The material of a wire has a density of 1.4 g per cm³. If it is not wetted by a liquid of surface tension 44 dyne per cm, then the maximum radius of the wire which can float on the surface of the liquid is

The correct answer is 1/7 cm.

Let us analyze the physical conditions for a wire floating on the surface of a liquid.
When a long wire of radius _r, length _l, and density _ρ is placed on the surface of a liquid, it does not sink if the upward force due to surface tension balances the weight of the wire.

For a wire that is not wetted by the liquid, the surface of the liquid supports the wire from below. The maximum support due to surface tension occurs when the contact angle is such that the surface tension force acts vertically upward on both sides of the wire.
Therefore, the maximum upward force due to surface tension on the wire of length _l is:
$F_T=2\cdot T\cdot l$
where $T$ is the surface tension of the liquid.

The downward force acting on the wire is its weight, which is given by:
$W=m\cdot g=V\cdot \rho \cdot g$
Since the wire is cylindrical, its volume is $V=\pi \cdot r^2\cdot l$. Thus, the weight is:
$W=\pi \cdot r^2\cdot l\cdot \rho \cdot g$

For the wire to float, the upward surface tension force must be greater than or equal to the weight of the wire. For the maximum radius $r$, these forces are equal:
$2\cdot T\cdot l=\pi \cdot r^2\cdot l\cdot \rho \cdot g$

Dividing both sides by the length $l$, we get:
$2\cdot T=\pi \cdot r^2\cdot \rho \cdot g$

Solving for the radius $r$:
$r^2=\frac{2\cdot T}{\pi \cdot \rho \cdot g}$

Given data in CGS units:
Density, $\rho =1.4{\text{g/cm}}^3$
Surface tension, $T=44\text{dyne/cm}$
Acceleration due to gravity, $g\approx 980{\text{cm/s}}^2$
Using $\pi \approx \frac{22}{7}$, let's substitute these values into the equation:

$r^2=\frac{2\cdot 44}{\frac{22}{7}\cdot 1.4\cdot 980}$

Simplifying the denominator:
$\frac{22}{7}\cdot 1.4=22\cdot 0.2=4.4$
Now substitute this back:
$r^2=\frac{88}{4.4\cdot 980}$

Since $\frac{88}{4.4}=20$, we have:
$r^2=\frac{20}{980}=\frac{2}{98}=\frac{1}{49}$

Taking the square root on both sides to find the maximum radius:
$r=\sqrt{\frac{1}{49}}=\frac{1}{7}\text{cm}$

Thus, the maximum radius of the wire that can float on the surface of the liquid is 1/7 cm.