Question

Consider a rectangular block of wood moving with a velocity $ {v_0} $ in a gas at temperature $ T $ and mass density $ \rho $. Assume the velocity is along the x-axis and the area of cross-section of the block perpendicular to $ {v_0} $ is $ A $. The drag force on the block is (where $ m $ is the mass of the gas molecule)
A) $ 4\rho A{v_0}\sqrt {\dfrac{{kT}}{m}} $
B) $ 2\rho A{v_0}\sqrt {\dfrac{{kT}}{{3m}}} $
C) $ \dfrac{{\rho A}}{{2{v_0}}}\sqrt {\dfrac{{kT}}{m}} $
D) $ \dfrac{{{v_0}}}{{\rho A}}\sqrt {\dfrac{{kT}}{{2m}}} $

Answer 1

Hint: The rectangular block of wood travelling in a gas will be imparted momentum from the molecules in front of them and then by the molecules on the back end. The gas will have a velocity associated with the RMS velocity of the gas.

Formula used: In this solution, we will use the following formula:
 $\Rightarrow {v_{RMS}} = \sqrt {\dfrac{{kT}}{m}} $ where $ {v_{RMS}} $ is the root mean square velocity of gas molecules at a temperature $ T $ that has a molecular mass $ m $
Force acting on a body $ F = \dfrac{{\Delta P}}{{\Delta t}} $ where $ \Delta P $ is the change in momentum and $ \Delta t $ is the change in time.

Complete step by step answer
Let $ n = $ no. of molecules per unit volume $ V $
The root mean square velocity or the RMS velocity is the average velocity with which gas molecules travel in a gas.
When the block is moving with speed $ {v_0} $, the relative velocity of the gas molecules on its front face will be $ {v_{RMS}} + {v_0} $. The momentum transferred to the block per a head-on collision will be
 $\Rightarrow p = 2m({v_{RMS}} + {v_0}) $
Now, only half of the molecules in front of the block will be moving towards the block. So, the number of collisions experienced in time $ \Delta t $ denoted by $ N $ will be the product of the relative velocity of the block and half the number of gas molecules that will be experienced by the cross-sectional area of the block. So,
 $\Rightarrow N = \dfrac{1}{2}({v_{RMS}} + {v_0})n\Delta tA $
Hence the total momentum transferred to the block in time $ \Delta t $ will be
 $\Rightarrow {P_1} = np $
 $\Rightarrow {P_1} = m{({v_{RMS}} + {v_0})^2}nA\Delta t $
This is the drag force experienced by the block in the front. At the back of the block, where the relative velocity of the gas molecules will be $ {v_{RMS}} + {v_0} $, the momentum transferred in time $ \Delta t $ will be
 $\Rightarrow {P_2} = m{({v_{RMS}} - {v_0})^2}nA\Delta t $
Hence the force experienced by the block will be
 $\Rightarrow F = \dfrac{{\Delta P}}{{\Delta t}} = \dfrac{{{P_1} - {P_2}}}{{\Delta T}} $
So, we can write
 $\Rightarrow F = m\left[ {{{\left( {{v_{RMS}} + {v_0}} \right)}^2} - {{\left( {{v_{RMS}} - {v_0}} \right)}^2}} \right]nA $
 $ \Rightarrow F = \,4mnA{v_{RMS}}{v_0} $
Since $ mn = \rho $, we can write
 $\Rightarrow F = \,4\rho A{v_{RMS}}{v_0} $
Now, we know that the RMS velocity of the gas will be
 $\Rightarrow {v_{RMS}} = \sqrt {\dfrac{{kT}}{m}} $,
Hence the net force acting on the block will be
 $\Rightarrow F = 4\rho A\sqrt {\dfrac{{kT}}{m}} {v_0} $
Hence the correct choice is option (A).

Note
Here the force that we have calculated is the average drag force experienced by the block as not all the gas molecules will be moving with $ {v_{RMS}} $ but it is an average velocity of the gas. Also, we can rule out option (C) and (D) directly as they do not have the correct dimensional formula of force.