Sunday, June 5, 2016

Physics 9702 Doubts | Help Page 239

  • Physics 9702 Doubts | Help Page 239



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

For springs P and Q (each having spring constant k),
k1eff = k + k = 2k

For the effective spring constant of the whole system,
1 / keff = 1/2k + 1/3k                (since spring R has spring constant 3k)
keff = [1/2k + 1/3k]-1 = 6k / 5

Hooke’s law: F = keff e
Extension e = W / keff = 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.




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