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JEE Advanced 2010Paper 2
Exam
JEE Advanced
Year
2010
Paper
Paper 2
Subject
Physics
Chapter
Mechanical Properties of Fluids

JEE Advanced 2010 · Paper 2 · Physics · Mechanical Properties of Fluids

Paragraph: When liquid medicine of density is to be put in the eye,...

Paragraph: When liquid medicine of density ρ\rho is to be put in the eye, it is done with the help of a dropper. As the bulb on the top of the dropper is pressed, a drop forms at the opening of the dropper. We wish to estimate the size of the drop. We first assume that the drop formed at the opening is spherical because that requires a minimum increase in its surface energy. To determine the size, we calculate the net vertical force due to the surface tension TT when the radius of the drop is RR. When this force becomes smaller than the weight of the drop, the drop gets detached from the dropper.Question: After the drop detaches, its surface energy is

Options

  1. A1.4×10−6 J1.4 \times 10^{-6} \mathrm{~J}
  2. B2.7×10−6 J2.7 \times 10^{-6} \mathrm{~J}
  3. C5.4×10−6 J5.4 \times 10^{-6} \mathrm{~J}
  4. D8.1×10−9 J8.1 \times 10^{-9} \mathrm{~J}

Correct answer

B. 2.7×10−6 J2.7 \times 10^{-6} \mathrm{~J}

Step-by-step solution

Surface energy,E=(4\piR2)T=(4π)(1.4×10−3)2(0.11)=2.7×10−6 J\begin{aligned}E & =\left(4 \piR^2\right) T \\ & =(4 \pi)\left(1.4 \times10^{-3}\right)^2(0.11) \\ & =2.7 \times10^{-6} \mathrm{~J} \end{aligned}\therefore The correct option is (b).
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