MHT CET · Physics · Mechanical Properties of Fluids

The work done in blowing a soap bubble of radius \(R\) is \(W_1\) at room temperature. Now the soap solution is heated. From the heated solution another soap bubble of radius \(2 R\) is blown and the work done is \(W_2\). Then

A \(W_2=0\)
B \(W_2=4 W_1\)
C \(W_2<4 W_1\)
D \(W_2=W_1\)

Verified Solution

Answer & Solution

Correct Answer

(C) \(W_2<4 W_1\)

Step-by-step Solution

Detailed explanation

The work done in forming a soap in forming a soap bubble of radius \(r\) is \(W_1=8 \pi r^2 T_1\), where \(T_1\) is the surface tension at room temperature and the work done to form a soap bubble of radius \(2 r\) is
\(\therefore \frac{W_2}{W_1}=\frac{328 \pi r^2 T_2}{8 \pi r^2 T_1}=\frac{4 T_2}{T_1}\)
If \(T_1=T_2\), then \(W_2=4 W_1\)
But it is given that the solution is heated. As the temperature is increased, the surface tension of the soap solution is decreases
\(\therefore T_2

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Explore more questions on app

From MHT CET

Explore more questions on app