A small sphere falling through a viscous liquid reaches terminal velocity v_t = (2r²·g·(ρ_s − ρ_l))/(9η). Drop spheres of known radius and density into a tall cylinder of glycerine; time the descent between two marked levels. Solve for η (coefficient of viscosity). Apply correction for finite cylinder diameter.
-- NCERT Physics Lab Manual Class 11, Part 6, p. 99Experiment 07 Viscosity Terminal
8 MCQs9-step worked example
Source: NCERT Experimental SkillsOfficial key: NTA-verifiedLast updated: 24 Sep 2026
Experiment 07 Viscosity Terminal, explained for NEET
The timing in this experiment starts too early. A ball dropped into the liquid does not fall at terminal velocity from the instant it is released — it accelerates first, and only after the drag has grown to match the net downward force does the speed settle. A stopwatch started at the liquid surface therefore measures a stretch of motion that is slower than terminal, and since the coefficient of viscosity comes out inversely proportional to the measured speed, the reported value of the coefficient comes out too large. That is why the upper mark on the tube is set well below the surface, as the procedure in the NCERT Physics Lab Manual Class 11, Part 6, page 99, specifies.
The physics behind the balance is short. A sphere falling through a viscous liquid carries three forces: its weight downward, the upthrust of the displaced liquid upward, and the viscous drag upward, which by Stokes' law grows in proportion to the speed. Drag rises as the ball speeds up until the three forces sum to zero. From then on the ball falls at a constant speed — the terminal velocity — and the balance condition rearranges into the working relation for the coefficient of viscosity in terms of the ball's radius, the two densities, and that constant speed.
For NEET, the experiment is worth about 0.4 questions a year, and the questions are usually one of three kinds: the force balance itself, a scaling question on how terminal velocity responds to a change of radius, and a procedural question about where the timing marks belong. The scaling is the one that catches repeaters — the speed depends on the square of the radius, not the first power.
Watch-out: never drop the upthrust term. Using the ball's own density in place of the density difference inflates the coefficient by a large factor, and the wrong answer usually appears as an option.
Can you answer these Experiment 07 Viscosity Terminal MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
A small sphere descends at its terminal velocity through a viscous liquid. Which force balance holds at that instant?
Show answer and why every option is right or wrong
Answer: C. Option C is correct: at terminal velocity the net force is zero, so the downward weight is matched by the two upward forces together — upthrust plus viscous drag — which is the balance the working relation of this experiment is built on (NCERT Physics Lab Manual Class 11, Part 6, page 99).
Why A is wrong: A is wrong because upthrust alone cannot balance the weight of a sphere denser than the liquid — if it did, the sphere would not sink at all.
Why B is wrong: B is wrong because it drops the upthrust term, the same omission that inflates the computed coefficient of viscosity when students use the sphere's density instead of the density difference.
Why D is wrong: D is wrong because it points the net force upward; an upthrust exceeding weight plus drag would drive the sphere back towards the surface, not let it fall steadily.
MCQ 2Easy RecallPractice
By Stokes' law, the viscous force on a small sphere moving slowly through a viscous liquid is proportional to which of the following?
Show answer and why every option is right or wrong
Answer: A. Option A is correct: Stokes' law makes the drag directly proportional to the speed (and to the radius, to the first power in each), which is exactly why the drag can grow to meet the net downward force and then hold the sphere at one steady speed (NCERT Physics Lab Manual Class 11, Part 6, page 99).
Why B is wrong: B is wrong because a speed-squared drag belongs to fast, turbulent flow; Stokes' law is the slow-flow, streamline result this experiment assumes.
Why C is wrong: C is wrong because the weight of the sphere goes as the cube of the radius, not the drag; confusing the two destroys the radius-squared scaling of terminal velocity.
Why D is wrong: D is wrong because a drag independent of speed could never rise to balance the weight, and no terminal velocity would ever be reached.
MCQ 3Easy RecallPractice
The coefficient of viscosity determined in this experiment is expressed in which SI unit?
Show answer and why every option is right or wrong
Answer: D. Option D is correct: the coefficient of viscosity has the SI unit pascal second (Pa s, equivalently kg m⁻¹ s⁻¹), which is the unit the result of this experiment is reported in (NCERT Physics Lab Manual Class 11, Part 6, page 99).
Why A is wrong: A is wrong because N m⁻¹ is force per unit length — the unit of surface tension, a different liquid property measured by a different experiment in this unit.
Why B is wrong: B is wrong because kg m⁻¹ s⁻² is pressure; viscosity carries one extra power of time, giving kg m⁻¹ s⁻¹.
Why C is wrong: C is wrong because the poise is the CGS unit, not the SI unit; the question asks for SI, and 1 Pa s = 10 poise.
MCQ 4Direct ApplicationPractice
Two spheres of the same material, of radii r and 2r, are allowed to fall through the same viscous liquid. The ratio of the terminal velocity of the smaller sphere to that of the larger one is:
Show answer and why every option is right or wrong
Answer: B. Option B is correct: terminal velocity is proportional to the square of the radius (the densities, g and the coefficient of viscosity are unchanged here), so doubling the radius multiplies the terminal velocity by 4, giving 1 : 4 (NCERT Physics Lab Manual Class 11, Part 6, page 99).
Why A is wrong: A is wrong because it treats terminal velocity as proportional to the first power of radius, reading the radius out of Stokes' drag alone and forgetting that the weight also scales with radius.
Why C is wrong: C is wrong because it uses the cube of the radius — that is how the weight of the sphere scales, not how the terminal velocity does.
Why D is wrong: D is wrong because it inverts the dependence: the larger sphere falls faster, not slower, in the same liquid.
MCQ 5Direct ApplicationPractice
A sphere of radius 1.0 × 10⁻³ m and density 7.8 × 10³ kg m⁻³ falls at a terminal velocity of 5.0 × 10⁻³ m s⁻¹ through a liquid of density 1.26 × 10³ kg m⁻³. Taking g = 9.8 m s⁻², the coefficient of viscosity of the liquid is closest to:
Show answer and why every option is right or wrong
Answer: D. Option D is correct: substituting into the terminal-velocity relation gives 2 × (1.0 × 10⁻³)² × (7.8 × 10³ − 1.26 × 10³) × 9.8 ÷ (9 × 5.0 × 10⁻³) ≈ 2.8 Pa s (NCERT Physics Lab Manual Class 11, Part 6, page 99).
Why A is wrong: A is wrong because it drops the factor 2 in the numerator, halving the correct result.
Why B is wrong: B is wrong because it uses the sphere's density 7.8 × 10³ kg m⁻³ in place of the density difference — the upthrust has been forgotten, and the answer is inflated.
Why C is wrong: C is wrong because it is a factor-of-ten slip in placing the decimal point while dividing by 9 × 5.0 × 10⁻³.
MCQ 6Direct ApplicationPractice
In this experiment the upper timing mark on the tube is placed a good distance below the liquid surface. The reason is:
Show answer and why every option is right or wrong
Answer: A. Option A is correct: the sphere is still accelerating just after it enters the liquid, and the terminal-velocity relation applies only to uniform motion, so the timed interval must begin after the speed has settled (NCERT Physics Lab Manual Class 11, Part 6, page 99).
Why B is wrong: B is wrong because the upthrust depends on the volume of liquid displaced by the sphere, which does not change with depth in this experiment.
Why C is wrong: C is wrong because the liquid is treated as incompressible and of uniform density throughout the tube; depth does not change the drag coefficient.
Why D is wrong: D is wrong because the weight of the sphere is fixed by its own mass and does not vary with how far it has fallen.
MCQ 7CalculationPractice
A student starts the stopwatch at the moment the sphere touches the liquid surface and stops it at a mark lower down, then uses that average speed as the terminal velocity. Compared with the true value, the coefficient of viscosity obtained is:
Show answer and why every option is right or wrong
Answer: B. Option B is correct: the early part of the fall is slower than terminal, so the average speed over the timed stretch comes out below the terminal velocity, and since the coefficient of viscosity varies inversely with that speed, the reported value is too large (NCERT Physics Lab Manual Class 11, Part 6, page 99).
Why A is wrong: A is wrong on the first step: including the accelerating phase lowers the average speed, it does not raise it, so the error runs the other way.
Why C is wrong: C is wrong because the distance is fixed by the marks while the time is lengthened by the slow start; they do not scale together, so the speed genuinely changes.
Why D is wrong: D is wrong because this is a one-sided procedural bias that shifts every reading in the same direction, not a random scatter.
MCQ 8CalculationPractice
A sphere falling through a viscous liquid crosses three successive marks that divide a 0.300 m stretch into three equal intervals of 0.100 m each. The measured times for the three intervals are 6.0 s, 5.0 s and 5.0 s in order. The terminal velocity that should be used is:
Show answer and why every option is right or wrong
Answer: C. Option C is correct: two equal times over two equal intervals is the signature of uniform motion, so the sphere had reached terminal velocity by the second mark, and 0.100 m ÷ 5.0 s = 2.0 × 10⁻² m s⁻¹ (NCERT Physics Lab Manual Class 11, Part 6, page 99).
Why A is wrong: A is wrong because the first interval is the one still contaminated by the accelerating phase — its longer time is the evidence, and it is the interval to discard.
Why B is wrong: B is wrong because averaging the accelerating stretch in with the uniform one drags the value below terminal velocity, which is the very bias the lower marks exist to avoid.
Why D is wrong: D is wrong because speeds over successive intervals are not added; the two equal readings are repeats of the same steady speed, not two contributions to it.
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How do you solve a Experiment 07 Viscosity Terminal question? A worked example
- 1
Given
• Radius of the sphere, r = 1.5 × 10⁻³ m• Density of the sphere material, ρ = 7.8 × 10³ kg m⁻³• Density of the liquid, σ = 1.3 × 10³ kg m⁻³• Distance between the two timing marks, both well below the liquid surface, s = 0.200 m• Time taken between the marks, t = 12.5 s• Acceleration due to gravity, g = 9.8 m s⁻² (problem-defined exact value)
- 2
Required
The coefficient of viscosity of the liquid.
- 3
Concept
Between the two marks the sphere is already moving uniformly, so the three forces on it sum to zero: weight downward, upthrust upward, and the Stokes' drag upward, which is proportional to the speed. Setting the net force to zero at that steady speed gives the working relation used in this experiment (NCERT Physics Lab Manual Class 11, Part 6, page 99).
- 4
Formula
Terminal velocity from the timed interval: v = s / t.
Force balance rearranged for the coefficient of viscosity: η = 2r²(ρ − σ)g ÷ (9v). - 5
Substitution
v = 0.200 m ÷ 12.5 s
η = 2 × (1.5 × 10⁻³ m)² × (7.8 × 10³ − 1.3 × 10³) kg m⁻³ × 9.8 m s⁻² ÷ (9 × v) - 6
Calculation
v = 1.60 × 10⁻² m s⁻¹
(1.5 × 10⁻³)² = 2.25 × 10⁻⁶ m²
ρ − σ = 6.5 × 10³ kg m⁻³
Numerator = 2 × 2.25 × 10⁻⁶ × 6.5 × 10³ × 9.8 = 2.87 × 10⁻¹
Denominator = 9 × 1.60 × 10⁻² = 1.44 × 10⁻¹
η = 2.87 × 10⁻¹ ÷ 1.44 × 10⁻¹ = 1.99
The numbers 2 and 9 in the relation are exact counting constants of the formula, and g = 9.8 m s⁻² is given as an exact problem-defined value; none of them limits the significant figures. The precision is set by the two-figure data (r and the density difference). - 7
Final answer
η ≈ 2.0 Pa s.
- 8
Common trap
Using ρ instead of (ρ − σ) — that is, forgetting the upthrust — would give 2 × 2.25 × 10⁻⁶ × 7.8 × 10³ × 9.8 ÷ 1.44 × 10⁻¹ ≈ 2.4 Pa s, about 20% high. The second trap is starting the stopwatch at the liquid surface rather than at the upper mark: the accelerating stretch lowers the measured speed and pushes η upward.
- 9
Similar NEET-style question
A sphere of radius 2.0 × 10⁻³ m and density 8.0 × 10³ kg m⁻³ takes 8.0 s to fall between two marks 0.240 m apart in a liquid of density 1.0 × 10³ kg m⁻³, the upper mark lying well below the surface. Taking g = 10 m s⁻² (exact), find the coefficient of viscosity of the liquid.
What to remember before solving Experiment 07 Viscosity Terminal questions
More in Experimental Skills: 6 exam traps and mistakes · 3 formulas · 1 question pattern from its other lessons.
Experiment 07 Viscosity Terminal questions from past NEET papers
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Sources
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