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.