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Chapter 8: Problem 31

The dimensional formula for coefficient of viscosity (A) \(\left[\mathrm{MLT}^{-1}\right]\) (B) \(\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right]\) (C) \(\left[\mathrm{MLT}^{-2}\right]\) (D) \(\left[\mathrm{ML}^{2} \mathrm{~T}^{-1}\right]\)

Short Answer

Expert verified
The correct dimensional formula for the coefficient of viscosity is \(\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right]\). Thus, the correct answer is (B).

Step by step solution

01

Recall the equation of viscous force

The equation for viscous force acting on a fluid is given by F = 畏 * A * (dv/dy), where F is the viscous force, 畏 is the coefficient of viscosity, A is the area of contact between fluid layers, and dv/dy is the velocity gradient (change in velocity with respect to the change in the distance between fluid layers).
02

Write down the dimensions for each term

We can now write down the dimensions for each term in the equation: 1. Force (F) has dimensions of \(\left[\mathrm{MLT}^{-2}\right]\), since force is mass times acceleration. 2. Area (A) has dimensions of \(\left[\mathrm{L}^{2}\right]\), since it is a length squared. 3. Velocity gradient (dv/dy) has dimensions of \(\left[\mathrm{LT}^{-1} \mathrm{~L}^{-1}\right] = \left[\mathrm{T}^{-1}\right]\), as velocity has dimensions of \(\left[\mathrm{LT}^{-1}\right]\) divided by distance with dimensions of \(\left[\mathrm{L}\right]\).
03

Substitute the dimensions into the equation

We can substitute the dimensions for the terms in the equation F = 畏 * A * (dv/dy): \(\left[\mathrm{MLT}^{-2}\right] = \left[\eta\right] \cdot \left[\mathrm{L}^{2}\right] \cdot \left[\mathrm{T}^{-1}\right]\)
04

Solve for the dimensions of the coefficient of viscosity

Now, we need to find the dimensions of 畏 (coefficient of viscosity). To do this, rearrange the above equation to isolate 畏: \(\left[\eta\right] = \frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{L}^{2}\right] \cdot \left[\mathrm{T}^{-1}\right]}\)
05

Simplify the equation

Next, simplify the equation by performing dimension cancellation and obtaining the final expression for the coefficient of viscosity: \(\left[\eta\right] = \frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{L}^{2} \mathrm{~T}^{-1}\right]}\) \(\left[\eta\right] = \left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right]\) The correct dimensional formula for the coefficient of viscosity is \(\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right]\). Thus, the correct answer is (B).

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Most popular questions from this chapter

A tank, which is open at the top, contains a liquid up to a height \(H\). A small hole is made in the side of the tank at a distance \(y\) below the liquid surface. The liquid emerging from the hole lands at a distance \(x\) from the \(\operatorname{tank}\) (A) If \(y\) is increased from zero to \(H, x\) will first increase and then decrease. (B) \(x\) is maximum for \(y=H / 2\). (C) The maximum value of \(x\) is \(H\). (D) The maximum value of \(x\) will depend on the density of the liquid.

A tank is filled with water and an orifice is made in the wall so that the horizontal range \(x\) of water rushing out is maximum. If \(H\) is the height of water above the orifice in the tank, then (A) \(x=H\) (B) \(x=2 H\) (C) \(2 x=H\) (D) \(4 x=H\)

The surface tension of a liquid is \(5 \mathrm{~N} / \mathrm{m}\). If a film is held on a ring of area \(0.02 \mathrm{~m}^{2}\), its total surface energy is about (A) \(5 \times 10^{-2} \mathrm{~J}\) (B) \(2.5 \times 10^{-2} \mathrm{~J}\) (C) \(2 \times 10^{-1} \mathrm{~J}\) (D) \(3 \times 10^{-1} \mathrm{~J}\)

If Young's modulus of iron is \(2 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}\) and the interatomic spacing between two molecules is \(3 \times 10^{-10} \mathrm{~m}\), the interatomic force constant is (A) \(60 \mathrm{~N} / \mathrm{m}\) (B) \(120 \mathrm{~N} / \mathrm{m}\) (C) \(3 \mathrm{~N} / \mathrm{m}\) (D) \(180 \mathrm{~N} / \mathrm{m}\)

A uniform cylinder of length \(\ell\) and mass \(M\) having cross-sectional area \(A\) is suspended, with its length vertical, from a fixed point by a massless spring, such that it is half submerged in a liquid of density at equilibrium position. The extension of the spring when it is in equilibrium is: (A) \(\frac{M g}{k}\left(1-\frac{\ell A \sigma}{M}\right)\) (B) \(\frac{M g}{k}\left(1-\frac{\ell A \sigma}{2 M}\right)\) (C) \(\frac{M g}{k}\left(1+\frac{\ell A \sigma}{M}\right)\) (D) \(\frac{M g}{k}\)
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