A metal wire is attached at one end to a fixed point and a load is hung from the other end so that the wire hangs vertically.

Question 3

A metal wire is attached at one end to a fixed point and a load is hung from the other end so that the wire hangs vertically. The load is increased from zero to 20 N. This causes the wire to extend elastically by 5.0 mm. The load is then reduced to 12 N and the extension decreases to 3.0 mm.

How much strain energy is released during the unloading process?

A 0.8×10^-2 J             B 1.8×10^-2 J               C 2.4×10^-2 J             D 3.2×10^-2 J

Reference: Past Exam Paper – June 2016 Paper 12 Q21

Solution:

Answer: D.

Strain energy released under the unloading process is obtained by the area under the load-extension graph between 3.0 mm and 5.0 mm.

Strain energy   = area under graph

                        = area of trapezium

                        = ½ × (12+20) × (5.0 – 3.0) × 10^-3

                        = 0.032 J = 3.2×10^–2 J

 
Alternatively, the area under the graph could be split into the area of a triangle and that of a rectangle instead of calculating the area of a trapezium.

The diagram shows the reading on an analogue ammeter. Which digital ammeter reading is the same as the reading on the analogue ammeter?

Question 6

The diagram shows the reading on an analogue ammeter.

Which digital ammeter reading is the same as the reading on the analogue ammeter?

display units                display reading

A                μA                        1600

B                μA                        160

C                mA                       16.0

D                 A                          1.60

Reference: Past Exam Paper – November 2014 Paper 11 & 12 Q3

Solution:

Answer: A.

This question is on the conversion of units.

Reading on analogue ammeter = 1.6 mA = 1.6 × 10^-3 A

1 μA = 10^-6 A

So, 1 A = 10⁶ uA

So, digital ammeter reading = 1.6 × 10^-3 A = 1.6 × 10^-3 × 10⁶ μA 

Digital ammeter reading= 1.6 × 10³ μA = 1600 μA

A number of identical springs are joined in four arrangements. Which arrangement has the same spring constant as a single spring?

Question 2

A number of identical springs are joined in four arrangements.

Which arrangement has the same spring constant as a single spring?

Reference: Past Exam Paper – June 2016 Paper 11 Q20

Solution:

Note that more notes on springs in series and parallel can be obtained here.

Answer: C.

Let the spring constant of a single spring = k

When 2 springs are joined end-to-end (in series),

Effective spring constant = (1/k + 1/k)^-1 = k/2                        [Choice A]

When 2 springs are joined side-by-side (in parallel),

Effective spring constant = k + k = 2k                                   [Choice B]

Choice C may be considered to be as if the arrangement in B is in series with another similar one [or equivalently, the arrangement in A is in parallel with another similar one]. 

Considering the former case,

Effective spring constant = k/2 + k/2 = k

This has the same spring constant as a single spring.

Considering the latter case,

Effective spring constant = (1/2k + 1/2k)^-1 = 2k/2 = k

Again, it has the same spring constant as a single spring.