DP Physics (first assessment 2025)

Question EXE.2.HL.TZ0.2

Date	Example questions Example questions	Marks available	[Maximum mark: 20]	Reference code	EXE.2.HL.TZ0.2
Level	HL	Paper	2	Time zone	TZ0
Command term	Calculate, Draw, Explain, Outline, Show that, State, Suggest	Question number	2	Adapted from	N/A

[Maximum mark: 20]

EXE.2.HL.TZ0.2

A geophone is an instrument designed to measure the movement of ground rocks.

When the ground moves, the magnet-spring system oscillates relative to the coil. An emf is generated in the coil. The magnitude of this emf is proportional to the speed of the magnet relative to the coil.

(a.i)

State the movement direction for which the geophone has its greatest sensitivity.

[1]

Markscheme

Vertical direction / parallel to springs ✓

(a.ii)

Outline how an emf is generated in the coil.

[2]

Markscheme

The magnetic field moves relative to the coil ✓

As field lines cut the coil, forces act on (initially stationary) electrons in the wire (and these move producing an emf) ✓

(a.iii)

Explain why the magnitude of the emf is related to the amplitude of the ground movement.

[3]

Markscheme

The springs have a natural time period for the oscillation ✓

A greater amplitude of movement leads to higher magnet speed (with constant time period) ✓

So field lines cut coil more quickly leading to greater emf ✓

(a.iv)

In one particular event, a maximum emf of 65 mV is generated in the geophone. The geophone coil has 150 turns.

Calculate the rate of flux change that leads to this emf.

[2]

Markscheme

Use of $ε = - \frac{Δ (N Φ)}{Δ t}$ ✓

$\frac{ΔΦ}{Δ t} = (-) \frac{65 \times 10^{- 3}}{150}$

$= 0.43$ mWb s⁻¹ ✓

(b)

Suggest two changes to the system that will make the geophone more sensitive.

[4]

Markscheme

Any two suggestions from:

Increase number of turns in coil ✓
Because more flux cutting per cycle ✓

Increase field strength of magnet ✓
So that there are more field lines ✓

Change mass-spring system so that time period decreases ✓
So magnet will be moving faster for given amplitude of movement ✓

The geophone is mounted on the ground at point Z and an explosion is produced at point W some distance away. Sound from the explosion travels to the geophone via the clay layer in the ground.

Diagram not to scale

The speed of sound in clay is 3.00 km s⁻¹; the speed of sound in sandstone is 4.70 km s⁻¹

(c.i)

Show that, when sound travels from clay to sandstone, the critical angle is approximately 40°.

[2]

Markscheme

_cn_s $= \frac{3.0}{4.7} = 1.57$ ✓

Critical angle $= \sin^{- 1} (\frac{1}{1.57}) = 39.6 °$ ✓

(c.ii)

The angle between the clay–air surface and path 1 is 80°.

Draw, on the diagram, the subsequent path of a sound wave that travels initially in the clay along path 1.

[2]

Markscheme

ray shown reflected back into the clay (and then to Z) at (by eye) the incidence angle ✓

ray shown refracted into the sandstone with angle of refraction greater than angle of incidence (by eye) ✓

Another explosion is produced at X. The sound from this explosion is detected twice at the geophone at Z. Some sound travels directly from X to Z through clay along path 2. Other sound travels through clay via Y along path 3.

The vertical thickness of the clay layer is d. The distance XZ is 80.0 m.

The time between the arrival of the sounds due to the path difference is 6.67 ms.

(d)

Calculate d.

[4]

Markscheme

distance difference $= (3000 \times 0.00667 =) 19.8$ m ✓

½ distance difference $= 9.9$ m so YZ $= 49.9$ m ✓

$d = (\sqrt{{YZ}^{2} - {(\frac{XZ}{2})}^{2}}) = \sqrt{49 . 9^{2} - 40^{2}}$ ✓