# 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 = √(3

^{2}+ 4^{2}) = 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}= k

_{1}+ k

_{2}+ ….

**In series:**effective spring constant, 1/k

_{eff}= 1/k

_{1}+ 1/k

_{2}+ ….

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.

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