Answer

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**Hint:**

To solve this question we use Bernoulli’s theorem which is ${P_1} + \dfrac{1}{2}\rho {v_1}^2 + \rho g{h_1} = {P_2} + \dfrac{1}{2}\rho {v_2}^2 + \rho g{h_2}$

And we know the gauge pressure is pressure above that of the atmospheric pressure at that point can be represented as $P - {P_{atm}} = \rho gh$ .

Where $P \Rightarrow $ is the total pressure at that point or at that level

**Step by step solution:**

First we mark another point Q at the same level of P let as assume the point P and Q are $h$ height below the free surface of the water tank.

As shown in figure.

From figure we can write the gauge pressure at point Q is ${P_Q} - {P_{atm}} = \rho gh$

Where $\rho \Rightarrow $ density of water

${P_Q} \Rightarrow $ Total pressure at point Q

$h \Rightarrow $ Height from free surface

So from this we can Wright the pressure at point Q is

$ \Rightarrow {P_Q} = {P_{atm}} + \rho gh$ ......... (1)

Pressure at point P is equal to the atmospheric pressure

${P_P} = {P_{atm}}$ ......... (2)

Velocity at point Q is approximately equal to zero

${v_Q} = 0$....... (3)

Let us assume the velocity at point P is with which water comes out is ${v_P}$

Step 2

Now we apply Bernoulli’s theorem for horizontal points P and Q.

$ \Rightarrow {P_P} + \dfrac{1}{2}\rho {v_P}^2 + \rho g{h_P} = {P_Q} + \dfrac{1}{2}\rho {v_Q}^2 + \rho g{h_Q}$

Because point P and Q are at same horizontal level ${h_P} = {h_Q}$ so Bernoulli theorem reduced to

$ \Rightarrow {P_P} + \dfrac{1}{2}\rho {v_P}^2 = {P_Q} + \dfrac{1}{2}\rho {v_Q}^2$

Put all values in this equation from (1) (2) and (3)

\[ \Rightarrow {P_{atm}} + \dfrac{1}{2}\rho {v_P}^2 = \left( {{P_{atm}} + \rho gh} \right) + \dfrac{1}{2}\rho {\left( 0 \right)^2}\]

\[ \Rightarrow \dfrac{1}{2}\rho {v_P}^2 = \rho gh\]

Now ${P_Q} - {P_{atm}} = \rho gh = 5 \times {10^5}N/{m^2}$ given in question

Put $\rho gh = 5 \times {10^5}$ and density of water $\rho = 1000kg/{m^3}$

\[ \Rightarrow \dfrac{1}{2} \times 1000 \times {v_P}^2 = 5 \times {10^5}\]

\[ \Rightarrow {v_P}^2 = \dfrac{{5 \times {{10}^5} \times 2}}{{1000}}\]

Further solving

\[ \Rightarrow {v_P} = \sqrt {\dfrac{{5 \times {{10}^5} \times 2}}{{1000}}} \]

\[ \Rightarrow {v_P} = \sqrt {1000} \]

\[ \Rightarrow {v_P} = 31.62m/\sec \]

Hence the velocity to exit the water tank at point P is approximately $32m/\sec $

**therefore Option B is correct**

**Note:**

We can solve this question by another short method which is given below

We know gauge pressure $P = \rho gh$ from this we can find height of hole from free surface

$

\Rightarrow 5 \times {10^5} = 1000 \times 10 \times h \\

\Rightarrow h = 50m \\

$

And now apply Torricelli’s theorem formula velocity of Efflux $v = \sqrt {2gh} $

$

\Rightarrow v = \sqrt {2 \times 10 \times 50} \\

\Rightarrow v = \sqrt {1000} \\

\therefore v = 31.62m/\sec \\

$

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