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NEET PHYSICSMechanical Properties of FluidMedium

Question

A small sphere of radius rr falls from rest in a viscous liquid. As a result, heat is produced due to the viscous force. The rate of production of heat when the sphere attains its terminal velocity is proportional to:

A

r3r^3

B

r2r^2

C

r5r^5

D

r4r^4

Step-by-Step Solution

  1. Terminal Velocity (vtv_t): When a sphere of radius rr falls through a viscous fluid, it attains a constant terminal velocity given by the formula vt=2r2(ρσ)g9ηv_t = \frac{2r^2(\rho - \sigma)g}{9\eta} . This implies that vtr2v_t \propto r^2.
  2. Viscous Force (FF): According to Stokes' Law, the viscous drag force acting on the sphere moving with terminal velocity is F=6πηrvtF = 6\pi \eta r v_t . Substituting the proportionality for vtv_t, we get Fr(r2)r3F \propto r(r^2) \propto r^3.
  3. Rate of Heat Production (Power): The rate of heat production is equal to the power (PP) dissipated by the viscous force, which is the product of the force and velocity: P=F×vtP = F \times v_t.
  4. Calculation: Substituting the proportionalities: P(r3)×(r2)P \propto (r^3) \times (r^2) Pr5P \propto r^5 Thus, the rate of heat production is proportional to the fifth power of the radius.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Mechanical Properties of Fluid. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSMechanical Properties of Fluidsphereradiusviscousliquidresult

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