First exercise: Determination of the volume of a solid Consider a solid (S) of density ρs = 1 g/cm3. (S) is immersed in a liquid of density ρ.(S) is in equilibrium and the volume of the immersed part is Vi (adjacent figure).

1-(S) floats on the surface of a liquid. a) Name the two forces acting on (S).

:Buoyant Force (Archimedes up thrust)

: weight of the object

b) Tell, for each of the two forces, whether it is a contact force or an action from a distance force.

Weight : action from a distance. Archimedes up thrust force: contact force.

c) Give the line of action and the direction of each of these two forces.

𝑊+ h𝑇 = 0 Condition of equilibrium

𝑊=− h𝑇

d) Write down the vector relation between these two forces.

i. Line of action. ( vertical) ii. Direction. (downward) 𝑊 i. Line of action. ( vertical)

ii. Direction. (upward)

e) Reproduce the figure and represent, without a scale, these two forces.

2- We repeat the experiment by putting (S) successively in different liquids. The adjacent graph represents the Variation of Vi as a function of ρ.

a) According to the graph, does the volume of the immersed part increase or decrease when the density of the liquid increases?

According to the graph, we notice that the immersed volume decreases when the density of the liquid increases.

𝐹 h𝑇 =𝜌𝑉 𝑖𝑚𝑚𝑔

𝑉 𝑖𝑚𝑚=𝐹 h𝑇

𝜌𝑔 are inversely proportional

b) For ρ = 1 g/cm3, (S) is totally immersed in the liquid. Why?

𝜌𝑙𝑖𝑞𝑢𝑖𝑑=𝜌𝑠𝑜𝑙𝑖𝑑=1 g / cm 3 Floating object in the liquid

If it was not written in the question “According to the graph”, we can solve it by this way.

The immersed volume decreases when the density of the liquid increases.

c) Deduce graphically the volume of (S).

For =1g/cm3 , we find Vi = 200 cm3 .

Vi = Vs = 200 cm3 (S) is totally immersed

d) Calculate the mass m of (S).

𝜌=𝑚𝑣

𝑚=𝜌×𝑣𝑚=(1×103)×(200×10−6)

𝑚=0.2𝑘𝑔

𝑂𝑅𝜌=𝑚𝑣

𝑚=𝜌×𝑣𝑚=1×200

𝑚=0.2𝑘𝑔𝑚=200𝑔×10−3

e) Calculate the weight W of (S).

W=m×𝑔W=0.2×10W=2𝑁

3- (S) is immersed in another liquid of density ρL . We observe that 25 % of the object is immersed.

a) Show that the density of this liquid is ρL = 4 g/cm3.

𝑽 𝑖𝑚𝑚=25

100 𝑽 𝑜𝑏𝑗𝑒𝑐𝑡

𝑽 𝑖𝑚𝑚=0.25×200¿50 cm 3According to the graph , for Vi = 50 cm3 we find =4g/cm3

b) Deduce the value F of the Archimedes up thrust exerted by liquid on (S).

𝐹 h𝑇 =𝜌𝑉 𝑖𝑚𝑚𝑔𝐹 h𝑇 =(4×103)×(50×10−6)×10

𝐹 h𝑇 =2𝑁

c) By comparing W and F, deduce that the solid floats at the surface of the liquid.

𝐹 h𝑇 =𝑊

Second exercise Normal functioning of a lamp The object of this exercise is to study the functioning of a lamp (L) that carries the inscriptions (3V; 3W).I- Resistance of the lamp (L)The lamp (L) is connected in a convenient circuit so as to function normally.

1) a) What is the voltage across (L)?

UL = 3 V

b) What is the power consumed by (L)?

c) Deduce the value of the current I0 carried by (L).

2) (L) may be considered as a resistor of resistance r. Show that r = 3 Ω.

P = 3 W

𝑃= 𝐼𝑜𝑈 𝐼𝑜=𝑃𝑈 𝐼 𝑜=

33=1𝐴

Ohm’s Law

𝑟=31=3Ω𝑟=

𝑈 𝐿

𝐼 𝑜

II- Functioning of the lamp (L) We connect (L) in series with a resistor (D) of resistance R = 17 Ω across the poles of a generator delivering a constant voltage UPN = 12V. A current I passes then in the circuit. 1) a) Determine the value of the resistance equivalent to the combination of R and r.

Re = R + r ( series grouping ) Re = 17 + 3 = 20

b) Determine the value of I.

Ohm’s Law

𝐼=𝑈𝑔

𝑅𝑒𝑞𝐼=12

20 𝐼=0.6 𝐴 c) (L) does not function normally. Why?

(L) will not function normally because I = 0.6 A < = 1 A

2) To make (L) function normally, we replace (D) by another resistor (D’) of resistance R’. R’ must be smaller than R. Why?

𝑈𝑔=𝐼×𝑅𝑒𝑞

𝑅𝑒𝑞=𝑈𝑔

𝐼

According to Ohm’s law I and are inversely proportionalFor (L) to function normally I should increase Re should decrease R should decrease R’ < R

THIRD EXERCISE The magnifierThe following figure shows:The lens (L), its optical axis, its object focus F and its image focus F’.The image A’B’ of object AB given by (L).A particular incident ray issued from B, in a direction passing through F and meeting the lens at point I.- An emergent ray KF’ corresponding to a particular ray issued from B.

1- Characteristics of the image a) What is the nature of A’B’? Why ?

A’B’ is virtual since it is on the same side of the object focus w.r.t the lens.

b) Determine graphically the size of the image and its distance from the lens

A ’B’=𝑠𝑐𝑎𝑙𝑒×𝑑𝑖𝑣𝑖𝑠𝑖𝑜𝑛A ’B’=1×3A ’B’=3 𝑐𝑚

O

O 𝐴 ′=1×6O 𝐴 ′=6 𝑐𝑚

2- Construction of the object AB a) Trace on your question sheet:1. The path of the emergent ray corresponding to the incident ray FI

2. The incident ray corresponding to the emergent ray KF’.

b) Construct the object AB

B

A

The point object B is the intersection of the two incident rays.A is the orthogonal projection of B on the optical axis.

b) The distance of the object AB from the lens.

B

A

3- Characteristics of the objectDetermine graphicallya) The size of the object AB.

AB=𝑠𝑐𝑎𝑙𝑒×𝑑𝑖𝑣𝑖𝑠𝑖𝑜𝑛AB=1×1AB=1𝑐𝑚

O

O A=1×2O A=2𝑐𝑚

4- Role of a magnifierThe lens L acts, in this case, as a magnifier. Why?

The lens acts as a a magnifier .Sinceis virtual, erect, and larger than the objectand (L) is converging