Consider a rectangular block of wood moving with a velocity `v_(0)` in a gas at temperature T and mass density p. Assume the velocity is along x-axis and the are of cross-section of the block perpendicular to `v_(0)` is A. show that the drag force on the block is `4rAv_(0)sqrt((kT)/(m))` where, m is the mass of the gas molecule.

Options

A. `4pAv_(0)sqrt((kT)/(m))`
B. `2pAv_(0)sqrt((kT)/(3m))`
C. `(pA)/(2v_(0))sqrt((kT)/(m))`
D. `(v_(0))/(pA)sqrt((kT)/(2m))`

Answer

A

Explanation

Speed of wooden block `=v_(0)`
Cross-sectional area of block = A
Temperature of the gas= T
Let n be the number of molecules per unit volume. Rms speed of gas molecules `=v_(rms).`
Relative speed of the molecule with respect to front face of the block `= v+v_(0)`
In head on collision, momentum transferred to block per collision `=2m (v+v_(0))` when m is the mass of molecule.
Number of collision in time `Deltat,=1/2(v+v_(0))nDeltatA.` (factor of `1/2` appears due to particles moving towards block)
hence, momentum transferred in time `Deltat,=m (v+v_(0))^(2)nADeltat`
Net force `(Deltap)/(Deltat)=m[v+v_(0))^(2)-(v-v_(0))^(2)"]"nA`
`=mnA(4v v_(0))=(4rhoAv)_(0)=(4rhoAv)v_(0)(becausemn=rho)" "...(i)`
Also, `1/2mv^(2)=1/2kT" "("v along x-axis")`
`thereforev=sqrt((KT)/(m))`
`therefore"Net force"=4rhoAsqrt((KT)/(m))v_(0)." "["Using (i)"]`
This is required drag force.