Fluid Mechanics

JEE Main level, single correct, medium

75 of 192

Question 75 of 192

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)

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Answer: (A) 5 m s−15 \mathrm{~m} \mathrm{~s}^{-1}

According to Bernoulli's theorem
P1+12ρv12=P2+12ρv22…(i)P_1+\frac{1}{2} \rho v_1^2=P_2+\frac{1}{2} \rho v_2^2 \ldots(\mathrm{i})
From question,
P1−P2=3×105,A1A2=5P_1-P_2=3 \times 10^5, \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 \times 10^5=\frac{1}{2} \times 1000\left(5 v_1^2-v_1^2\right. \\ & \Rightarrow \quad 600=6 v_1 \times 4 v_1 \\ & \Rightarrow \quad v_1^2=25 \\ & \therefore \quad v_1=5 \mathrm{~m} / \mathrm{s} \end{aligned}

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