Question

A square plate of 10 cm side moves parallel to another plate with a relative velocity of 10 cm $s^{-1}$, both plates immersed in water. If the viscous force is 2 x ${10}^{-3}$ N, calculate the perpendicular distance between the plates.

• A. 0.07 cm
• B. 0.04 cm
• C. 0.06 cm
• D. 0.05 cm

Answer 1

The correct option is D 0.05 cm

side length $=10cm={10}^{-1}m$

Area of plate $=100\times {cm}^2={10}^{-2}{m}^2$

viscosity constant $=8.9\times {10}^{-4}Pa\cdot s$

viscosity constant $\left(\mu \right)\approx 9\times {10}^{-4}Pa\cdot S$

Force $\left(F\right)=2\times {10}^{-3}N$

now $=$

$F=\mu A\frac{\Delta V}{\Delta x}\Delta V=$ relative velocity

putting value

$2\times {10}^{-3}N=(9\times {10}^{-4}Pa\cdot S)\times \left({10}^{-2}{m}^2\right)\times \frac{\left({10}^{-1}\right)}{\Delta x}$

$\Delta x=0.05cm$

Hence distance $=0.05cm$

Correct option $\left(D\right)$