Q

Confused! kindly explain, 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-1)

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-1)

• Option 1)

$4\pi \: mJ$

• Option 2)

$0.2\pi \: mJ$

• Option 3)

$2\pi \: mJ$

• Option 4)

$0.4\pi \: mJ$

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As we learnt in

Surface Energy -

$W=T\times \Delta A$

$T= \frac{W}{\Delta A}$

- wherein

$\Delta A\rightarrow increase\: in\: area$

$W=T \Delta A=0.03\times 2\left(4\pi \left(5^{2}-3^{2} \right )\times 10^{-4} \right )$

$W=24\pi \times16\times10^{-4}=0.384\pi \times 10^{-3}$

$\therefore\ \; W=0.4\pi mJ$

Correct option is 4.

Option 1)

$4\pi \: mJ$

This is an incorrect option.

Option 2)

$0.2\pi \: mJ$

This is an incorrect option.

Option 3)

$2\pi \: mJ$

This is an incorrect option.

Option 4)

$0.4\pi \: mJ$

This is the correct option.

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