The force required to take away a flat circular plate of radius $2 \ cm$ from the surface of water is $[$ Surface tension of water $= 70 \times 10^{-3} \ Nm^{-1}, \pi = \frac{22}{7} ]$

  • A
    $4.4 \times 10^{-4} \ N$
  • B
    $8.8 \times 10^{-3} \ N$
  • C
    $6.6 \times 10^{-4} \ N$
  • D
    $11 \times 10^{-3} \ N$

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$A$ paper disc of radius $R$ from which a circular hole of radius $r$ is cut out,is floating in a liquid of surface tension $T$. The force on the disc due to surface tension will be

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 $3.$

The work done in increasing the size of a soap film from $10 \, cm \times 6 \, cm$ to $10 \, cm \times 11 \, cm$ is $3 \times 10^{-4} \, J$. The surface tension of the film is:

Figure $(a)$ shows a thin liquid film supporting a small weight $= 4.5 \times 10^{-2} \, N$. What is the weight supported by a film of the same liquid at the same temperature in Figure $(b)$ and $(c)$? Explain your answer physically.

$A$ square frame of each side $L$ is dipped in a soap solution and taken out. The force acting on the film formed is ($T =$ surface tension of soap solution). (in $TL$)

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