Question

Dimension of coefficient of viscosity

Answer 1

Answer:

• SI unit of coefficient of viscosity is Newton-second per square metre (N−s/m^2)  

• It is equivalent to Pascal - seconds.

• Its dimensional formula is [M^1 L^−2 T^−1]

• CGS unit is ‘poise’.  

• Mathematically,
• Coefficient of viscosity = Eta(n) = (Tangential Force∗Distance between          Layers/(Area∗Velocity gradient)

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Answer 2

$\maltese \: \underline{ \bf Answer} \: \maltese$

Drag force/Viscous force is calculated by the given below formula :

$:\implies \sf F_v = 6\pi \eta rv$

• Fᵥ = Viscous force/ Drag force
• η = Coefficient of viscosity
• r = radius
• v = velocity

We are asked to find the Dimensions of Coefficient of viscosity.

But we know that, '6' and 'π' both are dimensionaless quantity beacuse they are constant.

$\dag\:\underline{\tt Dimensions \: of \: Viscous \: force \: (F_v) : }$

$:\implies \sf [F_v] = [MLT^{-2}]$

$\dag\:\underline{\tt Dimensions \: of \: Radius\: (r) : }$

$:\implies \sf [r] = [L]$

$\dag\:\underline{\tt Dimensions \: of \: Velocity\: (v) : }$

$:\implies \sf [v] = [LT^{-1}]$

Now, let's find the Dimensions of Coefficient of viscosity :

$\longrightarrow\:\:\sf [F_v] = [\eta] \: [ r] \: [v ]$

$\longrightarrow\:\:\sf [MLT^{-2}] = [\eta] \: [L] \: [LT^{-1} ]$

$\longrightarrow\:\:\sf [MLT^{-2}] = [\eta] \: [L^{2} T^{-1} ]$

$\longrightarrow\:\:\sf \dfrac{ [MLT^{-2}]}{[L^{2} T^{-1} ]} = [\eta]$

$\longrightarrow\:\:\sf [\eta] = \dfrac{ [MLT^{-2}]}{[L^{2} T^{-1} ]}$

$\longrightarrow\:\:\sf [\eta] = [MLT^{-2}][L^{ - 2} T^{1} ]$

$\longrightarrow\:\: \pink{\underline{ \orange{\boxed{\red{\bf [\eta] = [ML^{ - 1} T^{ - 1} ] }}}}\:\:\bigstar}$