A metal part weighing 390 g has a void. The part was hung on a thread and completely lowered into the water.

A metal part weighing 390 g has a void. The part was hung on a thread and completely lowered into the water. What volume does the void have if the part stretches the thread from a force of 3.35 N

390 g = 0.39 kg
Р = мg, where m – mass, g – gravitational acceleration, Р – weight
P1 = 0.39 kg * 10 n / kg = 3.9 n
P2 = 3.35 n
∆Р = Р1-Р2
∆Р = 3.9 n – 3.35 n = 0.55 n
0.55 n = F (Arch), Archimedes force
F (Arkh) = p (w) * V (mn) * g
Where p (w) is the density of the liquid, V (tn) is the volume of the submerged part of the body
0.55 = 1000 * V (TP) * 10
0.55 = 10000 V (tp)
V (tp) = 0.000055 m3
Density of metal (steel) p (c) = 7800 kg / m3
V (s) = m (s) / p (s)
V (s) = 0.39 kg / 7800 kg / m3 = 0.00005 m3
∆V = V (p) = V (tp) -V (c) = 0.000055 m3 – 0.000050 m3 = 0.000005 m3 = 5 cm3
Answer: V (n) = 0.000005 m3 = 5 cm3