Question
For the arrangement shown in the figure, find the time interval in seconds after which the water jet ceases to cross the wall. Area of the cross section of the tank $$A = \sqrt 5 {m^2}$$ and area of the orifice $$a = 4\,c{m^2}.$$ [Assume that the container remaining fixed]
For the arrangement shown in the figure, find the time interval in seconds after which the water jet ceases to cross the wall. Area of the cross section of the tank $$A = \sqrt 5 {m^2}$$ and area of the orifice $$a = 4\,c{m^2}.$$ [Assume that the container remaining fixed]
A.
$$1000s$$
B.
$$2000s$$
C.
$$1500s$$
D.
$$500s$$
Answer :
$$1000s$$
Solution :
Let $$x$$ be the height of the water level from the hole $$v = \sqrt {2gx} = \frac{\ell }{t} = \frac{1}{t}\,......\left( {\text{i}} \right)$$
For vertical motion $$x = \frac{1}{2}g{t^2}\,......\left( {{\text{ii}}} \right)$$
Now solving (i) & (ii) we get
$$ \Rightarrow x = 0.25\,m\left( {{\text{i}}{\text{.e}}{\text{. level goes down from }}0.81\,m{\text{ to}}\,0.25\,m} \right)$$
Using equation of continuity
$$\sqrt 5 .\frac{{dx}}{{dt}} = \sqrt {2gx} .4 \times {10^{ - 4}}$$
On solving $$t = 1000\,s$$
Let $$x$$ be the height of the water level from the hole $$v = \sqrt {2gx} = \frac{\ell }{t} = \frac{1}{t}\,......\left( {\text{i}} \right)$$
For vertical motion $$x = \frac{1}{2}g{t^2}\,......\left( {{\text{ii}}} \right)$$
Now solving (i) & (ii) we get
$$ \Rightarrow x = 0.25\,m\left( {{\text{i}}{\text{.e}}{\text{. level goes down from }}0.81\,m{\text{ to}}\,0.25\,m} \right)$$
Using equation of continuity
$$\sqrt 5 .\frac{{dx}}{{dt}} = \sqrt {2gx} .4 \times {10^{ - 4}}$$
On solving $$t = 1000\,s$$
