Question

The work done in increasing the radius of a soap bubble from 4cm to 5cm is........ Joule (given surface tension of soap water to be 25×20–3).

Answer 1

The work done is $0.565\times10^{-3}\ J$

Explanation:

Given that,

Surface tension $T=25\times10^{-3}$

Radius of bubble =4 cm

Radius of bubble = 5 cm

We need  to calculate the work done

Using formula of work done

$W=8\pi(r_{2}^2-r_{1}^2)T$

Where, T = surface tension

Put the value into the formula

$W=8\pi((5\times10^{-2})^2-(4\times10^{-2})^2)\times25\times10^{-3}$

$W=0.000565\ J$

$W=0.565\times10^{-3}\ J$

Hence, The work done is $0.565\times10^{-3}\ J$

Learn more :

Topic : surface tension

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