Search any question & find its solution
Question:
Answered & Verified by Expert
Water is flowing through a horizontal tube having cross-sectional areas of its two ends being $A$ and $A^{\prime}$ such that the ratio $A / A^{\prime}$ is 5 . If the pressure difference of water between the two ends is $3 \times$ $10^5 \mathrm{~N} \mathrm{~m}^{-2}$, the velocity of water with which it enters the tube will be (neglect gravity effects)
Options:
Solution:
1358 Upvotes
Verified Answer
The correct answer is:
$5 \mathrm{~m} \mathrm{~s}^{-1}$
$5 \mathrm{~m} \mathrm{~s}^{-1}$
According to Bernoulli's theorem
$$
P_1+\frac{1}{2} \rho v_1^2=P_2+\frac{1}{2} \rho v_2^2 \ldots(\mathrm{i})
$$
From question,
$$
P_1-P_2=3 \times 10^5, \frac{A_1}{A_2}=5
$$
According to equation of continuity
$$
\begin{aligned}
& A_1 v_1=A_2 v_2 \\
& \text { or, } \frac{A_1}{A_2}=\frac{v_2}{v_1}=5 \\
& \Rightarrow \quad v_2=5 v_1
\end{aligned}
$$
From equation (i)
$$
\begin{aligned}
& P_1-P_2=\frac{1}{2} \rho\left(v_2^2-v_1^2\right) \\
& \text { or } 3 \times 10^5=\frac{1}{2} \times 1000\left(5 v_1^2-v_1^2\right. \\
& \Rightarrow \quad 600=6 v_1 \times 4 v_1 \\
& \Rightarrow \quad v_1^2=25 \\
& \therefore \quad v_1=5 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
$$
P_1+\frac{1}{2} \rho v_1^2=P_2+\frac{1}{2} \rho v_2^2 \ldots(\mathrm{i})
$$
From question,
$$
P_1-P_2=3 \times 10^5, \frac{A_1}{A_2}=5
$$
According to equation of continuity
$$
\begin{aligned}
& A_1 v_1=A_2 v_2 \\
& \text { or, } \frac{A_1}{A_2}=\frac{v_2}{v_1}=5 \\
& \Rightarrow \quad v_2=5 v_1
\end{aligned}
$$
From equation (i)
$$
\begin{aligned}
& P_1-P_2=\frac{1}{2} \rho\left(v_2^2-v_1^2\right) \\
& \text { or } 3 \times 10^5=\frac{1}{2} \times 1000\left(5 v_1^2-v_1^2\right. \\
& \Rightarrow \quad 600=6 v_1 \times 4 v_1 \\
& \Rightarrow \quad v_1^2=25 \\
& \therefore \quad v_1=5 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.