Q&A - Ask Doubts and Get Answers
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

 
Answers (1)
705 Views

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.

Exams
Articles
Questions