A horizontal force of 0.5N is required to move a metal plate of area 10⁻² m² with a velocity of 3×10⁻² m/s, when it rests on 0.5×10⁻³ m thick layer of glycerin. Find the coefficient of viscosity of glycerin.

A horizontal force of 0.5N is required to move a metal plate of area 10⁻² m² with a velocity of 3×10⁻² m/s, wh

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Q. A horizontal force of 0.5N is required to move a metal plate of area 10⁻² m² with a velocity of 3\times10⁻² m/s, when it rests on 0.5\times10⁻³ m thick layer of glycerin. Find the coefficient of viscosity of glycerin.

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Concept understanding — Viscosity

Imagine a thick, golden stream of honey being poured from a jar. It moves slowly, in a smooth, lazy ribbon. Now picture water from a tap — it rushes out, splashes, and flows freely. The difference you see is viscosity. It is the fluid's internal friction, its reluctance to let one layer slide past another. Honey has high viscosity; water has low viscosity.

At the microscopic level, a fluid is made of molecules that are constantly jostling and colliding. In a more viscous fluid, the molecules are either larger, more tangled, or have stronger attractive forces between them. When you try to make one layer of the fluid move relative to the layer next to it, these molecular interactions resist that motion. That resistance is what you feel as "thickness" or "stickiness."

Now, let's make this precise. Consider a fluid trapped between two large parallel plates, one stationary and one moving at a steady speed. The layer of fluid right next to the moving plate sticks to it and moves with the same speed. The layer next to the stationary plate stays at rest. In between, the speed of the fluid changes gradually from zero to the plate's speed. This is called a velocity gradient.

The force required to keep the top plate moving is the viscous force. Experiments show that this force $F$ is proportional to two things: the area $A$ of the plate in contact with the fluid, and the velocity gradient $\frac{dv}{dx}$ (how quickly the speed changes as you move perpendicular to the flow). The constant of proportionality is called the coefficient of viscosity, denoted by $\eta$ (eta).

F = \eta \, A \, \frac{dv}{dx}

Here, $F$ is the viscous force that opposes relative motion between adjacent fluid layers. $A$ is the area of contact between those layers. $\frac{dv}{dx}$ is the velocity gradient — the rate at which velocity changes with distance perpendicular to the flow. And $\eta$ is the viscosity of the fluid, a property that depends on the fluid itself and its temperature (heating a fluid usually lowers its viscosity). …

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Newton's law of viscosity applied to a plate sliding over a thin liquid layer.

By Newton's law of viscous flow, $F = \eta A\dfrac{v}{l}$ (velocity gradient $\approx v/l$ across the thin layer of thickness $l$ ): …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 19 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. Which among the following forces is developed due to resistance of a fluid flow?$
(a) Viscous force
(b) Inertial force
(c) Gravitational force
(d) Pressure force
›Reveal solutionSolution

Viscosity is the fluid property responsible for a dragging/frictional force between adjacent layers of fluid moving at different speeds -- this is called the viscous force, and it directly resists flow.

Viscosity arises due to internal friction between adjacent layers of a fluid that are moving with different velocities relative to each other. The tangential force per unit area needed to maintain this relative motion (or the resisting force it produces) is the viscous force, given by Newton's law of viscosity:

F = eta A (dv/dx)

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. What is the dimensional formula of the coefficient of viscosity?$
(a) ML^-2T^-1
(b) ML^-1T^-2
(c) ML^-1T^-1
(d) ML^-2T^-2
›Reveal solutionSolution

The coefficient of viscosity has dimensional formula ML^-1T^-1 (SI unit: Pa·s = kg m^-1 s^-1).

From Newton's law of viscous flow, the viscous force between two layers of fluid is F = η A (dv/dx), where A is the area of contact and dv/dx is the velocity gradient. Solving for η: η = F / [A · (dv/dx)]. Now substitute dimensions: [F] = MLT^-2, [A …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The dimensions of the coefficient of viscosity are$
(a) [MLT^-1]
(b) [ML^2T^-2]
(c) [M^-1LT^-1]
(d) [ML^-1T^-1]
›Reveal solutionSolution
Coefficient of viscosity has dimensions [ML^-1 T^-1]. Answer (D).

Newton's law of viscosity: F = eta A (dv/dx), so eta = F/(A (dv/dx)).

Dimensions:

F -> [MLT^-2]

A -> [L^2] …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. Viscous force$
(a) is a non-conservative force
(b) depends on the velocity
(c) acts opposite to the fluid flow
(d) all of these
›Reveal solutionSolution
All three properties hold, so the answer is (D) all of these.

Viscous force between fluid layers (F = eta A dv/dx):

Non-conservative: it converts mechanical energy into heat and depends on the path, so it is non-conservative.

Velocity-dependent: it is proportional to the velocity gradient (relative velocity of the layers). …
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CBSE 2025Set ANNUAL1 markMCQ
$Q. With increase of temperature viscosity of maximum liquids (A) decreases (B) increases (C) remains same (D) none of these$
›Reveal solutionSolution

Viscosity of most liquids decreases as temperature increases.

Viscosity in a liquid arises mainly from intermolecular cohesive forces between neighbouring layers of the liquid. As temperature rises, molecules gain kinetic energy, move about more freely, and the average intermolecular spacing increases — weakening the cohesive forces that resist relative motion between layers. This reduces the liquid's visc …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The frictional resistance for fluids in motion is$
(a) inversely proportional to the square of the surface area of contact
(b) inversely proportional to the surface area of contact
(c) proportional to the square of the surface area of contact
(d) proportional to the surface area of contact
›Reveal solutionSolution

Newton's law of viscosity gives the viscous force as F = eta A (dv/dx) -- directly proportional to the area of contact between the fluid layers, not inversely.

For a fluid in motion, adjacent layers moving at different speeds experience an internal frictional (viscous) force opposing relative sliding, given by Newton's law of viscosity:

F = eta * A * (dv/dx)

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Match the following - Column A item: Coefficient of viscosity. Pick the matching relation from Column B.$
(a) F.V (Force . Velocity)
(b) T is proportional to sqrt(l)
(c) eta is proportional to 1/(dv/dx)
(d) mu_s = tan(theta)
(e) Y is proportional to 1/l
(f) v^2/r
›Reveal solutionSolution

Newton's law of viscous flow gives the coefficient of viscosity eta in terms of the velocity gradient dv/dx.

By Newton's law of viscosity, the tangential (shearing) stress F/A between adjacent fluid layers is proportional to the velocity gradient perpendicular to the flow: F/A = eta (dv/dx), so eta = (F/A)/(dv/dx). For a fixed shearing stress, a fluid with larger viscosity eta produces a smaller velocity gradient dv/dx between its layers — …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. If two small spheres of radii r/2 and r are moving in fluid with equal constant velocity, then the ratio of viscous force acting on them will be (A) 1:2 (B) 2:1 (C) 1:4 (D) 4:1$
›Reveal solutionSolution

Viscous force ratio for radii $r/2$ and $r$ is $1:2$ , since $F\propto r$ .

Stokes' law: $F=6\pi\eta rv$ , so at the same velocity $v$ and viscosity $\eta$ , $F\propto r$ .

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Dimension of coefficient of viscosity is (A) [ML^-1T^-1] (B) [MLT] (C) [MLT^-2] (D) [ML^-2T^-2]$
›Reveal solutionSolution

Coefficient of viscosity has dimension $[ML^{-1}T^{-1}]$ .

From $F=\eta A\dfrac{dv}{dx}$ : $\eta=\dfrac{F}{A\,(dv/dx)}$ .

Dimensions: $[F]=[MLT^{-2}]$ , $[A]=[L^2]$ , $[dv/dx]=[LT^{-1}]/[L]=[T^{-1}]$ .

…

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CBSE 2024Set SET-AP55001 mark
$Q. The viscosity of gases ________ on increasing temperature.$
›Reveal solutionSolution

Gas viscosity increases with temperature (opposite to the trend in liquids), because it arises from momentum transfer between molecules moving between adjacent layers, which intensifies as the molecules move faster.

In a gas, viscosity is caused by molecules diffusing between layers moving at different velocities relative to each other, carrying momentum with them and thereby exerting a dragging (frictional) effect between the layers. As temperature increases, molecular speeds increase (kinetic theory: average speed ∝ √T), so this momentum-transfer mechanism becomes more effective, and viscosity increases — roughly η ∝ √ …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. With an increase in temperature, the viscosity of liquid and gas, respectively will:$
(a) decrease and increase
(b) increase and increase
(c) decrease and decrease
(d) increase and decrease
›Reveal solutionSolution

Liquid viscosity is dominated by intermolecular cohesive forces, which weaken as temperature rises, so liquid viscosity decreases with temperature. Gas viscosity is dominated by molecular momentum transfer, which increases with temperature, so gas viscosity increases with temperature.

In a liquid, viscosity is caused mainly by the cohesive forces between closely-packed molecules. As temperature increases, molecules gain kinetic energy and move more freely, weakening these cohesive forces, so the liquid flows more easily — viscosity decreases.

…

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CBSE 2023Set ANNUAL1 markMCQ
$Q. The dimensional formula for coefficient of viscosity is:$
(a) ML^-2T^-2
(b) MLT^-2
(c) ML^-1T^-2
(d) ML^-1T^-1
›Reveal solutionSolution

Viscosity is defined by Newton's law of viscous flow, F = eta A (dv/dx); solving for eta and substituting dimensions gives ML^-1T^-1.

By Newton's law of viscosity, the viscous force between two liquid layers is

F = eta A (dv/dx)

where A is the area of contact and dv/dx is the velocity gradient (rate of change of velocity with distance).

Rearranging,

eta = F / [A x (dv/dx)]

Now write the dimensions of each quantity: …

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Q.A horizontal force of 0.5N is required to move a metal plate of area 10⁻² m² with a velocity of 3×10⁻² m/s, when it rests on 0.5×10⁻³ m thick layer of glycerin. Find the coefficient of viscosity of glycerin.

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