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JEE Main 202329 JanuaryMorning Shift
Exam
JEE Main
Year
2023
Date
29 January
Shift
Morning Shift
Subject
Physics
Chapter
Mechanical Properties of Fluids

JEE Main 2023 · 29 January · Morning Shift · Physics · Mechanical Properties of Fluids

Surface tension of a soap bubble is 2 . 0 × 10 ^ - 2 N m ^ - 1 ....

Surface tension of a soap bubble is 2.0×10−2Nm−12 . 0 \times {10}^{- 2}N{m}^{- 1} . Work done to increase the radius of soap bubble from 3.5cm3 . 5cm to 7cm7cm will be : [Take π=227\pi = \dfrac{22}{7} ]Surface tension of a soap bubble is 2 . 0 × 10 ^ - 2 N m ^ - 1 . Work done to increase the radius of soap bubble from 3

Options

  1. A0.72×10−4J0 . 72 \times {10}^{- 4}J
  2. B5.76×10−4J5 . 76 \times {10}^{- 4}J
  3. C18.48×10−4J18 . 48 \times {10}^{- 4}J
  4. D9.24×10−4J9 . 24 \times {10}^{- 4}J

Correct answer

C. 18.48×10−4J18 . 48 \times {10}^{- 4}J

Step-by-step solution

Surface tension of a soap bubble is 2 . 0 × 10 ^ - 2 N m ^ - 1 . Work done to increase the radius of soap bubble from 3

As we know,

Work done = change in surface energy of soap bubble

Therefore,

Work done =2T(4πr22−4πr12)= 2T\left(4\pi{r}_{2}^{2} - 4\pi{r}_{1}^{2}\right)

=8\piT(r22−r12)= 8\piT\left({r}_{2}^{2} - {r}_{1}^{2}\right)

=8×227×0.02(72−3.52)×10−4J= 8 \times \dfrac{22}{7} \times 0 . 02\left({7}^{2} - 3 . {5}^{2}\right) \times {10}^{- 4}J

=18.48×10−4J= 18 . 48 \times {10}^{- 4}J

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