Question

The work done W is required by an agent to form a bubble of radius R. An extra amount of work △W is required to increase the radius by △R. If \(\frac{△R}{R}\)=1%,\(\frac{△W}{W}\) is:

• A. 2%
• B. 1%
• C. 4%
• D. 3%
• E. 0.50%

Answer 1

The Correct Option is A

Step 1: Initial Work to Form Bubble

Work required to form bubble of radius R:

\[ W = 8\pi \gamma R^2 \]

(where γ is surface tension)

Step 2: Work for Increased Radius

For radius R + ΔR:

\[ W + \Delta W = 8\pi \gamma (R + \Delta R)^2 \]

Step 3: Expand and Find ΔW

\[ \Delta W = 8\pi \gamma [(R + \Delta R)^2 - R^2] \]

\[ \Delta W ≈ 8\pi \gamma [2R\Delta R] \] (for small ΔR)

Step 4: Calculate Ratio

Given \( \frac{\Delta R}{R} = 1\% \):

\[ \frac{\Delta W}{W} = \frac{16\pi \gamma R\Delta R}{8\pi \gamma R^2} = 2\frac{\Delta R}{R} = 2\% \]

Answer 2

1. Relate work done to surface tension and radius:

The work done (W) to form a bubble of radius R is proportional to the change in surface area and is given by:

\[W = S \times \Delta A\]

where S is the surface tension. A bubble has two surfaces (inner and outer), so the change in surface area is:

\[\Delta A = 2(4\pi R^2 - 0) = 8\pi R^2\]

Therefore:

\[W = 8\pi R^2 S\]

2. Calculate the extra work done (ΔW):

The extra work (ΔW) required to increase the radius by ΔR is:

\[\Delta W = S \times \Delta A' = S \times 2(4\pi (R + \Delta R)^2 - 4\pi R^2)\]

\[\Delta W = 8\pi S((R + \Delta R)^2 - R^2) = 8\pi S(R^2 + 2R\Delta R + (\Delta R)^2 - R^2)\]

Since ΔR is very small compared to R, we can neglect the (\Delta R)² term:

\[\Delta W \approx 8\pi S(2R\Delta R) = 16\pi SR\Delta R\]

3. Find the ratio ΔW/W:

\[\frac{\Delta W}{W} = \frac{16\pi SR\Delta R}{8\pi R^2 S} = \frac{2\Delta R}{R}\]

We are given \(\frac{\Delta R}{R} = 0.01 = 1\%\). Therefore:

\[\frac{\Delta W}{W} = 2(0.01) = 0.02 = 2\%\]

Final Answer: The final answer is \(\boxed{A}\)