Answer
$\approx 1.1\;\rm mm/min$
Work Step by Step
a) We know that the tank is losing water from the hole which we chose to be point 2 and the surface of the water to be point 1 (See the red notations in the figure below).
And since the tank is emptying, the velocity of water at point 1 is not zero.
To find the volume flow rate, we need to use Bernoulli’s equation.
$$\color{red}{\bf\not}P_1+\frac{1}{2}\rho v_1 ^2+y_1\rho g=\color{red}{\bf\not}P_2+\frac{1}{2}\rho v_2^2+y_2\rho g$$
Noting that the water surface and the hole are at atmospheric pressure since both are exposed to air.
$$ \frac{1}{2}\color{red}{\bf\not}\rho v_1 ^2+y_1\color{red}{\bf\not}\rho g= \frac{1}{2}\color{red}{\bf\not}\rho v_2^2+y_2\color{red}{\bf\not}\rho g$$
We chose point 2 to be our origin at which $y=0$, so $y_2=0$ and $y_1=d$
$$ \frac{1}{2} v_1 ^2+d g= \frac{1}{2} v_2^2+0\tag{Multiplying by $ 2$}$$
$$ v_1 ^2+2d g= v_2^2 \tag 1$$
Now we need to find $v_1$ in terms of $v_2$ and we know that the continuity equation states that
$$v_1A_1=v_2A_2$$
where $A$ is the cross-sectional area which is given by $\pi r^2$
Hence,
$$v_1=\dfrac{v_2A_2}{A_1}=\dfrac{v_2(\color{red}{\bf\not}\pi r_2^2)}{(\color{red}{\bf\not}\pi r_1^2)}=\dfrac{v_2 r_2^2 }{ r_1^2}$$
where $r_1=R$ and $r_2=r$, as shown in the figure below.
$$v_1= \dfrac{v_2r^2 }{ R^2 }\tag 2$$
Plugging into (1);
$$ \left[\dfrac{v_2r^2 }{ R^2 }\right]^2+2d g= v_2^2 $$
$$ \dfrac{v_2^2r^4 }{ R^4 } +2d g= v_2^2 $$
$$ v_2^2-\dfrac{v_2^2r^4 }{ R^4 } =2d g $$
$$ v_2^2\left[1 -\dfrac{r^4 }{ R^4 }\right] =2d g $$
$$ v_2 =\sqrt{\dfrac{2d g}{1 -\dfrac{r^4 }{ R^4 }} } \tag 3$$
Therefore, the volume flow rate of water at the hole (point 2) is given by
$$\left(\dfrac{\Delta V}{\Delta t}\right)_2=v_2A_2=v_2 (\pi r^2)$$
Plugging from (3);
$$\left(\dfrac{\Delta V}{\Delta t}\right)_2 =\sqrt{\dfrac{2d g}{1 -\dfrac{r^4 }{ R^4 }} } (\pi r^2)$$
$$\left(\dfrac{\Delta V}{\Delta t}\right)_2 =\sqrt{\dfrac{2\pi^2 r^4d g }{1 -\dfrac{r^4 }{ R^4 }} } $$
$$\left(\dfrac{\Delta V}{\Delta t}\right)_2 =\sqrt{\dfrac{2\pi^2 r^4d g }{ \dfrac{R^4-r^4 }{ R^4 }} } $$
$$\left(\dfrac{\Delta V}{\Delta t}\right)_2 =\sqrt{\dfrac{2\pi^2 r^4 R^4 d g }{ R^4-r^4 } } $$
$$\boxed{\left(\dfrac{\Delta V}{\Delta t}\right)_2 =\pi r^2R^2\sqrt{\dfrac{2 d g }{ R^4-r^4 } } } $$
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b) Now we need to find the rate at which the water level will initially drop in mm/min, which is the speed of the water surface.
Using (2);
$$v_1= \dfrac{v_2r^2 }{ R^2 } $$
Plugging from (3);
$$v_1= \dfrac{ r^2 }{ R^2 }\sqrt{\dfrac{2d g}{1 -\dfrac{r^4 }{ R^4 }} } $$
$$v_1= \dfrac{ r^2 }{ R^2 }\sqrt{\dfrac{2d g}{ \dfrac{R^4-r^4 }{ R^4 }} } $$
$$v_1= \dfrac{ r^2 }{ R^2 }\sqrt{\dfrac{2d gR^4}{ R^4-r^4 } } $$
$$v_1= \dfrac{ r^2R^2 }{ R^2 }\sqrt{\dfrac{2d g }{ R^4-r^4 } } $$
$$v_1= r^2 \sqrt{\dfrac{2d g }{ R^4-r^4 } } $$
Plugging the given in mm and minutes.
$$v_1= (2)^2 \sqrt{\dfrac{2(1000) (9800\times 60^2)}{ (1000)^4-(2)^4 } } $$
$$v_1=\color{red}{\bf 1.062}\;\rm mm/min$$
