Question

Water from a tap emerges vertically downwards with an initial speed of $1\text{m/s}$. If the cross-sectional area of tap is ${10}^{-4}{\text{m}}^2$ then find the cross sectional area of stream $0.15\text{m}$ below the tap.

• A. $5\times {10}^{-4}{\text{m}}^2$
• B. $1\times {10}^{-4}{\text{m}}^2$
• C. $5\times {10}^{-5}{\text{m}}^2$
• D. $2\times {10}^{-5}{\text{m}}^2$

Answer 1

The correct option is C $5\times {10}^{-5}{\text{m}}^2$

Let us suppose cross-sectional area of stream is $A_b$ and speed of water is $v_f$ at $0.15\text{m}$ below the tap.
Given that,
Initial speed of water $\left(v_i\right)=1\text{m/s}$
Cross-sectional area of tap $\left(A_t\right)={10}^{-4}{\text{m}}^2$

To find the speed of water at $0.15\text{m}$ below of tap:
Let us consider the mass of water be $m$
On applying, law of conservation of mechanical energy
$M.E_i=M.E_f$
$\frac{1}{2}m(v_i{)}^2+mg{h}_1=\frac{1}{2}m(v_f{)}^2+mg{h}_2.........\left(1\right)$
From the figure, the reference level has been chosen at $0.15\text{m}$ below the tap, $\therefore$ $h_1=0.15\text{m}$ and $h_2=0\text{m}$
$\Rightarrow \frac{1}{2}\times m\times 1^2+m\times 10\times 0.15=\frac{1}{2}\times m\times (v_f{)}^2+m\times 10\times (0)$

$\Rightarrow v_f=\sqrt{1^2+2\times 10\times 0.15}=2\text{m/s}...\left(1\right)$
On applying equation of continuity
$A_t{v}_i=A_b{v}_f$
$\Rightarrow {10}^{-4}\times 1=A_b\times 2$ [From (1)] $\Rightarrow A_b=5\times {10}^{-5}{\text{m}}^2$