Question

Water flows steadily from an open tank as shown in figure. Assuming area of cross-section of the hole is much less than the area of cross-section of the tank, which of the following option(s) is/are correct? (Take value of $g=10{\text{m/s}}^2$).

• A. Velocity of water coming out of the pipe is $4\sqrt{10}{\text{ms}}^{-1}$.
• B. Velocity of water coming out of the pipe is $\sqrt{10}{\text{ms}}^{-1}$.
• C. Value of $x$ is $7.97\text{m}$.
• D. Value of $x$ is $9.97\text{m}$

Answer 1

Answer: (A), (C)

Value of $x$ is $7.97\text{m}$.

Let the height of the pipe opening from the surface of water in the tank be $h$.
From the figure, we can deduce that the value of $h=10-2=8\text{m}$.
Then, speed of efflux from the pipe
$v_x=\sqrt{2gh}=\sqrt{2\times 10\times 8}=4\sqrt{10}{\text{ms}}^{-1}$

Using kinematic equation in the vertical direction for water coming out of the pipe,
$y=ut+\frac{1}{2}g{t}^2$
Since $u=0$, we get $y=\frac{1}{2}g{t}^2$
Given, $y=2\text{m}$
$\Rightarrow 2=\frac{1}{2}g{t}^2$
or $t=\sqrt{0.4}=0.63\text{s}$
Horizontal range of water
$x=v_{x}t=4\sqrt{10}\times \left(0.63\right)$
$=7.97\text{m}$
Thus, options (a) and (c) are correct answers.