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Question

If the surface tension of the soap bubble is 0.035 N/m, then the work done in blowing the soap bubble of radius 5 cm in the air is.

The correct answer is

 2.2 mJ

Calculating Work Done Blowing Soap Bubble

This problem involves calculating the work done to increase the surface area of a soap bubble. The key concept here is surface tension, which is the force acting per unit length along the surface of a liquid. Work must be done against this surface tension to expand the surface area.

Understanding Surface Tension and Work

Surface tension ($\gamma$) is defined as the energy required to increase the surface area of a liquid by a unit amount. Mathematically, it's the ratio of the excess pressure inside a bubble to the change in radius, or the force per unit length. When blowing a soap bubble, we are essentially stretching the soap film, which requires energy. Since a soap bubble has two surfaces (an inner and an outer surface), the work done is calculated considering both these surfaces.

Formula for Work Done

The work done ($W$) in increasing the surface area of a liquid film is given by the product of the surface tension ($\gamma$) and the change in surface area ($\Delta A$):

$W = \gamma \times \Delta A$

For a spherical soap bubble of radius r, the surface area of one surface is $A_1 = 4 \pi r^2$. Since a soap bubble has two surfaces (inner and outer), the total surface area is $A_{total} = 2 \times A_1 = 2 \times (4 \pi r^2) = 8 \pi r^2$.

Therefore, the work done to blow a soap bubble of radius r is:

$W = \gamma \times (8 \pi r^2)$

Step-by-Step Calculation

  1. Identify Given Values:
    • Surface tension ($\gamma$) = 0.035 N/m
    • Radius of the soap bubble (r) = 5 cm
  2. Convert Units: The radius must be converted from centimeters to meters for consistency with the surface tension unit (N/m).
    • $r = 5 \text{ cm} = 5 \times 10^{-2} \text{ m} = 0.05 \text{ m}$
  3. Apply the Formula: Substitute the given values into the formula $W = 8 \pi \gamma r^2$.
    • $W = 8 \times \pi \times (0.035 \text{ N/m}) \times (0.05 \text{ m})^2$
  4. Calculate the Square of the Radius:
    • $(0.05 \text{ m})^2 = 0.0025 \text{ m}^2$
  5. Perform the Multiplication:
    • $W = 8 \times \pi \times 0.035 \times 0.0025 \text{ J}$
    • $W \approx 8 \times 3.14159 \times 0.035 \times 0.0025 \text{ J}$
    • $W \approx 0.002199 \text{ J}$
  6. Convert Joules to MilliJoules (mJ): Since the options are in milliJoules, convert the result.
    • $1 \text{ J} = 1000 \text{ mJ}$
    • $W \approx 0.002199 \text{ J} \times 1000 \text{ mJ/J}$
    • $W \approx 2.199 \text{ mJ}$

Final Result

Rounding the result to one decimal place, the work done is approximately 2.2 mJ.

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Important Questions from Surface Tension

  1. Which of the following statements is NOT correct about surface tension?

  2. If a liquid droplet and a soap bubble are formed from the same liquid and have the same radius $R$, how does the excess pressure inside the soap bubble ($\Delta P_{bubble}$) compare to the excess pressure inside the liquid droplet ($\Delta P_{droplet}$)?

  3. Mercury does NOT wet the glass. This is due to the property of the liquid known as

  4. ________ is a surface phenomenon.

  5. A liquid drop is spherical in shape due to

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