JEE PYQ: Mechanical Properties Of Fluids Question 28
Question 28 - 2019 (10 Apr 2019 Shift 2)
Water from a tap emerges vertically downwards with an initial speed of 1.0 ms$^{-1}$. The cross-sectional area of the tap is $10^{-4}$ m$^2$. Assume that the pressure is constant throughout the stream of water and that the flow is streamlined. The cross-sectional area of the stream, 0.15 m below the tap would be: [Take $g = 10$ ms$^{-2}$]
(1) $2 \times 10^{-5}$ m$^2$
(2) $5 \times 10^{-5}$ m$^2$
(3) $5 \times 10^{-4}$ m$^2$
(4) $1 \times 10^{-5}$ m$^2$
Show Answer
Answer: (2) $5 \times 10^{-5}$ m$^2$
Solution
Using Bernoulli’s equation: $P + \frac{1}{2}(v_1^2 - v_2^2) + \rho g h = P$ $\Rightarrow v_2^2 = v_1^2 + 2gh$ $\Rightarrow v_2 = \sqrt{v_1^2 + 2gh}$ Equation of continuity: $A_1 v_1 = A_2 v_2$ $(1 \text{ cm}^2)(1 \text{ m/s}) = (A_2)\sqrt{(1)^2 + 2 \times 10 \times \frac{15}{100}}$
$10^{-4} \times 1 = A_2 \times 2$ $\therefore A_2 = \frac{10^{-4}}{2} = 5 \times 10^{-5}$ m$^2$
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