The sphere of diameter 0.02 m falls in a fluid of kinematic viscosity of 10 stokes with the terminal velocity of 0.02 m/s. What is the value of the coefficient of drag on the falling sphere?

This question was previously asked in
SSC JE CE Previous Year Paper 7 (Held On: 27 Jan 2018 Morning)
View all SSC JE CE Papers >
  1. 40
  2. 60
  3. 80
  4. 100

Answer (Detailed Solution Below)

Option 2 : 60
Free
Building Materials for All AE/JE Civil Exams Mock Test
15 K Users
20 Questions 20 Marks 20 Mins

Detailed Solution

Download Solution PDF

Concept:

Drag Force (Fd) is applied to the body when it moves through and liquid media.

Drag is defined as forceful pull experienced by the flat plate while the fluid flows over it.

Pressure drag comes from the eddying motions that are set up in the fluid by the passage of the body; This drag is associated with the formation of the wake in the flow.

Frictional drag comes from friction between the fluid and the surfaces over which it is flowing.

\({{\rm{F}}_{\rm{d}}} = \frac{{{{\rm{C}}_{\rm{d}}} \times {\rm{\rho }} \times {\rm{A}} \times {{\rm{V}}^2}}}{2}\)

\({\rm{Drag\;Coefficient}},{\rm{\;}}{{\rm{C}}_{\rm{d}}} = \frac{{24}}{{{{\rm{R}}_{\rm{e}}}}}\) when Re < 0.2

Drag Coefficient, Cd = \({24\over R_e}({1+{3\over 16R_e}})\) when 0.2 < Re < 5

Where Cd is the drag coefficient and Re is Reynold’s Number.

Calculation:

\({{\rm{R}}_{\rm{e}}} = \frac{{{\rm{V}} \times {\rm{d}}}}{{\rm{\nu }}} = \frac{{0.02 \times 0.02}}{{10 \times {{10}^{ - 4}}}} = 0.4\)

\({{\rm{C}}_{\rm{d}}} = \frac{{24}}{{0.4}} = 60\)

Note:

the The given answer is as per official  exam of SSC JE.

Latest SSC JE CE Updates

Last updated on May 28, 2025

-> SSC JE notification 2025 for Civil Engineering will be released on June 30. 

-> Candidates can fill the SSC JE CE application from June 30 to July 21.

-> The selection process of the candidates for the SSC Junior Engineer post consists of Paper I, Paper II, Document Verification, and Medical Examination.

-> Candidates who will get selected will get a salary range between Rs. 35,400/- to Rs. 1,12,400/-.

-> Candidates must refer to the SSC JE Previous Year Papers and SSC JE Civil Mock Test, SSC JE Electrical Mock Test, and SSC JE Mechanical Mock Test to understand the type of questions coming in the examination.

-> The Staff Selection Commission conducts the SSC JE exam to recruit Junior Engineers in different disciplines under various departments of the Central Government.

Get Free Access Now
Hot Links: teen patti gold downloadable content teen patti rich teen patti boss