An object is solid throughout. When the object is completely submerged in ethyl

Question

An object is solid throughout. When the object is completely submerged in ethyl alcohol, its apparent weight is 15.0 N. When completely submerged in water, its apparent weight is 12.3 N. What is the volume of the object?

A solid cylinder (radius = 0.200 m, height = 0.170 m) has a mass of 19.8 kg. This cylinder is floating in water. Then oil (the specific gravity of the oil = 0.813) is poured on top of the water until the situation shown in the drawing results. How much of the height of the cylinder is in the oil?

Explanation / Answer

Let the volume of the solid object be V

We know,

Apparent weight= real weight- Buyuont force

So, In ethyl alcohol,

15= Weight(W) - (V*et*g) [where et is volume of ethyl alcohol]

In water,

12.3 = W-V*w*g [ w is density of water]

As W is equal in both cases,

15-12.3=(V*w*g) - (V*et*g)

2.7 =V*g*(w-et)

w=1000kg/m3 and et=789kg/m3

g=9.8m/s2

Putting the values we get V=1.306 * 10-3 m3

Second part:

Well to solve this I need the drawing which is not given. However, I can give an idea of how to solve this type of a question.

Assuming the entire cylinder is submerged into oil and water.

Let the height inside the oil be x, so height inside water=(0.17-x)

So, since the cylinder is floating, apparent weight=0

So, Weight= Buyont force by oil + buyont force by water

so, 19.8*g= V1 * 813*g + V2*1000*g

V1=x*(r2)

V2=(.17-x)*(r2)

So, 813x + 1000(0.17-x)=19.8/(r2)=157.56

x=0.0665 m

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