Question

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

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

Answer 1

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

As we know

$P+\frac{1}{2}\rho v^2+\rho gh=$ costant

Given, initial speed of the water,

$v_1=1m/s$

Using Bernoulli's equation

$P+\frac{1}{2}\rho \left(v_1^2\right)+\rho gh=P+\frac{1}{2}\rho v_2^2$

$v_2^2=v_1^2+2gh$

$v_2=\sqrt{v_1^2+2gh}$

$v_2=\sqrt{1+2\times 10\times 0.15}$

$=2m/s$

Given, area of cross section of the tap, $A_1={10}^{-4}{m}^2$

Applying continuity equation at tap position and at a distance $0.15m$ below the tap.

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

${10}^{-4}\times 1=A_2{v}_2$

$A_2{v}_2={10}^{-4}$

$A_2\left(2\right)={10}^{-4}$

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