Terminal Velocity

Answer: B. Terminal velocity is reached when the sphere stops accelerating, meaning the net force is zero — gravity is exactly balanced by buoyancy plus viscous drag (NCERT Class 11 Physics, Chapter 9, page 192).

Why A is wrong: A is wrong because weight is only one of three forces acting. At terminal velocity, weight is balanced by buoyancy + drag, not unbalanced.

Why C is wrong: C is wrong because buoyant force alone does not balance weight — viscous drag also contributes.

Why D is wrong: D is wrong because drag alone does not balance weight; buoyancy also acts upward. The net of all three forces is zero.

Answer: A. A is correct. At terminal velocity the weight (less buoyancy) balances the viscous drag: (4/3)πr³(ρ − σ)g = 6πηrv, so v = 2r²(ρ − σ)g/(9η). v goes as r², and doubling r quadruples it.

Why B is wrong: B is wrong because 2v assumes terminal velocity goes as r. The weight grows as r³ while the drag grows only as r, leaving v ∝ r².

Why C is wrong: C is wrong because 8v tracks only the weight, which goes as r³, and forgets that the drag also grows, in proportion to r.

Why D is wrong: D is wrong because v/2 has larger drops falling SLOWER. Larger drops fall faster, which is why big raindrops hit harder than drizzle.

Answer: C. Stokes' law (F = 6πηrv) applies to slow, laminar flow around a sphere — the low Reynolds number regime. At high speeds, turbulence sets in and the drag law changes (NCERT Class 11 Physics, Chapter 9, page 188).

Why A is wrong: A is wrong because at high speeds the flow becomes turbulent and Stokes' linear drag law breaks down. The law requires slow (low Reynolds number) flow.

Why B is wrong: B is wrong because Stokes' law assumes an incompressible Newtonian fluid in laminar flow, the opposite of compressible turbulent conditions.

Why D is wrong: D is wrong because Stokes' law is derived specifically for a sphere. Non-spherical bodies have different drag coefficients.

Answer: C. Terminal velocity is proportional to r². Doubling r gives (2r)²/r² = 4 times the original terminal velocity, so the answer is 4v_t (NCERT Class 11 Physics, Chapter 9, page 192).

Why A is wrong: A is wrong because this assumes v_t ∝ r (linear). The correct dependence is v_t ∝ r², so doubling r quadruples v_t, not doubles it. This is a common mistake — Stokes drag is linear in r, but gravity minus buoyancy scales as r³, making v_t ∝ r².

Why B is wrong: B is wrong because 8v_t would require v_t ∝ r³. Since v_t = (2r²g(ρ_s − ρ_f))/(9η), the dependence is r², not r³.

Why D is wrong: D is wrong because terminal velocity increases (not decreases) with radius. A larger sphere has more gravitational force relative to drag.

Answer: B. From v_t = (2r²g(ρ_s − ρ_f))/(9η), terminal velocity is inversely proportional to η. With all other quantities identical, v_A/v_B = η_B/η_A = 3η/η = 3. So v_A : v_B = 3 : 1 (NCERT Class 11 Physics, Chapter 9, page 192).

Why A is wrong: A is wrong because this reverses the inverse relationship. Higher viscosity means lower terminal velocity, so the sphere in the less viscous liquid (A) is faster, giving 3 : 1 not 1 : 3.

Why C is wrong: C is wrong because v_t ∝ 1/η (inverse first power), not 1/η². Squaring the viscosity ratio is not justified by the terminal velocity formula.

Why D is wrong: D is wrong because v_t ∝ 1/η gives a ratio of 3 : 1, not 9 : 1. The ratio 9 : 1 would apply if v_t ∝ 1/η².

Answer: A. v_t ∝ (ρ_s − ρ_f). Originally (ρ_s − ρ_f) = (2.5 − 1.5) × 10³ = 1.0 × 10³ kg/m³. New: (2.5 − 2.0) × 10³ = 0.5 × 10³ kg/m³. Ratio = 0.5/1.0 = 1/2. Terminal velocity halves (NCERT Class 11 Physics, Chapter 9, page 192).

Why B is wrong: B is wrong because increasing the fluid density reduces (ρ_s − ρ_f), which decreases terminal velocity, not increases it.

Why C is wrong: C is wrong because terminal velocity depends on the density difference (ρ_s − ρ_f). Changing ρ_f changes this difference and hence v_t.

Why D is wrong: D is wrong because the density difference halved (from 1.0 × 10³ to 0.5 × 10³), so v_t halves — not reduces to one-fifth.

Answer: B. Initially, drag is zero and the sphere accelerates. As speed increases, drag (∝ v) increases, reducing net force and hence acceleration. The v–t curve is concave-down, asymptotically approaching terminal velocity — a curve with decreasing slope leveling off (NCERT Class 11 Physics, Chapter 9, page 192).

Why A is wrong: A is wrong because a straight line implies constant acceleration. Here, viscous drag increases with velocity, continuously reducing acceleration. The slope decreases over time.

Why C is wrong: C is wrong because velocity does not decrease after reaching a maximum. Once terminal velocity is reached, the sphere continues at that constant speed — there is no deceleration phase.

Why D is wrong: D is wrong because the sphere starts from rest (v = 0 at t = 0) and must accelerate to reach terminal velocity. A horizontal line would mean it was already at terminal velocity from the start.

Answer: D. v_t = (2r²g(ρ_s − ρ_f))/(9η) = (2 × (1.0 × 10⁻³)² × 9.8 × (7.0 × 10³))/(9 × 9.8) = (2 × 10⁻⁶ × 9.8 × 7.0 × 10³)/(9 × 9.8). The 9.8 cancels: = (2 × 10⁻⁶ × 7.0 × 10³)/9 = (14.0 × 10⁻³)/9 ≈ 1.56 × 10⁻³ m/s ≈ 1.5 × 10⁻³ m/s. Here g = 9.8 m/s² is treated as exact (problem-defined) and does not limit significant figures (NCERT Class 11 Physics, Chapter 9, page 192).

Why A is wrong: A is wrong because it is off by a factor of 10. A careful substitution gives the numerator as 14.0 × 10⁻³ and denominator 9, yielding ~1.56 × 10⁻³, not 10⁻².

Why B is wrong: B is wrong because 10⁻¹ m/s is far too large. With η = 9.8 Pa·s (a highly viscous oil) and a millimeter-sized sphere, the terminal velocity is of order 10⁻³ m/s.

Why C is wrong: C is wrong because it is off by a factor of 10 in the other direction. The r² term gives 10⁻⁶, multiplied by density difference 7.0 × 10³ gives 7.0 × 10⁻³ in the numerator — leading to ~10⁻³ order, not 10⁻⁴.