Question

A tube has variable cross-section as shown in figure. The diameter of the tube inlet and exit are $12\text{mm}$ and $4\text{mm}$ respectively. Find the range of water falling on the horizontal surface if the piston is moving with a constant velocity of $1\text{m/s}$. Given, $h=4.9\text{m},g=9.8{\text{ms}}^{-2}$.

• A. $4.5\text{m}$
• B. $9\text{m}$
• C. $18\text{m}$
• D. $15\text{m}$

Answer 1

The correct option is B $9\text{m}$

Given that
$d_1=12\text{mm}=12\times {10}^{-3}\text{m}$
$d_2=4\text{mm}=4\times {10}^{-3}\text{m}$
$V_1=1\text{m/s}$
Now applying continuity equation between inlet and outlet
$A_1{V}_1=A_2{V}_2$
$\Rightarrow \frac{\pi }{4}{d}_1^2{V}_1=\frac{\pi }{4}{d}_2^2{V}_2$

$\Rightarrow {(12\times {10}^{-3})}^2\times 1=(4\times {10}^{-3})\times V_2$

$\Rightarrow V_2=9\text{m/s}$

For height $h$,
$h=\frac{1}{2}g{t}^2$
$\Rightarrow t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times 4.9}{9.8}}=1\text{second}$
So,
Horizontal range $\left(R\right)=V_2\times t=9\times 1=9\text{m}$