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

Answer 1

The correct option is A $x=r{\left(\frac{H}{H+h}\right)}^{\frac{1}{4}}$

By energy conservation,

$\frac{\frac{1}{2}\rho v_1^2}{\frac{1}{2}\rho v_2^2}=\frac{\rho gH}{\rho g(H+h)}$

Hence, $\frac{v_2}{v_1}=\sqrt{\frac{H+h}{H}}$,

Also, since by mass conservation, $A_1{v}_1=A_2{v}_2$,

$r^2{v}_1=x^2{v}_2$

From the two relations, we get, $\frac{x}{r}={\left(\frac{H}{H+h}\right)}^{\frac{1}{4}}$.

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