# Physics 9702 Doubts | Help Page 117

__Question 592: [Matter > Hooke’s law]__
Two springs, one with spring
constant k

_{1}= 4 kN m^{–1}and other with spring constant k_{2}= 2 kN m^{–1}, are connected as shown.
What is total extension of the
springs when supporting a load of 80 N?

A 1.3 cm B 4 cm C
6 cm D 60 cm

**Reference:**

*Past Exam Paper – November 2014 Paper 13 Q25*

__Solution 592:__**Answer: C.**

This system consists of more than 1
spring attached in series to each other. The 2 strings each has its own spring
constant. However, to calculate the extension, we need to find out the
effective spring constant in the system.

For springs attached in series, the effective
spring constant is given by

1 / k

_{eff}= 1/k_{1}+ 1/k_{2}
Spring constant, k

_{eff}= [1/k_{1}+ 1/k_{2}]^{-1}= [1/4 + 1/2]^{-1}= 4/3 kN m^{-1}
Spring constant, k

_{eff}= 4/3 kN m^{-1}= 4000/3 Nm^{-1}
Hooke’s law: F = ke

Extension e = F / k

_{eff}= 80 / (4000/3) = 0.06m = 6cm

__Question 593: [Electromagnetism]__**(a)**Define the tesla

**(b)**Two long straight vertical wires X and Y are separated by distance of 4.5 cm, as illustrated in Fig.1.

Wires pass through a horizontal card
PQRS.

Current in wire X is 6.3 A in the
upward direction. Initially, there is no current in wire Y.

(i) On Fig.1, sketch, in plane PQRS,
the magnetic flux pattern due to the current in wire X. Show at least four flux
lines.

(ii) Magnetic flux density B at a
distance x from a long straight current-carrying wire is given by the
expression

B = Î¼

_{0}I / 2Ï€x
where I is the current in the wire
and Î¼

_{0}is the permeability of free space.
Calculate magnetic flux density at
wire Y due to the current in wire X.

(iii) A current of 9.3 A is now
switched on in wire Y. Use answer in (ii) to calculate the force per unit
length on wire Y.

**(c)**Currents in the two wires in (b)(iii) are not equal.

Explain whether force per unit
length on the two wires will be the same, or different.

**Reference:**

*Past Exam Paper – June 2013 Paper 42 Q5*

__Solution 593:__**(a)**The uniform magnetic flux normal to a long straight wire carrying a current of 1A creates

**a force per unit length of 1Nm**

^{-1}. (The tesla is a measure of the magnetic flux density.)

**(b)**

(i)

Sketch:

concentric circles

increasing separation (must show
more than 3 circles)

correct direction (anticlockwise,
looking down)

(right hand grip rule)

(ii)

B = (Î¼

_{o}I)/(2Ï€x) = (4Ï€ x 10^{-7}x 6.3) / (2Ï€ x 4.5x10^{-2}) = 2.8x10^{-5}T
(iii) F = BILsin(Î¸) Î¸ = 90

^{o}
F/L = 2.8x10

^{-5}x 9.3 x 1 = 2.6x10^{-4}Nm^{-1}**(c)**The force per unit length depends on the product of the two currents.

__OR__From Newton’s third law, the action and reaction are equal and opposite. So, the force per unit length is the same for both.

__Question 594: [Forces > Weight]__
What is meant by weight of an
object?

A the gravitational field acting on
the object

B the gravitational force acting on
the object

C the mass of the object multiplied
by gravity

D the object’s mass multiplied by
its acceleration

**Reference:**

*Past Exam Paper – June 2007 Paper 1 Q9*

__Solution 594:__**Answer: B.**

The weight of an object is the
gravitational FORCE acting on the object. Weight is a force and its unit is
newton. The gravitational field is a region in space where a mass experiences a
force. [A is incorrect]

Usually, when we say gravity, we
refer to the force of gravity. Weight is not the mass of the object multiplied
by ‘gravity’, but ‘acceleration due to gravity’ or ‘gravitational field
strength’. [C is incorrect]

A moving object (e.g. a car) may
accelerate forwards, but this does not make it its weight. Weight is due to acceleration
due to GRAVITY, not any other acceleration. [D is
incorrect]

__Question 595: [Current of Electricity]__
A network of resistors, each of resistance
R, is shown in Fig.1.

**(a)**Calculate total resistance, in terms of R, between points

(i) A and C

(ii) B and X

(iii) A and Z

**(b)**Two cells of e.m.f. E

_{1}and E

_{2}and negligible internal resistance are connected into network in (a).

Currents in the network are as
indicated in Fig.2.

Use Kirchhoff’s laws to state the relation

(i) between currents I

_{1}, I_{2}and I_{3}
(ii) between E

_{2}, R, I_{2}and I_{3}in loop BCXYB
(iii) between E

_{1}, E_{2}, R, I_{1}and I_{2}in loop ABCXYZA**Reference:**

*Past Exam Paper – June 2009 Paper 21 Q7*

__Solution 595:__**(a)**

(i) {Point B
and C may be considered to be the same since there is no component connected
between them.}

R

(ii) {B and
Y are junctions at which current would change when going through the components
connected to those junctions. So, the components between those junctions are in
parallel.

[For components connected
in series, the same current flows through them]

If another resistor is
added in parallel across BY, then the total resistance between B and Y would
change but the resistance between AB, across which a single resistor is
present, remains the same (=R)}

{It should also be noted
that points B and C can be considered to be the same point since there is no
component between B and C. The same is true for points Y and X [that is the can
be considered to be the same]. So, when considering the

__whole circuit__, the resistance between BY, BX or CX are the same. But if we consider the resistor between B and Y or that between C and X__separately__, then the resistance between BY or CX are just R each.}
([1/R + 1/R]

It is then clear that the resistance BY, BX and CX are the same. Note,
however, that this resistance between them is ‘0.5R’ (as calculate), and NOT ‘R’.}

^{-1}=) 0.5R
{

**Why is it the case that the resistance between BY, BX and CX is the same? I can understand about BY and CX, but I don't understand about BX. In BX, why are we only considering the resistor on the line BCX? Why are we excluding the branch BY? I'm confused because: 1. The current branches out at the point B, so why exclude BY? 2. In other questions regarding voltage and resistance between two points, we usually include all branches in between.**__Question:__**First of all, we are not excluding BY. Secondly, it is important to think of it as part of a complete circuit. If we think of them separately, the resistance of BY is R, that between CX is R, resistance between BC is zero and resistance between XY is also zero. But this is not how we think of them. Think of it as part of a complete circuit, as will be explained below. Everything below will be about a complete circuit.**

__Explanation:__
Consider the vertical wire
between B and C. No component (resistor) is connected between B and C – so B
and C can be considered to be the same point. The same goes for the vertical
wire between X and Y.

Thus, the section BCXY can
be drawn as follows:

(iii) {The
current flowing through the resistor between A and B is the same as that flowing
through the resistor between Y and W. But this current is not the same flowing
through the resistor between B and Y, nor the same as that flowing through C
and X. So, between B and Y (as shown in the diagram), the resistors are in
parallel.

So, the resistor between A
and B is not in series with the resistor between B and Y, nor is it in series
with the resistor between C and X. However, since the total current flowing the
2 resistors in parallel is the same as that flowing between AB and YZ, then the
resistors between AB and YZ are in series with EQUIVALENT resistance between BY}

(R + [1/R + 1/R]

^{-1}+ R =) 2.5R**(b)**

(i) {From
Kirchhoff’s law, the sum of currents entering a junction should be equal to the
sum of currents leaving the junction. I

_{1}and I_{2}are entering junction B and I_{3}is leaving it.}
I

_{1}+ I_{2}= I_{3}
(ii) {We
need to consider loop BCXYB. From Kirchhoff’s law, the sum of p.d.’s in a loop
should be equal to the e.m.f. in the loop. The e.m.f. in the loop is E

_{2}. The p.d. across B and Y is I_{3}R and the p.d. across the resistor between C and X is I_{2}R.}
E

_{2}= I_{3}R + I_{2}R
(iii) {We
need to consider loop ABCXYZA. Again from Kirchhoff’s law, the sum of p.d.’s in
a loop should be equal to the e.m.f. in the loop. The currents from E

_{1}and E_{2}opposes each other. So, the overall e.m.f in the loop is obtained by subtracting the two (here we are assuming that E_{1}is greater than E_{2}, that’s why we take E_{1}– E_{2}). Between A and B, the p.d. is I_{1}R. Between the resistor in CX, the p.d. is I_{2}R. Across Y and Z, p.d. is I_{1}R (since current I_{1}is flowing from the positive terminal of the battery E_{1}, the same current should return to its negative terminal). So, the sum of p.d. is I_{1}R + I_{2}R + I_{1}R.}
E

_{1}– E_{2}= 2I_{1}R – I_{2}R

__Question 596: [Electrostatic]__
A small charge q is placed in the
electric field of large charge Q.

Both charges experience a force F.

What is electric field strength of
the charge Q at the position of the charge q?

A F / Qq B F / Q C
FqQ D F / q

**Reference:**

*Past Exam Paper – November 2009 Paper 11 Q27 & Paper 12 Q26 & November 2013 Paper 13 Q31*

__Solution 596:__**Answer: D.**

A small charge q is placed in the electric
field of a large charge Q. Both charges experience force F.

Electric force F at position of
charge q = Eq

Note that F also account
for the magnitudes of the charges. The electric force can also be given by
another formula (Coulomb’s law: F = Qq / 4Ï€Ïµ

_{0}r^{2}) but we don’t need to calculate this here. The electric force is already given to be F.
Electric field strength of charge Q
at position of charge q, E = F/q

For solution 595, I think there's a typo here in part (iii):

ReplyDelete{The current flowing through the resistor between A and B is the same as that flowing through the resistor between Y and W.} - because there is no W on the diagram.

But I have another question.

Why is it the case that the resistance between BY, BX and CX is the same?

I can understand about BY and CX, but I don't understand about BX.

In BX, why are we only considering the resistor on the line BCX? Why are we excluding the branch BY? I'm confused because:

1. The current branches out at the point B, so why exclude BY?

2. In other questions regarding voltage and resistance between two points, we usually include all branches in between.

THanks for the correction.

DeleteAs for your doubt, I have added a few lines above. See if it helps.

where is mayjune 2009 # P22 Question 7 solution?.

DeleteGo to

Deletehttp://physics-ref.blogspot.com/2014/11/9702-june-2009-paper-22-worked.html

what if 595(a)(ii) there is a resistor between BC? what would be the answer?

ReplyDeleteThen, this resistor would be in series in the bottom-most resistor. The equivalent resistance would be [1/R + 1/2R]^-1 = 2R / 3

Delete