Surface tension of a soap solution is $1.9 \times 10^{-2} \text{ N/m}$. The work done in blowing a soap bubble of $2.0 \text{ cm}$ diameter will be:

  • A
    $7.6 \times 10^{-6} \pi \text{ J}$
  • B
    $15.2 \times 10^{-6} \pi \text{ J}$
  • C
    $1.9 \times 10^{-6} \pi \text{ J}$
  • D
    $1 \times 10^{-4} \text{ J}$

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If $W_1$ is the work done in increasing the radius of a soap bubble from $r$ to $2r$ and $W_2$ is the work done in increasing the radius of the soap bubble from $2r$ to $3r$, then $W_1: W_2=$

Work done in increasing the size of a soap bubble from a radius of $3\, cm$ to $5\, cm$ is nearly (Surface tension of soap solution $= 0.03\, N/m$)

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$A$ soap bubble of radius $r$ is blown up to form a bubble of radius $2r$ under isothermal conditions. If $T$ is the surface tension of the soap solution, the energy spent in the blowing is: (in $\pi T r^2$)

When a mercury drop of radius $R$ splits up into $1000$ droplets of radius $r$,the change in surface energy is ($\pi R^2 T$). ($T=$ surface tension of mercury).

The surface tension of the soap water solution is $\frac{1}{10 \pi} \text{ N m}^{-1}$. The free energy of the surface layer of a soap bubble of diameter $5 \text{ mm}$ will be:

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