JEE Main 2014 · 19 Apr · Shift 1 · Physics Q9

← JEE Main 2014 · 19 Apr · Shift 1

PhysicsSingle correct+4 / -1Question 9 of 90

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

A horizontal pipe with water flowing to the right, narrower at A and wider at B; vertical manometer tubes rise from A and B, and the water level in the tube at A is 5 cm lower than at B.

7.5 cc/s

8.0 cc/s

10.0 cc/s

12.5 cc/s

Answer and worked solution

Official answer: A

The level is lower over the narrow section A, so $P_B-P_A=\rho g(0.05)=500$ Pa. Bernoulli: $v_A^2-v_B^2=\dfrac{2\times500}{1000}=1$ .

Continuity: $v_A=\frac{10}6v_B$ . Then $\left(\frac{100}{36}-1\right)v_B^2=1$ , so $v_B=0.75$ m/s.
$Q=10\times10^{-6}\times0.75=7.5\times10^{-6}\text{ m}^3\text{/s}=7.5\text{ cc/s}.$

Topics:Bernoulli's principle Continuity equation

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JEE Main 2014 · 19 Apr · Shift 1, Physics question 9