Question

13. 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 oil 1.5 x 10^-3 m thick. Find the
coefficient of viscosity of oil.
[Ans. 7.5 Ns/m?)​

Answer 1

$\color{darkblue}\underline{\underline{\sf Given-}}$

• Force (F) = 1N
• Area (A) = ${\sf 10^{-2}m^2}$
• Velocity (v) =${\sf 2×10^{-2}m/s}$
• thickness of layer ${\sf (x)=1.5×10^{-3}m}$

$\color{darkblue}\underline{\underline{\sf To \: Find-}}$

• Coefficient of viscosity ${\sf (\eta)}$

$\color{green}\bullet\underline{\boxed{\sf Newton's\:Law\;of\: viscosity-}}$

$\Large\color{red}\blacksquare\boxed{\sf F=\eta A\dfrac{dv}{dx}}$

$\implies{\sf 1=\eta × 10^{-2}\left(\dfrac{2×10^{-2}}{1.5×10^{-3}}\right)}$

$\implies{\sf 1=\eta×10^{-2}×(1.33×10)}$

$\implies{\sf 1=\eta × 10^{-2}×13.3 }$

$\implies{\sf \eta =\dfrac{1}{13.3×10^{-2}} }$

$\color{red}\implies{\sf \eta = 7.5\:poise }$

$\color{darkblue}\underline{\underline{\sf Answer-}}$

Coefficient of viscosity of oil is $\color{red}{\sf 7.5\:poise}$.

Answer 2

Answer

• Coefficient of viscosity of oil = 7.5

Given

• A horizontal force of 1 N is required to move a metal plate of area  10⁻² m² with a velocity of  2 * 10⁻² m/s, when it rests on a layer of oil  1.5 * 10⁻³ m thick

To Find

• Coefficient of viscosity of oil

Formula Used

$\bf F=\eta A\dfrac{dv}{dx}$

where ,

F denotes force

η denotes coefficient of viscosity

A denotes area

dv denotes velocity

dx denotes thickness

Solution

$\bf F=\eta A\dfrac{dv}{dx}\\\\\implies 1=\eta \times 10^{-2}\times \dfrac{2 \times 10^{-2}}{1.5 \times 10^{-3}}\\\\ \implies \bf \eta = \dfrac{1.5 \times 10^{-3}}{2 \times 10^{-4}}\\\\\implies \bf \eta = \dfrac{1.5 \times 10}{2}\\\\\implies \bf \eta =\dfrac{15}{2}\\\\\implies \bf \eta =7.5$

So , Coefficient of viscosity of oil , η = 7.5 poise