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NEET
Exam
NEET
Subject
Physics
Chapter
Mechanical Properties of Fluids

NEET · Physics · Mechanical Properties of Fluids

Match List-I with List-II regarding a solid sphere of radius r...

Match List-I with List-II regarding a solid sphere of radius rr moving through a fluid.

List-IList-II
(A) Viscous force on a sphere moving at a constant given velocity(I) Proportional to r3r^3
(B) Terminal velocity of a solid sphere falling under gravity(II) Zero
(C) Viscous force on a solid sphere falling at its terminal velocity(III) Proportional to rr
(D) Viscous drag on a sphere moving in an ideal fluid(IV) Proportional to r2r^2

Choose the correct answer from the options given below:

Options

  1. A(A)-(III), (B)-(I), (C)-(IV), (D)-(II)
  2. B(A)-(III), (B)-(IV), (C)-(I), (D)-(II)
  3. C(A)-(IV), (B)-(III), (C)-(I), (D)-(II)
  4. D(A)-(I), (B)-(IV), (C)-(III), (D)-(II)

Correct answer

B. (A)-(III), (B)-(IV), (C)-(I), (D)-(II)

Step-by-step solution

According to Stokes' law, the viscous force on a spherical body is F=6πηrvF = 6\pi\eta r v.
For a constant given velocity vv, the viscous force F∝rF \propto r. Thus, (A) matches with (III).

The terminal velocity of a sphere falling under gravity is given by vt=2r2(ρ−σ)g9ηv_t = \dfrac{2r^2(\rho - \sigma)g}{9\eta}.
Therefore, terminal velocity vt∝r2v_t \propto r^2. Thus, (B) matches with (IV).

The viscous force at terminal velocity is F=6πηrvtF = 6\pi\eta r v_t.
Substituting vt∝r2v_t \propto r^2, we get F∝r(r2)=r3F \propto r(r^2) = r^3. Thus, (C) matches with (I).

An ideal fluid is non-viscous (η=0\eta = 0). Therefore, the viscous drag is zero. Thus, (D) matches with (II).

The correct matching is (A)-(III), (B)-(IV), (C)-(I), (D)-(II).

Answer: (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
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