**ANSWERS:**

**QN1:**

From the law of floatation: A floating object displaces its own weight in the fluid in which it floats. So the amount of water displaced by the boat is likewise **10000 tonnes**

**QN2:**

The important thing to note here is that, in order for the piece made of both iron and wood (attached together) to float in water, the THE COMBINATION OF THEIR DENSITY must be less than that of water, i.e. assuming their combined total volume is V, then their combined density will be

combined density,(CD)=(mass of iron (MI)+mass of wood)/V................(i)

Also,

The total volume, V=Volume of iron+Volume of wood

V=(Mass of iron,(MI)/density of iron)+(Mass of wood/density of wood)..........................................................(ii)

So the above two equations will provide us with the required solution, bearing in mind that the density of water is 1.0g/cc

With the following information is provided,

Density of iron=0.8g/cc

Density of wood=0.3g/cc

Mass of wood=100g

Volume of iron=?

then in order to get the required volume (of iron), we need first to know the mass of iron (MI), after which we will get its volume by taking the ratio of its mass to its density (given above as 0.8g/cc)

Therefore from (i) above:

CD=(MI+100)/V

and at the moment let's consider this quantity to be equal to the density of water, but bearing in mind that it is supposed to be less that that of water for floating conditions for the pieces of woood and iron to be satisfied.

Therefore:

(MI+100)/V=1

from which we get

V=MI+100

Again from (ii) above

(MI/0.8)+(100/0.3)=V

From these two equations therefore, we finally find that:

(MI/0.8)+(100/0.3)=MI+100 or simply

30MI+8000=24MI+2400

from which upon solving, we get a **NEGATIVE VALUE OF MI.** Now since the mass of iron cannot be negative,** it implies that IT IS THEREFORE IMPOSSIBLE for such a combination of masses of wood and iron to be attached together so that it floats in water!**

On the other hand however, assuming the MI were a positive value, we could simply divide MI to the density of iron to get the THRESHOLD volume i.e. the value we were to get would be the maximum volume above which there would be no possibility for the two attached pieces to flow. This is true bearing in mind the assumption we made when we equated to 1.0g/cc, the combined density(CD) of the two pieces (i.e to the density of water, 1.0g/cc)

**QN3:**

(a) UPTHRUST=weight in air-weight in water=40N-30N=**10N**

(b) From the law of flotation, Upthrust=weight of water displaced

UPTHRUST=10N=1kgf=**1000gf**, assuming the acceleration due to gravity is 10 meters per second per second!

(c)Mass of displaced water is **1000g**

(d)Volume of displaced water is 1000g/(1.0g/cc)=**1000cc**

(e) Mass and volume of the object

Weight of object in air is 40N. The mass of the object is therefore 4kg or **4000g**

Volume of object=volume of water displaced=mass of water displaced/density of water=1**000cc**

(f)Relative density (RD) =(Mass/volume)/(density of water)=(4000g/1000cc)/1.0g/cc)=**4.0**

(g)Density=RDx density of water=4.0x1g/cc=**4.0g/cc**

**QN4:**

Given, weight in air 60N=6kgf=6000gf

weight in water 40N

Therefore:

Upthrust=20N=Weight of water displaced=2kgf=2000gf

mass of water displaced=2000g

Volume of the object=2000gxdensity of water=2000gx1.0g/cc=2000cc

Hence:

(a) **RD=(6000g/2000cc)/1.0g/cc)=3.0**

(b)** Density=3.0x1.0g/cc=3.0g/cc**

Kama kuna sehemu nimekosea, naomba mnirekebishe, ni siku nyingi. Natumia copyrigt ya ubongo wa mwaka 1987 (More than 30 years!)