Answer

Verified

393k+ views

**Hint:**The velocity of the water stream at exit can be given by Torricelli's equation. Consider the maximum height attained by the water stream.

$\Rightarrow {v^2} = 2gh$ where $v$ is the velocity of stream at exit, $g$ is the acceleration due to gravity and $h$ is the height of the free surface of the liquid from exit,

$\Rightarrow H = \dfrac{{{v^2}{{\sin }^2}\alpha }}{{2g}}$ where $H$ is the maximum height of the water stream from exit point and $\alpha $ is the angle the velocity makes with the horizontal.

**Complete step by step answer**

Torricelli’s equation gives us the velocity of the water stream at exit due to the pressure of the liquid above the exit.

To solve, let us call the height of exit above line XX’ equal to $h'$.

Then, $h' = L\sin \alpha $.

Substituting the known values we get,

$\Rightarrow h' = 2\sin 30^\circ $

$\Rightarrow 2 \times \dfrac{1}{2} $

$\Rightarrow 1m $

The height of exit above the free surface is

$\Rightarrow h - h' = 10 - 1 $

$\Rightarrow 9m $

Where $h$ is the height of the free surface of liquid from line XX’.

Then, velocity at the exit is given by

$\Rightarrow {v^2} = 2g(h - h')$

where $v$ is the velocity of stream at exit, $g$ is the acceleration due to gravity.

Therefore, substituting the known values, we get

$\Rightarrow {v^2} = 2\left( {9.8} \right)\left( 9 \right) $

$\Rightarrow 176.4 $

Square rooting both sides we get,

$\Rightarrow v = \sqrt {176.4} $

$\Rightarrow 13.3m/s $

Therefore, the velocity at the exit is $13.3m/s$.

The maximum height attained by the water stream will be

$\Rightarrow H = \dfrac{{{v^2}{{\sin }^2}\alpha }}{{2g}} $

$\Rightarrow \dfrac{{176.4\left( {{{\sin }^2}30} \right)}}{{2\left( {9.8} \right)}} $

Solving further we get,

$\Rightarrow H = \dfrac{{176.4\left( {{{\left( {\dfrac{1}{2}} \right)}^2}} \right)}}{{2\left( {9.8} \right)}} $

$\Rightarrow \dfrac{{176.4\left( {\dfrac{1}{4}} \right)}}{{2\left( {9.8} \right)}} $

$\Rightarrow 2.25m $

Therefore, the total height from the line XX’ is

$\Rightarrow {H_t} = H + h' $

$\Rightarrow 2.25 + 1 $

$\Rightarrow 3.25m $

The closest to the answer is option D.

**Hence, our correct option is (D).**

**Note**

A common point of error is to incorrectly leave $H = 2.25m$ as the final answer. Note that the maximum height $H$ of any projectile as calculated from $H = \dfrac{{{v^2}{{\sin }^2}\alpha }}{{2g}}$ is only the height of the body from the point where it has the velocity $v$. Hence, for the water stream, this is the exit point which is 1m above line XX’.

Recently Updated Pages

Draw a labelled diagram of DC motor class 10 physics CBSE

A rod flies with constant velocity past a mark which class 10 physics CBSE

Why are spaceships provided with heat shields class 10 physics CBSE

What is reflection Write the laws of reflection class 10 physics CBSE

What is the magnetic energy density in terms of standard class 10 physics CBSE

Write any two differences between a binocular and a class 10 physics CBSE

Trending doubts

Difference Between Plant Cell and Animal Cell

Give 10 examples for herbs , shrubs , climbers , creepers

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Name 10 Living and Non living things class 9 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Fill the blanks with proper collective nouns 1 A of class 10 english CBSE

Select the word that is correctly spelled a Twelveth class 10 english CBSE

Write the 6 fundamental rights of India and explain in detail