Question

calculate the work done on blowing a soap bubbles of radius 0.5cm surface tension of soap solution is 25×10^-3 N/m​

Answer 1

Answer:

$\boxed{\mathfrak{Work \ done \ (W) = 3.14 \ mJ \ \ \ or \ \ \ \pi \ mJ}}$

Given:

Radius of soap bubble (r) = 0.5 cm = $\sf 5 \times 10^{-3}$ m

Surface tension of soap solution (S) = $\sf 25 \times 10^{-3}$ N/m

To Find:

Work done (W) on blowing a soap bubbles

Explanation:

In case of soap bubble work done is given as:

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

Where:

$\sf \Delta A \rightarrow$ Change in surface area

So,

$\sf \implies W = 2S(4\pi {r}^{2} - 0) \\ \\ \sf \implies W = 2S \times 4\pi {r}^{2} \\ \\ \sf \implies W = 8S\pi {r}^{2} \\ \\ \sf \implies W =8 \times 25 \times {10}^{ - 3} \times 3.14 \times 5 \times {10}^{ - 3} \\ \\ \sf \implies W =3140 \times {10}^{ - 6} \\ \\ \sf \implies W =3.14 \times {10}^{ - 3} \: J \\ \\ \sf \implies W =3.14 \: mJ$

$\therefore$

Work done (W) on blowing a soap bubbles = 3.14 mJ or π mJ

Answer 2

✔ Explanation :-

✔ Given :-

• Radius(r) of Soap bubbles is 0.5cm ➡ 5 × 10^{-3}m.

• Surface tension (s) of soap is solution ➡ 25 × 10^{-3}N/m.

✔ To Find :-

• The work done(w) on blowing a soap bubbles.

✔ Solution :-

We know that,

W = 2S × ΔA

[ Putting values ]

=> W = 2S × ( 4πr² - 0 )

=> W = 2S × 4πr²

=> W = 8Sπr²

=> W = 8 × 25 × 10^{-3} × 3.14 × 5 × 10^{-3}

=> W = 3140 × 10^{-6}

=> W = 3.14 mJ

Hence,

• The work done on blowing a soap bubbles is 3.14mJ.