A stream at \( M_1 = 2 \) and \( 50 \) kPa passes through a normal shock. The downstream static pressure is
Answer: A
The pressure ratio at Mach 2 is 4.5, so \( 50 \times 4.5 = 225 \) kPa. This abrupt jump across a vanishingly thin front is what makes shock waves so destructive structurally and what a supersonic intake must manage rather than simply endure.
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Assertion (A): An expansion shock cannot exist in a gas.
Reason (R): The entropy of the gas would have to decrease across such a discontinuity.
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Answer: B
Both are true and the second explains the first. The conservation equations formally admit an expansion discontinuity, but evaluating its entropy change gives a decrease, which the second law forbids. Supersonic expansions must therefore be spread through a fan of Mach waves instead of concentrated into a front.
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Assertion (A): A supersonic intake is built to compress the flow through several weak oblique shocks.
Reason (R): Entropy rise across a weak shock varies as the cube of its strength.
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Answer: C
Both statements are true and the second is the reason for the first. Because the loss grows so steeply with strength, splitting a compression among several weak shocks costs far less stagnation pressure than one strong shock achieving the same thing. Recovered stagnation pressure translates directly into engine thrust.
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Assertion (A): Stagnation pressure falls across an oblique shock.
Reason (R): An oblique shock raises the entropy of the gas passing through it.
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Answer: D
Both are true and the second explains the first. Any irreversibility destroys stagnation pressure, and a shock of finite strength is irreversible whatever its inclination. The loss is smaller than for a normal shock at the same upstream Mach number, because only the normal component is processed.
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Assertion (A): A blunt-nosed re-entry vehicle carries a detached bow shock.
Reason (R): A blunt nose would require a flow deflection larger than any attached shock can provide.
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Answer: A
Both statements are true and the reason accounts for the assertion. At the stagnation streamline the required turning is effectively ninety degrees, far beyond the maximum deflection, so the shock must stand off. The bluntness is deliberate: the detached shock holds the hottest gas away from the surface.
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Assertion (A): The tangential velocity component is unchanged across an oblique shock.
Reason (R): No pressure gradient acts along the shock surface.
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Answer: B
Both are true and the second explains the first. Applying momentum conservation in the direction along the shock, where the pressure difference has no component, leaves the tangential velocity untouched. Only the normal component is processed, which is what makes the oblique shock reducible to a normal shock problem.
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Assertion (A): The downstream Mach number behind a weak oblique shock is always subsonic.
Reason (R): The velocity component normal to an oblique shock is always subsonic downstream.
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Answer: C
The assertion is false while the reason is true. The normal component does drop below sonic speed, but the tangential component survives intact, so the resultant is usually still supersonic. This is precisely what allows a supersonic vehicle to carry several oblique shocks in succession.
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Assertion (A): Stagnation temperature falls across a strong normal shock.
Reason (R): A normal shock destroys a large fraction of the stagnation pressure.
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Answer: D
The assertion is false and the reason is true. A shock is adiabatic, so the total energy and with it the stagnation temperature are unchanged however strong the shock. It is the stagnation pressure that is lost, and at Mach 2 nearly twenty-eight per cent of it goes.
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Assertion (A): A shock wave may be treated as a discontinuity in any engineering calculation.
Reason (R): Its thickness is of the order of a few molecular mean free paths.
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Answer: A
Both statements are true and the second is the reason for the first. A front well under a micron thick is negligible beside any body dimension, so the jump conditions may be applied across it without resolving its interior. Resolving that interior would in any case require kinetic theory rather than continuum mechanics.
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Assertion (A): The density ratio across a normal shock cannot exceed six in air.
Reason (R): The limiting ratio is the specific heat ratio plus one divided by the specific heat ratio minus one.
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Answer: B
Both are true and the second gives the reason. Substituting 1.4 gives 2.4 over 0.4, that is six. Pressure and temperature ratios grow without limit as the shock strengthens, but the density saturates, and dissociation at hypersonic speeds is what raises the ceiling in practice.
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