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JEE Main 2026 (21 Jan, Shift 1) · Bernoulli's Equation

Bernoulli's Equation question from JEE Main 2026 (21 Jan, Shift 1)

Q6JEE Main 2026 · 21 JAN · SHIFT 1Medium

Bernoulli's Equation

Water flows through a horizontal tube as shown in the figure. The difference in height between the water columns in vertical tubes is 5 cm and the area of cross-sections at AA and BB are 6 cm26\text{ cm}^2 and 3 cm23\text{ cm}^2 respectively. The rate of flow will be ‾ cm3/s\underline{\hspace{3em}}\ \text{cm}^3/\text{s}. (take g=10 m/s2g = 10\text{ m/s}^2)

Figure

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Answer: (d)

  1. Pressure difference from the columns: PA−PB=ρghP_A - P_B = \rho g h.
  2. Bernoulli: 12(vB2−vA2)=gh=10×0.05=0.5⇒vB2−vA2=1\frac{1}{2}(v_B^2 - v_A^2) = gh = 10\times0.05 = 0.5 \Rightarrow v_B^2 - v_A^2 = 1
  3. Continuity: vB=63vA=2vA⇒3vA2=1⇒vA=13 m/s=1003 cm/sv_B = \dfrac{6}{3}v_A = 2v_A \Rightarrow 3v_A^2 = 1 \Rightarrow v_A = \dfrac{1}{\sqrt{3}}\text{ m/s} = \dfrac{100}{\sqrt{3}}\text{ cm/s}
  4. Q=AAvA=6×1003=2003 cm3/sQ = A_A v_A = 6\times\dfrac{100}{\sqrt{3}} = 200\sqrt{3}\text{ cm}^3/\text{s}
  5. So the answer is (d).
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