Physics · Mechanical Properties of Matter
JEE Main 2026 — 4 April, Morning Shift — Question 20
The surface tension of a soap solution is 3.5×10⁻² N/m. The work required to increase the radius of a soap bubble from 1 cm to 2 cm is J. The value of is ______. ()
Answer: 264
Numerical answer — enter this value.
Step-by-step solution
Work = 2T×4π(r₂²-r₁²) = 2×3.5×10⁻²×4π×(4-1)×10⁻⁴ = 2×3.5×4π×3×10⁻⁶ = 84π×10⁻⁶ = 84×(22/7)×10⁻⁶ = 264×10⁻⁶ J ⇒ α=264
Answer key and solution verified before publishing.
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- Exam
- JEE Main 2026
- Subject
- Physics
- Chapter
- Mechanical Properties of Matter
- Topic
- Surface Tension and Surface Energy