Question

A horizontal force of 1 N is required to move a metal plate of area $\tt 10^{-2}$ m² with a velocity of $\tt 2 x 10^{-2}$ m/s, when it rests on a layer of oil $\tt 1.5 x 10^{-3}$ m thick. Find the coefficient of viscosity of oil.​

Answer 1

Explanation:

A horizontal force of 1 N is required to move a metal plate of area 10^-2 m² with a velocity of 2 x 10^-2m/s, when it rests on a layer of ... coefficient of viscosity of oil. [Ans. 7.5 Ns/m?). 

please mark Brainlist answer

Answer 2

Given :-

• Force (F) = 1 N
• Area (A) = $\tt 10^{-2}$ m²
• Velocity (V) = $\tt 2 x 10^{-2}$ m/s
• Thickness of layer = $\tt 1.5 x 10^{-3}$

To Find :-

• The (coefficient of viscosity) η of oil

Answer :-

✏ The coefficient of viscosity (η) of oil is 7.5 poise.

Solution :-

Applying , Newton law of Viscosity

✅$</em></strong><strong><em>F</em></strong><strong><em>=</em></strong><strong><em> </em></strong><strong><em>η</em></strong><strong><em>A</em></strong><strong><em> \</em></strong><strong><em>d</em></strong><strong><em>frac{dv}{dx}$

➡ $1 = </em></strong><strong><em>η</em></strong><strong><em> \times {10}^{ - 2} ( \</em></strong><strong><em>d</em></strong><strong><em>frac{2 \times 10 ^{ - 2} }{1.5 \times 10 ^{ - 3} } )$

➡ 1 = η×10−² ×(1.33×10)

➡ 1 = η×10−² ×13.3

➡ $η = \dfrac{1}{13.3 \times {10}^{ - 2} }$

✒η = 7.5 poise

________________________________