Question

Soap bubble is blown to a diameter of 7 cm. If36960 ergs of work is done in blowing it further, find the new radius, if
surface tension of the soap solution is 40 dynes/cm.​

Answer 1

Solution

Radius of the expanded bubble is 7 cm

Given

• Initial Diameter (D) = 7 cm

• Initial Radius (R) = 7/2 cm

• Work Done (W) = 36960 Ergs

• Surface Tension (S) = 40 Dyne/cm

To finD

New Radius of the Bubble

Let r be the new radius of the bubble

Using the Relation,

$\boxed{\boxed{\sf Work \ Done = Surface \ Tension \times Increase \ In \ Surface \ Area }}$

Substituting the values,

$\tt \: 36960 = 40 \times \bigg(4\pi {r}^{2} - 4\pi( { \dfrac{7}{2}) }^{2} \bigg) \times 2 \\ \\ \longrightarrow \: \tt \: 462 = 4\pi( {r}^{2} - \dfrac{49}{4} ) \\ \\ \longrightarrow \: \tt \: {r}^{2} - \dfrac{49}{4} = \dfrac{21 \times 7}{4} \\ \\ \longrightarrow \: \tt \: { r }^{2} = 49 \\ \\ \\ \longrightarrow \: \boxed{ \boxed{ \tt \: r = 7 \: cm}}$

Answer 2

Answer:

Solution

Radius of the expanded bubble is 7 cm

Given

Initial Diameter (D) = 7 cm

Initial Radius (R) = 7/2 cm

Work Done (W) = 36960 Ergs

Surface Tension (S) = 40 Dyne/cm

To finD

New Radius of the Bubble

Let r be the new radius of the bubble

Using the Relation,

\boxed{\boxed{\sf Work \ Done = Surface \ Tension \times Increase \ In \ Surface \ Area }}

Work Done=Surface Tension×Increase In Surface Area

Substituting the values,

\begin{gathered} \tt \: 36960 = 40 \times \bigg(4\pi {r}^{2} - 4\pi( { \dfrac{7}{2}) }^{2} \bigg) \times 2 \\ \\ \longrightarrow \: \tt \: 462 = 4\pi( {r}^{2} - \dfrac{49}{4} ) \\ \\ \longrightarrow \: \tt \: {r}^{2} - \dfrac{49}{4} = \dfrac{21 \times 7}{4} \\ \\ \longrightarrow \: \tt \: { r }^{2} = 49 \\ \\ \\ \longrightarrow \: \boxed{ \boxed{ \tt \: r = 7 \: cm}}\end{gathered}

36960=40×(4πr

2

−4π(

2

7

)

2

)×2

⟶462=4π(r

2

−

4

49

)

⟶r

2

−

4

49

=

4

21×7

⟶r

2

=49

⟶

r=7cm