Fig. 4.1 shows a metal cylinder of height 4.5 cm and base area 24 cm2.

Question 38

Fig. 4.1 shows a metal cylinder of height 4.5 cm and base area 24 cm².

Fig. 4.1

The density of the metal is 7900 kg m^-3.

(a) Show that the mass of the cylinder is 0.85 kg. [2]

(b) The cylinder is placed on a plank, as shown in Fig. 4.2.

Fig. 4.2

The plank is at an angle of 40° to the horizontal.

Calculate the pressure on the plank due to the cylinder. [3]

(c) The cylinder then slides down the plank with a constant acceleration of 3.8 m s^-2.

A constant frictional force f acts on the cylinder.

Calculate the frictional force f. [3]

Reference: Past Exam Paper – November 2015 Paper 21 Q4

Solution:

(a)

density = mass / volume        

{mass = density × volume

Volume = height × base area

Volume = 4.5×10^-2 × 24×10^-4 = 4.5 × 24 × 10^-6 m³}

mass = 7900 × 4.5 × 24×10^-6 = 0.85 (0.853) kg

(b)

pressure = force / area

{The force is the component of the weight acting perpendicularly to the surface of the plank.

Force = Weight × cos 40° = W cos 40°}

force = W cos 40°

{pressure = force / area

Force = W cos 40° = mg cos 40°}

pressure = (0.85 × 9.81 cos 40°) / (24×10^-4)

pressure = 2.7 (2.66) × 10³ Pa

(c)

{Resultant force F = ma}

F = ma

{Consider the forces acting along the surface of the plank.

Component of weight (downward) along plank = W sin 40° = mg cos 40°

The frictional force f opposes motion and so, acts upwards along the plank.

The resultant force is downwards as thr cylinder slides down the plank.

Resultant force = W sin 40° – f = ma}

W sin 40° – f = ma

0.85×9.81×sin 40° – f = 0.85 × 3.8

{5.36 – f = 3.23}

f (= 5.36 – 3.23) = 2.1 N

• 9702 June 2019 Paper 22 Worked Solution | A-Level Physics

Question 5
A battery of electromotive force (e.m.f.) 5.6 V and internal resistance r is connected to two external resistors, as shown in Fig. 5.1.

A long solenoid has an area of cross-section of 28 cm2, as shown in Fig. 5.1.

Question 7

(a) Define the tesla. [2]

(b) A long solenoid has an area of cross-section of 28 cm², as shown in Fig. 5.1.

Fig. 5.1

A coil C consisting of 160 turns of insulated wire is wound tightly around the centre of

the solenoid.

The magnetic flux density B at the centre of the solenoid is given by the expression

B = μ0n I

where I is the current in the solenoid, n is a constant equal to 1.5 × 10³ m^-1 and μ0 is

the permeability of free space.

Calculate, for a current of 3.5 A in the solenoid,

(i) the magnetic flux density at the centre of the solenoid, [2]

(ii) the flux linkage in the coil C. [2]

(c) (i) State Faraday’s law of electromagnetic induction. [2]

(ii) The current in the solenoid in (b) is reversed in direction in a time of 0.80 s.

Calculate the average e.m.f. induced in coil C. [2]

Reference: Past Exam Paper – June 2013 Paper 41 & 43 Q5

Solution:

(a) The tesla is the (uniform magnetic) flux is normal to a long (straight) wire carrying a current of 1 A, creates a force per unit length of 1 N m^–1.

(b)

(i)

{Magnetic flux density B = μ₀n I}

Magnetic flux density B = (4π×10^–7) × (1.5×10³) × 3.5

Magnetic flux density B = 6.6 × 10^–3 T

(ii)

{Flux linkage = Magnetic Flux × Number of turns

Flux linkage = BA × N

Cross-sectional area A should be in SI units.}

Flux linkage = (6.6 × 10^–3) × (28 × 10^–4) × 160

Flux linkage = 3.0 × 10^–3 Wb

(c) (i) Faraday’s law of electromagnetic induction states that the (induced) e.m.f. is proportional to rate of change of (magnetic) flux (linkage).

(ii)

{To reverse the direction of the current, the current must first be reduced to zero, and then increased to the same value in the opposite direction. So, the flux linkage calculated above should be  considered twice.

Average e.m.f. induced = change in flux linkage / time}

Average e.m.f. = (2 × 3.0 × 10^–3) / 0.80

Average e.m.f. = 7.4 × 10^–3 V

What happens when waves pass through a gap equal to their wavelength?

Question 30

What happens when waves pass through a gap equal to their wavelength?

A There is diffraction and the wavelength decreases.

B There is diffraction and the wavelength stays the same.

C There is no diffraction and the wavelength decreases.

D There is no diffraction and the wavelength stays the same.

Reference: Past Exam Paper – November 2018 Paper 12 Q28

Solution:

Answer: B.

When the gap is equal in width to the wavelength of the wave, diffraction occurs. [C and D incorrect]

During diffraction, the wavelength stays the same as the wave has not changed medium. It has only passed through a gap. The distance between wavefronts remains constant. [B is correct]