• 9702 March 2016 Paper 42 Worked Solutions | A-Level Physics

Question 3
A student carries out an experiment to determine the specific heat capacity of a liquid using the apparatus illustrated in Fig. 3.1.




Question 6
A parallel beam of ultrasound is incident normally on the surface of a layer of fat of thickness 1.1 cm, as shown in Fig. 6.1.




Question 9
A particle of charge +q and mass m is travelling with a constant speed of 1.6 × 105 m s-1 in a vacuum.

Which row best defines elastic and plastic behaviour of a material?

Question 7

Which row best defines elastic and plastic behaviour of a material?

Reference: Past Exam Paper – November 2009 Paper 11 Q20 & Paper 12 Q19

Solution:

Answer: D.

An elastic material returns to its original shape and size when the deforming force is removed.

Up to the limit of proportionality, the extension of the material is proportional to the force. The curve of the force-extension graph is linear. The material is said to obey Hooke’s law up to the limit of proportionality.

Beyond the limit of proportionality P is a point called the elastic limit. From extension = 0 to the P, the extension is proportional to the force and the material regains its original shape when the force is removed.

Beyond P up to the elastic limit, the extension is no longer proportional to the force (Hooke’s is no longer obeyed) BUT the material still regain its original shape when the force is removed.

Thus, as long as the elastic limit is not exceeded, the material have an elastic behaviour. For the extensions up to the limit of proportionality P, the extension is proportional to the force (i.e. it obeys Hooke’s law – the graph is a straight line). Even beyond P, the material would behave elastically (regain its original size and shape) as long as the elastic limit is not reached. [A incorrect]

As for a plastic material, as the force is removed, it does not regain its original size and shape – it suffers permanent deformation. If a large enough force is applied, it could break. [C incorrect]

A horizontal force-extension curve would mean that for the extension does not depend on the force applied. Meaning that for any given force, the material would extend continuously. [B incorrect]

A metal wire in a circuit is damaged. The resistivity of the metal is unchanged but the cross-sectional area of the wire is reduced over a length of 3.0 mm, as shown in Fig. 6.2.

Question 10

(a) State what is meant by an electric current. [1]

(b) A metal wire has length L and cross-sectional area A, as shown in Fig. 6.1.

Fig. 6.1

I is the current in the wire,

n is the number of free electrons per unit volume in the wire,

v is the average drift speed of a free electron and

e is the charge on an electron.

(i) State, in terms of A, e, L and n, an expression for the total charge of the free electrons in the wire. [1]

(ii) Use your answer in (i) to show that the current I is given by the equation

I = nAve.

[2]

(c) A metal wire in a circuit is damaged. The resistivity of the metal is unchanged but the cross-sectional area of the wire is reduced over a length of 3.0 mm, as shown in Fig. 6.2.

Fig. 6.2

The wire has diameter d at cross-section X and diameter 0.69 d at cross-section Y.

The current in the wire is 0.50 A.

(i) Determine the ratio

average drift speed of free electrons at cross-section Y

average drift speed of free electrons at cross-section X

 [2]

(ii) The main part of the wire with cross-section X has a resistance per unit length of

1.7 × 10^-2 Ω m^-1.

For the damaged length of the wire, calculate

1. the resistance per unit length, [2]

2. the power dissipated. [2]

(iii) The diameter of the damaged length of the wire is further decreased. Assume that the current in the wire remains constant.

State and explain qualitatively the change, if any, to the power dissipated in the damaged length of the wire. [2]

 [Total: 12]

Reference: Past Exam Paper – November 2017 Paper 22 Q6

Solution:

(a) Electric current is the flow of charge carriers.

(b)

(i)

{n is the number of electrons per unit volume

A×L = volume

So, nAL is the number of electrons

Charge of 1 electron = e

Total charge = number of electrons × charge of 1 electron}

nALe

(ii)

(t is time taken for electrons to move length L)

Current I = Q / t

I = nALe / t

or

{Speed v = distance / time                  v = L / t                        so, t = L / v}

I = nALe / (L / v)

or

I = nAvte / t and I = nAve

(c)

(i)

{ I = nAve        so, v = I / nAe

The drift velocity v is inversely proportional to the cross-sectional area A

Ratio    = v at cross-section Y / v at cross-section X

            = area at X / area at Y}

ratio     = area at X / area at Y

= [πd² / 4] / [π(0.69d)² / 4]       or d² / (0.69d)²                or 1 / 0.69²

= 2.1

(A = πd²          so, A is proportional to d²)

(ii)

1.

{R = ρL / A      giving R/L = ρ / A        Since ρ is the same for wires of the same material.}

R = ρL / A                   or R / L ∝ 1 / A          

{Resistance per unit length: R / L ∝ 1 / A

For wire X: 1.7 × 10^-2 Ω m^-1 ∝ 1 / area X                 eqn (1)

For wire Y: resistance per unit length  ∝ 1 / area Y   eqn (2)

Divide (2) by (1),}

resistance per unit length        = 1.7 × 10^-2 × (area at X / area at Y)

= 1.7 × 10^-2 × 2.1

= 3.6 × 10^-2 Ω m^-1

2.

P = I²R            or P = V²/ R

{Damaged length = 3.0 mm

Resistance per unit length = 3.6 × 10^-2 Ω m^-1

Resistance of damaged length = Resistance per unit length × Damaged length}

R = 3.6 × 10^-2 × 3.0 × 10^-3 (= 1.08 × 10^-4 Ω)

{P = I²R                                   or P = V²/ R }

P = 0.50² × 1.08 × 10^-4               or P = (5.4 × 10^-5)² / 1.08 × 10^-4

P = 2.7 × 10^-5 W

(iii)

{R = ρL / A}

The cross-sectional area decreases, so the resistance increases

(P = I²R) So, the power increases