$A$ thread is tied slightly loose to a wire frame as shown in the figure. The frame is dipped into a soap solution and taken out,such that the frame is completely covered with a soap film. When the portion $A$ is punctured with a pin,the thread:

  • A
    Becomes concave toward $A$
  • B
    Becomes convex towards $A$
  • C
    Remains in the initial position
  • D
    Either $(a)$ or $(b)$ depending on the size of $A$ w.r.t. $B$

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What is the ratio of the surface tension force acting on the curved part to that on the flat part?

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When liquid medicine of density $\rho$ is to be put in the eye, it is done with the help of a dropper. As the bulb on the top of the dropper is pressed, a drop forms at the opening of the dropper. We wish to estimate the size of the drop. We first assume that the drop formed at the opening is spherical because that requires a minimum increase in its surface energy. To determine the size, we calculate the net vertical force due to the surface tension $T$ when the radius of the drop is $R$. When the force becomes smaller than the weight of the drop, the drop gets detached from the dropper.
$1.$ If the radius of the opening of the dropper is $r$, the vertical force due to the surface tension on the drop of radius $R$ (assuming $r \ll R$) is
$(A)$ $2 \pi r T$ $(B)$ $2 \pi R T$ $(C)$ $\frac{2 \pi r^2 T}{R}$ $(D)$ $\frac{2 \pi R^2 T}{r}$
$2.$ If $r=5 \times 10^{-4} \, m, \rho=10^3 \, kg \, m^{-3}, g=10 \, m/s^2, T=0.11 \, Nm^{-1}$, the radius of the drop when it detaches from the dropper is approximately
$(A)$ $1.4 \times 10^{-3} \, m$ $(B)$ $3.3 \times 10^{-3} \, m$
$(C)$ $2.0 \times 10^{-3} \, m$ $(D)$ $4.1 \times 10^{-3} \, m$
$3.$ After the drop detaches, its surface energy is
$(A)$ $1.4 \times 10^{-6} \, J$ $(B)$ $2.7 \times 10^{-6} \, J$
$(C)$ $5.4 \times 10^{-6} \, J$ $(D)$ $8.1 \times 10^{-6} \, J$
Give the answer for questions $1, 2$ and $3.$

Small droplets of a liquid are usually more spherical in shape than larger drops of the same liquid because

$A$ square frame of side $L$ is dipped in a liquid. On taking out,a membrane is formed. If the surface tension of the liquid is $T$,the force acting on the frame will be (in $TL$)

The force necessary to pull a circular plate of $5\, cm$ radius from a water surface,for which the surface tension is $75\, dynes/cm$,is

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