The experiment of measuring the density of brine Here is a scale, a cuboid block, water and beaker. Please design an experiment to measure the density of brine

The experiment of measuring the density of brine Here is a scale, a cuboid block, water and beaker. Please design an experiment to measure the density of brine

1. Put water into the beaker, and then put the wood block into the beaker. Measure the underwater depth of the wood block with a scale, which is h, and record the bottom area of the wood block as S2. Similarly, put the wood block into the brine, and then measure the underwater depth of the wood block with a scale, which is H3

Second grade physics salt water density operation problem fast!
There is a solution: 1. Measure the mass of the empty beaker M1 with a balance; 2. Pour an appropriate amount of salt water into the empty beaker and weigh their total mass m2; 3. Pour the salt water in the beaker into the measuring cylinder and measure the volume V of salt water in the measuring cylinder
4. Calculate the density = (m2-m1) / v
It is said that when the salt water in the beaker is poured into the measuring cylinder, the wall of the beaker will be stained with salt water, and the weight of the salt water is too large
But I don't understand. Although there is salt water on the beaker wall, what part of salt water is poured into the measuring cylinder. Even if there is salt water on the beaker wall, it is still in the total mass. Why is it too large

The reason is very simple. So, if you pour all the water into the measuring cylinder, is there still a little water in the beaker? It is the water that affects the later measurement. The volume of water you see in the measuring cylinder is not the volume of all brine, but the volume of brine left after the total brine has removed the brine left in the beaker, compared with the actual volume, The data you get will be a little smaller, so when you use ρ = m / V, m you use the total mass of brine and the volume of brine left after removing the brine in the beaker, the denominator will become smaller, and the molecular invariant quotient will naturally become larger. Your misunderstanding is that you do not connect the physical quantity with the corresponding number, which is the source of your confusion
The correct way is: 1. Pour an appropriate amount of salt water into the beaker, and measure its mass (the total mass of beaker + salt water) as m; 2. Pour part of salt water into the measuring cylinder (Note: do not pour out, leave part of salt water) and measure its volume as V, and the mass of salt water in the remaining beaker is m; 3. Calculate the density ρ = (M-M) / v
By using the difference, we can avoid the problem you said and improve the accuracy

Xiaoming wound a little lead wire around one end of a uniform wooden pole, so that the wooden pole can float vertically in the liquid, so as to make a densimeter. Put it into the water, the distance between the liquid level and the lower end of the wooden pole is 16.5cm; then put it into the salt water, the distance between the liquid level and the lower end of the wooden pole is 14.5cm. If the volume of the lead wire is very small, it can be ignored, what is the salt water density measured by Xiaoming?
---------Please explain in detail - immortal uncles - thank you!

P water, P salt and P salt water are used to represent the density of water salt and salt water respectively, and t is used to represent the salt content
(by the way, the hope table tells me to use numbers directly)

T = m salt / M total = m salt / (M Salt + m water)
M salt = P brine * V * t
Mwater = pwater * V
T = P brine * V * t / (P brine * V * t + P water * V)
Eliminate V
T = P brine * t / (P brine * t + P water)
There's no digital tape. I'm looking forward to it
PS: "P salt" is absolutely useless. P salt can only be used in solid state. In this case, P salt can not be used when it is completely dissolved