Answer
See the detailed answer below.
Work Step by Step
First, we need to find the volume flow rate, so we can find the two velocities.
Noting that this problem is the inverse of the previous one.
b)
Since the author told us to assume that the air is an ideal fluid, we can use Bernoulli’s equation.
We chose point 1 at the left-wider pipe and point 2 at the same height (on the same horizontal line) at the right-narrower pipe. This means that $y_2-y_1=0$.
$$P_1+\frac{1}{2}\rho v_1^2+\rho gy_1=P_2+\frac{1}{2}\rho v_2^2+\rho gy_2$$
Hence,
$$P_1+\frac{1}{2}\rho v_1^2+\rho gy_1=P_2+\frac{1}{2}\rho v_2^2+\rho gy_2$$
$$P_1-P_2= \frac{1}{2}\rho v_2^2-\frac{1}{2}\rho v_1^2+\rho gy_2-\rho gy_1$$
$$P_1-P_2= \frac{1}{2}\rho (v_2^2- v_1^2)+\overbrace{\rho g(y_2- y_1)}^{=0}$$
$$\overbrace{P_1-P_2}^{-h\rho_{Hg} g}= \frac{1}{2}\rho (v_2^2- v_1^2) $$
Noting that the pressure difference $P_1-P_2=h\rho_{Hg}g$ since the pressure above the mercury surface in the left tube is $P_1$ and the pressure above the mercury surface in the right tube is $P_2$.
$$ h\rho_{Hg} g =- \frac{1}{2}\rho (v_2^2- v_1^2) $$
$$ h\rho_{Hg} g = \frac{1}{2}\rho (v_1^2- v_2^2)\tag 1 $$
Now we need to find $v_1$ and $v_2$, and we know that the volume rate is given by
$$Q=A_1v_1=A_2v_2$$
Hence,
$$v_1=\dfrac{Q}{A_1}=\dfrac{Q}{\pi r_1^2}=\dfrac{Q}{\pi\left[\frac{d_1}{2}\right]^2}$$
$$v_1=\dfrac{4Q}{\pi d_1^2}\tag 2$$
And by the same approach,
$$v_2 =\dfrac{4Q}{\pi d_2^2}\tag 3$$
Plugging (2) and (3) into (1);
$$ h\rho_{Hg} g = \frac{1}{2}\rho \left(\left[\dfrac{4Q}{\pi d_1^2}\right]^2- \left[\dfrac{4Q}{\pi d_2^2}\right]^2\right) $$
$$ h\rho_{Hg} g = \frac{16Q^2}{2\pi^2}\rho \left( \dfrac{1}{ d_1^4} - \dfrac{1}{ d_2^4} \right) $$
Solving for $Q$;
$$ \dfrac{ h\rho_{Hg} g\pi^2 }{8\rho\left( \dfrac{1}{ d_1^4} - \dfrac{1}{ d_2^4} \right) }= Q^2 $$
$$Q=\sqrt{\dfrac{ h\rho_{Hg} g\pi^2 }{8\rho\left( \dfrac{1}{ d_1^4} - \dfrac{1}{ d_2^4} \right) }}$$
Plugging the known and remember to convert the units to SI units.
$$Q=\sqrt{\dfrac{ (10\times 10^{-2})(13600) (9.8)\pi^2 }{8(1.28)\left( \dfrac{1}{ (2\times 10^{-3})^4} - \dfrac{1}{ (1\times 10^{-2})^4} \right) }}$$
$$Q= \color{red}{\bf 4.537\times 10^{-4}}\;\rm m^3/s$$
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b) To find the two velocities, we just need to plug the previous result into (2) and (3).
$$v_1=\dfrac{4(4.537\times 10^{-4})}{\pi (2\times 10^{-3})^2}=\color{red}{\bf 144}\;\rm m/s$$
And
$$v_2=\dfrac{4(4.537\times 10^{-4})}{\pi (1\times 10^{-2})^2}=\color{red}{\bf 5.78}\;\rm m/s$$
