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BITSAT 2010
Exam
BITSAT
Year
2010
Subject
Physics
Chapter
Mechanical Properties of Fluids

BITSAT 2010 · Physics · Mechanical Properties of Fluids

Two circular plates of radius 5 cm each, have a 0.01 mm thick water...

Two circular plates of radius 5cm5 \mathrm{ cm} each, have a 0.01 mm thick water film between them. Then what will be the force required to separate these plate (S.T. of water =73=73 dyne/cm)?

Options

  1. A125N125 \mathrm{ N}
  2. B95N95 \mathrm{ N}
  3. C115N115 \mathrm{ N}
  4. D105N105 \mathrm{ N}

Correct answer

C. 115N115 \mathrm{ N}

Step-by-step solution

F=2ATd=2×π×(0.05)2×73×10−30.01×10−3=36.5π≈115 newton \begin{aligned} \mathrm{F}=& \frac{2 \mathrm{AT}}{\mathrm{d}}=\frac{2 \times \pi \times(0.05)^{2} \times 73 \times 10^{-3}}{0.01 \times 10^{-3}} \\=& 36.5 \pi \approx 115 \text{ newton } \end{aligned}
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