Q135 / 160 · AP EAMCET 2022 (06 Jul Shift 1) · Physics

A cylindrical metal box whose flat surface has an area of $0.01 \mathrm{~m}^2$ rests on liquid of $0.3 \mathrm{~mm}$ thickness. If upon applying a horizontal force of magnitude $\frac{1}{3} \mathrm{~N}$, the box slides with a constant speed of $0.09 \mathrm{~ms}^{-1}$, the coefficient of viscosity of the liquid is nearly

  • A
    $2.5 \times 10^{-2} \mathrm{Pa.s}$
  • Correct answer
    B
    $1.1 \times 10^{-1} \mathrm{Pa.s}$
  • C
    $1.1 \times 10^{-2} \mathrm{Pa.s}$
  • D
    $2.5 \times 10^{-1} \mathrm{~Pa} . \mathrm{s}$

Solution

We know that $$ \mathrm{F}_{\mathrm{drag}}=\eta \mathrm{A} \frac{\mathrm{dv}}{\mathrm{dx}} $$ For metal box to be moving with constant velocity $$ \begin{aligned} & \mathrm{F}_{\mathrm{drag}}=\mathrm{F} \\ & \Rightarrow \eta \mathrm{A} \frac{\mathrm{dv}}{\mathrm{dx}}=\mathrm{F} \\ & \Rightarrow \eta=\mathrm{F} \frac{\mathrm{dx}}{\mathrm{Adx}}=\frac{\frac{1}{3} \times 0.3 \times 10^{-3}}{0.01 \times(0.09-0)} \\ & =1.1 \mathrm{PaS} \end{aligned} $$

Hence, the correct answer is option: 'B'.

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