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 4ρ 0 kT Av m , where m is the mass of the gas molecule

This is a long answer type question as classified in NCERT Exemplar

Consider the diagram

let n =number of molecules per unit volume

V_rms= rms speed of gas molecule

When block is moving with speed v_o, relative speed of molecules w.r.t front face =v+v_o

Coming head on, momentum transferred to block per collision =2m (v+v_o)

Number of collisions in time $?t$ = $\frac{1}{2}$ (v+v_o)n $?t$ A  where A is the area of cross section.

So momentum transferred in time $?t$ =m (v+v_o)²nA $?t$   this is from front surface

Similarly momentum transferred in time $?t$ = m (v-v_o)²nA $?t$ ) this is from back surface

Drag force = mnA (v+v_o)²- (v-v_o)²)

= mnA (4w_o)=4mnAvv_o

= 4 $\rho A$ vv_o

So $\rho$ =mn/V=M/

...more

This is a long answer type question as classified in NCERT Exemplar

Consider the diagram

let n =number of molecules per unit volume

V_rms= rms speed of gas molecule

When block is moving with speed v_o, relative speed of molecules w.r.t front face =v+v_o

Coming head on, momentum transferred to block per collision =2m (v+v_o)

Number of collisions in time $?t$ = $\frac{1}{2}$ (v+v_o)n $?t$ A  where A is the area of cross section.

So momentum transferred in time $?t$ =m (v+v_o)²nA $?t$   this is from front surface

Similarly momentum transferred in time $?t$ = m (v-v_o)²nA $?t$ ) this is from back surface

Drag force = mnA (v+v_o)²- (v-v_o)²)

= mnA (4w_o)=4mnAvv_o

= 4 $\rho A$ vv_o

So $\rho$ =mn/V=M/V

If v = velocity along x axis

Then we can write KE= ½ mv²=1/2 K_BT

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<p><span data-teams="true">This is a long answer type question as classified in NCERT Exemplar</span></p><p>Consider the diagram</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1745555811phpT1CtOp_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1745555811phpT1CtOp.jpeg" alt="" width="319" height="139"></picture></div></div><p>let n =number of molecules per unit volume</p><p>V_rms= rms speed of gas molecule</p><p>When block is moving with speed v_o,relative speed of molecules w.r.t front face =v+v_o</p><p>Coming head on, momentum transferred to block per collision =2m (v+v_o)</p><p>Number of collisions in time<span contenteditable="false"> <math> <mo>? </mo> <mi>t</mi> </math> </span> =<span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> </math> </span> (v+v_o)n<span contenteditable="false"> <math> <mo>? </mo> <mi>t</mi> </math> </span>A&nbsp; where A is the area of cross section.</p><p>So momentum transferred in time <span contenteditable="false"> <math> <mo>? </mo> <mi>t</mi> </math> </span>=m (v+v_o)²nA<span contenteditable="false"> <math> <mo>? </mo> <mi>t</mi> </math> </span>&nbsp; this is from front surface</p><p>Similarly momentum transferred in time <span contenteditable="false"> <math> <mo>? </mo> <mi>t</mi> </math> </span>= m (v-v_o)²nA<span contenteditable="false"> <math> <mo>? </mo> <mi>t</mi> </math> </span>) this is from back surface</p><p>Drag force = mnA (v+v_o)²- (v-v_o)²)</p><p>= mnA (4w_o)=4mnAvv_o</p><p>= 4<span contenteditable="false"> <math> <mi>ρ</mi> <mi>A</mi> </math> </span>vv_o</p><p>So <span contenteditable="false"> <math> <mi>ρ</mi> </math> </span>=mn/V=M/V</p><p>If v = velocity along x axis</p><p>Then we can write KE= ½ mv²=1/2 K_BT</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1745558247phpGbWMad_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1745558247phpGbWMad.jpeg" alt="" width="285" height="86"></picture></div></div>

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