A metal wire of cross-sectional area 0.20 mm2 hangs vertically from a fixed point. A load of 84 N is then attached to the lower end of the wire.

Question 13

A metal wire of cross-sectional area 0.20 mm2 hangs vertically from a fixed point. A load of 84 N is then attached to the lower end of the wire. The wire obeys Hooke’s law and increases in length by 0.30%.

What is the Young modulus of the metal of the wire?

A 1.4 × 10⁵ Pa

B 1.4 × 10⁸ Pa

C 1.4 × 10⁹ Pa

D 1.4 × 10¹¹ Pa

Reference: Past Exam Paper – June 2016 Paper 13 Q21

Solution:

Answer: D.

Young modulus = stress / strain

Stress = force / area

Stress = 84 / (0.2×10^-6)

Recall that 1 mm² = 1 mm × 1 mm = 10^-3 m × 10^-3 m = 10^-6 m²

The wire increases in length by 0.30 %. This represents the strain.

Strain = extension / original length = 0.30 % = 0.30 / 100 = 0.0030              (the percentage needs to be expressed as a decimal)

Young modulus = stress / strain

Young modulus = 84 / [(0.2×10^-6) × 0.0030]

Young modulus = 1.4×10¹¹ Pa

A coil C having 120 turns of wire is wound on to one end of the core. The area of cross-section of coil C is 1.5 cm2.

Question 8

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

(b) A solenoid S is wound on a soft-iron core, as shown in Fig. 9.1.

Fig. 9.1

A coil C having 120 turns of wire is wound on to one end of the core. The area of cross-section of coil C is 1.5 cm².

A Hall probe is close to the other end of the core.

When there is a constant current in solenoid S, the flux density in the core is 0.19 T. The

reading on the voltmeter connected to the Hall probe is 0.20 V.

The current in solenoid S is now reversed in a time of 0.13 s at a constant rate.

(i) Calculate the reading on the voltmeter connected to coil C during the time that the current is changing. [2]

(ii) Complete Fig. 9.2 for the voltmeter readings for the times before, during and after the

direction of the current is reversed.

Fig. 9.2

[4]

 [Total: 8]

Reference: Past Exam Paper – November 2018 Paper 41 & 43 Q9

Solution:

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

(b)

(i)

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

Change in flux linkage = BAN}

induced e.m.f. = (ΔB)AN / Δt

{As the current is reversed, the flux density B is also reversed. That is, it changes from 0.19 T to {to zero and from zero to} – 0.19 T. The change in flux density = 2 × 0.19 T}

induced e.m.f. = (2 × 0.19 × 1.5×10^-4 × 120) / 0.13

induced e.m.f. = 0.053 V                                           

(ii)

reading on voltmeter connected to coil C / V:            0          0.053               0

{For an e.m.f. to be induced, there should be a change in flux density. So, before the current changes and after the current changes, there is no change in current and thus, no change in flux density. So, no e.m.f. is induced.

Note that there IS a current in these cases, but the current is NOT changing.

During the change in current, the induced e.m.f. is 0.053 V (as calculated above).}

reading on voltmeter connected to Hall probe / V: zero in middle column

final column correct sign (negative)                                                              

final column correct magnitude (0.20)                                                          

{Hall voltage = BI / ntq

When the current I is zero, the Hall voltage is also zero. (middle column).

When the current is reversed, its direction changes. So, the Hall voltage will be in the reversed direction – a negative sign should be included.

Hall voltage after current changes = - 0.20 V}

A solid rubber ball has a diameter of 8.0 cm. It is released from rest with the top of the ball 80 cm above a horizontal surface.

Question 40

A solid rubber ball has a diameter of 8.0 cm. It is released from rest with the top of the ball 80 cm above a horizontal surface. It falls vertically and then bounces back up so that the maximum height reached by the top of the ball is 45 cm, as shown.

If the kinetic energy of the ball is 0.75 J just before it strikes the surface, what is its kinetic energy just after it leaves the surface?
A 0.36 J                       B 0.39 J                       C 0.40 J                       D 0.42 J

Reference: Past Exam Paper – June 2013 Paper 11 Q17

Solution:

Answer: B.


This is a tricky question. It is necessary to deal with the bottom of the ball at 72 cm at the start and 37 cm at the end.

From the conservation of energy, as the ball falls, its potential energy is being converted to kinetic energy.


Just before it strikes the surface, the kinetic energy is 0.75 J (all potential energy has been converted to KE). The bottom of the ball has zero PE and a KE of 0.75 J. This value of KE = 0.75 J is for the bottom of the ball. So, throughout our calculation, we need to consider the bottom of the ball instead of the top.

PE of bottom of the ball at the point of release = KE of bottom of ball just before striking the surface
mgh = KE
m × 9.81 × 0.72 = 0.75
Mass m of the ball = 0.11 kg


Now, for the rebound, all of the kinetic energy just after the ball leaves the surface is converted to potential energy at the maximum height reached. The maximum of height of the bottom of the ball after rebound is 45 – 8 = 37 cm.

KE just after leaving the surface = PE of bottom of ball at a height of 37 cm
KE = mgh
KE = 0.11 × 9.81 × 0.37 = 0.3854 J


Note that 0.75 J is the KE of the ball just before hitting the surface while 0.3854 J is the KE of the ball just after leaving the surface. This indicates that some energy (0.75 – 0.3854) J has been lost.