Question

a circular loop of wire of radius 5/Π cm is suspended from 1 arm of a balance the plane of the loop is in contact with the surface of soap solution the pull on the loop due to surface tension is found to be 0.8x10^-3 kg weight calculate surface tension of soap solution​

Answer 1

Given:

A circular loop of wire of radius  $\mathbf{\frac{5}{\pi } \ cm}$ is suspended from 1 arm of a balance the plane of the loop is in contact with the surface of soap solution the pull on the loop due to surface tension is found to be $\mathbf{0.8\times 10^{-3}\ \ \frac{N}{m}}$.

To Find:

What is weight calculated due to surface tension of soap solution?

Solution:

$\mathbf{\textrm{Surface tension of soap solution}(\sigma )=0.8\times 10^{-3} \ \frac{N}{m}}$

$\mathbf{\textrm{Radius of soap solution}(R )=\dfrac{5}{\pi } \ \ \ \ \ \ \ (cm)}$

We can also write:

$\mathbf{\textrm{Radius of soap solution}(R )=\dfrac{5\times 10^{-2}}{\pi } \ \ \ \ (m)}$

We know the formula of relation between force and surface tension:

Force = Surface tension× Length

Here length is circumference of soap wire circular loop:

$\mathbf{\textrm{Force due to surface tension}(F )=\sigma \times (2\times \pi \times R) }$

On putting respective value in above equation:

$\mathbf{\textrm{Force due to surface tension}(F )=0.8\times 10^{-3} \times 2\times \pi \times \dfrac{5\times 10^{-2}}{\pi } }$

$\mathbf{\textrm{Force due to surface tension}(F )=8\times 10^{-5} \ \ \ \ \ (N) }$

Force due to surface tension is same as weight due to surface tension which value is $\mathbf{8\times 10^{-5} \ (N) }$.

Answer 2

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Force due to surface tension is same as weight due to surface tension which value is 8 × 10^-5 N