Verified Solution • 2025–26 Session moderate

QJEEM23Ph1cf3: JEE Main 2023 Physics - Class 11 JEEM23Ph1cf3 Solution: Surface tension of a soap bubble is . Work done to increase the …

Surface tension of a soap bubble is 2.0×10−2Nm−12.0 \times 10^{-2} \mathrm{Nm}^{-1} . Work done to increase the radius of soap bubble from 3.5 cm3.5 \mathrm{~cm} to 7 cm7 \mathrm{~cm} will be:

Take [π=227]\left[\pi=\frac{22}{7}\right]

Multiple Choice Options

  • A 18.48×10−4 J18 .48 \times 10^{-4} \mathrm{~J}
  • B 5.76×10−4 J5.76 \times 10^{-4} \mathrm{~J}
  • C 0.72×10−4 J0.72 \times 10^{-4} \mathrm{~J}
  • D 9.24×10−4 J9.24 \times 10^{-4} \mathrm{~J}

Step-by-Step Verified Solution

Surface area of soap bubble =2×4πR2=2 \times 4 \pi \mathrm{R}^{2} Work done == change in surface energy ×TS\times \mathrm{T}_{\mathrm{S}}

=TS×8π×(R22−R12)=\mathrm{T}_{\mathrm{S}} \times 8 \pi \times\left(\mathrm{R}_{2}^{2}-\mathrm{R}_{1}^{2}\right)

=2×10−2×8×227×49×34×10−4=2 \times 10^{-2} \times 8 \times \frac{22}{7} \times 49 \times \frac{3}{4} \times 10^{-4}

=18.48×10−4 J=18.48 \times 10^{-4} \mathrm{~J}

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