Mercury is poured into a U-shaped manometric tube with a cross section of 1 cm2; the mercury levels in both knees

Mercury is poured into a U-shaped manometric tube with a cross section of 1 cm2; the mercury levels in both knees are the same. The air volume in the sealed elbow is 10 cm3. How much mercury do you need to put into an open knee to fill it? Atmospheric pressure is 1.01 kPa, both legs of the pressure gauge are the same.

Find the excess pressure using the following formula:
h = V / S;
h = 10 * 10 ^ (- 6) / (1 * 10 ^ (- 4));
h = 0.1 m;
Find the pressure above the sealed end using the formula:
P = Po * V / V1 = Po * h / (h – h1);
Let’s transform:
Po * h / (h – h1) = Po + ro * g * (h – h1);
ro * g * (h – h1) ^ 2 + Po * (h – h1) – Po * h = 0;
D = Po * (Po + 4 * ro * g * h);
(h – h1) = (- Po + √ (Po * (Po + 4 * ro * g * h))) / (2 * ro * g);
h1 = h – (- Po + √ (Po * (Po + 4 * ro * g * h))) / ro * g;
Then we get the following:
ho = Po / (ro * g);
ho = 101 * 10 ^ 3 / (13600 * 9.8);
ho = 0.758 m;
h1 = h – (- ho + √ (ho * (ho + 4 * h)) / 2);
h1 = h – ho * (√ (1 + 4h / ho) – 1) / 2;
h1 = 0.1 – 0.758 * (√ (1 + 4 * 0.1 / 0.758) – 1) / 2;
h1 = 0.0106;
Then we find the total height of the mercury column:
H = h + h1;
H = 2h – ho * (√ (1 + 4h / ho) – 1) / 2;
H = 0.1 + 0.0106;
H = 0.1106 m;
Let’s find the total volume:
V / hg = H * S = 2 * V – ho * (√ (1 + 4h / ho) – 1) * S;
V / hg = 2 * V – ho * (√ (1 + 4 * V / (ho * S)) – 1) * S;
V / hg = 11.06 cm3;
Let’s find the total mass of the poured mercury:
m = V / hg * ro;
m = 11.06 * 13.6;
m = 150 gr.
Answer: m = 150 gr.