Q.A metal block of area $0.10\ \text{m}^{2}$ is connected to a $0.010\ \text{kg}$ mass via a string that passes over an ideal pulley (considered massless and frictionless). A liquid with a film thickness of $0.30\ \text{mm}$ is placed between the block and the table. When released the block moves to the right with a constant speed of $0.085\ \text{m s}^{-1}$. Find the coefficient of viscosity of the liquid.

Showing the 12 most recent of 19 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.A bubble rises in a viscous liquid. If radius of bubble doubles, terminal velocity becomes (A) Double (B) Four times (C) Half (D) Same

   ›Reveal solutionSolution

   Terminal velocity in Stokes' regime scales as $r^2$, so doubling the bubble's radius quadruples its terminal (rise) velocity.

   Concept and Intuition

   A bubble rising through a viscous liquid reaches terminal velocity when the net upward buoyant force balances the downward viscous drag (given by Stokes' law, $F = 6\pi\eta r v$). Balancing these forces shows the terminal speed depends on the square of the radius, because the buoyant/weight-difference force scales with volume ($r^3$) while drag scales with $r$ (at fixed $v$) — the ratio gives $v_t \propto r^2$.

   Step-by-Step Solution

   1. At terminal velocity, net force = 0: buoyant force (upward, on the lighter bubble in the denser liquid) equals viscous drag.
   2. $\dfrac{4}{3}\pi r^3(\rho_{liquid}-\rho_{gas})g = 6\pi\eta r v_t$.
   3. Solve for $v_t$: $v_t = \dfrac{2r^2(\rho_{liquid}-\rho_{gas})g}{9\eta}$.
   4. All quantities except $r$ are unchanged, so $v_t \propto r^2$. …

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2. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.A small sphere of radius 'r' is dropped in a viscous liquid. When it is moving with terminal velocity in the liquid, the relation between the rate of heat produced $\left(\frac{dQ}{dt}\right)$ and 'r' is (A) $\frac{dQ}{dt} \propto r^2$ (B) $\frac{dQ}{dt} \propto r^5$ (C) $\frac{dQ}{dt} \propto \frac{1}{r^2}$ (D) $\frac{dQ}{dt} \propto \frac{1}{r^5}$

   ›Reveal solutionSolution

   At terminal velocity, all the gravitational PE lost is converted to heat via viscous drag; combining $v_t \propto r^2$ (Stokes' law) with the $r^3$-scaling effective weight gives $\frac{dQ}{dt} \propto r^5$.

   Concept and Intuition

   At terminal velocity, the sphere moves at constant speed, so its kinetic energy doesn't change — all the gravitational potential energy it loses per second is dissipated as heat by the viscous drag force. The rate of heat production is therefore just the power delivered by gravity (net of buoyancy), i.e., force times velocity, both of which have their own dependence on the sphere's radius.

   Step-by-Step Solution

   1. At terminal velocity, net force = 0: weight = buoyant force + viscous drag. $\frac{4}{3}\pi r^3 \rho g = \frac{4}{3}\pi r^3 \sigma g + 6\pi\eta r v_t$ where $\rho$ = density of sphere, $\sigma$ = density of liquid.
   2. This gives $v_t \propto r^3 / r = r^2$ (Stokes' law result: $v_t = \dfrac{2r^2(\rho-\sigma)g}{9\eta}$).
   3. The viscous drag force at terminal velocity equals the effective weight (weight − buoyancy), which is proportional to the sphere's volume, i.e., $F_{visc} \propto r^3$. …

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3. AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
   Q.An air bubble of radius 0.5 mm rises in a long vertical column of liquid of coefficient of viscosity 0.2 $\mathrm{N\ s\ m^{-2}}$ and density 900 $\mathrm{kg\ m^{-3}}$. If the density of air is neglected, then the terminal velocity of the air bubble is (Acceleration due to gravity = 10 $\mathrm{m\ s^{-2}}$) (A) 2.5 $\mathrm{mm\ s^{-1}}$ (B) 5 $\mathrm{mm\ s^{-1}}$ (C) 3.5 $\mathrm{mm\ s^{-1}}$ (D) 7 $\mathrm{mm\ s^{-1}}$

   ›Reveal solutionSolution

   Applying Stokes' law terminal-velocity formula (with air's density neglected) gives a bubble rise speed of 2.5 mm/s.

   Concept and Intuition

   An air bubble rising through a viscous liquid experiences an upward net buoyant force (because it's much less dense than the surrounding fluid) balanced by a downward viscous drag (Stokes' drag, proportional to velocity) once it reaches terminal velocity. The magnitude of this terminal velocity has the same form as for a sinking dense sphere, but driven by the density difference $(\rho_{liquid}-\rho_{bubble})\approx\rho_{liquid}$ since the bubble's own density is negligible.

   Step-by-Step Solution

   1. Stokes' terminal velocity formula: $v = \dfrac{2r^2g(\rho_{liquid}-\rho_{air})}{9\eta}$. Since $\rho_{air}\approx0$ is neglected, this simplifies to $v = \dfrac{2r^2g\rho_{liquid}}{9\eta}$.
   2. Convert radius: $r = 0.5\ \text{mm} = 5\times10^{-4}\ \text{m}$, so $r^2 = 2.5\times10^{-7}\ \text{m}^2$.
   3. Substitute values ($g=10\ \mathrm{ms^{-2}}$, $\rho_{liquid}=900\ \mathrm{kg\,m^{-3}}$, $\eta=0.2\ \mathrm{N\,s\,m^{-2}}$): …

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4. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
   Q.Two drops of same radius are falling through air with steady velocity of $5\,\mathrm{cms^{-1}}$. If the two drops coalesce, the terminal velocity would be (A) $10\,\mathrm{cms^{-1}}$ (B) $5 \times 4^{1/3}\,\mathrm{cms^{-1}}$ (C) $5 \times 4^{2/3}\,\mathrm{cms^{-1}}$ (D) $2.5\,\mathrm{cms^{-1}}$

   ›Reveal solutionSolution

   Using $v_t\propto r^2$ and conservation of volume when two drops merge, the new terminal velocity is $5\times4^{1/3}\,\mathrm{cm/s}$.

   Concept and Intuition

   A small sphere falling through a viscous fluid (Stokes' law regime) reaches a terminal velocity $v_t=\dfrac{2r^2(\rho-\sigma)g}{9\eta}\propto r^2$. When two identical drops coalesce, mass (and hence volume, since density is unchanged) is conserved, which fixes the new radius, and from there the new terminal velocity follows via the $r^2$ scaling.

   Step-by-Step Solution

   1. Volume conservation: $\dfrac{4}{3}\pi R^3 = 2\times\dfrac{4}{3}\pi r^3 \Rightarrow R^3=2r^3 \Rightarrow R=r\cdot2^{1/3}$.
   2. Terminal velocity scaling: $v_t\propto r^2$, so $\dfrac{V}{v_t}=\left(\dfrac{R}{r}\right)^2=(2^{1/3})^2=2^{2/3}$. …

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5. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.If water flows with a velocity of $20\ \mathrm{cm\,s^{-1}}$ in a pipe of radius 2 cm, then the flow is (The coefficient of viscosity of water is $10^{-3}\ \mathrm{kg\,m^{-1}\,s^{-1}}$ and density of water is $10^3\ \mathrm{kg\,m^{-3}}$) (A) turbulant (B) steady flow (C) non-viscous (D) unsteady

   ›Reveal solutionSolution

   Computing the Reynolds number for water flowing through a pipe shows $Re \approx 8000$, far above the critical range, so the flow is turbulent.

   Concept and Intuition

   Whether a flow is smooth (streamline/laminar) or chaotic (turbulent) is determined by the dimensionless Reynolds number $Re = \dfrac{\rho v D}{\eta}$, which compares inertial forces to viscous forces. Below roughly 1000–2000 the flow stays laminar; well above it, turbulence sets in.

   Step-by-Step Solution

   1. Given $v = 20\ \mathrm{cm\,s^{-1}} = 0.2\ \mathrm{m\,s^{-1}}$, radius $= 2\ \mathrm{cm}$ so diameter $D = 4\ \mathrm{cm} = 0.04\ \mathrm{m}$.
   2. $\eta = 10^{-3}\ \mathrm{kg\,m^{-1}\,s^{-1}}$, $\rho = 10^3\ \mathrm{kg\,m^{-3}}$.
   3. $Re = \dfrac{\rho v D}{\eta} = \dfrac{10^3\times0.2\times0.04}{10^{-3}} = \dfrac{8}{10^{-3}} = 8000$. …

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6. AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
   Q.A small solid sphere of mass 10 g and density 2600 kg m$^{-3}$ is dropped into a long vertical column of glycerine. When the sphere attains terminal velocity, the magnitude of the viscous force acting on the sphere is (Acceleration due to gravity = 10 m s$^{-2}$ and density of glycerine = 1300 kg m$^{-3}$) (A) $20 \times 10^{-3}$ N (B) $25 \times 10^{-3}$ N (C) $75 \times 10^{-3}$ N (D) $50 \times 10^{-3}$ N

   ›Reveal solutionSolution

   At terminal velocity, the net force is zero, so the viscous drag exactly balances the sphere's effective weight (weight minus buoyant upthrust) — giving $50\times10^{-3}$ N.

   Concept and Intuition

   When a sphere falls through a viscous fluid, three forces act on it: gravity (down), buoyancy (up), and viscous drag (up, opposing motion, growing with speed). At terminal velocity the sphere no longer accelerates, so these three forces balance: $F_v + F_{buoyancy} = mg$, i.e. $F_v = mg - F_{buoyancy}$.

   Step-by-Step Solution

   1. Mass of sphere $m=10\text{ g}=0.01$ kg; weight $mg = 0.01\times10=0.1$ N.
   2. Buoyant force $=\rho_{fluid}\,V\,g = \dfrac{\rho_{fluid}}{\rho_{sphere}}\times mg$ (since $V=m/\rho_{sphere}$). …

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7. AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
   Q.If two rain drops of radii $r_1$ and $r_2$ reach the ground with terminal velocities $v_1$ and $v_2$ and linear momenta p and 32p respectively, then $r_1 : r_2 =$ (A) 1 : 16 (B) 1 : 2 (C) 2 : 1 (D) 16 : 1

   ›Reveal solutionSolution

   Since terminal velocity scales as $r^2$ and mass as $r^3$, momentum scales as $r^5$; matching the given momentum ratio of $1:32 = 1:2^5$ gives $r_1:r_2 = 1:2$.

   Concept and Intuition

   A falling raindrop reaches terminal velocity when viscous drag (Stokes' law, $\propto r$) balances the net downward weight (buoyancy-corrected gravity, $\propto r^3$), giving $v_t \propto r^2$. Combined with mass $m \propto r^3$ (constant density), the momentum $p=mv_t$ scales as $r^5$ — a strong dependence on radius that makes ratios of momentum a sensitive probe of the ratio of radii.

   Step-by-Step Solution

   1. Terminal velocity: $v_t \propto r^2$ (from balancing Stokes drag against effective weight).
   2. Mass: $m \propto r^3$ (density constant, volume $\propto r^3$).
   3. Momentum: $p = mv_t \propto r^3 \times r^2 = r^5$.
   4. Given $p_1 : p_2 = p : 32p = 1 : 32 = 1 : 2^5$. …

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8. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.A spherical ball of radius $1 \times 10^{-4}$ m and of density $10^4\ \mathrm{kgm^{-3}}$ falls freely under gravity through a distance 'h' before entering a tank of water. After entering water if the velocity of the ball does not change, then 'h' is (The coefficient of viscosity of water $9.8 \times 10^{-6}\ \mathrm{Nsm^{-2}}$) (A) 20.4 cm (B) 20.4 mm (C) 20.4 m (D) 10.2 m

   ›Reveal solutionSolution

   "Velocity unchanged on entry" means the free-fall speed already equals the water's terminal velocity; computing that terminal velocity and inverting $v^2=2gh$ gives $h\approx20.4$ m.

   Concept and Intuition

   A small sphere falling through a viscous fluid speeds up only until viscous drag balances the net weight (weight minus buoyancy) — beyond that it moves at a constant "terminal velocity." The problem tells us the ball enters the water already moving at exactly this terminal speed (since its velocity doesn't change on entry), so the free-fall height before entering water must be exactly what's needed to reach that speed under gravity alone.

   Step-by-Step Solution

   1. Terminal velocity (Stokes' law): $v_t=\dfrac{2r^2(\rho-\sigma)g}{9\eta}$, with $r=1\times10^{-4}$ m, $\rho=10^4\ \mathrm{kgm^{-3}}$, $\sigma=1000\ \mathrm{kgm^{-3}}$ (water), $\eta=9.8\times10^{-6}\ \mathrm{Nsm^{-2}}$, $g=9.8\ \mathrm{ms^{-2}}$.
   2. Numerator: $2\times(1\times10^{-4})^2\times9000\times9.8=2\times10^{-8}\times9000\times9.8=1.764\times10^{-3}$.
   3. Denominator: $9\times9.8\times10^{-6}=8.82\times10^{-5}$. …

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9. AP EAPCET 2023Set ap-2023-05-23-FN1 markMCQ
   Q.A liquid of density of $3000\ \mathrm{kg\ m^{-3}}$ and coefficient of viscosity 0.1 Pas flows through a pipe with diameter 0.05 m with a velocity of $0.2\ \mathrm{ms^{-1}}$. The Reynolds number of the fluid is (A) 300 (B) 400 (C) 500 (D) 600

   ›Reveal solutionSolution

   Direct substitution into the Reynolds number formula gives $Re=300$. Answer: (A).

   Concept and Intuition

   The Reynolds number is a dimensionless quantity that predicts whether fluid flow will be laminar or turbulent, defined as the ratio of inertial forces to viscous forces in the flow: $Re = \dfrac{\rho v D}{\eta}$, where $\rho$ is fluid density, $v$ is flow speed, $D$ is the pipe diameter, and $\eta$ is the coefficient of viscosity.

   Step-by-Step Solution

   1. Given: $\rho = 3000\ \mathrm{kg\,m^{-3}}$, $\eta = 0.1\ \mathrm{Pa\,s}$, $D=0.05\ \mathrm{m}$, $v=0.2\ \mathrm{ms^{-1}}$.
   2. Compute numerator: $\rho v D = 3000 \times 0.2 \times 0.05 = 30$. …

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10. AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
    Q.A steel ball of radius 0.05 cm and density $7.8\ \mathrm{g\,cm^{-3}}$ is dropped into a tank of water. The terminal velocity of the steel ball is (Density of water = $1\ \mathrm{g\,cm^{-3}}$ and viscosity of water = 0.001 Pa s) (A) $3.42\ \mathrm{ms^{-1}}$ (B) $1.81\ \mathrm{ms^{-1}}$ (C) $5.11\ \mathrm{ms^{-1}}$ (D) $3.77\ \mathrm{ms^{-1}}$

    ›Reveal solutionSolution

    Stokes' law gives the terminal velocity of a small sphere falling through a viscous fluid; plugging in the given numbers yields $\approx 3.77\ \text{m/s}$.

    Concept and Intuition

    A small sphere falling through a viscous fluid quickly reaches a terminal (constant) velocity once the upward viscous drag (Stokes' force $6\pi\eta r v$) plus buoyancy exactly balance the downward weight. Solving that balance for $v$ gives Stokes' terminal-velocity formula, which depends on the square of the radius (small spheres reach terminal velocity fast and at low speed) and linearly on the density difference between the sphere and the fluid.

    Step-by-Step Solution

    1. Stokes' terminal velocity formula: $v = \dfrac{2r^2(\rho-\sigma)g}{9\eta}$, where $\rho$ = density of the ball, $\sigma$ = density of the fluid.
    2. Convert given values to SI: $r = 0.05\ \text{cm} = 5\times10^{-4}$ m, so $r^2 = 2.5\times10^{-7}\ \text{m}^2$; $\rho = 7800\ \text{kg/m}^3$, $\sigma=1000\ \text{kg/m}^3$, so $\rho-\sigma=6800\ \text{kg/m}^3$; $\eta = 0.001\ \text{Pa·s}$; $g=10\ \text{m/s}^2$. …

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11. AP EAPCET 2022Set ap-2022-07-12-AN1 markMCQ
    Q.Two capillary tubes A and B are connected in series. The length and radius of the bore of tube A are twice those of tube B. The ratio of the pressure difference across the tubes A and B is (A) 8:1 (B) 1:8 (C) 4:1 (D) 1:4

    ›Reveal solutionSolution

    In series flow, the same volume flow rate $Q$ passes through both capillaries, so the pressure drop across each follows Poiseuille's law: $\Delta P \propto L/r^4$.

    Concept and Intuition

    Poiseuille's law for laminar flow through a capillary states $Q = \dfrac{\pi \Delta P r^4}{8\eta L}$, i.e. $\Delta P = \dfrac{8\eta L Q}{\pi r^4}$. When two capillaries are connected in series, the same liquid (same $\eta$) flows through both at the same rate $Q$, so the pressure drop across each tube is proportional to $L/r^4$.

    Step-by-Step Solution

    1. $\Delta P \propto \dfrac{L}{r^4}$ for fixed $Q$ and $\eta$. …

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12. AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
    Q.An object of mass 10 kg is released from rest in a liquid. If the object moves a distance of 2 m while sinking in time duration of 1 s, then the mass of the liquid displaced by the submerged object is (Acceleration due to gravity = $10\ \mathrm{m\,s^{-2}}$) (A) 5 kg (B) 6 kg (C) 3 kg (D) 4 kg

    ›Reveal solutionSolution

    Find the net downward acceleration from the given kinematics, then use Newton's second law (weight minus buoyant force) to find the mass of displaced liquid.

    Concept and Intuition

    As the object sinks, two forces act (ignoring viscous drag, which the problem implies by giving constant acceleration): gravity pulling down, and the buoyant force (weight of displaced liquid) pushing up. From the observed uniformly accelerated motion, we can back out the net acceleration and hence the buoyant force.

    Step-by-Step Solution

    1. The object starts from rest and covers $s=2\,\text{m}$ in $t=1\,\text{s}$, so using $s=\tfrac12 a t^2$: $2 = \tfrac12 a (1)^2 \Rightarrow a=4\,\text{m/s}^2$.
    2. Newton's second law (downward positive): $mg - F_{\text{buoyant}} = ma$.
    3. $F_{\text{buoyant}} = m(g-a) = 10(10-4) = 60\,\text{N}$. …

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