}}{{6\left( {2\pi r} \right)}}\end{array}\)

Substituting the values in the above expression

\(\begin{array}{c}0.072\;{\rm{N/m}} = \frac{{{F_t}}}{{6 \times 2\pi \times 3.0 \times {{10}^{ - 5}}\;{\rm{m}}}}\\{F_t} = 8.143 \times {10^{ - 5}}\;{\rm{N}}\end{array}\)

Also, the expression for the weight of the insect acting on the surface of the water is given by,

\(\begin{array}{c}{F_g} = mg\cos \theta \\{F_g} = 0.016 \times {10^{ - 3}}\;{\rm{kg}} \times 9.81\;{\rm{m/}}{{\rm{s}}^2} \times \cos 0^\circ \times \frac{{1\;{\rm{N}}}}{{1\;{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^2}}}\\{F_g} = 1.5696 \times {10^{ - 4}}\;{\rm{N}}\end{array}\)

Comparing both force values,

\({F_t} < {F_g}\)

The surface tension force is less than the weight of the insect acting on top of the water surface.

Thus, the six-legged insect will not remain on top of the water.