Two square metal plates, each of side 10 cm are immersed in water. One plate moves parallel to the other with a velocity of 5 cm `s^(-1)`. If the viscous force is 150 dyne, what is their distance of separation ? Given `eta_("water")` = .001 PI

Explanation

To solve the problem step by step, we will use the formula for viscous force and the given data.

### Step 1: Identify the given values
- Side of the square plates, \( L = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Area of one plate, \( A = L^2 = (0.1 \, \text{m})^2 = 0.01 \, \text{m}^2 \)
- Viscous force, \( F_v = 150 \, \text{dyne} = 150 \times 10^{-5} \, \text{N} = 1.5 \times 10^{-3} \, \text{N} \)
- Relative velocity, \( v = 5 \, \text{cm/s} = 0.05 \, \text{m/s} \)
- Viscosity of water, \( \eta = 0.001 \, \text{poise} = 0.001 \times 0.1 \, \text{N s/m}^2 = 10^{-4} \, \text{N s/m}^2 \)
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