The soap's holding it back

Classical Mechanics Level 4

The diagram on the left shows a soap film formed between two square figures--side lengths $4a$ and $a$ each--made of uniform wire. Now, the bigger square is being held in a horizontal plane while the smaller is slowly let drop vertically, which reaches an equilibrium state after dropping a height of $h$ .

Let $T$ be the surface tension of the soap, $\lambda$ the mass per unit length of the wire, and $g$ the acceleration due to gravity.

Given that $h = \dfrac{n \lambda ga}{2 \sqrt{4T^2 - \lambda^2 g^2}},$ find the integer value $n.$

Assumptions:

Assume that the soap film forms a surface exactly as shown in the diagram (i.e., without any curvature).

The answer is 3.

1 solution

A Former Brilliant Member
Jan 17, 2018

Balancing the forces in the vertical direction on the smaller square yields :

$mg=F\cos \theta$

where $\theta$ is the angle formed by line joining boundary of smaller square and bigger square with the vertical and $F$ is net force acting on the smaller square due to surface tension.

Using a bit of geometry:

$\cos \theta =\dfrac{h}{\sqrt{h^2+\dfrac{9a^2}{4}}}$

$F=(2)(T)(4a)$

$\therefore \dfrac{8aTh}{\sqrt{h^2+\dfrac{9a^2}{4}}} = 4a\lambda g$

Squaring and solving , we get :

$h = \dfrac{3 \lambda ga}{2 \sqrt{4T^2 - \lambda^2 g^2}}$

@Tapas Mazumdar . Where did you get get the question paper ? Has Fiitjee uploaded the paper?

A Former Brilliant Member - 3 years, 4 months ago

Yes. At reg.fiitjee.com

Tapas Mazumdar - 3 years, 4 months ago

@Tapas Mazumdar How many marks are you expecting ?

A Former Brilliant Member - 3 years, 4 months ago

@A Former Brilliant Member – Results are out

Arunava Das - 3 years, 4 months ago

The soap's holding it back

The answer is 3.

1 solution

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