Sarthaks Test
0 votes
349 views
in Physics by (8.6k points)

Consider a rectangular block of wood moving with a velocity v0 in a gas at temperature T and mass density ρ. Assume the velocity is along x-axis and the area of cross-section of the block perpendicular to v0 is A. Show that the drag force on the block is 4ρAv0√(kT/m), where m is the mass of the gas molecule.

1 Answer

0 votes
by (13.6k points)
selected by
 
Best answer

n = no. of molecules per unit volume

vrms = rms speed of gas molecules

When block is moving with speed vo, relative speed of molecules w.r.t.

front face = v + vo

Coming head on, momentum transferred to block per collission

= 2 m (v + vo), where m = mass of molecule.

No. of collission in time Δt = 1/2(v + vo)nΔtA, where A = area of cross section of block and factor of 1/2 appears due to particles moving towards block.

Momentum transferred in time Δt = m(v + vo)2AΔt  from front surface

Similarly momentum transferred in time Δt = m(v -vo)2nAΔt  from back surface

Net force (drag force) = mnA [(v + vo) - (v - vo)2] from front

= mnA (4vvo) = (4mnAv)vo)

= (4ρAv)vo 

We also have (1/2)mv2 = (1/2)kT (v - is the velocity along c-axis)

Therefore, v = √kT/m

Thus drag = 4ρA√(kT/m)vo 

Related questions

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...