Question

Let $W_1$ be the work done in blowing a soap bubble of radius $r$ from a soap solution at room temperature. The soap solution is now heated and a second soap bubble of radius $2r$ is blown from the heated soap solution. If $W_2$ is the work done in forming this second bubble, then

• A. $W_2 = 2W_1$
• B. $W_2 = 4W_1$
• C. $W_2 > 4W_1$
• D. $W_2 < 4W_1$

Hint:

If the temperature had remained constant, doubling the radius would increase the required work by a factor of $2^2 = 4$ due to the area scaling rule ($W \propto r^2$). However, because heating the solution decreases its surface tension coefficient, the required work must end up being strictly less than this unheated value, pointing directly to $W_2 < 4W_1$.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The problem compares the work done to create two different soap bubbles. The first bubble has a radius $r$ and is blown at room temperature. The second bubble has its radius doubled to $2r$, but is blown from a solution that has been heated. We need to determine the correct inequality relation between the work values $W_1$ and $W_2$.

Step 2: Key Formula or Approach:
1. A soap bubble has two free surfaces (an inner and an outer surface). The work done ($W$) to form a bubble of radius $R$ against surface tension ($T$) is equal to the increase in surface energy: $$W = 2 \times T \times \Delta A = 2 \times T \times (4\pi R^2) = 8\pi R^2 T$$ 2. Surface tension is temperature-dependent: heating a liquid decreases its cohesive molecular forces, causing its surface tension coefficient to drop ($T_{\text{hot}} < T_{\text{cold}}$).

Step 3: Detailed Explanation:
Let's write down the expression for the first bubble blown at room temperature ($T_1$): $$W_1 = 8\pi r^2 T_1$$ Now write down the expression for the second bubble blown from the heated solution, where radius $R = 2r$ and surface tension drops to $T_2$: $$W_2 = 8\pi (2r)^2 T_2 = 8\pi (4r^2) T_2 = 4 \cdot (8\pi r^2 T_2)$$ Let's take the ratio of $W_2$ to $W_1$: $$\frac{W_2}{W_1} = \frac{4 \cdot (8\pi r^2 T_2)}{8\pi r^2 T_1} = 4 \left(\frac{T_2}{T_1}\right)$$ $$W_2 = 4W_1 \left(\frac{T_2}{T_1}\right)$$ Since the soap solution was heated for the second bubble, its temperature increased. Because surface tension decreases with rising temperature, we know that: $$T_2 < T_1 \implies \frac{T_2}{T_1} < 1$$ Substituting this inequality boundary constraint into our equation reveals that: $$W_2 < 4W_1$$

Step 4: Final Answer:
The work done for the second bubble satisfies the condition $W_2 < 4W_1$, matching option (D).