Question

Consider a water jar of radius $R$ that has water filled up to height $H$ and is kept on a stand of height $h$ (see figure). Through a hole of radius $r (r << R)$ at its bottom, the water leaks out and the stream of water coming down towards the ground has a shape like a funnel as shown in the figure. If the radius of the cross-section of water stream when it hits the ground is $x$. Then :

• A. $x = r \left(\frac{H}{H+h}\right)$
• B. $x = r \left(\frac{H}{H+h}\right)^{\frac{1}{2}}$
• C. $x = r \left(\frac{H}{H+h}\right)^{\frac{1}{4}}$
• D. $x = r \left(\frac{H}{H+h}\right)^{2}$

Answer 1

Answer: (C)

$x = r \left(\frac{H}{H+h}\right)^{\frac{1}{4}}$

Answer 2

$\frac{1}{2}\rho v^{2}_{1}+\rho gh=\frac{1}{2}\rho v^{2}_{2}$
$v^{2}_{1}+2gh=v^{2}_{2}$
$2gH+2gh=v^{2}_{2}$
$a_{1}v_{1}=a_{2}v_{2}$
$\pi r^{2}\sqrt{2gh}=\pi\lambda^{2}v_{2}$
$\frac{r^{2}}{x^{2}}\sqrt{2gh}=v_{2}$
$2gH+2gh=\frac{r^{4}}{x^{4}}2gh$
$x=r\left[\frac{H}{H+h}\right]^{\frac{1}{4}}$