A thin metal wire of density rho floats on th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) For the wire to float,the downward gravitational force must be balanced by the upward force due to surface tension.The gravitational force (weight

Vedclass · Accepted Answer

$\sqrt{\frac{2 T}{\pi \rho g}}$

Vedclass · Answer

(A) For the wire to float,the downward gravitational force must be balanced by the upward force due to surface tension.The gravitational force (weight) acting on the wire is $W = mg = (	ext{volume} 	imes ho) g = (\pi r^2 l) ho g$,where $r$ is the radius and $l$ is the length of the wire.The surface tension force acts along the two sides of the wire on the water surface,so the upward force is $F_s = 2Tl$.Equating the forces for the limiting case of floating: $Mg = 2Tl$.Substituting the values: $(\pi r^2 l) ho g = 2Tl$.Canceling $l$ from both sides: $\pi r^2 ho g = 2T$.Solving for $r$: $r^2 = \frac{2T}{\pi ho g}$,which gives $r = \sqrt{\frac{2T}{\pi ho g}}$.Thus,the maximum radius is proportional to $\sqrt{\frac{2T}{\pi ho g}}$.

$A$ thin metal wire of density $\rho$ floats on the water surface horizontally. If it is $\text{NOT}$ to sink in water,then the maximum radius of the wire is proportional to $(T = \text{surface tension of water}, g = \text{gravitational acceleration})$.

Similar Questions

$A$ liquid is coming out from a vertical tube. The relation between the weight of the drop $W$,surface tension of the liquid $T$,and radius of the tube $r$ is given by,if the angle of contact is zero.

The spherical shape of a rain-drop is due to

If the work done in increasing the size of a soap film from $10\;cm \times 6\;cm$ to $10\;cm \times 11\;cm$ is $2 \times 10^{-4}\;J$,then the surface tension is:

An iron needle slowly placed on the surface of water floats because

Vedclass Test Series

Exam Paper Generator

Online Exam Module