Physics Q33 – JEE-MAIN 2026

Home Physics Mechanical Properties of FluidsQuestion #256

The surface tension of a soap bubble is

0.03 N/m

. The work done in increasing the diameter of bubble from

2 cm

6 cm

α π \times 10^{- 4} J

. The value of

α

is _______. (Take

π = 3.14

)

Clue 1 of 3: Understand the nature of a soap bubble

A soap bubble has two free surfaces, which means the total surface area involved in work done calculations is twice the geometric surface area of a sphere.

🧠 Full Solution Path

Step 1: Identify the formula for work done on a soap bubble✦ Active

A soap bubble has two free surfaces. The work done ( $W$ ) in changing its surface area is given by $W = T \times Δ A^{'}$ , where $T$ is the surface tension and $Δ A^{'}$ is the change in the total surface area. The surface area of a sphere is $A = π D^{2}$ , so for a soap bubble, the total surface area is $A^{'} = 2 π D^{2}$ .

Step 2: Calculate the change in total surface area○ Expand

The initial diameter is $D_{1} = 2 cm = 0.02 m$ . The final diameter is $D_{2} = 6 cm = 0.06 m$ . Initial total surface area: $A_{1}^{'} = 2 π D_{1}^{2} = 2 π (0.02)^{2} = 2 π (0.0004) = 0.0008 π m^{2}$ . Final total surface area: $A_{2}^{'} = 2 π D_{2}^{2} = 2 π (0.06)^{2} = 2 π (0.0036) = 0.0072 π m^{2}$ . Change in total surface area: $Δ A^{'} = A_{2}^{'} - A_{1}^{'} = 0.0072 π - 0.0008 π = 0.0064 π m^{2}$ .

Step 3: Calculate the work done and determine

α

○ Expand

Given surface tension $T = 0.03 N/m$ . Work done $W = T \times Δ A^{'} = 0.03 \times 0.0064 π = 0.000192 π J$ . The problem states $W = α π \times 10^{- 4} J$ . Equating the two expressions for $W$ : $0.000192 π = α π \times 10^{- 4}$ $0.000192 = α \times 10^{- 4}$ $α = \frac{0.000192}{10^{- 4}} = 0.000192 \times 10^{4} = 1.92$ .

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Physics Question 33 – JEE-MAIN 2026

Clue 1 of 3: Understand the nature of a soap bubble

🧠 Full Solution Path