A spherical body of radius r and density σ falls freely through a viscous liquid having density ρ and viscosity η and attains a terminal velocity v₀. Estimated maximum error in the quantity η is: (Ignore errors associated with σ, ρ and g, gravitational acceleration) (a) 2[(Δ r)/(r) - (Δ v₀)/(v₀)] (b) 2[(Δ r)/(r) + (Δ v₀)/(v₀)] (c) (2Δ r)/(r) + (Δ v₀)/(v₀) (d) 2(Δ r)/(r) - (Δ v₀)/(v₀)

Answer: (c). Terminal velocity (Stokes' law): v₀ = (2r²(σ - ρ)g)/(9η). So η = (2r²(σ - ρ)g)/(9v₀) ∝ (r²)/(v₀) (ignoring errors in σ, ρ, g). (Δ η)/(η) = 2(Δ r)/(r) + (Δ v₀)/(v₀) So the answer is (c) (2Δ r)/(r) + (Δ v₀)/(v₀).

A spherical body of radius r and density σ falls freely through a… | JEE Main 2026 (21 Jan, Shift 2)

Q32JEE Main 2026 · 21 JAN · SHIFT 2Medium

Errors in Measurement

A spherical body of radius $r$ and density $\sigma$ falls freely through a viscous liquid having density $\rho$ and viscosity $\eta$ and attains a terminal velocity $v_0$ . Estimated maximum error in the quantity $\eta$ is: (Ignore errors associated with $\sigma,\ \rho$ and $g$ , gravitational acceleration)

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Answer: (c)

Terminal velocity (Stokes' law): $v_0 = \frac{2r^2(\sigma - \rho)g}{9\eta}$ .
So $\eta = \frac{2r^2(\sigma - \rho)g}{9v_0} \propto \frac{r^2}{v_0}$ (ignoring errors in $\sigma, \rho, g$ ).
$\frac{\Delta \eta}{\eta} = 2\frac{\Delta r}{r} + \frac{\Delta v_0}{v_0}$
So the answer is (c) $\frac{2\Delta r}{r} + \frac{\Delta v_0}{v_0}$ .

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