A water drop of radius 1 μm falls in a situation where the effect of buoyant force is negligible. Co-efficient of viscosity of air is 1.8 × 10–5 Nsm–2 and its density is negligible as compared to that of water (106 gm–3). Terminal velocity of the water drop is
(Take acceleration due to gravity = 10 ms–2)
(Take acceleration due to gravity = 10 ms–2)
145.4 × 10–6 ms–1
118.0 × 10–6 ms–1
132.6 × 10–6 ms–1
123.4 × 10–6 ms–1
The Correct Option is D
Approach Solution - 1
To find the terminal velocity of the water drop, we can use Stokes' Law. According to Stokes' Law, the terminal velocity \(v_t\) of a spherical object moving under the influence of gravity through a fluid is given by:
\(v_t = \frac{{2r^2g(\rho - \sigma)}}{{9\eta}}\)
where:
- \(r\) is the radius of the drop.
- \(g\) is the acceleration due to gravity.
- \(\rho\) is the density of the drop (water).
- \(\sigma\) is the density of the fluid (air) - negligible in this case.
- \(\eta\) is the coefficient of viscosity of the fluid (air).
Given data:
- \(r = 1 \, \mu m = 1 \times 10^{-6} \, m\)
- \(g = 10 \, \text{m/s}^2\)
- \(\rho = 10^6 \, \text{gm/m}^3 = 10^3 \, \text{kg/m}^3\)
- \(\eta = 1.8 \times 10^{-5} \, \text{Ns/m}^2\)
- \(\sigma = 0 \, \text{kg/m}^3\) (negligible)
Substituting these values into the formula:
\(v_t = \frac{{2 \times (1 \times 10^{-6})^2 \times 10 \times (10^3 - 0)}}{{9 \times 1.8 \times 10^{-5}}}\)
Simplifying further:
\(v_t = \frac{{2 \times 10^{-12} \times 10^4}}{{9 \times 1.8 \times 10^{-5}}}\)
\(v_t = \frac{{2 \times 10^{-8}}}{{16.2 \times 10^{-5}}}\)
\(v_t = \frac{{2 \times 10^{-8}}}{{16.2 \times 10^{-5}}} \, \approx 1.234 \times 10^{-4} \, \text{m/s}\)
Converting it to the format given in the options:
\(v_t = 123.4 \times 10^{-6} \, \text{ms}^{-1}\)
Therefore, the correct answer is 123.4 × 10–6 ms–1.
Approach Solution -2
The correct answer is (D) : 123.4 × 10–6 ms–1
\(6πηrv=mg\)
\(6πηrv=\frac{4}{3}πr^3ρg\)
or
\(v=\frac{2}{9}\frac{ρr^2g}{η}\)
\(=\frac{2}{9}×\frac{10^3×(10^{−6})^2×10}{1.8×10^{−5}}\)
= 123.4×10–6 m/s
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Concepts Used:
Viscosity
Viscosity is a measure of a fluid’s resistance to flow. The SI unit of viscosity is poiseiulle (PI). Its other units are newton-second per square metre (N s m-2) or pascal-second (Pa s.) The dimensional formula of viscosity is [ML-1T-1].
Viscosity: Formula
Viscosity is measured in terms of a ratio of shearing stress to the velocity gradient in a fluid. If a sphere is dropped into a fluid, the viscosity can be determined using the following formula:
η = [2ga2(Δρ)] / 9v
Where ∆ρ is the density difference between fluid and sphere tested, a is the radius of the sphere, g is the acceleration due to gravity and v is the velocity of the sphere.
Viscosity: Types
- Dynamic viscosity: When the viscosity is measured directly by measuring force. It is defined as the ratio of shear stress to the shear strain of the motion. Dynamic viscosity is used to calculate the rate of flow in liquid.
- Kinematic viscosity: There is no force involved. It can be referred to as the ratio between the dynamic viscosity and density of the fluid. It can be computed by dividing the dynamic viscosity of the fluid with fluid mass density.
- Laminar flow: Laminar flow is the type of flow in which the fluid moves smoothly or in a regular path from one layer to the next. Laminar flow occurs in lower velocities.