Question

A soap bubble has a radius of 2cm the work done in doubling it's radius is (T = 30dyne/cm)​

Answer 1

Answer:

$\boxed{\mathfrak{Work \ done \ (W) = 2880\pi \ erg}}$

Given:

Initial radius of soap bubble ($\rm r_i$) = 2 cm

Final radius of soap bubble ($\rm r_f$) = 4 cm

Surface tension of soap solution (T) = 30 dyne/cm

Explanation:

In case of soap bubble work done (W) is given as:

$\boxed{ \bold{W = 2T\Delta A}}$

$\sf \Delta A \longrightarrow$ Change in surface area of soap bubble

So,

$\rm \implies W = 2T(4\pi {r_f}^{2} - 4\pi {r_i}^{2}) \\ \\ \rm \implies W = 2 \times 30(4\pi({r_f}^{2} -{r_i}^{2} )) \\ \\ \rm \implies W =60(4\pi( {4}^{2} - {2}^{2} )) \\ \\ \rm \implies W =240\pi(16 - 4) \\ \\ \rm \implies W =240\pi \times 12 \\ \\ \rm \implies W =2880\pi \: erg$