benjaminlimbu454
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The viscosity [tex]$\eta[tex]$[/tex] of the liquid flowing through a capillary tube of length [tex]$[/tex]L[tex]$[/tex] and radius [tex]$[/tex]r$[/tex] is given by

[tex]
\frac{v}{t}=\pi \frac{\left(P_1-P_2\right) r^4}{8 \eta L}
[/tex]

Where [tex]$P_1[tex]$[/tex] and [tex]$[/tex]P_2[tex]$[/tex] are the pressures at the ends of the tube, [tex]$[/tex]t[tex]$[/tex] is the time taken, [tex]$[/tex]r[tex]$[/tex] is the radius, and [tex]$[/tex]v$[/tex] is the volume of the liquid.

(i) Find the expression for the fractional error in [tex]$\eta$[/tex].



Answer :

GinnyAnswer
Sure, let's find the expression for the viscosity \(\eta\) from the given equation. The given formula is:

[tex]\[ \frac{v}{t} = \pi \frac{(P1 - P2)}{8 \eta L} \][/tex]

We need to isolate \(\eta\) on one side of the equation:

### Step 1: Multiply both sides of the equation by \(8 \eta L\)
Doing this will help get \(\eta\) out of the denominator on the right-hand side.

[tex]\[ 8 \eta L \cdot \frac{v}{t} = \pi (P1 - P2) \][/tex]

### Step 2: Divide both sides by \(8 \left(\frac{v}{t}\right) L\)
To isolate \(\eta\), we divide both sides by \(8 \left(\frac{v}{t}\right) L\):

[tex]\[ \eta = \frac{\pi (P1 - P2)}{8 \left(\frac{v}{t}\right) L} \][/tex]

### Final Expression
We have successfully isolated \(\eta\):

[tex]\[ \eta = \pi \frac{(P1 - P2)}{8 \left(\frac{v}{t}\right) L} \][/tex]

So the expression for the viscosity \(\eta\) in terms of the given variables is:

[tex]\[ \eta = \pi \frac{(P1 - P2)}{8 \left(\frac{v}{t}\right) L} \][/tex]

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