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A soap bubble of surface tension $0.04 \text{ N/m}$ is blown to a diameter of $7 \text{ cm}$. If $(15000 - x) \text{ }\mu\text{J}$ of work is done in blowing it further to make its diameter $14 \text{ cm}$, then the value of $x$ is _______.
($\pi = 22/7$)

Work Done Soap Bubble Inflation Calculation

This section provides a step-by-step calculation for the work done in increasing the diameter of a soap bubble, based on surface tension.

Physics Concepts for Work Done

The work done ($W$) to change the surface area of a liquid film is related to surface tension ($T$) and the change in area ($\Delta A$). For a soap bubble, which has two surfaces (inner and outer), the formula is:

$W = 2 \times T \times \Delta A$

The surface area ($A$) of a sphere with radius ($r$) is given by $A = 4 \pi r^2$. Therefore, the change in area is $\Delta A = A_{final} - A_{initial} = 4 \pi r_{final}^2 - 4 \pi r_{initial}^2$.

Calculation Steps

Step 1: Define Given Parameters

  • Surface Tension, $T = 0.04 \text{ N/m}$
  • Initial Diameter, $d_1 = 7 \text{ cm} = 0.07 \text{ m}$
  • Final Diameter, $d_2 = 14 \text{ cm} = 0.14 \text{ m}$
  • Work Done Formula: $W = (15000 - x) \text{ }\mu\text{J}$
  • Value of Pi, $\pi = 22/7$

Step 2: Determine Initial and Final Radii

  • Initial Radius, $r_1 = d_1 / 2 = 0.07 / 2 = 0.035 \text{ m}$
  • Final Radius, $r_2 = d_2 / 2 = 0.14 / 2 = 0.07 \text{ m}$

Step 3: Calculate Change in Surface Area ($\Delta A$)

The change in area depends on the squares of the radii:

$\Delta A = 4 \pi (r_2^2 - r_1^2)$

Substituting the values:

$\Delta A = 4 \times \frac{22}{7} \times \left( (0.07)^2 - (0.035)^2 \right) \text{ m}^2$

$\Delta A = \frac{88}{7} \times (0.0049 - 0.001225) \text{ m}^2$

$\Delta A = \frac{88}{7} \times 0.003675 \text{ m}^2$

$\Delta A = 0.0462 \text{ m}^2$

Step 4: Calculate Total Work Done ($W$)

Use the work formula $W = 2 \times T \times \Delta A$:

$W = 2 \times (0.04 \text{ N/m}) \times (0.0462 \text{ m}^2)$

$W = 0.08 \times 0.0462 \text{ J}$

$W = 0.003696 \text{ J}$

Convert the work done to microJoules ($\mu J$):

$W = 0.003696 \times 10^6 \text{ }\mu\text{J} = 3696 \text{ }\mu\text{J}$

Step 5: Determine the value of $x$

Equate the calculated work ($W$) with the given expression for work done:

$15000 - x = 3696$

Solve for $x$:

$x = 15000 - 3696$

$x = 11304$

The value of $x$ is 11304.

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Important Questions from Mechanics

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