Question

An ideal liquid flowing through a pipe $\text{A}$ of cross-section $0.2{\text{m}}^2$ with velocity $10\text{m/s}$ enters a $\text{T}$ - junction. One side of the $\text{T}$ -junction $\text{B}$ has cross-section area $0.1{\text{m}}^2$ and the other side $\text{C}$ has cross-section area $0.05{\text{m}}^2$. If the velocity of water in $\text{C}$ is $15\text{m/s},$ then in $\text{B}$ the velocity is

• A. $1\text{m/s}$
• B. $10\text{m/s}$
• C. $12.5\text{m/s}$
• D. $1\text{cm/s}$

Answer 1

The correct option is C $12.5\text{m/s}$

We have from the continuity equation,

$v_A{A}_A=v_B{A}_B+v_C{A}_C......\left(1\right)$

Where $v$ represents the velocity and $A$ represents the area of cross section.



Given: $v_A=10\text{m/s}$, $A_A=0.2{\text{m}}^2$, $A_B=0.1{\text{m}}^2$, $v_C=15\text{m/s}$, $A_C=0.05{\text{m}}^2$

Putting these values in equation $\left(1\right)$

$10\times 0.2=(v_B\times 0.1)+(15\times 0.05)$

$2=v_B\times 0.1+0.75$

$v_B=\frac{2-0.75}{0.1}=\frac{1.25}{0.1}$

$v_B=12.5{\text{ms}}^{-1}$