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JIPMER 2013
Exam
JIPMER
Year
2013
Subject
Physics
Chapter
Mechanical Properties of Fluids

JIPMER 2013 · Physics · Mechanical Properties of Fluids

The work done in increasing the size of a soap film for 10 ~cm × 6...

The work done in increasing the size of a soap film for 10 cm×6 m10 \mathrm{~cm} \times 6 \mathrm{~m} to 10 cm×11 cm10 \mathrm{~cm} \times 11 \mathrm{~cm} is 3×10−4 J3 \times 10^{-4} \mathrm{~J}. The surface tension of the film is

Options

  1. A1.0×10−2 N/m1.0 \times 10^{-2} \mathrm{~N} / \mathrm{m}
  2. B6.0×10−2 N/m6.0 \times 10^{-2} \mathrm{~N} / \mathrm{m}
  3. C3.0×10−2 N/m3.0 \times 10^{-2} \mathrm{~N} / \mathrm{m}
  4. D1.5×10−2 N/m1.5 \times 10^{-2} \mathrm{~N} / \mathrm{m}

Correct answer

C. 3.0×10−2 N/m3.0 \times 10^{-2} \mathrm{~N} / \mathrm{m}

Step-by-step solution

\begin{aligned} & \text{ Work done }=\text{ Area increased } \\ & \text{ So, surface tension } \\ & \qquad \begin{aligned} S_T & =\frac{3 \times 10^{-4}}{2 \times(10 \times 11-10 \times 6) \times 10^{-4}} \\ & =3 \times 10^{-2} \mathrm{ N} / \mathrm{m}\end{aligned}\end{aligned}$
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