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NCERT Exemplar · Q31

Q.Consider a rectangular block of wood moving with a velocity v0v_0 in a gas at temperature TT and mass density ρ\rho. Assume the velocity is along xx-axis and the area of cross-section of the block perpendicular to v0v_0 is AA. Show that the drag force on the block is 4ρAv0kTm4\rho A v_0 \sqrt{\dfrac{kT}{m}}, where mm is the mass of the gas molecule.

Punjab PsebLong· 5mImportance★★★★★est
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Concept: Kinetic Theory Explanation

Model the gas using the same simplified kinetic-theory picture used to derive gas pressure: treat the molecules as moving only along the three coordinate axes, with n/2n/2 of the molecules moving in the +x+x direction and n/2n/2 moving in the −x-x direction (where n=ρ/mn=\rho/m is the number density), each with the characteristic 1-D thermal speed vxv_x defined by

12mvx2=12kT⇒vx=kTm.\tfrac12 m v_x^2 = \tfrac12 k T \quad\Rightarrow\quad v_x = \sqrt{\frac{kT}{m}}.

Step 1 — Collisions on the front face. In the block's rest frame (block moving at v0v_0 along +x+x), the molecules that were moving in the −x-x direction (density n/2n/2, speed vxv_x) approach the front face with relative speed (vx+v0)(v_x+v_0). In an elastic collision with the (much heavier) block, each such molecule transfers momentum 2m(vx+v0)2m(v_x+v_0). The number of such collisions per unit area per unit time is (n/2)(vx+v0)(n/2)(v_x+v_0), so the pressure on the front face is

Pfront=n2(vx+v0)⋅2m(vx+v0)=nm(vx+v0)2.P_{\text{front}} = \frac{n}{2}(v_x+v_0)\cdot 2m(v_x+v_0) = nm(v_x+v_0)^2.

Step 2 — Collisions on the back face. Molecules originally moving in the +x+x direction (density n/2n/2, speed vxv_x) strike the back face with relative speed (vx−v0)(v_x-v_0), transferring momentum 2m(vx−v0)2m(v_x-v_0) each:

Pback=n2(vx−v0)⋅2m(vx−v0)=nm(vx−v0)2.P_{\text{back}} = \frac{n}{2}(v_x-v_0)\cdot 2m(v_x-v_0) = nm(v_x-v_0)^2. …

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