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JEE Main 2020 (9 Jan, Shift 1) · Bernoulli's Equation

Bernoulli's Equation question from JEE Main 2020 (9 Jan, Shift 1)

Q24JEE Main 2020 · 9 JAN · SHIFT 1Medium

Bernoulli's Equation

Water flows in a horizontal tube (see figure).

The pressure of water changes by 700 Nm–2

between A and B where the area of cross section

are 40 cm2 and 20 cm2, respectively. Find the

rate of flow of water through the tube.

(density of water = 1000 kgm–3)

Figure

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Answer: (d)

  1. Continuity: vB=4020vA=2vAv_B = \dfrac{40}{20}v_A = 2v_A
  2. Bernoulli: PA−PB=12ρ(4vA2−vA2)=1500 vA2=700P_A - P_B = \frac{1}{2}\rho(4v_A^2 - v_A^2) = 1500\,v_A^2 = 700
  3. vA2=0.467⇒vA≈0.683 m/s=68.3 cm/sv_A^2 = 0.467 \Rightarrow v_A \approx 0.683\text{ m/s} = 68.3\text{ cm/s}
  4. Q=AAvA=40×68.3≈2732 cm3/sQ = A_A v_A = 40\times68.3 \approx 2732\text{ cm}^3/\text{s}
  5. So the answer is (d) 2720 cm3/s2720\text{ cm}^3/\text{s}.
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