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Turpentine oil is flowing through a tube...

Turpentine oil is flowing through a tube of length `L` and radius `r`. The pressure difference between the two ends of the tube is `p` , the viscosity of the coil is given by `eta = (p (r^(2) - x^(2)))/(4 vL)`, where `v` is the velocity of oil at a distance `x` from the axis of the tube. From this relation, the dimensions of viscosity `eta` are

A

`[M^(0)L^(0)T^(0)]`

B

`[MLT^(-1)]`

C

`[ML^(2)T^(-2)]`

D

`[ML^(-1)T^(-1)]`

Text Solution

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To find the dimensions of viscosity \(\eta\) given the relation: \[ \eta = \frac{p (r^2 - x^2)}{4 v L} \] we will follow these steps: ### Step 1: Identify the dimensions of each variable 1. **Pressure \(p\)**: The dimension of pressure is given by: \[ [p] = \frac{F}{A} = \frac{MLT^{-2}}{L^2} = ML^{-1}T^{-2} \] 2. **Radius \(r\)**: The dimension of radius is: \[ [r] = L \] 3. **Distance \(x\)**: The dimension of distance is the same as radius: \[ [x] = L \] 4. **Velocity \(v\)**: The dimension of velocity is: \[ [v] = \frac{L}{T} \] 5. **Length \(L\)**: The dimension of length is: \[ [L] = L \] ### Step 2: Substitute the dimensions into the viscosity formula Now, substituting the dimensions into the viscosity formula: \[ \eta = \frac{p (r^2 - x^2)}{4 v L} \] ### Step 3: Calculate the dimensions of the numerator and denominator 1. **Numerator**: The term \(p (r^2 - x^2)\): - The dimension of \(r^2\) and \(x^2\) is: \[ [r^2] = [x^2] = L^2 \] - Thus, the dimension of \(r^2 - x^2\) is: \[ [r^2 - x^2] = L^2 \] - Therefore, the dimension of the numerator becomes: \[ [p (r^2 - x^2)] = [p] \cdot [r^2 - x^2] = (ML^{-1}T^{-2})(L^2) = ML^{1}T^{-2} \] 2. **Denominator**: The term \(4 v L\): - The dimension of \(v\) is: \[ [v] = \frac{L}{T} \] - The dimension of \(L\) is \(L\), so: \[ [4vL] = \left[\frac{L}{T}\right] \cdot [L] = \frac{L^2}{T} \] ### Step 4: Combine the dimensions to find viscosity Now we can find the dimensions of viscosity \(\eta\): \[ [\eta] = \frac{[p (r^2 - x^2)]}{[4 v L]} = \frac{ML^{1}T^{-2}}{\frac{L^2}{T}} = \frac{ML^{1}T^{-2} \cdot T}{L^2} = \frac{ML^{1}T^{-1}}{L^2} = ML^{-1}T^{-1} \] ### Conclusion Thus, the dimensions of viscosity \(\eta\) are: \[ [\eta] = ML^{-1}T^{-1} \]
To find the dimensions of viscosity \(\eta\) given the relation: \[ \eta = \frac{p (r^2 - x^2)}{4 v L} \] we will follow these steps: ...
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