Start Learning Free

Before You Begin

Question #42121

Difficulty: 8/10 | 🟢 Free

The Question

A solid sphere of initial radius r0r_0 and density ρs\rho_s is falling in a viscous liquid of density ρl\rho_l (where ρs>ρl\rho_s > \rho_l). The system is slowly heated such that the sphere's radius expands as r(t)=r0(1+αt)r(t) = r_0(1 + \alpha t) and the liquid's viscosity decreases as η(t)=η0(1−βt)\eta(t) = \eta_0(1 - \beta t), where α\alpha and β\beta are small positive constants. Assuming the sphere continuously maintains its quasi-steady terminal velocity at each instant, what is the acceleration of the sphere at t=0t = 0?

Options

  • A.

    2gr029η0[β(ρs−ρl)−α(ρs+2ρl)]\frac{2g r_0^2}{9 \eta_0} \left[ \beta (\rho_s - \rho_l) - \alpha (\rho_s + 2\rho_l) \right]

  • B.

    2gr029η0(ρs−ρl)(2α+β)\frac{2g r_0^2}{9 \eta_0} (\rho_s - \rho_l) (2\alpha + \beta)

  • C.

    2gr029η0[β(ρs−ρl)−α(ρs−2ρl)]\frac{2g r_0^2}{9 \eta_0} \left[ \beta (\rho_s - \rho_l) - \alpha (\rho_s - 2\rho_l) \right]

  • D.

    2gr029η0[β(ρs−ρl)−3αρs]\frac{2g r_0^2}{9 \eta_0} \left[ \beta (\rho_s - \rho_l) - 3\alpha \rho_s \right]

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Viscosity and Stokes' Law|Thermal Expansion of Area and Volume|Differential Calculus

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to apply dynamic equilibrium to a system with time-varying parameters.

The lethal trap lies in blindly differentiating the standard terminal velocity formula vT=2r2g9η(ρs−ρl)v_T = \frac{2r^2 g}{9\eta}(\rho_s - \rho_l).

Students who do this implicitly assume the sphere's density remains constant while its volume expands, which violates the conservation of mass and leads directly to Trap Option B.

By forcing the mass to be constant (M=43πr03ρsM = \frac{4}{3}\pi r_0^3 \rho_s) while the displaced fluid volume changes (V=43πr(t)3V = \frac{4}{3}\pi r(t)^3), the buoyancy force becomes a dynamic function of time, leading to a completely different transient acceleration.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterMechanical Properties of Fluids - Class 11 Part 2
ConceptViscosity and Stokes' Law
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree