(a) Pressure decreases as one ascends the atmosphere. If the density of air is ρ, what is the change in pressure dp over a differential height dh?

(b) Considering the pressure p to be proportional to the density, find the pressure p at a height h if the pressure on the surface of the earth is p0.

(c) If Po = 1.03×105 N m-2, ρo = 1.29 kg m-3 and g = 9.8 m s-2, at what height will the pressure drop to (1/10) the value at the surface of the earth?

(d) This model of the atmosphere works for relatively small distances. Identify the underlying assumption that limits the model.

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    Answered by

    Payal Gupta | Contributor-Level 10

    4 months ago

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

    (a) consider a horizontal parcel of air with cross section A and height dh

    Let the pressure on the top surface and bottom surface be P and p+dp. If the parcel is in equilibrium , then the net upward force must be balanced by the weight

    (P+dP)-PA=- ρ A d h × g

    dP= - ρ g d h

    negative sign shows that pressure decreases with height.

    (b) let ρ o be the density of air on the surface of earth.

    As per question , pressure  density

    P P o = ρ ρ 0

    ρ = ρ o P o P

    dP= - ρ o g P o P d h

    d P p = - ρ o g d h P o

    P o P d P P = - ρ o g P o 0 h d h

    In P P o = - ρ o g h P o

    P=Poe(- ρ o g h P o )

    (c) as P =Po e - ρ o g h P o

    in P P o = - ρ o g h P o

    p=1/10 Po

    in( 1 10 P o P o ) =- - ρ o g h P o

    in1/10 =- ρ o g P o h ρ 0

    h=- P o ρ o g in1/10= - P o P o g i n ( 10 ) -1= P o P o g i n ( 10 )

    = P o P o g × 2.303

    = 1.013 × 10 5 1.22 × 9.8 × 2.303 = 0.16 × 10 5 m

    = 16 × 103m

    (d) we know that

    P ρ  , temperature remain

    ...more

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