Question

Water from a tap emerges vertically downwards with an initial speed of $1.0m{s}^{-1}$. The cross-sectional area of the tap is ${10}^{-4}{m}^2$. Assume that the pressure is constant throughout the stream of water and that the flow is steady. The cross-sectional area of the stream $0.15m$ below the tap is ($g=10m{s}^{-2}$)

• A. $5.0\times {10}^{-4}{m}^2$
• B. $1.0\times {10}^{-5}{m}^2$
• C. $5.0\times {10}^{-5}{m}^2$
• D. $2.0\times {10}^{-5}{m}^2$

Answer 1

Answer: (C)

$5.0\times {10}^{-5}{m}^2$

Volume flow rate $={10}^{-4}\frac{m^3}{sec}$

$\Delta P=\rho gh={10}^3\frac{kg}{m^3}\times 10\frac{m^2}{s}\times 0.15$

$\Rightarrow \Delta P=1500P_a$

$\Rightarrow \Delta P=\frac{1}{2}\rho ({v_2}^2-{v_1}^2)$

$\Rightarrow 1500=\frac{1}{2}\times {10}^3(v^2-1)$

$\Rightarrow v=2\frac{m}{s}$

By equation of continuity,

$A_1{v}_1=A_2{v}_2$

$\Rightarrow {10}^{-4}\times 1=A_2\times 2$

$\Rightarrow A_2=5\times {10}^{-5}{m}^2$