JEE Main 2026 — Jan 21, Shift 1 — Physics solution

JEE Main2026Jan 21, Shift 1Physics

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 A and B are 6 cm $^{2}$ and 3 cm $^{2}$ respectively.

The rate of flow will be ____ cm $^{3}$ /s. (take g = 10 m/s $^{2}$ )

A
$\frac{200}{3}$
B
$2006$
C
$2003$
D
$1003$

Solution

Continuity: $A_{A} v_{A} = A_{B} v_{B} \Rightarrow v_{B} = 2 v_{A}$ (since $A_{A} / A_{B} = 2$ ).

The height difference of water columns equals the pressure difference between the two sections divided by $ρ g$ : $ρ g h = P_{A} - P_{B}$ .

Bernoulli: $P_{A} - P_{B} = \frac{1}{2} ρ (v_{B}^{2} - v_{A}^{2}) = \frac{3}{2} ρ v_{A}^{2}$ . Combining: $g h = \frac{3}{2} v_{A}^{2} \Rightarrow v_{A} = \frac{2 \cdot 10 \cdot 0.05}{3} = \frac{1}{3}$ m/s $= \frac{100}{3}$ cm/s.

Volume flow rate $= A_{A} v_{A} = 6 \cdot \frac{100}{3} = \frac{600}{3} = 2003$ cm $^{3}$ /s.

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