Soap has surface tension σ. We make a plane soap film and on the soap film we placed a loop of rubber band of total length l. The cross sectional area of the rubber band is S and its Young's modulus of elasticity is E. Now the film inside the loop is pierced. The rubber loop acquires a circular shape. Find the tension in the rubber loop.

Question

Soap has surface tension σ. We make a plane soap film and on the soap film we placed a loop of rubber band of total length l. The cross sectional area of the rubber band is S and its Young's modulus of elasticity is E. Now the film inside the loop is pierced. The rubber loop acquires a circular shape. Find the tension in the rubber loop. (A) 2 σ l (B) (E σ l s/π E s+σ l) (C) (E σ l s/2 π Es -σ l) (D) Es -l σ

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/177a1a998c382dd8804d0f46beaafc0e-.png" /> 2 σ R θ=2 T (θ/2) T =σ R ( T / s )= Y × ((2 π R -l)/l) =( Y /1)((2 π T /σ)-1) Y =(π TY /σ l)-( T / s ) ⇒ T =( Y σ s /2 π Ys -σ l )

$2 σ l$

$\frac{E σ l s}{π E s + σ l}$

$\frac{E σ l s}{2 π E s - σ l}$

$E s - l σ$

$2 σ Rθ = 2 T \frac{θ}{2}$
$T = σ R$
$\frac{T}{s} = Y \times \frac{( 2 π R - l )}{l}$
$= \frac{Y}{1} (\frac{2 π T}{σ} - 1)$
$Y = \frac{π T Y}{σ l} - \frac{T}{s}$
$\Rightarrow T = \frac{Yσ s}{2 πY s - σ l}$

2σl

πEs+σlEσls​

2πEs−σlEσls​

Es−lσ

2σRθ=2T2θ​ T=σR sT​=Y×l(2πR−l)​ =1Y​(σ2πT​−1) Y=σlπTY​−sT​ ⇒T=2πYs−σlYσs​

$2 σ l$

$\frac{E σ l s}{π E s + σ l}$

$\frac{E σ l s}{2 π E s - σ l}$

$E s - l σ$

$2 σ Rθ = 2 T \frac{θ}{2}$
$T = σ R$
$\frac{T}{s} = Y \times \frac{( 2 π R - l )}{l}$
$= \frac{Y}{1} (\frac{2 π T}{σ} - 1)$
$Y = \frac{π T Y}{σ l} - \frac{T}{s}$
$\Rightarrow T = \frac{Yσ s}{2 πY s - σ l}$