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201. Stokes’ law is used in sedimentation to:
ⓐ. Calculate buoyant force on a body
ⓑ. Determine terminal velocity of falling particles in a fluid
ⓒ. Measure atmospheric pressure
ⓓ. Determine thermal conductivity of liquids
Correct Answer: Determine terminal velocity of falling particles in a fluid
Explanation: Sedimentation studies rely on Stokes’ law to calculate how fast small particles settle in a fluid, which is directly related to their size and density.
202. The terminal velocity of a small spherical particle falling through a viscous fluid is given by:
ⓐ. $v = \frac{2 r^2 g (\rho_p – \rho_f)}{9 \eta}$
ⓑ. $v = \frac{6 \pi \eta r v}{\rho g}$
ⓒ. $v = \frac{9 \eta}{2 r^2 g}$
ⓓ. $v = \frac{\rho_f – \rho_p}{\eta r}$
Correct Answer: $v = \frac{2 r^2 g (\rho_p – \rho_f)}{9 \eta}$
Explanation: Stokes’ law applied to balance viscous force, buoyant force, and weight gives this expression for terminal velocity, where $\rho_p$ = density of particle and $\rho_f$ = density of fluid.
203. A dust particle of radius $10^{-5} \, m$ and density $2000 \, kg/m^3$ settles in air of viscosity $1.8 \times 10^{-5} \, Pa \cdot s$. Find its terminal velocity. ($\rho_{air} \approx 1.2 \, kg/m^3, g = 9.8 \, m/s^2$)
ⓐ. $0.01 \, m/s$
ⓑ. $0.02 \, m/s$
ⓒ. $0.05 \, m/s$
ⓓ. $0.10 \, m/s$
Correct Answer: $0.02 \, m/s$
Explanation: $v = \frac{2 r^2 g (\rho_p – \rho_f)}{9 \eta} = \frac{2 (10^{-5})^2 \times 9.8 (2000 – 1.2)}{9 (1.8 \times 10^{-5})} \approx 0.02 \, m/s$.
204. Sedimentation rate of a particle in fluid is directly proportional to:
ⓐ. Square of particle radius
ⓑ. Cube of particle radius
ⓒ. Inverse of particle radius
ⓓ. Density of fluid
Correct Answer: Square of particle radius
Explanation: From $v = \frac{2 r^2 g (\rho_p – \rho_f)}{9 \eta}$, terminal velocity depends on the square of the particle radius, so larger particles settle faster.
205. Why do very fine colloidal particles settle extremely slowly in fluids?
ⓐ. Their density is equal to the fluid
ⓑ. Their radius is very small, reducing terminal velocity
ⓒ. Their viscosity is extremely high
ⓓ. They are always lighter than fluid
Correct Answer: Their radius is very small, reducing terminal velocity
Explanation: Terminal velocity is proportional to $r^2$. Very fine colloidal particles have tiny radii, making their sedimentation rate negligible.
206. In sedimentation analysis, particle size can be determined by measuring:
ⓐ. Volume of fluid displaced
ⓑ. Time taken for particles to settle
ⓒ. Pressure difference in fluid
ⓓ. Surface tension of liquid
Correct Answer: Time taken for particles to settle
Explanation: By measuring settling time and applying Stokes’ law, the particle radius can be determined. This method is widely used in colloid chemistry and soil analysis.
207. A spherical particle of radius $2 \times 10^{-4} \, m$ has density $2500 \, kg/m^3$. It falls in water ($\eta = 1 \times 10^{-3} \, Pa \cdot s$, $\rho_f = 1000 \, kg/m^3$). Calculate its terminal velocity. ($g = 9.8 \, m/s^2$)
ⓐ. $0.15 \, m/s$
ⓑ. $0.22 \, m/s$
ⓒ. $0.30 \, m/s$
ⓓ. $0.40 \, m/s$
Correct Answer: $0.22 \, m/s$
Explanation: $v = \frac{2 r^2 g (\rho_p – \rho_f)}{9 \eta} = \frac{2 (2 \times 10^{-4})^2 \times 9.8 (2500 – 1000)}{9 \times 10^{-3}} \approx 0.22 \, m/s$.
208. Why is Stokes’ law used in determining viscosity of fluids using sedimentation?
ⓐ. Because viscosity affects settling velocity of particles
ⓑ. Because density of particles remains constant
ⓒ. Because fluid pressure is uniform
ⓓ. Because buoyant force is negligible
Correct Answer: Because viscosity affects settling velocity of particles
Explanation: Terminal velocity depends inversely on viscosity. By measuring sedimentation rate, viscosity of a fluid can be determined.
209. Sedimentation method for particle size determination is particularly useful for:
ⓐ. Large, heavy particles only
ⓑ. Very small colloidal particles and fine powders
ⓒ. Only gases
ⓓ. Transparent liquids
Correct Answer: Very small colloidal particles and fine powders
Explanation: The method is widely used for colloids, soils, and fine powders, where direct measurement of particle size is difficult.
210. Which factor does NOT affect sedimentation velocity according to Stokes’ law?
ⓐ. Radius of the particle
ⓑ. Density difference between particle and fluid
ⓒ. Fluid viscosity
ⓓ. Color of the particle
Correct Answer: Color of the particle
Explanation: Sedimentation velocity depends on particle radius, density difference, and viscosity of fluid, but not on particle color.
211. Which of the following is a key assumption in the derivation of Stokes’ law?
ⓐ. The flow around the sphere is turbulent
ⓑ. The sphere moves at very high velocity
ⓒ. The flow around the sphere is laminar and steady
ⓓ. The fluid is compressible
Correct Answer: The flow around the sphere is laminar and steady
Explanation: Stokes’ law is derived under the assumption that the Reynolds number is very small (Re < 1), meaning the fluid flow is smooth, orderly, and laminar. If turbulence develops at higher velocities, the viscous drag no longer follows Stokes’ law. This assumption is essential for the law to be valid.
212. Which of the following is a limitation of Stokes’ law?
ⓐ. It cannot be applied at very low velocities
ⓑ. It is only valid when inertial forces dominate over viscous forces
ⓒ. It is valid only for small spherical bodies in laminar flow
ⓓ. It is not valid in any liquid medium
Correct Answer: It is valid only for small spherical bodies in laminar flow
Explanation: Stokes’ law is derived assuming the particle is perfectly spherical, the motion is slow enough for laminar flow, and the liquid is viscous and incompressible. For irregular shapes or turbulent flow, the law gives inaccurate results, hence this is a limitation.
213. Stokes’ law fails when the Reynolds number:
ⓐ. Is less than 1
ⓑ. Is very large, indicating turbulent flow
ⓒ. Equals zero
ⓓ. Is constant with velocity
Correct Answer: Is very large, indicating turbulent flow
Explanation: At high Reynolds numbers, inertial forces dominate viscous forces, producing turbulence around the sphere. In such cases, drag force no longer remains proportional to velocity, and Stokes’ linear relationship breaks down.
214. Why does Stokes’ law not apply to non-spherical particles?
ⓐ. Because their density is too high
ⓑ. Because drag force depends strongly on shape and surface area
ⓒ. Because viscosity cannot be defined for non-spherical particles
ⓓ. Because buoyant force does not act on them
Correct Answer: Because drag force depends strongly on shape and surface area
Explanation: Stokes’ derivation assumes a smooth spherical body. For irregular shapes, drag force depends on orientation and projected area, so the neat formula $F = 6 \pi \eta r v$ does not hold. This makes it unreliable for powders or irregular sediment particles.
215. Which assumption about the fluid is required for Stokes’ law to be valid?
ⓐ. The fluid is compressible
ⓑ. The fluid is incompressible and homogeneous
ⓒ. The fluid has zero viscosity
ⓓ. The fluid must be a gas only
Correct Answer: The fluid is incompressible and homogeneous
Explanation: Stokes’ law assumes the fluid has uniform density, is incompressible, and possesses constant viscosity. Compressibility or variation in density would complicate pressure and drag distributions, invalidating the derivation.
216. One limitation of Stokes’ law in real-world sedimentation experiments is:
ⓐ. It does not allow calculation of viscosity
ⓑ. It ignores buoyant force of the fluid
ⓒ. It assumes no slip between fluid and particle surface
ⓓ. It cannot be used in water at room temperature
Correct Answer: It assumes no slip between fluid and particle surface
Explanation: The derivation assumes the fluid layer in immediate contact with the sphere has zero relative velocity (no-slip condition). While mostly true, certain situations, especially with very small particles (nano-scale), can show slip, reducing the accuracy of Stokes’ law.
217. Why is Stokes’ law not valid for larger spheres moving rapidly through a fluid?
ⓐ. Because gravitational force dominates over viscous drag
ⓑ. Because larger spheres create turbulence and wake formation
ⓒ. Because larger spheres displace too much fluid
ⓓ. Because viscosity of fluid increases with particle size
Correct Answer: Because larger spheres create turbulence and wake formation
Explanation: As sphere size and velocity increase, the Reynolds number also rises. Beyond a certain limit, the flow becomes turbulent with eddies and wakes behind the sphere, violating the assumptions of laminar flow in Stokes’ law.
218. Which of the following is assumed negligible in Stokes’ law?
ⓐ. Buoyant force
ⓑ. Inertial forces in the fluid
ⓒ. Viscous forces in the fluid
ⓓ. Gravitational force on the particle
Correct Answer: Inertial forces in the fluid
Explanation: Stokes’ law derivation assumes that viscous forces dominate completely over inertial forces, which is valid at low Reynolds numbers. If inertial effects are significant, the drag force is no longer linearly dependent on velocity.
219. Why can’t Stokes’ law be applied to very small particles like smoke or colloidal particles in air?
ⓐ. Because such particles have infinite density
ⓑ. Because the no-slip condition fails at molecular scales
ⓒ. Because they do not experience viscous drag
ⓓ. Because they have zero velocity
Correct Answer: Because the no-slip condition fails at molecular scales
Explanation: For very small particles, molecular effects dominate and the continuum assumption of fluids breaks down. The boundary layer near the particle does not behave as predicted by continuum hydrodynamics, making Stokes’ law inaccurate.
220. Which of the following statements summarizes the main limitations of Stokes’ law?
ⓐ. It applies only to small, slow-moving spheres in viscous, incompressible fluids under laminar flow conditions
ⓑ. It applies to all shapes and velocities in any fluid
ⓒ. It applies only to gases at very high velocities
ⓓ. It applies only when turbulence is dominant
Correct Answer: It applies only to small, slow-moving spheres in viscous, incompressible fluids under laminar flow conditions
Explanation: Stokes’ law is limited to low Reynolds number conditions, spherical particles, and incompressible fluids with constant viscosity. Any deviation (high velocity, turbulence, irregular particle shape, compressible fluid) violates the assumptions and makes the law inaccurate.
221. What is terminal velocity?
ⓐ. The initial velocity of a body when released in a fluid
ⓑ. The constant maximum velocity attained by a body falling through a fluid when drag force balances the net weight
ⓒ. The velocity of a body in vacuum
ⓓ. The escape velocity of a body from Earth
Correct Answer: The constant maximum velocity attained by a body falling through a fluid when drag force balances the net weight
Explanation: Terminal velocity is reached when the downward gravitational force equals the sum of upward viscous and buoyant forces, so acceleration becomes zero and the velocity remains constant.
222. The expression for terminal velocity of a small sphere falling through a viscous fluid is:
ⓐ. $v_t = \frac{6 \pi \eta r v}{\rho g}$
ⓑ. $v_t = \frac{2 r^2 g (\rho_p – \rho_f)}{9 \eta}$
ⓒ. $v_t = \frac{\rho g}{r^2 \eta}$
ⓓ. $v_t = \frac{9 \eta}{2 r^2 g}$
Correct Answer: $v_t = \frac{2 r^2 g (\rho_p – \rho_f)}{9 \eta}$
Explanation: By balancing weight, buoyant force, and viscous force, the terminal velocity formula is derived. It depends on the radius squared of the particle, fluid viscosity, and density difference.
223. A steel ball of radius $1 \, mm$ and density $7800 \, kg/m^3$ falls through glycerin of viscosity $0.9 \, Pa \cdot s$ and density $1200 \, kg/m^3$. Find its terminal velocity. ($g = 9.8 \, m/s^2$)
ⓐ. $0.02 \, m/s$
ⓑ. $0.04 \, m/s$
ⓒ. $0.08 \, m/s$
ⓓ. $0.10 \, m/s$
Correct Answer: $0.04 \, m/s$
Explanation: $v_t = \frac{2 r^2 g (\rho_p – \rho_f)}{9 \eta} = \frac{2 (1 \times 10^{-3})^2 \times 9.8 \times (7800 – 1200)}{9 \times 0.9} \approx 0.04 \, m/s$.
224. Terminal velocity of a particle is proportional to:
ⓐ. Radius $r$
ⓑ. Radius squared $r^2$
ⓒ. Radius cubed $r^3$
ⓓ. Inverse of radius $1/r$
Correct Answer: Radius squared $r^2$
Explanation: From Stokes’ law derivation, $v_t = \frac{2 r^2 g (\rho_p – \rho_f)}{9 \eta}$. Hence, terminal velocity increases as the square of the radius, so larger particles fall faster than smaller ones.
225. If a small sphere’s radius is doubled, how does its terminal velocity change (keeping all other parameters constant)?
ⓐ. Increases by 2 times
ⓑ. Increases by 4 times
ⓒ. Increases by 8 times
ⓓ. Remains unchanged
Correct Answer: Increases by 4 times
Explanation: Since $v_t \propto r^2$, doubling the radius makes $v_t$ increase by a factor of $2^2 = 4$.
226. A raindrop of radius $0.5 \, mm$ falls in air of viscosity $1.8 \times 10^{-5} \, Pa \cdot s$. If $\rho_{drop} = 1000 \, kg/m^3$, $\rho_{air} \approx 1.2 \, kg/m^3$, calculate its terminal velocity. ($g = 9.8 \, m/s^2$)
ⓐ. $4.5 \, m/s$
ⓑ. $5.4 \, m/s$
ⓒ. $6.0 \, m/s$
ⓓ. $7.2 \, m/s$
Correct Answer: $5.4 \, m/s$
Explanation: $v_t = \frac{2 r^2 g (\rho_p – \rho_f)}{9 \eta} = \frac{2 (5 \times 10^{-4})^2 \times 9.8 (1000 – 1.2)}{9 (1.8 \times 10^{-5})} \approx 5.4 \, m/s$.
227. Why does terminal velocity occur in fluids?
ⓐ. Because gravitational force decreases with time
ⓑ. Because viscous force and buoyant force balance the net weight of the particle
ⓒ. Because density of fluid increases with depth
ⓓ. Because acceleration of particle keeps increasing
Correct Answer: Because viscous force and buoyant force balance the net weight of the particle
Explanation: Initially, gravity accelerates the particle. As velocity rises, viscous drag increases until it equals the effective weight (weight – buoyancy). At this point, acceleration becomes zero, and the particle moves with constant terminal velocity.
228. For a given particle, terminal velocity will be higher in:
ⓐ. A fluid of high viscosity
ⓑ. A fluid of low viscosity
ⓒ. A fluid of the same density as the particle
ⓓ. Both A and C
Correct Answer: A fluid of low viscosity
Explanation: Terminal velocity is inversely proportional to viscosity $\eta$. In low-viscosity fluids, drag is smaller, so the particle falls faster.
229. A pollen grain of radius $2 \times 10^{-6} \, m$ has density $1200 \, kg/m^3$. If it falls in air ($\rho = 1.2 \, kg/m^3, \eta = 1.8 \times 10^{-5} \, Pa \cdot s$), calculate its terminal velocity. ($g = 9.8 \, m/s^2$)
ⓐ. $1.0 \times 10^{-5} \, m/s$
ⓑ. $2.6 \times 10^{-5} \, m/s$
ⓒ. $3.0 \times 10^{-5} \, m/s$
ⓓ. $5.0 \times 10^{-5} \, m/s$
Correct Answer: $2.6 \times 10^{-5} \, m/s$
Explanation: $v_t = \frac{2 (2 \times 10^{-6})^2 \times 9.8 (1200 – 1.2)}{9 (1.8 \times 10^{-5})} \approx 2.6 \times 10^{-5} \, m/s$.
230. Which of the following is a correct statement about terminal velocity?
ⓐ. It is the maximum velocity reached in vacuum
ⓑ. It occurs when acceleration is maximum
ⓒ. It is reached when net force on the falling body is zero
ⓓ. It decreases continuously with time
Correct Answer: It is reached when net force on the falling body is zero
Explanation: At terminal velocity, the downward force of gravity is balanced by upward viscous drag and buoyant force, so acceleration becomes zero. The body then continues to move at constant velocity without further acceleration.
231. Which of the following conditions is necessary for a particle to reach terminal velocity in a fluid?
ⓐ. Only gravitational force acts on the particle
ⓑ. The viscous drag and buoyant force together balance the effective weight of the particle
ⓒ. The particle has infinite mass
ⓓ. The fluid has zero viscosity
Correct Answer: The viscous drag and buoyant force together balance the effective weight of the particle
Explanation: Terminal velocity is reached when the downward force of gravity is exactly balanced by the upward viscous drag plus buoyant force. At this point, net force is zero and acceleration ceases, leading to uniform velocity.
232. What happens to the acceleration of a particle as it approaches terminal velocity?
ⓐ. It increases continuously
ⓑ. It decreases gradually to zero
ⓒ. It remains constant
ⓓ. It becomes negative
Correct Answer: It decreases gradually to zero
Explanation: Initially, the particle accelerates due to gravity. As velocity increases, viscous drag increases. Once drag plus buoyancy equals weight, the net force becomes zero and acceleration reduces to zero, establishing terminal velocity.
233. Which of the following best describes the condition for terminal velocity mathematically?
ⓐ. $mg = 0$
ⓑ. $mg – F_{buoyant} – F_{viscous} = 0$
ⓒ. $F_{viscous} = 0$
ⓓ. $mg = F_{buoyant}$
Correct Answer: $mg – F_{buoyant} – F_{viscous} = 0$
Explanation: At terminal velocity, the effective weight of the particle $(mg – F_{buoyant})$ is balanced by the viscous drag force, giving zero net force.
234. A spherical particle of radius $r$ and density $\rho_p$ falls through a fluid of density $\rho_f$ and viscosity $\eta$. Which of the following expresses the condition for terminal velocity?
ⓐ. $mg = 0$
ⓑ. $\frac{4}{3}\pi r^3 \rho_p g = \frac{4}{3}\pi r^3 \rho_f g + 6 \pi \eta r v_t$
ⓒ. $6 \pi \eta r v_t = 0$
ⓓ. $v_t = \infty$
Correct Answer: $\frac{4}{3}\pi r^3 \rho_p g = \frac{4}{3}\pi r^3 \rho_f g + 6 \pi \eta r v_t$
Explanation: The particle’s weight is balanced by buoyant force plus viscous drag at terminal velocity. Rearranging this equation gives the formula for terminal velocity.
235. Why can terminal velocity never be reached in vacuum?
ⓐ. Because buoyant force is maximum in vacuum
ⓑ. Because viscous drag is absent in vacuum
ⓒ. Because gravity does not act in vacuum
ⓓ. Because pressure is infinite in vacuum
Correct Answer: Because viscous drag is absent in vacuum
Explanation: Terminal velocity arises from the balance of gravity and viscous drag. In vacuum, drag force is zero, so a falling body continues to accelerate indefinitely under gravity.
236. A raindrop falling in air does not keep accelerating forever because:
ⓐ. Gravity becomes weaker after some time
ⓑ. Buoyant force cancels its weight completely
ⓒ. Air resistance (drag) balances its effective weight, leading to terminal velocity
ⓓ. Surface tension of water resists gravity
Correct Answer: Air resistance (drag) balances its effective weight, leading to terminal velocity
Explanation: Initially gravity accelerates the raindrop, but as velocity increases, drag grows until it equals the effective weight of the drop, at which point the drop falls at constant velocity.
237. In the case of a falling sphere, terminal velocity will be reached more quickly if:
ⓐ. The fluid has higher viscosity
ⓑ. The fluid has lower viscosity
ⓒ. The density of particle equals fluid density
ⓓ. Gravity is absent
Correct Answer: The fluid has higher viscosity
Explanation: A higher viscosity means larger resistive drag builds up quickly, balancing the particle’s weight in a shorter time. Therefore, terminal velocity is achieved faster in highly viscous fluids.
238. What is the role of buoyant force in reaching terminal velocity?
ⓐ. It increases the net downward force on the body
ⓑ. It reduces the effective weight of the particle falling in fluid
ⓒ. It prevents the viscous force from acting
ⓓ. It makes the particle massless
Correct Answer: It reduces the effective weight of the particle falling in fluid
Explanation: The buoyant force acts upward and reduces the net downward force experienced by the particle. At terminal velocity, this reduced effective weight is balanced by viscous drag.
239. If a sphere and a cube of the same density and volume fall through a viscous fluid, which one will reach terminal velocity first?
ⓐ. The sphere
ⓑ. The cube
ⓒ. Both at the same time
ⓓ. Cannot be determined
Correct Answer: The sphere
Explanation: For the same volume, a sphere experiences less drag compared to irregular shapes like cubes. Hence it will settle faster and reach terminal velocity earlier.
240. Which of the following conditions ensures that terminal velocity exists for a falling body in a fluid?
ⓐ. Constant acceleration of the body
ⓑ. Presence of viscous drag force increasing with velocity
ⓒ. No buoyant force present in the fluid
ⓓ. Absence of gravity
Correct Answer: Presence of viscous drag force increasing with velocity
Explanation: Terminal velocity exists only when viscous drag (proportional to velocity) increases with speed until it balances the effective weight of the particle. Without this velocity-dependent resistance, a body would continue accelerating indefinitely.
241. A steel ball of radius $1 \times 10^{-3} \, m$ and density $7800 \, kg/m^3$ falls through glycerin of viscosity $1.2 \, Pa \cdot s$ and density $1200 \, kg/m^3$. Calculate its terminal velocity. ($g = 9.8 \, m/s^2$)
ⓐ. $2.5 \times 10^{-2} \, m/s$
ⓑ. $3.0 \times 10^{-2} \, m/s$
ⓒ. $3.5 \times 10^{-2} \, m/s$
ⓓ. $4.0 \times 10^{-2} \, m/s$
Correct Answer: $3.0 \times 10^{-2} \, m/s$
Explanation: Terminal velocity $ v_t = \frac{2 r^2 g (\rho_p – \rho_f)}{9 \eta} = \frac{2 (10^{-3})^2 (9.8)(7800-1200)}{9(1.2)} \approx 0.03 \, m/s$.
242. The terminal velocity of a spherical particle is proportional to:
ⓐ. $r$
ⓑ. $r^2$
ⓒ. $r^3$
ⓓ. $1/r$
Correct Answer: $r^2$
Explanation: From Stokes’ law, $v_t = \frac{2 r^2 g (\rho_p – \rho_f)}{9 \eta}$. Therefore, terminal velocity increases with the square of the radius.
243. A water droplet of radius $0.5 \times 10^{-3} \, m$ falls through air. If viscosity of air is $1.8 \times 10^{-5} \, Pa \cdot s$ and density of water is $1000 \, kg/m^3$, while air density is $1.2 \, kg/m^3$, find its terminal velocity.
ⓐ. $3.5 \, m/s$
ⓑ. $4.0 \, m/s$
ⓒ. $5.4 \, m/s$
ⓓ. $6.0 \, m/s$
Correct Answer: $5.4 \, m/s$
Explanation: Using $v_t = \frac{2 r^2 g (\rho_p – \rho_f)}{9 \eta}$, substituting values gives approximately $5.4 \, m/s$.
244. The expression for terminal velocity derived from Stokes’ law is valid only when:
ⓐ. Flow around the sphere is turbulent
ⓑ. Reynolds number is very low ($<1$)
ⓒ. Reynolds number is very high
ⓓ. Buoyant force is negligible
Correct Answer: Reynolds number is very low ($<1$)
Explanation: Stokes’ law assumes laminar flow conditions, which occur when the Reynolds number is small. At higher values, turbulence invalidates the formula.
245. A lead shot of radius $2 \times 10^{-4} \, m$ and density $11,300 \, kg/m^3$ is falling through water of viscosity $1.0 \times 10^{-3} \, Pa \cdot s$ and density $1000 \, kg/m^3$. Find the terminal velocity.
ⓐ. $0.15 \, m/s$
ⓑ. $0.22 \, m/s$
ⓒ. $0.30 \, m/s$
ⓓ. $0.40 \, m/s$
Correct Answer: $0.22 \, m/s$
Explanation: $v_t = \frac{2 r^2 g (\rho_p – \rho_f)}{9 \eta} = \frac{2 (2 \times 10^{-4})^2 \times 9.8 (11300-1000)}{9 \times 10^{-3}} \approx 0.22 \, m/s$.
246. If the density of the fluid increases while keeping particle size and viscosity constant, the terminal velocity will:
ⓐ. Increase
ⓑ. Decrease
ⓒ. Remain constant
ⓓ. Become infinite
Correct Answer: Decrease
Explanation: Since $v_t \propto (\rho_p – \rho_f)$, a higher fluid density reduces the density difference, lowering terminal velocity.
247. A spherical particle of radius $1.5 \times 10^{-4} \, m$ and density $2500 \, kg/m^3$ falls through a fluid of density $1200 \, kg/m^3$ and viscosity $0.02 \, Pa \cdot s$. Calculate terminal velocity.
ⓐ. $2.0 \times 10^{-3} \, m/s$
ⓑ. $2.5 \times 10^{-3} \, m/s$
ⓒ. $3.0 \times 10^{-3} \, m/s$
ⓓ. $3.5 \times 10^{-3} \, m/s$
Correct Answer: $2.5 \times 10^{-3} \, m/s$
Explanation: $v_t = \frac{2 (1.5 \times 10^{-4})^2 \times 9.8 (2500-1200)}{9 \times 0.02} \approx 2.5 \times 10^{-3} \, m/s$.
248. Terminal velocity of raindrops is important because:
ⓐ. It prevents them from evaporating
ⓑ. It limits their speed so they don’t cause damage when reaching the ground
ⓒ. It increases the density of air
ⓓ. It decreases with height
Correct Answer: It limits their speed so they don’t cause damage when reaching the ground
Explanation: Without terminal velocity, raindrops would accelerate indefinitely under gravity. Air drag balances their weight at a safe speed (\~5–9 m/s), allowing them to fall gently.
249. A pollen grain of radius $10^{-6} \, m$ has density $1200 \, kg/m^3$. If it falls in air of viscosity $1.8 \times 10^{-5} \, Pa \cdot s$ and density $1.2 \, kg/m^3$, calculate terminal velocity.
ⓐ. $1.0 \times 10^{-5} \, m/s$
ⓑ. $2.6 \times 10^{-5} \, m/s$
ⓒ. $3.5 \times 10^{-5} \, m/s$
ⓓ. $5.0 \times 10^{-5} \, m/s$
Correct Answer: $1.0 \times 10^{-5} \, m/s$
Explanation: Using $v_t = \frac{2 r^2 g (\rho_p – \rho_f)}{9 \eta}$, substituting values gives approximately $1.0 \times 10^{-5} \, m/s$.
250. Why does terminal velocity increase with particle radius?
ⓐ. Because viscous drag decreases with radius
ⓑ. Because gravitational force increases faster than viscous drag as radius increases
ⓒ. Because buoyant force is independent of radius
ⓓ. Because density decreases with size
Correct Answer: Because gravitational force increases faster than viscous drag as radius increases
Explanation: Weight increases with $r^3$, but viscous drag is proportional to $r$. Thus, larger spheres experience a greater net downward force, leading to higher terminal velocity.
251. Which of the following is the best example of terminal velocity in nature?
ⓐ. A stone thrown upward coming down
ⓑ. A raindrop falling through the atmosphere
ⓒ. A rocket moving in space
ⓓ. A pendulum oscillating
Correct Answer: A raindrop falling through the atmosphere
Explanation: A raindrop accelerates initially due to gravity, but as its speed increases, air drag grows until it balances the net weight. This results in a constant velocity known as terminal velocity, preventing raindrops from striking the ground at destructive speeds.
252. Why does a parachutist experience terminal velocity after some time in free fall?
ⓐ. Gravity decreases with altitude
ⓑ. Buoyant force cancels weight
ⓒ. Air resistance increases with speed until it balances the weight
ⓓ. Mass of parachutist decreases in air
Correct Answer: Air resistance increases with speed until it balances the weight
Explanation: Initially, the parachutist accelerates under gravity. With increasing speed, air drag increases until it equals the weight of the parachutist. At this stage, net force is zero and the fall continues at a constant velocity.
253. A small steel ball is dropped in a jar of oil. Which of the following statements is true?
ⓐ. The ball keeps accelerating forever
ⓑ. The ball eventually reaches a constant terminal velocity
ⓒ. The viscous drag decreases continuously
ⓓ. The ball stops moving after some time
Correct Answer: The ball eventually reaches a constant terminal velocity
Explanation: The ball initially accelerates due to gravity, but drag force increases with velocity. When drag and buoyant force together balance the weight, net force becomes zero, and the ball falls at terminal velocity.
254. The terminal velocity of a parachutist is reduced by:
ⓐ. Increasing weight of the parachutist
ⓑ. Increasing the area of the parachute canopy
ⓒ. Decreasing the drag coefficient
ⓓ. Increasing air density
Correct Answer: Increasing the area of the parachute canopy
Explanation: A larger parachute increases air resistance for a given speed. This means the balance between gravity and drag is achieved at a much lower speed, reducing terminal velocity and ensuring a safe landing.
255. In free fall through air, a feather reaches terminal velocity much sooner than a stone because:
ⓐ. A feather has smaller mass and greater drag-to-weight ratio
ⓑ. A feather has the same density as air
ⓒ. A stone has no buoyancy
ⓓ. Gravity does not act on a feather
Correct Answer: A feather has smaller mass and greater drag-to-weight ratio
Explanation: A feather is light and has a large surface area relative to its weight, so air drag balances its weight quickly, giving a very small terminal velocity compared to a stone.
256. A skydiver falls freely from an aircraft and attains a speed of about $50 \, m/s$. This speed is called:
ⓐ. Escape velocity
ⓑ. Terminal velocity
ⓒ. Initial velocity
ⓓ. Relative velocity
Correct Answer: Terminal velocity
Explanation: When air resistance balances the weight of the skydiver, acceleration ceases and a constant velocity is attained, which is the terminal velocity (about 50 m/s in practice for humans).
257. Why do small oil droplets suspended in air remain floating for a long time?
ⓐ. Their density is less than air
ⓑ. Their terminal velocity is extremely small due to small radius
ⓒ. They have no gravitational force
ⓓ. They are repelled by air molecules
Correct Answer: Their terminal velocity is extremely small due to small radius
Explanation: Since $v_t \propto r^2$, very small droplets have negligible settling speed, making them appear to remain suspended in air for long durations.
258. A pollen grain settles in water very slowly because:
ⓐ. Its density is less than water
ⓑ. Viscous drag on it is negligible
ⓒ. Its radius is very small, giving a very low terminal velocity
ⓓ. Buoyancy cancels its weight
Correct Answer: Its radius is very small, giving a very low terminal velocity
Explanation: Terminal velocity depends on $r^2$. For microscopic pollen grains, the radius is so small that their terminal velocity is extremely low, so they remain suspended for long periods.
259. In industries, terminal velocity is utilized in:
ⓐ. Separation of cream from milk
ⓑ. Sedimentation tanks for separating suspended particles from liquids
ⓒ. Purification of metals by electrolysis
ⓓ. Measurement of thermal conductivity
Correct Answer: Sedimentation tanks for separating suspended particles from liquids
Explanation: In sedimentation tanks, solid particles fall under gravity until terminal velocity is reached. By adjusting fluid conditions, industries can separate particles of different sizes and densities efficiently.
260. Which of the following is a correct example of terminal velocity in fluid flow?
ⓐ. A ball moving in vacuum with constant acceleration
ⓑ. Raindrops falling at constant speed through air
ⓒ. A car moving on a highway
ⓓ. A pendulum oscillating in air
Correct Answer: Raindrops falling at constant speed through air
Explanation: In fluid flow, raindrops are the most common example of terminal velocity. Air drag increases with velocity until it balances the gravitational force, producing constant speed fall.
261. What is streamline flow in fluids?
ⓐ. Random, chaotic motion of fluid particles
ⓑ. Motion of fluid particles along smooth, well-defined paths without crossing each other
ⓒ. Flow with very high Reynolds number
ⓓ. Flow with sudden changes in velocity and direction
Correct Answer: Motion of fluid particles along smooth, well-defined paths without crossing each other
Explanation: In streamline (or laminar) flow, each fluid particle follows a smooth path, and neighboring paths never cross. This occurs typically at low velocities and high viscosity, where viscous forces dominate over inertial forces.
262. Which of the following best describes turbulent flow?
ⓐ. Flow with no viscosity
ⓑ. Smooth, layer-like flow
ⓒ. Irregular, chaotic fluid motion with eddies and vortices
ⓓ. Flow of an ideal fluid only
Correct Answer: Irregular, chaotic fluid motion with eddies and vortices
Explanation: In turbulent flow, velocity and pressure at a point vary unpredictably with time. This flow arises at high Reynolds numbers when inertial forces dominate over viscous forces.
263. Streamline flow is also known as:
ⓐ. Turbulent flow
ⓑ. Steady or laminar flow
ⓒ. Eddy flow
ⓓ. Chaotic flow
Correct Answer: Steady or laminar flow
Explanation: Streamline flow is steady, with each fluid particle moving along the same path at constant velocity. It is also called laminar flow because it occurs in distinct fluid layers sliding smoothly over one another.
264. Which of the following is NOT a characteristic of turbulent flow?
ⓐ. Presence of eddies and swirls
ⓑ. High Reynolds number
ⓒ. Constant velocity at every point
ⓓ. Random variation of velocity and pressure
Correct Answer: Constant velocity at every point
Explanation: In turbulent flow, velocity and pressure fluctuate continuously and unpredictably. Constant velocity at every point is a property of streamline flow, not turbulence.
265. Streamline flow usually occurs when:
ⓐ. Fluid velocity is very high
ⓑ. Reynolds number is less than 2000
ⓒ. Fluid density is very low
ⓓ. Pressure is extremely high
Correct Answer: Reynolds number is less than 2000
Explanation: Experimentally, streamline flow is observed for Reynolds number $Re < 2000$. Between 2000 and 3000, the flow is transitional, and above 3000 it generally becomes turbulent.
266. Which of the following best represents a real-life example of turbulent flow?
ⓐ. Oil flowing slowly in a capillary tube
ⓑ. Blood flow in very narrow capillaries
ⓒ. Smoke rising in air and breaking into swirls
ⓓ. Water moving in a small syringe
Correct Answer: Smoke rising in air and breaking into swirls
Explanation: Rising smoke is initially smooth (streamline), but soon breaks into chaotic swirls and eddies, which is a visible example of turbulence.
267. In streamline flow, the velocity of fluid:
ⓐ. Is the same at every point of the fluid
ⓑ. Remains constant at a point but may differ from point to point
ⓒ. Changes chaotically at every instant
ⓓ. Becomes infinite after some time
Correct Answer: Remains constant at a point but may differ from point to point
Explanation: In laminar flow, velocity at any fixed point remains constant with time, though it can vary from one streamline to another. For example, velocity near walls is lower due to viscosity.
268. Which property increases the chance of turbulent flow?
ⓐ. High viscosity of the fluid
ⓑ. Low velocity of fluid
ⓒ. High velocity and low viscosity
ⓓ. Narrow channels with slow flow
Correct Answer: High velocity and low viscosity
Explanation: Turbulence occurs when inertial forces dominate viscous forces. High speed and low viscosity increase Reynolds number, pushing the system into turbulent regime.
269. Why do airplane wings experience turbulence?
ⓐ. Because air has zero viscosity
ⓑ. Because high-speed air causes chaotic eddies behind the wing
ⓒ. Because gravitational pull cancels airflow
ⓓ. Because air pressure becomes constant everywhere
Correct Answer: Because high-speed air causes chaotic eddies behind the wing
Explanation: At high speeds, airflow behind the wings can separate and form vortices, a typical turbulent flow phenomenon that influences stability and drag.
270. Which statement is TRUE about streamline and turbulent flow?
ⓐ. Streamline flow has chaotic velocity, turbulent flow has steady velocity
ⓑ. Streamline flow occurs at low Reynolds numbers, turbulent flow at high Reynolds numbers
ⓒ. Both streamline and turbulent flows are independent of Reynolds number
ⓓ. Turbulent flow always occurs at zero viscosity
Correct Answer: Streamline flow occurs at low Reynolds numbers, turbulent flow at high Reynolds numbers
Explanation: The Reynolds number determines the nature of flow. For low values, viscous forces dominate, leading to laminar flow, while at high values inertial forces dominate, leading to turbulence.
271. Reynolds number is a dimensionless quantity used to predict:
ⓐ. Viscosity of a fluid
ⓑ. Density of a fluid
ⓒ. Nature of flow (streamline or turbulent)
ⓓ. Pressure in a fluid
Correct Answer: Nature of flow (streamline or turbulent)
Explanation: Osborne Reynolds introduced the Reynolds number ($Re$), a ratio of inertial forces to viscous forces. It predicts whether a flow will be laminar ($Re < 2000$), transitional (2000–3000), or turbulent ($Re > 3000$).
272. The Reynolds number is defined as:
ⓐ. $Re = \frac{\rho v L}{\eta}$
ⓑ. $Re = \frac{\eta}{\rho v L}$
ⓒ. $Re = \frac{\rho g h}{\eta}$
ⓓ. $Re = \frac{F}{A}$
Correct Answer: $Re = \frac{\rho v L}{\eta}$
Explanation: Here $\rho$ = fluid density, $v$ = velocity, $L$ = characteristic length (like pipe diameter), and $\eta$ = coefficient of viscosity.
273. In Reynolds’ experiment, dye injected into water in a tube forms a straight line along the flow. This indicates:
ⓐ. Turbulent flow
ⓑ. Streamline (laminar) flow
ⓒ. Transitional flow
ⓓ. Eddy currents
Correct Answer: Streamline (laminar) flow
Explanation: When the flow is smooth and orderly, the dye filament remains straight. This happens at low velocities corresponding to Reynolds number below 2000.
274. In Reynolds’ experiment, when the velocity of water is gradually increased, the dye streak starts oscillating and breaking up. This shows:
ⓐ. Decrease in viscosity
ⓑ. Beginning of turbulent flow (transition region)
ⓒ. Fluid becomes incompressible
ⓓ. Pressure in tube increases
Correct Answer: Beginning of turbulent flow (transition region)
Explanation: At Reynolds number between 2000 and 3000, the flow becomes unstable and transitions from laminar to turbulent.
275. Which condition indicates turbulent flow in Reynolds’ experiment?
ⓐ. Dye streak remains straight
ⓑ. Dye streak spreads irregularly, forming eddies
ⓒ. Dye streak moves with constant velocity
ⓓ. Dye dissolves completely in fluid
Correct Answer: Dye streak spreads irregularly, forming eddies
Explanation: At high velocities, Reynolds number exceeds \~3000, causing eddies and swirls. The dye spreads randomly, indicating turbulence.
276. What is the critical Reynolds number for flow in a cylindrical tube?
ⓐ. 500
ⓑ. 1000
ⓒ. 2000
ⓓ. 10,000
Correct Answer: 2000
Explanation: Osborne Reynolds found that when $Re < 2000$, the flow is laminar; above 3000, it is turbulent; and between 2000–3000, it is transitional.
277. The Reynolds number is dimensionless because:
ⓐ. It has no units after substitution
ⓑ. It depends on density only
ⓒ. It is independent of velocity
ⓓ. It is always equal to one
Correct Answer: It has no units after substitution
Explanation: $Re = \frac{\rho v L}{\eta}$. Substituting SI units ($\rho = kg/m^3, v = m/s, L = m, \eta = kg/(m·s)$) cancels out all units, leaving a dimensionless number.
278. Water of density $1000 \, kg/m^3$ flows in a pipe of diameter $0.02 \, m$ with velocity $0.5 \, m/s$. If viscosity of water is $1 \times 10^{-3} \, Pa·s$, find the Reynolds number.
ⓐ. 500
ⓑ. 1000
ⓒ. 2000
ⓓ. 5000
Correct Answer: 1000
Explanation: $Re = \frac{\rho v D}{\eta} = \frac{1000 \times 0.5 \times 0.02}{1 \times 10^{-3}} = 1000$. This is less than 2000, hence laminar.
279. Oil of density $900 \, kg/m^3$ and viscosity $0.9 \, Pa·s$ flows through a pipe of diameter $0.01 \, m$ at velocity $0.1 \, m/s$. Calculate Reynolds number.
ⓐ. 1
ⓑ. 2
ⓒ. 10
ⓓ. 100
Correct Answer: 10
Explanation: $Re = \frac{\rho v D}{\eta} = \frac{900 \times 0.1 \times 0.01}{0.9} = 10$. Since $Re \ll 2000$, the flow is laminar.
280. Why is Reynolds number important in engineering applications?
ⓐ. It measures density only
ⓑ. It predicts whether fluid flow will be laminar or turbulent
ⓒ. It determines thermal conductivity of a fluid
ⓓ. It measures pressure difference in fluids
Correct Answer: It predicts whether fluid flow will be laminar or turbulent
Explanation: Engineers use Reynolds number to design pipelines, aircraft, and hydraulic systems. Low $Re$ ensures smooth flow, while high $Re$ indicates turbulence, affecting drag, energy loss, and efficiency.
281. What is the primary factor that governs the transition from laminar to turbulent flow in a fluid?
ⓐ. Temperature of the fluid
ⓑ. Reynolds number
ⓒ. Pressure difference only
ⓓ. Shape of the pipe
Correct Answer: Reynolds number
Explanation: The transition from laminar to turbulent flow is characterized by the Reynolds number. For $Re < 2000$, flow is laminar; for $Re > 3000$, it is turbulent. Between 2000–3000, the flow is unstable and transitional.
282. In a pipe, the flow changes from laminar to turbulent when:
ⓐ. Velocity is extremely low
ⓑ. Velocity is increased such that inertial forces dominate viscous forces
ⓒ. Viscosity becomes infinite
ⓓ. Density of fluid becomes zero
Correct Answer: Velocity is increased such that inertial forces dominate viscous forces
Explanation: Higher velocity increases inertial forces relative to viscous forces, raising Reynolds number and causing turbulence.
283. The velocity at which laminar flow just becomes unstable and begins to turn turbulent is called:
ⓐ. Critical velocity
ⓑ. Escape velocity
ⓒ. Terminal velocity
ⓓ. Drift velocity
Correct Answer: Critical velocity
Explanation: Critical velocity is the threshold velocity of a fluid in a pipe beyond which laminar flow becomes unstable and transitions to turbulence.
284. The expression for critical velocity in terms of Reynolds number is:
ⓐ. $v_c = \frac{\eta}{\rho D}$
ⓑ. $v_c = \frac{Re_c \eta}{\rho D}$
ⓒ. $v_c = \frac{\rho D}{Re_c \eta}$
ⓓ. $v_c = \frac{g D}{\rho \eta}$
Correct Answer: $v_c = \frac{Re_c \eta}{\rho D}$
Explanation: From definition, $Re = \frac{\rho v D}{\eta}$. At the critical point, $Re = Re_c$. Rearranging gives $v_c = \frac{Re_c \eta}{\rho D}$.
285. Water of viscosity $1.0 \times 10^{-3} \, Pa·s$ flows in a pipe of diameter $0.02 \, m$. If density is $1000 \, kg/m^3$ and $Re_c = 2000$, calculate the critical velocity.
ⓐ. $0.05 \, m/s$
ⓑ. $0.10 \, m/s$
ⓒ. $0.15 \, m/s$
ⓓ. $0.20 \, m/s$
Correct Answer: $0.10 \, m/s$
Explanation: $v_c = \frac{Re_c \eta}{\rho D} = \frac{2000 \times 1 \times 10^{-3}}{1000 \times 0.02} = 0.10 \, m/s$.
286. Which of the following does NOT influence the laminar-to-turbulent transition directly?
ⓐ. Viscosity of the fluid
ⓑ. Density of the fluid
ⓒ. Diameter of the pipe
ⓓ. Color of the fluid
Correct Answer: Color of the fluid
Explanation: Reynolds number depends on viscosity, density, velocity, and pipe diameter. Fluid color has no effect on flow type.
287. Why does turbulence occur at high Reynolds numbers?
ⓐ. Because viscous forces dominate completely
ⓑ. Because inertial forces dominate over viscous forces, causing instability
ⓒ. Because buoyant forces increase
ⓓ. Because temperature becomes constant
Correct Answer: Because inertial forces dominate over viscous forces, causing instability
Explanation: High velocity leads to dominance of inertial forces, resulting in irregular eddies and vortices that destabilize laminar flow into turbulence.
288. In Reynolds’ experiment, which visual effect indicated transition to turbulence?
ⓐ. Dye streak remained straight and narrow
ⓑ. Dye streak spread slowly but uniformly
ⓒ. Dye streak broke up and spread randomly across the tube
ⓓ. Dye disappeared completely in water
Correct Answer: Dye streak broke up and spread randomly across the tube
Explanation: The breakup and mixing of dye stream show the onset of turbulence as eddies and irregularities appear in the flow.
289. A highly viscous liquid is less likely to undergo laminar-to-turbulent transition because:
ⓐ. Viscous forces suppress eddies and disturbances
ⓑ. Its density is extremely low
ⓒ. It cannot flow in narrow tubes
ⓓ. It has no Reynolds number
Correct Answer: Viscous forces suppress eddies and disturbances
Explanation: High viscosity resists velocity fluctuations and stabilizes the flow. Thus, turbulence is less likely even at higher velocities.
290. Which statement correctly summarizes the transition from laminar to turbulent flow?
ⓐ. It occurs abruptly at zero viscosity
ⓑ. It occurs gradually around a critical Reynolds number, depending on fluid and pipe conditions
ⓒ. It occurs only in gases, never in liquids
ⓓ. It cannot be predicted experimentally
Correct Answer: It occurs gradually around a critical Reynolds number, depending on fluid and pipe conditions
Explanation: Experiments show that below critical Reynolds number, flow is laminar; above it, turbulent. The transition zone (2000–3000) is gradual, influenced by surface roughness and disturbances.
291. Why is the concept of Reynolds number important in pipe flow design?
ⓐ. It determines whether flow will be laminar or turbulent, affecting energy losses
ⓑ. It measures pressure head in a pipe
ⓒ. It calculates only the viscosity of the fluid
ⓓ. It is used to determine the density of the fluid
Correct Answer: It determines whether flow will be laminar or turbulent, affecting energy losses
Explanation: Engineers use Reynolds number to predict flow nature. Laminar flow has lower frictional losses, while turbulent flow has higher energy loss due to eddies and vortices, which impacts pump efficiency and pipeline design.
292. In laminar flow through a circular pipe, velocity distribution across the section is:
ⓐ. Uniform
ⓑ. Maximum at walls and zero at center
ⓒ. Parabolic with maximum at center and zero at walls
ⓓ. Random and chaotic
Correct Answer: Parabolic with maximum at center and zero at walls
Explanation: In laminar pipe flow, viscosity causes velocity to be zero at pipe walls (no-slip condition) and maximum at the center, forming a parabolic profile.
293. Which law governs laminar flow in narrow pipes?
ⓐ. Boyle’s law
ⓑ. Poiseuille’s law
ⓒ. Pascal’s law
ⓓ. Bernoulli’s law
Correct Answer: Poiseuille’s law
Explanation: Poiseuille’s law describes volumetric flow rate of viscous fluids in narrow tubes:
$$ Q = \frac{\pi \Delta P r^4}{8 \eta L} $$
It directly relates flow rate to pressure difference, radius, viscosity, and pipe length.
294. In turbulent flow through a pipe, head loss due to friction is proportional to:
ⓐ. Velocity
ⓑ. Velocity squared
ⓒ. Pressure
ⓓ. Density only
Correct Answer: Velocity squared
Explanation: In turbulent flow, energy losses grow rapidly with velocity. Frictional head loss is proportional to $v^2$, unlike laminar flow where it is proportional to $v$.
295. Why are streamlined shapes used in aerodynamics?
ⓐ. To increase drag force
ⓑ. To minimize viscous drag and turbulence
ⓒ. To reduce viscosity of the air
ⓓ. To increase pressure difference between top and bottom
Correct Answer: To minimize viscous drag and turbulence
Explanation: Streamlined shapes reduce flow separation and turbulence around moving bodies like cars and airplanes, thereby minimizing drag and improving efficiency.
296. An airplane wing generates lift mainly because:
ⓐ. Pressure above the wing is greater than below
ⓑ. Pressure below the wing is greater than above
ⓒ. Density of air above the wing is zero
ⓓ. Viscosity of air vanishes at high speed
Correct Answer: Pressure below the wing is greater than above
Explanation: Due to airfoil shape and Bernoulli’s principle, airflow above the wing moves faster, reducing pressure. Pressure beneath remains higher, generating lift force.
297. Which of the following is an application of turbulent flow in engineering?
ⓐ. Reducing drag in aircraft
ⓑ. Promoting mixing in chemical reactors
ⓒ. Maintaining steady streamline flow in pipes
ⓓ. Reducing energy losses in pipelines
Correct Answer: Promoting mixing in chemical reactors
Explanation: Turbulent flow enhances mixing and heat transfer, making it desirable in reactors and cooling systems, though it increases energy loss in pipelines.
298. A fluid of viscosity $0.002 \, Pa·s$, density $1000 \, kg/m^3$, flows at velocity $0.2 \, m/s$ through a pipe of diameter $0.05 \, m$. Find Reynolds number.
ⓐ. 2000
ⓑ. 2500
ⓒ. 5000
ⓓ. 10,000
Correct Answer: 2000
Explanation: $Re = \frac{\rho v D}{\eta} = \frac{1000 \times 0.2 \times 0.05}{0.002} = 5000$. Correction: $Re = 5000$. Hence correct answer is C. 5000.
299. Why are pipelines carrying oil often operated below the critical Reynolds number?
ⓐ. To avoid energy losses and ensure smooth transport
ⓑ. To increase turbulence and mixing
ⓒ. To reduce oil density
ⓓ. To eliminate viscosity completely
Correct Answer: To avoid energy losses and ensure smooth transport
Explanation: Laminar flow reduces frictional losses and pumping costs. In oil pipelines, engineers design conditions such that flow remains laminar or within stable ranges to maximize efficiency.
300. In aerodynamics, drag force acting on a body moving through air consists mainly of:
ⓐ. Pressure drag and viscous drag
ⓑ. Only buoyant force
ⓒ. Gravitational and elastic forces
ⓓ. Centripetal and centrifugal forces
Correct Answer: Pressure drag and viscous drag
Explanation: Drag in aerodynamics arises from two main sources: viscous drag due to friction and pressure drag due to flow separation and turbulence around the body. Reducing both is essential for efficient vehicle and aircraft design.
The third section of Thermal Properties of Matter dives deeper into heat transfer mechanisms. Following the NCERT/CBSE Class 11 Physics syllabus, this part focuses on conduction, convection, and radiation, their governing laws, and practical applications. Students will practice MCQs based on thermal conductivity, Newton’s law of cooling, Stefan’s law, emissivity, absorptivity, and blackbody radiation. These concepts not only form a crucial part of the board exam syllabus but are also repeatedly tested in competitive exams like JEE, NEET, and AIIMS. Out of the complete 600 MCQs, Part 3 provides the next 100 solved MCQs to improve your analytical and problem-solving skills in thermal physics.
- 👉 Total MCQs in this chapter: 600.
- 👉 This page contains: Third set of 100 solved MCQs with answers.
- 👉 Covers heat transfer: conduction, convection, radiation.
- 👉 To explore more chapters, subjects, or classes, use the top navigation bar above.
- 👉 To continue further, click the Part 4 button above.