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Question 1112: [Vectors]

The vector diagram shows three coplanar forces acting on an object at P.

The magnitude of the resultant of these three forces is 1 N.

What is the direction of this resultant?

Reference: Past Exam Paper – June 2005 Paper 1 Q14

Solution 1112:

Answer: D.

First, consider the resultant of the 3 N vertical force and the 4 N horizontal force. This resultant is in the north-east direction, and its magnitude can be obtained from Pythagoras’ theorem.

Resultant = √(3² + 4²) = 5 N

The above resultant is anti-parallel to the 4 N force shown.

Thus, the direction of the resultant of all the three vectors is in north-east direction and its magnitude is (5 – 4 =) 1 N.

Question 1113: [Deformation > Hooke’s law]

Three springs are arranged vertically as shown.

Springs P and Q are identical and have spring constant k. Spring R has spring constant 3k.

What is the increase in the overall length of the arrangement when a force W is applied as shown?

A 5W / 6k                   B 4W / 3k                    C 7kW / 2                    D 4kW

Reference: Past Exam Paper – November 2012 Paper 11 Q23

Solution 1113:

Answer: A.

For springs in series and in parallel, the following formulae for the effective spring constant apply:

In parallel: effective spring constant, k_eff = k₁ + k₂ + ….

In series: effective spring constant, 1/k_eff = 1/k₁ + 1/k₂ + ….

For springs P and Q (each having spring constant k),

k_1eff = k + k = 2k

For the effective spring constant of the whole system,

1 / k_eff = 1/2k + 1/3k                (since spring R has spring constant 3k)

k_eff = [1/2k + 1/3k]^-1 = 6k / 5

Hooke’s law: F = k_eff e

Extension e = W / k_eff = W / (6k/5) = 5W / 6k  

Question 1114: [Current of Electricity]

When will 1 C of charge pass a point in an electrical circuit?

A when 1 A moves through a potential difference of 1 V

B when a power of 1 W is used for 1 s

C when the current is 5 mA for 200 s

D when the current is 10 A for 10 s

Reference: Past Exam Paper – June 2012 Paper 12 Q32

Solution 1114:

Answer: C.

Choice A: Ohm’s law: V = IR. Resistance of a conductor is 1 ohm when 1 A moves through a potential difference of 1 V

Choice B: Power = Energy / time. 1 J of energy is involved when a power of 1 W is used for 1 s

Choice C: Charge Q = It. Here, Q = (5×10^-3) × 200 = 1 C. 1 C of charge pass a point in an electrical circuit when the current is 5 mA for 200 s. [C is correct]

Choice D: Charge Q = It. Here, Q = 10 × 10 = 100 C.