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JEE Main 2014 (19 Apr, Shift 1) · Applications of Bernoulli's Theorem

Applications of Bernoulli's Theorem question from JEE Main 2014 (19 Apr, Shift 1)

Q9JEE Main 2014 · 19 APR · SHIFT 1Medium

Applications of Bernoulli's Theorem

Figure

In the diagram shown, if the difference in the two tubes of the manometer is 5 cm, the cross section of the tube at A and B is 6 mm2 and 10 mm2 respectively. The rate at which water flows through the tube ( g = 10 ms−2{}^{-2})

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Answer: (a)

  1. Manometer: ΔP=ρgh=1000(10)(0.05)=500 Pa\Delta P = \rho g h = 1000(10)(0.05) = 500\text{ Pa}
  2. Bernoulli: 12ρ(vA2−vB2)=ρgh⇒vA2−vB2=2gh=1\frac{1}{2}\rho(v_A^2 - v_B^2) = \rho g h \Rightarrow v_A^2 - v_B^2 = 2gh = 1
  3. Continuity: 6vA=10vB⇒vB=0.6vA6v_A = 10v_B \Rightarrow v_B = 0.6v_A, so 0.64vA2=1⇒vA=1.25 m/s0.64v_A^2 = 1 \Rightarrow v_A = 1.25\text{ m/s}
  4. Q=AAvA=(6×10−6)(1.25)=7.5×10−6 m3/s=7.5 cc/sQ = A_Av_A = (6\times10^{-6})(1.25) = 7.5\times10^{-6}\text{ m}^3\text{/s} = 7.5\text{ cc/s}
  5. So the answer is (a) 7.5 cc/s.
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