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JEE Main 201212 MayOnline
Exam
JEE Main
Year
2012
Date
12 May
Mode
Online
Subject
Physics
Chapter
Mechanical Properties of Fluids

JEE Main 2012 · 12 May · Online · Physics · Mechanical Properties of Fluids

Water is flowing through a horizontal tube having cross-sectional...

Water is flowing through a horizontal tube having cross-sectional areas of its two ends being AA and A′A^{\prime} such that the ratio A/A′A / A^{\prime} is 5 . If the pressure difference of water between the two ends is 3×3 \times 105 N m−210^5 \mathrm{~N} \mathrm{~m}^{-2}, the velocity of water with which it enters the tube will be (neglect gravity effects)

Options

  1. A5 m s−15 \mathrm{~m} \mathrm{~s}^{-1}
  2. B10 m s−110 \mathrm{~m} \mathrm{~s}^{-1}
  3. C25 m s−125 \mathrm{~m} \mathrm{~s}^{-1}
  4. D5010 m s−150 \sqrt{10} \mathrm{~m} \mathrm{~s}^{-1}

Correct answer

A. 5 m s−15 \mathrm{~m} \mathrm{~s}^{-1}

Step-by-step solution

According to Bernoulli's theorem P_1+12ρ\frac{1}{2} \rhov_1^2=P_2+12ρ\frac{1}{2} \rhov_2^2 \ldots(i)\mathrm{i})From question, P_1-P_2=3 ×\times10^5, A1A2=\frac{A_1}{A_2}=5 According to equation of continuity A1v1=A2v2 or, A1A2=v2v1=5⇒v2=5v1\begin{aligned}& A_1 v_1=A_2 v_2 \\ & \text{ or, } \frac{A_1}{A_2}=\frac{v_2}{v_1}=5 \\ & \Rightarrow \quad v_2=5 v_1 \end{aligned}From equation (i) P1−P2=12ρ(v22−v12) or 3×105=12×1000(5v12−v12⇒600=6v1×4v1⇒v12=25∴v1=5 m/s\begin{aligned}& P_1-P_2=\frac{1}{2} \rho\left(v_2^2-v_1^2\right) \\ & \text{ or }3 \times10^5=\frac{1}{2} \times1000\left(5 v_1^2-v_1^2\right. \\ & \Rightarrow \quad 600=6 v_1 \times4 v_1 \\ & \Rightarrow \quad v_1^2=25 \\ & \therefore \quad v_1=5 \mathrm{~m}/ \mathrm{s} \end{aligned}
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