Question

A tube has two areas of cross-section as shown in the figure. The diameter of the bigger section is double the diameter of the smaller section. Find the range of water falling on the horizontal surface, if piston is moving with a constant velocity $0.4\text{m/s}$ and $h=2.5\text{m}(g=10{\text{m/s}}^2)$

• A. $1.13\text{m}$
• B. $1.5\text{m}$
• C. $2\text{m}$
• D. $1.51\text{m}$

Answer 1

The correct option is A $1.13\text{m}$

Let $d$ be the diameter of smaller section.
$\therefore$ Area of smaller section $a=\frac{\pi }{4}{d}^2$ and Area of bigger section $A=\frac{\pi }{4}(2d{)}^2=\frac{\pi }{4}(4d^2)$
By continuity equation,
$A{v}_1=a{v}_2$
$\Rightarrow v_2=\frac{A}{a}{v}_1=\frac{\frac{\pi }{4}\left(4d^2\right)}{\frac{\pi }{4}{d}^2}\times 0.4=1.6\text{m/s}$
Here, $v_2$ is the horizontal velocity of water coming out of the tube.
Using $S=ut+\frac{1}{2}a{t}^2$ for vertical direction
$h=\frac{1}{2}g{t}^2[u_y=0]$
$\Rightarrow t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times 2.5}{10}}=\frac{1}{\sqrt{2}}\text{s}$
$\therefore$ Horizontal range of water
$R=v_2\times t$
$=v_2\times t$
$=1.6\times \frac{1}{\sqrt{2}}$
$=1.131\text{m}$