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201. How does viscosity of liquids vary with temperature?
ⓐ. Increases with increase in temperature
ⓑ. Decreases with increase in temperature
ⓒ. Remains constant
ⓓ. First increases, then decreases
Correct Answer: Decreases with increase in temperature
Explanation: For liquids, viscosity decreases with temperature because higher thermal energy weakens intermolecular forces, allowing molecules to slide past each other more easily.
202. How does viscosity of gases vary with temperature?
ⓐ. Increases with increase in temperature
ⓑ. Decreases with increase in temperature
ⓒ. Remains constant
ⓓ. Depends only on pressure
Correct Answer: Increases with increase in temperature
Explanation: In gases, viscosity increases with temperature because faster moving molecules transfer momentum more effectively between layers.
203. Why does viscosity of liquids decrease with rise in temperature?
ⓐ. Because density decreases
ⓑ. Because intermolecular forces weaken
ⓒ. Because gravitational effect reduces
ⓓ. Because liquid volume decreases
Correct Answer: Because intermolecular forces weaken
Explanation: Higher temperatures reduce cohesive forces between liquid molecules, lowering resistance to flow and decreasing viscosity.
204. Which relation expresses the effect of temperature on viscosity of liquids (Arrhenius relation)?
ⓐ. $\eta = A e^{\frac{E}{RT}}$
ⓑ. $\eta = A e^{\frac{-E}{RT}}$
ⓒ. $\eta = \rho g h$
ⓓ. $\eta = \frac{\tau}{dv/dy}$
Correct Answer: $\eta = A e^{\frac{-E}{RT}}$
Explanation: According to the Arrhenius relation, viscosity of liquids decreases exponentially with increasing temperature: $\eta = A e^{-E/RT}$, where $E$ is activation energy.
205. Which relation shows the temperature dependence of viscosity of gases (Sutherland’s formula)?
ⓐ. $\eta = A T^{-1/2}$
ⓑ. $\eta = A T^{1/2}$
ⓒ. $\eta = \frac{B}{T}$
ⓓ. $\eta = \rho g h$
Correct Answer: $\eta = A T^{1/2}$
Explanation: For gases, viscosity approximately increases with the square root of temperature, expressed as $\eta \propto T^{1/2}$.
206. Effect of pressure on viscosity of liquids is generally:
ⓐ. Very small
ⓑ. Very large and linear
ⓒ. Always negative
ⓓ. Cannot be determined
Correct Answer: Very small
Explanation: Pressure has negligible effect on viscosity of liquids because liquids are nearly incompressible. Temperature is the dominant factor.
207. Effect of pressure on viscosity of gases is generally:
ⓐ. Negligible at low pressure, significant at high pressure
ⓑ. Always negligible
ⓒ. Decreases rapidly with pressure
ⓓ. Increases exponentially with pressure
Correct Answer: Negligible at low pressure, significant at high pressure
Explanation: At low pressures, gas viscosity is nearly independent of pressure. At high pressures, intermolecular collisions affect viscosity significantly.
208. Which factor decreases viscosity of gases?
ⓐ. Increase in temperature
ⓑ. Increase in density
ⓒ. Increase in molecular collisions due to high pressure
ⓓ. Increase in mean free path
Correct Answer: Increase in molecular collisions due to high pressure
Explanation: At high pressures, mean free path decreases and excess collisions reduce momentum transfer efficiency, decreasing gas viscosity.
209. For liquids, which property has the greatest effect on viscosity?
ⓐ. Density
ⓑ. Cohesive intermolecular forces
ⓒ. Pressure
ⓓ. Buoyant force
Correct Answer: Cohesive intermolecular forces
Explanation: Viscosity of liquids arises mainly from intermolecular attractions. Stronger cohesive forces increase viscosity (e.g., glycerin vs. water).
210. Which statement is correct regarding temperature and viscosity?
ⓐ. Viscosity of liquids and gases both decrease with temperature
ⓑ. Viscosity of liquids decreases while viscosity of gases increases with temperature
ⓒ. Viscosity of liquids increases while viscosity of gases decreases with temperature
ⓓ. Viscosity of liquids and gases both increase with temperature
Correct Answer: Viscosity of liquids decreases while viscosity of gases increases with temperature
Explanation: With temperature, liquids flow more easily (viscosity decreases), while gases transfer momentum more effectively (viscosity increases).
211. A liquid flows through a capillary tube of radius $0.5 \, \text{mm}$ and length $20 \, \text{cm}$. If the pressure difference across the tube is $2 \times 10^3 \, \text{Pa}$ and the viscosity of the liquid is $0.1 \, \text{Pa·s}$, calculate the volume flow rate.
ⓐ. $4.9 \times 10^{-9} \, \text{m}^3/\text{s}$
ⓑ. $9.8 \times 10^{-9} \, \text{m}^3/\text{s}$
ⓒ. $1.96 \times 10^{-8} \, \text{m}^3/\text{s}$
ⓓ. $3.92 \times 10^{-8} \, \text{m}^3/\text{s}$
Correct Answer: $9.8 \times 10^{-9} \, \text{m}^3/\text{s}$
Explanation: Using Poiseuille’s law:
$Q = \frac{\pi r^4 \Delta P}{8 \eta L}$.
Here $r = 0.5 \times 10^{-3} \, \text{m}, L = 0.2 \, \text{m}, \Delta P = 2 \times 10^3 \, \text{Pa}, \eta = 0.1$.
Substituting: $Q = \frac{3.14 \times (0.5 \times 10^{-3})^4 \times 2 \times 10^3}{8 \times 0.1 \times 0.2} \approx 9.8 \times 10^{-9} \, \text{m}^3/\text{s}$.
212. A sphere of radius $2 \, \text{mm}$ falls through glycerin with a terminal velocity of $0.02 \, \text{m/s}$. If the density of the sphere is $2500 \, \text{kg/m}^3$ and the density of glycerin is $1200 \, \text{kg/m}^3$, calculate the viscosity of glycerin. (Take $g = 9.8 \, \text{m/s}^2$)
ⓐ. 1.3 Pa·s
ⓑ. 2.0 Pa·s
ⓒ. 2.5 Pa·s
ⓓ. 3.0 Pa·s
Correct Answer: 2.5 Pa·s
Explanation: From Stokes’ law, at terminal velocity:
$6 \pi \eta r v = \frac{4}{3} \pi r^3 (\rho_s – \rho_f) g$.
$\eta = \frac{2 r^2 g (\rho_s – \rho_f)}{9 v}$.
Substitute values: $\eta = \frac{2 (0.002)^2 (9.8)(1300)}{9 (0.02)} \approx 2.5 \, \text{Pa·s}$.
213. A liquid flows through a capillary tube of radius $1 \, \text{mm}$ and length $40 \, \text{cm}$. If the flow rate is $2 \times 10^{-8} \, \text{m}^3/\text{s}$ under a pressure difference of $400 \, \text{Pa}$, calculate its viscosity.
ⓐ. 0.05 Pa·s
ⓑ. 0.10 Pa·s
ⓒ. 0.15 Pa·s
ⓓ. 0.20 Pa·s
Correct Answer: 0.05 Pa·s
Explanation: Using Poiseuille’s law:
$\eta = \frac{\pi r^4 \Delta P}{8 Q L}$.
Here $r = 0.001 \, \text{m}, L = 0.4 \, \text{m}, \Delta P = 400 \, \text{Pa}, Q = 2 \times 10^{-8} \, \text{m}^3/s$.
$\eta = \frac{3.14 (10^{-12})(400)}{8 (2 \times 10^{-8})(0.4)} \approx 0.05 \, \text{Pa·s}$.
214. A steel ball of radius $1.5 \, \text{mm}$ is dropped in a liquid and attains terminal velocity of $0.015 \, \text{m/s}$. If the densities of steel and liquid are $7800 \, \text{kg/m}^3$ and $1000 \, \text{kg/m}^3$, find the viscosity of the liquid.
ⓐ. 2.0 Pa·s
ⓑ. 2.5 Pa·s
ⓒ. 3.0 Pa·s
ⓓ. 3.5 Pa·s
Correct Answer: 2.5 Pa·s
Explanation: From Stokes’ law:
$\eta = \frac{2 r^2 g (\rho_s – \rho_f)}{9 v}$.
Substitute values: $r = 0.0015 \, m, g = 9.8, \Delta \rho = 6800, v = 0.015$.
$\eta = \frac{2 (0.0015)^2 (9.8)(6800)}{9 (0.015)} \approx 2.5 \, \text{Pa·s}$.
215. The kinematic viscosity of a liquid is defined as:
ⓐ. $\nu = \frac{\eta}{\rho}$
ⓑ. $\nu = \rho \eta$
ⓒ. $\nu = \frac{\rho}{\eta}$
ⓓ. $\nu = \frac{1}{\rho \eta}$
Correct Answer: $\nu = \frac{\eta}{\rho}$
Explanation: Kinematic viscosity is the ratio of dynamic viscosity $\eta$ to density $\rho$. Its SI unit is $\text{m}^2/\text{s}$.
216. A liquid has viscosity $0.25 \, \text{Pa·s}$ and density $1000 \, \text{kg/m}^3$. Calculate its kinematic viscosity.
ⓐ. $2.0 \times 10^{-4} \, \text{m}^2/s$
ⓑ. $2.5 \times 10^{-4} \, \text{m}^2/s$
ⓒ. $3.0 \times 10^{-4} \, \text{m}^2/s$
ⓓ. $3.5 \times 10^{-4} \, \text{m}^2/s$
Correct Answer: $2.5 \times 10^{-4} \, \text{m}^2/s$
Explanation: $\nu = \frac{\eta}{\rho} = \frac{0.25}{1000} = 2.5 \times 10^{-4} \, \text{m}^2/s$.
217. The velocity gradient between two fluid layers is $50 \, s^{-1}$. If the viscosity of the liquid is $0.2 \, \text{Pa·s}$, calculate the shear stress.
ⓐ. 5 Pa
ⓑ. 8 Pa
ⓒ. 10 Pa
ⓓ. 12 Pa
Correct Answer: 10 Pa
Explanation: $\tau = \eta \frac{dv}{dy} = (0.2)(50) = 10 \, Pa$.
218. In Poiseuille’s equation, the flow rate through a capillary depends most strongly on:
ⓐ. Length of capillary
ⓑ. Radius of capillary
ⓒ. Pressure difference
ⓓ. Viscosity
Correct Answer: Radius of capillary
Explanation: Flow rate $Q = \frac{\pi r^4 \Delta P}{8 \eta L}$. Since $Q \propto r^4$, even a small increase in radius produces a very large increase in flow rate.
219. A glycerin column of height 0.5 m and density $1260 \, \text{kg/m}^3$ is filled in a capillary of radius $0.5 \, mm$. If viscosity is $1.5 \, Pa·s$, calculate the flow rate under gravity.
ⓐ. $2.5 \times 10^{-10} \, m^3/s$
ⓑ. $5.0 \times 10^{-10} \, m^3/s$
ⓒ. $7.5 \times 10^{-10} \, m^3/s$
ⓓ. $1.0 \times 10^{-9} \, m^3/s$
Correct Answer: $2.5 \times 10^{-10} \, m^3/s$
Explanation: Pressure difference = $\rho g h = 1260 \times 9.8 \times 0.5 \approx 6.2 \times 10^3 \, Pa$.
Using Poiseuille’s equation: $Q = \frac{\pi r^4 \Delta P}{8 \eta L}$. Substituting values gives about $2.5 \times 10^{-10} \, m^3/s$.
220. A fluid of viscosity $0.01 \, Pa·s$ flows through a capillary of length 0.1 m and radius 0.001 m under a pressure difference of 100 Pa. Find the volume flow rate.
ⓐ. $3.0 \times 10^{-9} \, m^3/s$
ⓑ. $3.9 \times 10^{-9} \, m^3/s$
ⓒ. $4.0 \times 10^{-9} \, m^3/s$
ⓓ. $4.5 \times 10^{-9} \, m^3/s$
Correct Answer: $3.9 \times 10^{-9} \, m^3/s$
Explanation: Using Poiseuille’s law: $Q = \frac{\pi r^4 \Delta P}{8 \eta L}$. Substituting values gives $3.9 \times 10^{-9} \, m^3/s$.
221. Stokes’ law is applicable to:
ⓐ. Motion of large bodies in fluids
ⓑ. Motion of very small spherical bodies at low velocities in viscous fluids
ⓒ. Motion of ideal fluids without viscosity
ⓓ. Fluids under high turbulence
Correct Answer: Motion of very small spherical bodies at low velocities in viscous fluids
Explanation: Stokes’ law is derived for laminar flow conditions around small spheres. It does not hold for turbulent flows or large objects.
222. According to Stokes’ law, the viscous drag force acting on a spherical body of radius $r$ moving with velocity $v$ in a fluid of viscosity $\eta$ is:
ⓐ. $F = \eta r v$
ⓑ. $F = 3 \pi \eta r v$
ⓒ. $F = 6 \pi \eta r v$
ⓓ. $F = 9 \pi \eta r^2 v$
Correct Answer: $F = 6 \pi \eta r v$
Explanation: Stokes derived that viscous drag is directly proportional to fluid viscosity, sphere radius, and velocity: $F = 6 \pi \eta r v$.
223. The dimensional formula of viscosity can be obtained from Stokes’ law as:
ⓐ. $[M^1 L^{-1} T^{-1}]$
ⓑ. $[M^0 L^0 T^0]$
ⓒ. $[M^1 L^0 T^{-2}]$
ⓓ. $[M L^2 T^{-2}]$
Correct Answer: $[M^1 L^{-1} T^{-1}]$
Explanation: From $F = 6 \pi \eta r v$, viscosity $\eta = \frac{F}{rv} $. Substituting dimensions: $[F] = MLT^{-2}, [r] = L, [v] = LT^{-1}$. Hence $[M L^{-1} T^{-1}]$.
224. Which of the following assumptions is made in deriving Stokes’ law?
ⓐ. The fluid flow is laminar
ⓑ. The fluid flow is turbulent
ⓒ. The sphere is moving with supersonic speed
ⓓ. The fluid has zero viscosity
Correct Answer: The fluid flow is laminar
Explanation: Stokes’ law assumes slow velocity and low Reynolds number ($Re < 1$), ensuring laminar flow around the sphere.
225. If a steel ball of radius 0.01 m falls through oil with viscosity 0.2 Pa·s at a velocity of 0.05 m/s, find the viscous drag force.
ⓐ. 0.018 N
ⓑ. 0.019 N
ⓒ. 0.020 N
ⓓ. 0.021 N
Correct Answer: 0.020 N
Explanation: $F = 6 \pi \eta r v = 6 \pi (0.2)(0.01)(0.05) \approx 0.020 \, N$.
226. Which of the following is NOT included in the derivation of Stokes’ law?
ⓐ. Viscous drag force
ⓑ. Buoyant force
ⓒ. Inertial drag force
ⓓ. Gravitational force on the sphere
Correct Answer: Inertial drag force
Explanation: Stokes’ law considers only viscous drag in laminar flow. Inertial drag dominates at higher Reynolds numbers, where Stokes’ law is invalid.
227. Stokes’ law is valid only when the Reynolds number is:
ⓐ. $Re < 1$
ⓑ. $Re < 100$
ⓒ. $Re > 1000$
ⓓ. $Re \to \infty$
Correct Answer: $Re < 1$
Explanation: For $Re < 1$, inertial forces are negligible compared to viscous forces, ensuring laminar flow around the sphere.
228. Derivation of Stokes’ law involves solving which type of fluid equations?
ⓐ. Continuity equation only
ⓑ. Euler’s equation
ⓒ. Navier–Stokes equations for viscous flow
ⓓ. Bernoulli’s equation
Correct Answer: Navier–Stokes equations for viscous flow
Explanation: Stokes’ law is derived from simplified Navier–Stokes equations for low Reynolds number laminar flows around a sphere.
229. Which term in Stokes’ law indicates dependence on velocity?
ⓐ. $6 \pi$
ⓑ. $\eta$
ⓒ. $r$
ⓓ. $v$
Correct Answer: $v$
Explanation: Stokes’ drag force $F = 6 \pi \eta r v$ is directly proportional to velocity. Faster motion increases drag linearly.
230. A ball of radius 1 mm is moving in a viscous liquid with velocity 0.01 m/s. If the viscous drag experienced is $1.88 \times 10^{-4} \, N$, calculate the viscosity of the liquid.
ⓐ. 0.5 Pa·s
ⓑ. 0.7 Pa·s
ⓒ. 1.0 Pa·s
ⓓ. 1.2 Pa·s
Correct Answer: 0.5 Pa·s
Explanation: Using $F = 6 \pi \eta r v$,
$\eta = \frac{F}{6 \pi r v} = \frac{1.88 \times 10^{-4}}{6 \pi (0.001)(0.01)} \approx 0.5 \, Pa·s$.
231. Stokes’ law is applied in sedimentation to:
ⓐ. Calculate pressure in deep fluids
ⓑ. Determine the velocity of a falling particle in a viscous medium
ⓒ. Measure surface tension
ⓓ. Measure density of gases
Correct Answer: Determine the velocity of a falling particle in a viscous medium
Explanation: Stokes’ law gives the drag force on a particle falling in a viscous fluid, which helps calculate its terminal velocity during sedimentation.
232. The terminal velocity of a spherical particle in a viscous medium is given by:
ⓐ. $v = \frac{2 r^2 (\rho_s – \rho_f) g}{9 \eta}$
ⓑ. $v = \frac{6 \pi \eta r}{\rho g}$
ⓒ. $v = \frac{9 \eta}{2 r^2 (\rho_s – \rho_f)}$
ⓓ. $v = \frac{\eta}{\rho g r}$
Correct Answer: $v = \frac{2 r^2 (\rho_s – \rho_f) g}{9 \eta}$
Explanation: By balancing viscous drag, buoyant force, and gravitational force, the expression for terminal velocity of a sphere is derived as above.
233. A steel ball of radius $0.1 \, \text{mm}$ falls in water. If densities of steel and water are $7800 \, \text{kg/m}^3$ and $1000 \, \text{kg/m}^3$, viscosity of water is $1.0 \times 10^{-3} \, \text{Pa·s}$, calculate terminal velocity. ($g = 9.8 \, \text{m/s}^2$)
ⓐ. 0.15 m/s
ⓑ. 0.25 m/s
ⓒ. 0.35 m/s
ⓓ. 0.45 m/s
Correct Answer: 0.25 m/s
Explanation: $v = \frac{2 r^2 g (\rho_s – \rho_f)}{9 \eta}$. Substituting: $v = \frac{2 (10^{-4})^2 (9.8)(6800)}{9 (10^{-3})} \approx 0.25 \, \text{m/s}$.
234. Sedimentation techniques using Stokes’ law are useful in:
ⓐ. Measuring viscosity of gases
ⓑ. Determining particle size in colloidal solutions
ⓒ. Measuring atmospheric pressure
ⓓ. Measuring buoyant force
Correct Answer: Determining particle size in colloidal solutions
Explanation: By observing sedimentation velocity of particles, Stokes’ law is used to calculate particle radius in colloids and suspensions.
235. In particle size analysis, smaller particles settle more slowly because:
ⓐ. They experience higher buoyant force
ⓑ. They experience smaller gravitational force and higher drag
ⓒ. Their density is lower
ⓓ. They have higher velocity
Correct Answer: They experience smaller gravitational force and higher drag
Explanation: Gravitational force is proportional to $r^3$ but viscous drag is proportional to $r$. Hence, small particles settle very slowly.
236. If two particles of radii $r_1$ and $r_2$ are suspended in a fluid, the ratio of their terminal velocities is:
ⓐ. $\frac{r_1}{r_2}$
ⓑ. $\frac{r_1^2}{r_2^2}$
ⓒ. $\frac{r_1^3}{r_2^3}$
ⓓ. $\frac{r_1^4}{r_2^4}$
Correct Answer: $\frac{r_1^2}{r_2^2}$
Explanation: From $v \propto r^2$, terminal velocity ratio depends on the square of particle radii.
237. A spherical pollen grain of radius $10^{-6} \, m$ settles in air with viscosity $1.8 \times 10^{-5} \, \text{Pa·s}$. If density difference is $500 \, \text{kg/m}^3$, calculate the terminal velocity. ($g = 9.8 \, \text{m/s}^2$)
ⓐ. $5.4 \times 10^{-8} \, m/s$
ⓑ. $1.1 \times 10^{-7} \, m/s$
ⓒ. $2.2 \times 10^{-7} \, m/s$
ⓓ. $3.5 \times 10^{-7} \, m/s$
Correct Answer: $2.2 \times 10^{-7} \, m/s$
Explanation: $v = \frac{2 r^2 g (\Delta \rho)}{9 \eta} = \frac{2 (10^{-6})^2 (9.8)(500)}{9 (1.8 \times 10^{-5})} \approx 2.2 \times 10^{-7} \, m/s$.
238. In sedimentation analysis, the viscosity of the medium must be known because:
ⓐ. Viscosity does not affect sedimentation
ⓑ. Sedimentation velocity depends on viscosity
ⓒ. Viscosity determines density difference
ⓓ. Viscosity affects only large particles
Correct Answer: Sedimentation velocity depends on viscosity
Explanation: Stokes’ law shows $v \propto 1/\eta$. Higher viscosity slows sedimentation, so accurate particle size determination requires knowing fluid viscosity.
239. A ball of radius $0.5 \, mm$ falls in oil of viscosity $0.1 \, Pa·s$ with a velocity $0.02 \, m/s$. If density difference between ball and oil is $1000 \, kg/m^3$, check if this velocity is terminal velocity using Stokes’ law. ($g = 9.8 \, m/s^2$)
ⓐ. Yes, matches terminal velocity
ⓑ. No, actual terminal velocity is higher
ⓒ. No, actual terminal velocity is lower
ⓓ. Cannot be determined
Correct Answer: Yes, matches terminal velocity
Explanation: $v = \frac{2 r^2 g \Delta \rho}{9 \eta} = \frac{2 (5 \times 10^{-4})^2 (9.8)(1000)}{9 (0.1)} \approx 0.02 \, m/s$. Matches given velocity, so it is terminal velocity.
240. Stokes’ law in sedimentation is particularly useful in industries for:
ⓐ. Designing turbines
ⓑ. Measuring particle sizes in paints, cement, and food colloids
ⓒ. Determining viscosity of oils
ⓓ. Measuring atmospheric pressure
Correct Answer: Measuring particle sizes in paints, cement, and food colloids
Explanation: Sedimentation analysis based on Stokes’ law helps control particle size in industrial processes like paints, cement, dairy, and pharmaceuticals.
241. Stokes’ law is valid only when the motion of the sphere takes place under:
ⓐ. High Reynolds number conditions
ⓑ. Turbulent flow conditions
ⓒ. Low Reynolds number conditions ($Re < 1$)
ⓓ. Vacuum conditions
Correct Answer: Low Reynolds number conditions ($Re < 1$)
Explanation: Stokes’ law assumes laminar flow around the sphere, which is valid only at very small Reynolds numbers where viscous forces dominate over inertial forces.
242. Which of the following is NOT an assumption in the derivation of Stokes’ law?
ⓐ. The sphere is rigid and smooth
ⓑ. Fluid flow is laminar
ⓒ. Fluid is incompressible
ⓓ. Fluid is turbulent
Correct Answer: Fluid is turbulent
Explanation: Stokes’ law assumes the fluid is incompressible, non-turbulent, and flow is laminar around the sphere. Turbulent flow violates these conditions.
243. Stokes’ law is not applicable when:
ⓐ. The sphere is very small and velocity is low
ⓑ. The fluid is viscous and flow is laminar
ⓒ. Reynolds number exceeds 1 and flow becomes inertial
ⓓ. Fluid is incompressible
Correct Answer: Reynolds number exceeds 1 and flow becomes inertial
Explanation: At higher Reynolds numbers, inertial forces dominate over viscous forces, causing deviations from Stokes’ law.
244. Which limitation is true about Stokes’ law in sedimentation experiments?
ⓐ. It neglects buoyant force
ⓑ. It assumes perfectly spherical particles
ⓒ. It applies to gases only
ⓓ. It works for all Reynolds numbers
Correct Answer: It assumes perfectly spherical particles
Explanation: Stokes’ law derivation assumes particles are spherical. In reality, many sedimenting particles are irregular in shape, making the law less accurate.
245. Why does Stokes’ law fail at high velocities?
ⓐ. Viscous forces dominate completely
ⓑ. Flow becomes turbulent and inertial drag dominates
ⓒ. Buoyancy vanishes at high speed
ⓓ. Density difference becomes negligible
Correct Answer: Flow becomes turbulent and inertial drag dominates
Explanation: At high velocities, turbulence appears, inertial forces exceed viscous drag, and Stokes’ drag formula no longer applies.
246. Stokes’ law assumes the liquid is:
ⓐ. Compressible and inviscid
ⓑ. Incompressible and viscous
ⓒ. Compressible and turbulent
ⓓ. Ideal without viscosity
Correct Answer: Incompressible and viscous
Explanation: The derivation of Stokes’ law requires the fluid to be incompressible and viscous with steady, laminar flow.
247. In applying Stokes’ law to very small particles like colloids, which factor limits its accuracy?
ⓐ. Brownian motion of particles
ⓑ. Excessive viscosity
ⓒ. Large size of particles
ⓓ. Surface tension effects
Correct Answer: Brownian motion of particles
Explanation: For microscopic colloidal particles, random Brownian motion disturbs sedimentation, making Stokes’ law inaccurate for predicting velocity.
248. Which of the following conditions is required for Stokes’ law to hold?
ⓐ. No slip condition at fluid–sphere boundary
ⓑ. Fluid compressibility must be large
ⓒ. Flow must be turbulent
ⓓ. Sphere must move at supersonic speed
Correct Answer: No slip condition at fluid–sphere boundary
Explanation: Stokes’ law derivation assumes the no-slip boundary condition: fluid molecules in contact with the sphere’s surface are stationary relative to it.
249. If particle shape deviates significantly from a sphere, Stokes’ law:
ⓐ. Still gives exact results
ⓑ. Overestimates or underestimates drag force
ⓒ. Becomes independent of viscosity
ⓓ. Is not affected
Correct Answer: Overestimates or underestimates drag force
Explanation: Stokes’ law assumes spherical particles. Irregular shapes alter drag coefficient, leading to inaccurate predictions.
250. Which of the following is NOT a limitation of Stokes’ law?
ⓐ. Assumes laminar flow around the sphere
ⓑ. Assumes Reynolds number less than 1
ⓒ. Assumes fluid is incompressible
ⓓ. Assumes fluid is superfluid with zero viscosity
Correct Answer: Assumes fluid is superfluid with zero viscosity
Explanation: Stokes’ law explicitly involves viscosity and is valid only for viscous fluids. It does not assume zero viscosity or superfluidity.
251. A small steel ball of radius $1 \, \text{mm}$ falls through glycerin with viscosity $1.5 \, \text{Pa·s}$. If the density of steel is $7800 \, \text{kg/m}^3$ and glycerin density is $1260 \, \text{kg/m}^3$, calculate the terminal velocity. ($g = 9.8 \, \text{m/s}^2$)
ⓐ. $1.2 \times 10^{-2} \, \text{m/s}$
ⓑ. $2.4 \times 10^{-2} \, \text{m/s}$
ⓒ. $3.6 \times 10^{-2} \, \text{m/s}$
ⓓ. $4.8 \times 10^{-2} \, \text{m/s}$
Correct Answer: $2.4 \times 10^{-2} \, \text{m/s}$
Explanation: Using $v = \frac{2 r^2 g (\rho_s – \rho_f)}{9 \eta}$. Substituting values:
$v = \frac{2 (0.001)^2 (9.8)(6540)}{9 (1.5)} \approx 0.024 \, \text{m/s}$.
252. A sphere of radius $0.5 \, \text{mm}$ is falling in a fluid of viscosity $0.25 \, \text{Pa·s}$. If its terminal velocity is $0.01 \, \text{m/s}$ and density difference between sphere and fluid is $2000 \, \text{kg/m}^3$, verify Stokes’ law. ($g = 9.8 \, \text{m/s}^2$)
ⓐ. Valid, calculated velocity matches given
ⓑ. Not valid, calculated velocity is higher
ⓒ. Not valid, calculated velocity is lower
ⓓ. Cannot be checked
Correct Answer: Valid, calculated velocity matches given
Explanation: $v = \frac{2 r^2 g \Delta \rho}{9 \eta} = \frac{2 (0.0005)^2 (9.8)(2000)}{9 (0.25)} \approx 0.01 \, \text{m/s}$. Matches given value.
253. A lead shot of radius $2 \, \text{mm}$ falls through oil of viscosity $0.8 \, \text{Pa·s}$. If density of lead = $11300 \, \text{kg/m}^3$ and oil = $900 \, \text{kg/m}^3$, calculate terminal velocity. ($g = 9.8 \, \text{m/s}^2$)
ⓐ. 0.20 m/s
ⓑ. 0.25 m/s
ⓒ. 0.30 m/s
ⓓ. 0.35 m/s
Correct Answer: 0.30 m/s
Explanation: $v = \frac{2 r^2 g (\rho_s – \rho_f)}{9 \eta} = \frac{2 (0.002)^2 (9.8)(10400)}{9 (0.8)} \approx 0.30 \, m/s$.
254. A spherical pollen grain of radius $5 \, \mu m$ settles in air ($ \eta = 1.8 \times 10^{-5} \, Pa·s$, $\rho_{air} = 1.2 \, kg/m^3$). If the pollen density is $1000 \, kg/m^3$, calculate terminal velocity.
ⓐ. $2.5 \times 10^{-5} \, m/s$
ⓑ. $4.5 \times 10^{-5} \, m/s$
ⓒ. $6.5 \times 10^{-5} \, m/s$
ⓓ. $8.5 \times 10^{-5} \, m/s$
Correct Answer: $4.5 \times 10^{-5} \, m/s$
Explanation: $v = \frac{2 (5 \times 10^{-6})^2 (9.8)(998.8)}{9 (1.8 \times 10^{-5})} \approx 4.5 \times 10^{-5} \, m/s$.
255. A ball of radius $0.2 \, cm$ is dropped in a fluid and experiences a viscous drag of $7.5 \times 10^{-4} \, N$. If its velocity is $0.02 \, m/s$, find the viscosity of the fluid.
ⓐ. 0.1 Pa·s
ⓑ. 0.2 Pa·s
ⓒ. 0.3 Pa·s
ⓓ. 0.4 Pa·s
Correct Answer: 0.2 Pa·s
Explanation: From Stokes’ law: $\eta = \frac{F}{6 \pi r v} = \frac{7.5 \times 10^{-4}}{6 \pi (0.002)(0.02)} \approx 0.2 \, Pa·s$.
256. A steel ball of radius $1.5 \, mm$ moves through glycerin at velocity $0.03 \, m/s$. If glycerin viscosity is $1.2 \, Pa·s$, calculate the viscous drag force.
ⓐ. $ 8.5 \times 10^{-4} \, N$
ⓑ. $ 1.0 \times 10^{-3} \, N$
ⓒ. $ 1.2 \times 10^{-3} \, N$
ⓓ. $ 1.5 \times 10^{-3} \, N$
Correct Answer: $ 1.2 \times 10^{-3} \, N$
Explanation: $F = 6 \pi \eta r v = 6 \pi (1.2)(0.0015)(0.03) \approx 1.2 \times 10^{-3} \, N$.
257. A ball of radius $0.01 \, m$ falls through a fluid and attains terminal velocity of $0.02 \, m/s$. If the viscosity of the fluid is $0.5 \, Pa·s$ and density of fluid = $1000 \, kg/m^3$, density of ball = ?
ⓐ. 1020 $kg/m^3$
ⓑ. 1100 $kg/m^3$
ⓒ. 1200 $kg/m^3$
ⓓ. 1300 $kg/m^3$
Correct Answer: 1200 $kg/m^3$
Explanation: From $v = \frac{2 r^2 g (\rho_s – \rho_f)}{9 \eta}$.
Rearrange: $\rho_s = \rho_f + \frac{9 \eta v}{2 r^2 g}$.
Substituting: $\rho_s = 1000 + \frac{9 (0.5)(0.02)}{2 (0.01)^2 (9.8)} \approx 1200 \, kg/m^3$.
258. A small sphere falls in oil and reaches a terminal velocity of $0.01 \, m/s$. If the radius of the sphere is doubled, keeping all other factors same, the new terminal velocity is:
ⓐ. Unchanged
ⓑ. Doubled
ⓒ. Four times
ⓓ. Eight times
Correct Answer: Four times
Explanation: From Stokes’ law: $v \propto r^2$. If radius is doubled, terminal velocity increases by factor of 4.
259. A particle of radius $0.002 \, m$ falls in a liquid of viscosity $0.3 \, Pa·s$. If density difference = $2000 \, kg/m^3$, calculate its terminal velocity. ($g = 9.8 \, m/s^2$)
ⓐ. 0.0087 m/s
ⓑ. 0.0097 m/s
ⓒ. 0.0107 m/s
ⓓ. 0.0117 m/s
Correct Answer: 0.0087 m/s
Explanation: $v = \frac{2 r^2 g \Delta \rho}{9 \eta} = \frac{2 (0.002)^2 (9.8)(2000)}{9 (0.3)} \approx 0.0087 \, m/s$.
260. A ball of radius $0.5 \, mm$ is falling in a viscous fluid of viscosity $0.15 \, Pa·s$. If density of sphere = $4000 \, kg/m^3$, density of fluid = $1000 \, kg/m^3$, calculate terminal velocity. ($g = 10 \, m/s^2$)
ⓐ. 0.005 m/s
ⓑ. 0.010 m/s
ⓒ. 0.015 m/s
ⓓ. 0.020 m/s
Correct Answer: 0.020 m/s
Explanation: $v = \frac{2 r^2 g (\rho_s – \rho_f)}{9 \eta} = \frac{2 (5 \times 10^{-4})^2 (10)(3000)}{9 (0.15)} \approx 0.02 \, m/s$.
261. Terminal velocity is defined as:
ⓐ. The maximum velocity a body can achieve in vacuum
ⓑ. The constant velocity attained by a body when viscous drag and buoyant force balance its weight in a fluid
ⓒ. The velocity at which gravitational force becomes zero
ⓓ. The initial velocity of a falling body
Correct Answer: The constant velocity attained by a body when viscous drag and buoyant force balance its weight in a fluid
Explanation: At terminal velocity, net force on the body is zero since downward weight = buoyant force + viscous drag. The body continues falling at a constant velocity.
262. When a small sphere falls in a viscous fluid, it eventually reaches terminal velocity because:
ⓐ. Weight increases with velocity
ⓑ. Buoyant force decreases
ⓒ. Viscous drag increases with velocity until equilibrium is reached
ⓓ. Viscosity of the fluid decreases
Correct Answer: Viscous drag increases with velocity until equilibrium is reached
Explanation: Drag force ($F = 6 \pi \eta r v$) increases with velocity and balances the net force, causing constant velocity (terminal velocity).
263. The expression for terminal velocity of a spherical particle in a viscous medium is:
ⓐ. $v_t = \frac{6 \pi \eta r v}{mg}$
ⓑ. $v_t = \frac{2 r^2 (\rho_s – \rho_f) g}{9 \eta}$
ⓒ. $v_t = \frac{\rho g h}{\eta}$
ⓓ. $v_t = \frac{9 \eta}{2 r^2 (\rho_s – \rho_f)}$
Correct Answer: $v_t = \frac{2 r^2 (\rho_s – \rho_f) g}{9 \eta}$
Explanation: Terminal velocity is derived by equating weight minus buoyant force with viscous drag.
264. At terminal velocity, the acceleration of the falling body is:
ⓐ. Zero
ⓑ. Maximum
ⓒ. Equal to $g$
ⓓ. Negative
Correct Answer: Zero
Explanation: Once terminal velocity is reached, net force is zero, hence acceleration becomes zero, and the body continues with constant velocity.
265. Which of the following does NOT affect terminal velocity?
ⓐ. Radius of the particle
ⓑ. Density difference between particle and fluid
ⓒ. Viscosity of the fluid
ⓓ. Gravitational acceleration being constant
Correct Answer: Gravitational acceleration being constant
Explanation: Terminal velocity depends on $r^2$, density difference, viscosity, and $g$. Since $g$ is constant near Earth’s surface, it does not change terminal velocity conditions.
266. If radius of a spherical particle is doubled, terminal velocity:
ⓐ. Remains unchanged
ⓑ. Increases by 2 times
ⓒ. Increases by 4 times
ⓓ. Increases by 8 times
Correct Answer: Increases by 4 times
Explanation: From $v_t \propto r^2$, doubling radius increases terminal velocity by factor of 4.
267. If viscosity of fluid increases, terminal velocity:
ⓐ. Increases
ⓑ. Decreases
ⓒ. Remains constant
ⓓ. First increases, then decreases
Correct Answer: Decreases
Explanation: Terminal velocity $v_t \propto \frac{1}{\eta}$. Increasing viscosity increases resistance, lowering terminal velocity.
268. Which of the following conditions must be satisfied for terminal velocity to be achieved?
ⓐ. Net force = 0
ⓑ. Weight = Buoyant force
ⓒ. Viscous force = 0
ⓓ. Density of sphere = density of fluid
Correct Answer: Net force = 0
Explanation: Terminal velocity is achieved when downward weight is exactly balanced by upward buoyant force and viscous drag, giving net force zero.
269. A ball falling in a viscous fluid attains terminal velocity faster if:
ⓐ. Fluid viscosity is low
ⓑ. Fluid viscosity is high
ⓒ. Ball density is low
ⓓ. Gravitational acceleration is smaller
Correct Answer: Fluid viscosity is low
Explanation: In a low viscosity fluid, resistance is smaller, and the ball accelerates quickly to reach its constant terminal velocity.
270. Which of the following best describes terminal velocity in everyday life?
ⓐ. A parachute slowing down and then moving at constant speed
ⓑ. A rocket leaving the Earth
ⓒ. A satellite orbiting the Earth
ⓓ. A pendulum oscillating
Correct Answer: A parachute slowing down and then moving at constant speed
Explanation: A parachute quickly reaches a constant fall speed when air resistance balances the downward weight, an everyday demonstration of terminal velocity.
271. A falling sphere in a viscous fluid reaches terminal velocity when:
ⓐ. Gravitational force becomes maximum
ⓑ. Buoyant force becomes zero
ⓒ. Viscous drag + buoyant force = gravitational force
ⓓ. Gravitational force = viscous drag only
Correct Answer: Viscous drag + buoyant force = gravitational force
Explanation: At terminal velocity, net force on the sphere is zero. The downward gravitational force is exactly balanced by the sum of upward viscous drag and buoyant force.
272. The condition for reaching terminal velocity depends strongly on:
ⓐ. Shape of the container
ⓑ. Balance of forces acting on the particle
ⓒ. Surface tension of the liquid
ⓓ. Capillary action
Correct Answer: Balance of forces acting on the particle
Explanation: Terminal velocity is achieved when all forces balance. The net acceleration becomes zero, and the particle moves with constant speed.
273. The equation of motion for a small sphere falling in a viscous fluid before reaching terminal velocity is:
ⓐ. $m \frac{dv}{dt} = mg$
ⓑ. $m \frac{dv}{dt} = mg – (F_b + F_v)$
ⓒ. $m \frac{dv}{dt} = F_b – mg$
ⓓ. $m \frac{dv}{dt} = F_v – mg$
Correct Answer: $m \frac{dv}{dt} = mg – (F_b + F_v)$
Explanation: The net downward force is weight minus buoyant force and viscous drag. When this net force = 0, terminal velocity is reached.
274. The time taken by a particle to reach near its terminal velocity is determined by:
ⓐ. Viscosity of fluid and particle size
ⓑ. Density of particle and density of fluid
ⓒ. Both A and B
ⓓ. None of these
Correct Answer: Both A and B
Explanation: The settling time depends on viscosity, particle radius, and density difference. These factors decide how quickly equilibrium is reached.
275. Which of the following indicates that terminal velocity has been reached?
ⓐ. Acceleration of particle becomes zero
ⓑ. Net downward force increases
ⓒ. Viscous force decreases
ⓓ. Buoyant force becomes larger than weight
Correct Answer: Acceleration of particle becomes zero
Explanation: At terminal velocity, net force = 0, hence acceleration = 0. The particle continues at constant velocity.
276. A raindrop attains terminal velocity while falling through air because:
ⓐ. Its mass decreases with time
ⓑ. Air resistance (drag) increases with velocity until balanced by weight
ⓒ. Gravity decreases with altitude
ⓓ. Air viscosity increases with velocity
Correct Answer: Air resistance (drag) increases with velocity until balanced by weight
Explanation: Drag force increases with velocity. Eventually, it balances the weight of the raindrop, resulting in constant terminal velocity.
277. The condition for terminal velocity in terms of forces is:
ⓐ. $F_g = F_b$
ⓑ. $F_g = F_v$
ⓒ. $F_g = F_b + F_v$
ⓓ. $F_v = F_b$
Correct Answer: $F_g = F_b + F_v$
Explanation: Gravity (downward) is exactly balanced by the sum of buoyant force and viscous drag (upward) at terminal velocity.
278. For a particle to reach terminal velocity, the Reynolds number should be:
ⓐ. Very high ($Re \gg 1000$)
ⓑ. Moderate ($Re \approx 100$)
ⓒ. Very low ($Re < 1$)
ⓓ. Independent of Re
Correct Answer: Very low ($Re < 1$)
Explanation: Stokes’ law and terminal velocity concepts apply under laminar flow (low Reynolds number) conditions.
279. When does a falling particle fail to reach terminal velocity in a fluid?
ⓐ. When fluid is incompressible
ⓑ. When viscosity is negligible
ⓒ. When density difference is high
ⓓ. When buoyant force is small
Correct Answer: When viscosity is negligible
Explanation: In absence of viscous drag, there is nothing to balance the weight, so the particle keeps accelerating indefinitely without terminal velocity.
280. The reason why larger raindrops fall faster than smaller ones is:
ⓐ. Their weight is larger and terminal velocity is proportional to radius squared
ⓑ. They experience less buoyant force
ⓒ. They have lower density
ⓓ. They disobey viscosity
Correct Answer: Their weight is larger and terminal velocity is proportional to radius squared
Explanation: From $v_t = \frac{2 r^2 (\rho_s – \rho_f) g}{9 \eta}$, terminal velocity increases with square of radius. Larger drops thus reach higher constant velocities before equilibrium.
281. A steel ball of radius $1 \, \text{mm}$ is falling through glycerin of viscosity $1.2 \, \text{Pa·s}$. The densities of steel and glycerin are $7800 \, \text{kg/m}^3$ and $1260 \, \text{kg/m}^3$. Calculate its terminal velocity. ($ g = 9.8 \, \text{m/s}^2$)
ⓐ. $1.5 \times 10^{-2} \, m/s$
ⓑ. $2.0 \times 10^{-2} \, m/s$
ⓒ. $2.5 \times 10^{-2} \, m/s$
ⓓ. $3.0 \times 10^{-2} \, m/s$
Correct Answer: $2.0 \times 10^{-2} \, m/s$
Explanation: Using formula $v_t = \frac{2 r^2 g (\rho_s – \rho_f)}{9 \eta}$.
Substitute $r = 0.001, \Delta \rho = 6540, g = 9.8, \eta = 1.2$.
$v_t = \frac{2 (10^{-6})(9.8)(6540)}{9 (1.2)} \approx 0.020 \, m/s$.
282. A lead shot of radius $2 \, \text{mm}$ falls through oil of viscosity $0.8 \, \text{Pa·s}$. Densities of lead and oil are $11300 \, \text{kg/m}^3$ and $900 \, \text{kg/m}^3$. Find its terminal velocity.
ⓐ. 0.15 m/s
ⓑ. 0.25 m/s
ⓒ. 0.35 m/s
ⓓ. 0.45 m/s
Correct Answer: 0.35 m/s
Explanation: $v_t = \frac{2 r^2 g (\rho_s – \rho_f)}{9 \eta}$.
Here $r = 0.002, \Delta \rho = 10400, g = 9.8, \eta = 0.8$.
So $v_t = \frac{2 (4 \times 10^{-6})(9.8)(10400)}{9 (0.8)} \approx 0.35 \, m/s$.
283. A sphere of radius $0.5 \, \text{mm}$ falls in water ($ \eta = 0.001 \, \text{Pa·s}, \rho_f = 1000 \, \text{kg/m}^3$). If its density is $8000 \, \text{kg/m}^3$, calculate terminal velocity.
ⓐ. 0.25 m/s
ⓑ. 0.50 m/s
ⓒ. 0.75 m/s
ⓓ. 1.0 m/s
Correct Answer: 1.0 m/s
Explanation: $v_t = \frac{2 (0.0005)^2 (9.8)(7000)}{9 (0.001)} \approx 1.0 \, m/s$.
284. A small glass bead of radius $10^{-4} \, m$ is falling in air ($ \eta = 1.8 \times 10^{-5} \, Pa·s$, $\rho_f = 1.2 \, kg/m^3$). If the bead density is $2500 \, kg/m^3$, calculate terminal velocity.
ⓐ. $1.2 \times 10^{-2} \, m/s$
ⓑ. $2.5 \times 10^{-2} \, m/s$
ⓒ. $3.5 \times 10^{-2} \, m/s$
ⓓ. $4.5 \times 10^{-2} \, m/s$
Correct Answer: $2.5 \times 10^{-2} \, m/s$
Explanation: $v_t = \frac{2 (10^{-8})(9.8)(2498.8)}{9 (1.8 \times 10^{-5})} \approx 0.025 \, m/s$.
285. A pollen grain of radius $5 \, \mu m$ settles in air. If $\rho_s = 1000 \, kg/m^3, \rho_f = 1.2 \, kg/m^3, \eta = 1.8 \times 10^{-5} \, Pa·s$, calculate terminal velocity.
ⓐ. $2.0 \times 10^{-5} \, m/s$
ⓑ. $4.0 \times 10^{-5} \, m/s$
ⓒ. $6.0 \times 10^{-5} \, m/s$
ⓓ. $8.0 \times 10^{-5} \, m/s$
Correct Answer: $4.0 \times 10^{-5} \, m/s$
Explanation: $v_t = \frac{2 (25 \times 10^{-12})(9.8)(998.8)}{9 (1.8 \times 10^{-5})} \approx 4.0 \times 10^{-5} \, m/s$.
286. A steel ball of radius $0.002 \, m$ falls in oil of viscosity $0.5 \, Pa·s$. If density of steel = $7800 \, kg/m^3$ and density of oil = $900 \, kg/m^3$, calculate terminal velocity.
ⓐ. 0.08 m/s
ⓑ. 0.10 m/s
ⓒ. 0.12 m/s
ⓓ. 0.15 m/s
Correct Answer: 0.12 m/s
Explanation: Using terminal velocity formula,
$v_t = \frac{2 (0.002)^2 (9.8)(6900)}{9 (0.5)} \approx 0.12 \, m/s$.
287. If the terminal velocity of a sphere of radius $r$ is $v$, then what will be the terminal velocity if radius is doubled, keeping all other conditions the same?
ⓐ. $2v$
ⓑ. $3v$
ⓒ. $4v$
ⓓ. $8v$
Correct Answer: $4v$
Explanation: From Stokes’ law, $v_t \propto r^2$. If $r$ doubles, velocity increases four times.
288. Two spheres of radii $r_1 = 1 \, mm$ and $r_2 = 2 \, mm$ are dropped in the same viscous liquid. What is the ratio of their terminal velocities?
ⓐ. 1 : 2
ⓑ. 1 : 3
ⓒ. 1 : 4
ⓓ. 1 : 8
Correct Answer: 1 : 4
Explanation: $v_t \propto r^2$. Thus, $v_1/v_2 = (r_1/r_2)^2 = (1/2)^2 = 1/4$.
289. A ball of radius $0.003 \, m$ falls through glycerin with viscosity $1.0 \, Pa·s$. If density difference between ball and glycerin is $5000 \, kg/m^3$, calculate terminal velocity.
ⓐ. 0.01 m/s
ⓑ. 0.05 m/s
ⓒ. 0.10 m/s
ⓓ. 0.15 m/s
Correct Answer: 0.05 m/s
Explanation: $v_t = \frac{2 r^2 g \Delta \rho}{9 \eta} = \frac{2 (9 \times 10^{-6})(9.8)(5000)}{9 (1.0)} \approx 0.05 \, m/s$.
290. A raindrop of radius $0.001 \, m$ falls through air ($ \eta = 1.8 \times 10^{-5} \, Pa·s$, $\rho_{air} = 1.2 \, kg/m^3$). If density of water = $1000 \, kg/m^3$, find its terminal velocity.
ⓐ. 3.0 m/s
ⓑ. 4.0 m/s
ⓒ. 5.0 m/s
ⓓ. 6.0 m/s
Correct Answer: 5.0 m/s
Explanation: $v_t = \frac{2 (10^{-6})(9.8)(998.8)}{9 (1.8 \times 10^{-5})} \approx 5.0 \, m/s$.
291. A raindrop falling through air does not keep accelerating but soon moves with a constant velocity because:
ⓐ. Air pressure becomes equal to water pressure
ⓑ. Its weight becomes zero after some time
ⓒ. Air drag force balances the gravitational force
ⓓ. Buoyant force increases with time
Correct Answer: Air drag force balances the gravitational force
Explanation: At terminal velocity, the downward weight of the raindrop is exactly balanced by the upward viscous drag and buoyant force of air, resulting in constant velocity.
292. The terminal velocity of fine dust particles in air is usually:
ⓐ. A few cm/s
ⓑ. A few m/s
ⓒ. Greater than 50 m/s
ⓓ. Equal to speed of sound
Correct Answer: A few cm/s
Explanation: Dust particles are extremely small, so their weight is very low, but viscous drag is significant. This results in very small terminal velocities, often only a few cm/s.
293. Why do parachutes slow down a falling person?
ⓐ. They reduce gravitational acceleration
ⓑ. They increase viscous drag by enlarging the surface area
ⓒ. They decrease the person’s density
ⓓ. They make the person weightless
Correct Answer: They increase viscous drag by enlarging the surface area
Explanation: A parachute increases air resistance (drag), lowering terminal velocity so the person descends slowly and safely.
294. In fluid mechanics, the constant velocity reached by an air bubble rising in water is also called:
ⓐ. Critical velocity
ⓑ. Terminal velocity
ⓒ. Escape velocity
ⓓ. Threshold velocity
Correct Answer: Terminal velocity
Explanation: Just like falling bodies, rising bubbles in a viscous medium reach a constant speed when buoyant force is balanced by viscous drag.
295. A steel ball of radius $1 \, \text{mm}$ is released in oil and attains a terminal velocity of $0.02 \, m/s$. If the radius is doubled, its new terminal velocity will be approximately:
ⓐ. 0.02 m/s
ⓑ. 0.04 m/s
ⓒ. 0.08 m/s
ⓓ. 0.16 m/s
Correct Answer: 0.08 m/s
Explanation: Terminal velocity $v_t \propto r^2$. Doubling radius increases $v_t$ four times: $0.02 \times 4 = 0.08 \, m/s$.
296. Which of the following is an example of terminal velocity in daily life?
ⓐ. A car moving at constant speed
ⓑ. A coin sinking in water at constant speed
ⓒ. A stone thrown upward
ⓓ. A pendulum at rest
Correct Answer: A coin sinking in water at constant speed
Explanation: The coin sinks until viscous drag balances its weight, after which it moves at constant terminal velocity.
297. For very small pollen grains in air, terminal velocity is extremely small because:
ⓐ. Their density is very high
ⓑ. Their radius is very small
ⓒ. Gravitational force is larger than drag
ⓓ. Air viscosity is zero
Correct Answer: Their radius is very small
Explanation: $v_t \propto r^2$. Tiny radius means extremely small terminal velocity, hence pollen grains float in air for long times.
298. Why do larger raindrops fall faster than smaller raindrops?
ⓐ. Larger drops have less density
ⓑ. Larger drops have higher buoyancy
ⓒ. Terminal velocity increases with square of radius
ⓓ. Viscous force decreases for large drops
Correct Answer: Terminal velocity increases with square of radius
Explanation: From Stokes’ law, $v_t \propto r^2$. Bigger drops fall faster and strike the ground with higher speed.
299. In blood flow, suspended cells settle at a constant speed in diagnostic centrifugation due to:
ⓐ. Terminal velocity in fluid medium
ⓑ. Archimedes’ principle
ⓒ. Pascal’s law
ⓓ. Bernoulli’s theorem
Correct Answer: Terminal velocity in fluid medium
Explanation: In centrifugation, red blood cells reach a terminal velocity when viscous drag balances the centrifugal force, aiding separation.
300. Which of the following demonstrates terminal velocity in gases?
ⓐ. A helium balloon rising in air at constant speed
ⓑ. An aircraft at cruising altitude
ⓒ. Sound propagation in air
ⓓ. Diffusion of oxygen in air
Correct Answer: A helium balloon rising in air at constant speed
Explanation: The balloon rises until buoyant force equals drag, and then moves upward at a constant terminal velocity, similar to falling bodies in air.
In Class 11 Physics, the chapter Mechanical Properties of Fluids is an important part of the NCERT/CBSE syllabus and provides the foundation for understanding fluid statics and dynamics. It carries significant weightage in board exams as well as competitive exams like JEE, NEET, and state-level tests. This collection offers a total of 700 MCQs with clear explanations, divided into 7 systematic parts. In this Part 3, you will attempt the next 100 multiple-choice questions with detailed answers, focusing on fluid pressure, manometry, and practical problem-solving.
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