Question

A water bug is suspended on the surface of a pond by surface tension (water does not wet the legs). The bug has six leg, and each leg is in contact with the water over a length of 5 mm. What is the maximum mass (in grams) of the bug if it is to avoid sinking

Answer 1

Answer:

m = 2.2 x 10⁻⁴ kg = 0.22 g

Explanation:

The surface tension of water is 0.072 N/m. So in order for the bug to avoid sinking, its weight per unit length of contact must be no more than the surface tension of water. Therefore,

$Weight\ of bug\ per\ unit\ length = Surface\ Tension\ of\ Water\\\frac{mg}{L} = Surface\ Tension\ of Water\\m = \frac{(Surface\ Tension\ of\ Water)(L)}{g}$

where,

m = mass of bug = ?

g = acceleration due to gravity = 9.81 m/s²

L = Contact length = (contact length of each leg)(No. of Legs) = (5 mm)(6)

L = 30 mm = 0.03 m

Therefore,

$m = \frac{(0.072\ N/m)(0.03\ m)}{9.81\ m/s^{2}}$

m = 2.2 x 10⁻⁴ kg = 0.22 g