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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

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  2. The property of a fluid which enables it to resist tensile stress is known as

  3. Above the critical micelle concentration (CMC), the option which correctly describes the variation of molar conductivity with increase in concentration of sodium dodecylsulphate în aqueous solution is

  4. Addition of detergent to liquid

  5. The mercury does not wet the glass tube. This is due to the property of liquid known as

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