Solveeit Logo
Login

Question

Question: Calculate the work done against surface tension in blowing a soap bubble from a radius of \(10cm\), ...

Calculate the work done against surface tension in blowing a soap bubble from a radius of 10cm10cm, if the surface tension of soap solution is 25×10−3N/m25 \times {10^{ - 3}}N/m.

Explanation

Solution

In blowing the bulb, some of the work is done for which energy is required. The relation between work done and the surface tension is used to solve this.

Formula used:
(1) Work done=T×= T \timesincrease in surface area
(2) Surface area of sphere =4πr2 = 4\pi {r^2}
Where TT is the surface tension and r is the radius of the sphere.

Complete step by step answer:
The surface of the soap bubble possess minimum area due to surface tension. To increases the radius of soap bubble, work is to be done against the force of attraction. Which is given by
Work done, W=T×W = T \times increase in area of the soap bubble
Where TT is the surface tension.
Since, the soap bubbles is in spherical shape
So, increase in area of soap bubble
=2×= 2 \times(Final area – initial area)
=2×(4πr22−4πr12)= 2 \times \left( {4\pi r_2^2 - 4\pi r_1^2} \right)
Where r, is the initial radius of soap bubble and r2{r_2} is the final radius of soap bubble.
So, increase in area =2×4π(r22−r12) = 2 \times 4\pi \left( {r_2^2 - r_1^2} \right)
Hence, work done =2×T×4π(r22−r12) = 2 \times T \times 4\pi \left( {r_2^2 - r_1^2} \right)
Hence, in the question.
Surface tension, T=25×10−3N/MT = 25 \times {10^{ - 3}}N/M
Initial radius of soap bubble, r1=10cm{r_1} = 10cm
=10×10−2m= 10 \times {10^{ - 2}}m
Final radius of soap bubble, r2=20cm{r_2} = 20cm
=20×10−2m= 20 \times {10^{ - 2}}m
So, work done =25×10−3×2×4π[(20×10−2)2−(10×10−2)2] = 25 \times {10^{ - 3}} \times 2 \times 4\pi \left[ {{{\left( {20 \times {{10}^{ - 2}}} \right)}^2} - {{\left( {10 \times {{10}^{ - 2}}} \right)}^2}} \right]
Work done =20×10−3×8×227×(400×10−4−100×10−4) = 20 \times {10^{ - 3}} \times 8 \times \dfrac{{22}}{7} \times \left( {400 \times {{10}^{ - 4}} - 100 \times {{10}^{ - 4}}} \right)
=200×10−3×3.14(400−100)×10−4 =200×3.14×10−3×300×10−4  = 200 \times {10^{ - 3}} \times 3.14\left( {400 - 100} \right) \times {10^{ - 4}} \\\ = 200 \times 3.14 \times {10^{ - 3}} \times 300 \times {10^{ - 4}} \\\
=18.84×10−3= 18.84 \times {10^{ - 3}} Joules
=1.88×10−2= 1.88 \times {10^{ - 2}}Joules

Note:
Here, the area of the soap bubble is taken to be 8πr28\pi {r^2}i.e. 2×4πr22 \times 4\pi {r^2} instead of only 4πr24\pi {r^2}. This is because a soap bubble has air both inside and outside. So, it has two free surfaces. Hence, the surface area of one soap bubble =2×= 2 \timessurface area of the sphere.