Question

Water from a tap emerges vertically downwards with an initial speed of $1.0m{s}^{-1}$. The crosssectional 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 streamlined. The cross-sectional area of the stream, $0.15m$ below the tap would be :
(Take $g=10m{s}^{-2}$)

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

Answer 1

Answer: (B)

$A_1{v}_1=A_2{v}_2$
$10^{-4}\times 1=A_2{v}_2$
$A_2{v}_2=10^{-4}$ .......(1)
$P+\frac{1}{2}\rho \left(v_1^2-v_2^2\right)+\rho gh=P$
$v_2^2=v_1^2+2gh$
$v_2=\sqrt{v_1^2+2gh}$
$=\sqrt{1+2\times 10\times 0.15}$
$\frac{10^{-4}}{A_2}=2$
$A_2=5\times 10^{-5}{m}^2$