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Figure shows a soap film formed between two square figures made of a uniform wire. The bigger square is held while keeping it in a horizontal plane and the smaller square is slowly allowed to drop vertically. It reaches an equilibrium state after dropping a height $h .$ Let surface tension of soap = $T .$ Mass per unit length of wire = $λ .$ Acceleration due to gravity = $g .$ Given that $h = \frac{n λ g a}{\sqrt{T^{2} - \frac{λ^{2} g^{2}}{4}}};$ find the value $n .$

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Detailed Solution

$\tan θ = \frac{3 a}{2 h}$

$2 \times 4 a T \times \cos θ = 4 a λ g$

$\cos θ = \frac{λ g}{2 T}$

$\Rightarrow h = \frac{3 λ g a}{2 \sqrt{4 T^{2} - λ^{2} g^{2}}}$

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