A metal block of volume 1 * 10^3 cm^3 is suspended from a thread as shown in the diagram. It is lowered into a beaker of liquid of unknown density until it is completely immersed but does not touch the bottom of the beaker. The beaker rests on a balance. The reading of the balance before the block is lowered in the liquid is 4.00 kg, and is 5.8 after the block is completely immersed in the liquid. The volume and mass of the thread may be neglected in all the following calculations

The density of the liquid is:

Question 2 options:
A) 9.8 x 103 kg/m3

B) 4.0 x 103 kg/cm3

C) 1.8 x 103 kg/m3

D) 5.8 x 103 kg/m3

If the thread is released so that the block sinks to the bottom of the beaker and rests there, then the pan balance reads 12 kg. The density of the material from which the block is made is:

Question 3 options:
A) 6.2 x 103 kg/m3

B) 1.2 x 103 kg/cm3

C) 5.8 x 103 kg/m3

D) 8.0 x 103 kg/m3

You know the increase in the balance is equal to bouyancy, or 1.8kg

but bouyancy is equal to the density of the liquid*volume displaced, so set them equal, and I think the density then must be equal to 1.8kg/1000cm^3, that converts to one of the answers.

in the last part, you know the total mass of the weight must be 8kg, and you know its volume, figure denstiy

To solve Question 2 and Question 3, we need to use the principles of buoyancy and Archimedes' principle.

Question 2: To find the density of the liquid, we can use Archimedes' principle, which states that the buoyant force acting on an object submerged in a fluid is equal to the weight of the fluid displaced by the object. We can calculate the volume of the liquid displaced by the submerged block, which is equal to the volume of the block itself.

Given that the volume of the block is 1 * 10^3 cm^3, we convert it to meters cubed by dividing by 10^6. So, the volume of the fluid displaced is 1 * 10^(-6) m^3.

Before the block is immersed, the reading on the balance is 4.00 kg. After immersion, the reading becomes 5.8 kg. The increase in weight (5.8 kg - 4.00 kg = 1.8 kg) is equal to the buoyant force acting on the block.

The buoyant force is given by the equation F_b = ρ_fluid * V_fluid * g, where ρ_fluid is the density of the liquid, V_fluid is the volume of the liquid displaced, and g is the acceleration due to gravity (approximately 9.8 m/s^2).

Substituting the values, we have 1.8 kg = ρ_fluid * 1 * 10^(-6) m^3 * 9.8 m/s^2.
Simplifying the equation, ρ_fluid = (1.8 kg) / (1 * 10^(-6) m^3 * 9.8 m/s^2).
Therefore, the density of the liquid is approximately 1.8 x 10^3 kg/m^3, which corresponds to option C.

Question 3: To find the density of the material from which the block is made, we need to consider the situation where the block sinks to the bottom of the beaker. In this case, the weight of the block is balanced by the buoyant force and the normal force exerted by the bottom of the beaker.

The balance reading when the block is at the bottom of the beaker is 12 kg. This weight includes the weight of the block and the weight of the displaced fluid. The weight of the fluid displaced is equal to the buoyant force acting on the block, which we calculated as 1.8 kg in the previous question.

Therefore, the weight of the block alone is 12 kg - 1.8 kg = 10.2 kg. This weight is equal to the mass of the block multiplied by the acceleration due to gravity.

The density of the block can be found using the equation ρ_block = mass_block / volume_block.
The volume of the block is given to be 1 * 10^3 cm^3, which is equal to 1 * 10^(-6) m^3 (after converting to cubic meters).

Substituting the values, we have ρ_block = (10.2 kg) / (1 * 10^(-6) m^3) = 1.02 x 10^7 kg/m^3.
Therefore, the density of the material from which the block is made is approximately 1.02 x 10^7 kg/m^3, which corresponds to option B.