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final_fall07a

# final_fall07a - PHY 206 FINAL EXAM Dec 10 2007 NAME...

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PHY 206 FINAL EXAM Dec. 10, 2007 NAME: SIGNATURE: ID: 1. 2. 3. 4. 5. Some relations you may need: p + ρgy + 1 2 ρv 2 = const. dQ = mCdT Δ V = βV 0 Δ T pV = nRT dQ = dU + dW dW = pdV e = W Q H dS = dQ T n a sin θ a = n b sin θ b n a s + n b s 0 = n b - n a R 1 s + 1 s 0 = 1 f γ = 1 p 1 - u 2 /c 2 For relative motion along x, x 0 : x 0 = γ ( x - ut ) t 0 = γ ( t - ux/c 2 ) v 0 x = v x - u 1 - uv x /c 2 Δ t = γ Δ t 0 L = L 0 γ f = r c + u c - u f 0 (source approaching)

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1. A large, open cylindrical tank is filled to height h with equal volumes of two liquids, having densities ρ 1 and ρ 2 ( > ρ 1 ), respectively. a) What is the gauge pressure at the interface between the liquids? b) Use Bernouilli’s equation to compute the velocity of efflux, v , through a small hole in the bottom of the tank. c) What does your expression in b) yield for v in the case ρ 1 = ρ 2 ? v ρ 1 ρ 2 h Solution: a) The gauge pressure (the difference from atmospheric pressure, p a ) at depth y in a liquid of density ρ is, p ( y ) - p a = ρgy . Here we have equal volumes of the two liquids, so the depth of the ρ 1 - ρ 2 interface is y = h/ 2 and the gauge pressure there is: p - p a = ρ 1 gh/ 2 . b) Compare points at the ρ 1 - ρ 2 interface and at the hole in the bottom of the tank where the fluid comes out. Since the tank is large, the velocity of the fluid at the interface is negligible and may be taken to be zero. The pressure there is that found in a). The pressure where the fluid comes out the hole is p a . Taking y = 0 at the hole and measuring positive y upward, Bernouilli says: p a + ρ 1 g h 2 + ρ 2 g h 2 = p a + 1 2 ρ 2 v 2 v = r ρ 1 + ρ 2 ρ 2 gh. c) For ρ 1 = ρ 2 = ρ our expression yields v = ρgh , i.e. the usual Torricelli’s theorem as we expect.
2. One mole of a monatomic gas is taken through the reversible cycle shown. At point

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final_fall07a - PHY 206 FINAL EXAM Dec 10 2007 NAME...

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