# Physics 9702 Doubts | Help Page 19

__Question 100: [Current of Electricity > Potential difference]__Circuit is set up as shown, supplied by 3 V battery. All resistances are 1 kΩ.

What will be the reading on voltmeter?

A 0 B 0.5 V C 1.0 V D 1.5 V

**Reference:**

*Past Exam Paper – June 2013 Paper 11 Q35*

__Solution 100:__**Answer: B.**

The circuit may be divided into 2 main loops. From Kirchhoff’s law, the potential difference across each loop is equal to the e.m.f. in the circuit.

For each loop, the potential at a point may be found by the

**potential divider equation**. The potential

__difference__across the voltmeter is equal to the difference in potential between points A and B indicated in the diagram.

**,**

__In loop 1__Total resistance = 1 + 1 = 2kΩ

Potential at point A = [1 / (1+1)] x 3 = 1.5V

**,**

__In loop2__Consider the 2 sections indicated.

Total resistance in section X = [1/1 + 1/1]

^{-1}+ 1 = 1.5kΩ

Total resistance in section Y = [1/3 + 1/1]

^{-1}= 0.75kΩ

Total resistance in loop 2 = 1.5 + 0.75 = 2.25kΩ

Potential at point B = [1.5 / (1.5 + 0.75)] x 3 = 2.0V

Potential difference across voltmeter = 2.0 – 1.5 =

**0.5V**

__Note 1:__{Note that for both loop 1 and 2, I have been considering the components on the left of A and B. If I had considered the components on the right of A and B, for the first loop, the potential would still be 0.5V since both resistances are the same. For loop 2, the potential at B would be [0.75 / (1.5 + 0.75)] x 3 = 1.0V. The potential difference across the voltmeter would still be (1.0 – 0.5 =) 0.5V. What’s important is that when you are considering the components on the left or the right in one loop, you need to do the same in the other loop. Potential difference is only a relative comparison. For example, the difference between 4 and 2 is 2 and the difference between 1000 and 1002 is still 2.}

__Note 2:__{In fact, the explanation I provided is not totally good (but it’s still good.) [I’m not referring to note 1.] This is because current flows from the positive terminal towards the negative terminal. So, a component’s junction which is closer to the positive terminal would have a greater potential while the other end of the component would have a lower potential. The potential difference will nevertheless be the same across the component. Consider for example the component on the left of point A in loop 1. Its left end would be at a potential of 3.0V and its right end would be at a potential of 1.5V such that the potential difference across it is still (3.0 – 1.5 =) 1.5V (as calculated). If you consider the solution I gave, the potential at A would actually be 1.5V (here, it’s the same because the resistances are the same). Similarly, for loop 2, the actual potential at B would have been 3.0 – 2.0 = 1.0V. Well, since in both loops, I needed to subtract the answer I obtained from 3.0V, this would not have affected the potential difference between the voltmeter since as I said, potential difference is only relative. What I have explained in note 2 is very essential and important, but I think very few teachers make this clear to students.}

__Question 101: [Kinematics > Conservation of energy]__Car travelling with speed 28 ms

^{-1}leaves motorway on an exit road. The end of the exit road is 22m higher than motorway.

If only the force of gravity is considered, what will be the speed of car at end of the exit road?

A. 7.3 ms

^{-1}B. 19 ms

^{-1}C 21 ms

^{-1}D 24 ms

^{-1}

**Reference:**

*Past Exam Paper – November 2012 Paper 13 Q18*

__Solution 101:__**Answer: B.**

From conservation of energy,

**Gain in Potential energy = Loss in Kinetic energy**

__Gain in potential energy__= mgh = m (9.81) (22) = 215.82(m)

Let speed of car at the end of the exit road = v

__Loss in kinetic energy__= ½m (28

^{2}) – ½mv2 = 0.5m(28

^{2}– v

^{2})

From conservation of energy,

0.5m(28

^{2}– v

^{2}) = 215.82(m)

v

^{2}= 28

^{2}– (215.82 / 0.5)

Speed, v = 18.8 = 19ms

^{-1}

__Question 102: [Momentum]__Stationary nucleus has nucleon number A.

Nucleus decays by emitting proton with speed v to form new nucleus with speed u. New nucleus and proton move away from one another in opposite directions.

Which equation gives v in terms of A and u?

A v = (A/4 – 1)u B v = (A – 1)u C v = Au D v = (A + 1)u

**Reference:**

*Past Exam Paper – June 2013 Paper 13 Q10*

__Solution 102:__**Answer: B.**

Nucleus is initially stationary, so its speed is zero. Its momentum (p = mv) is also zero initially.

The nucleus contains an amount of A nucleons.

Nucleus decays by emitting proton with speed v to form new nucleus with speed u.

Number of nucleons in new nucleus = A – 1

Let the mass of 1 nucleon = m

__After decay__,

Momentum of proton = mv {the proton is a nucleon}

Momentum of new nucleus = (A – 1)m(–u)

{negative is taken for the speed because the new nucleus and the proton travel in opposite direction. The direction of motion of the proton is taken as positive [v is taken as positive], so u should be taken as negative.}

Total momentum after decay = mv – (A – 1)mu

From the

**conservation of momentum**,

Sum of Momentum before decay = Sum of Momentum after decay

0 = mv – (A – 1)mu

v = (A – 1)u

__Question 103: [Vectors > Equilibrium]__Diagrams represent systems of coplanar forces acting at a point. Lengths of the force vectors represent magnitudes of the forces.

Which system of forces is in equilibrium?

**Reference:**

*Past Exam Paper – November 2002 Paper 1 Q15*

__Solution 103:__**Answer: A.**

For the forces to be in equilibrium, the resultant horizontal and vertical components of the force vectors should be zero.

The lengths of the vectors represent the magnitudes. Note that the diagonal vectors in the diagrams will have vertical and horizontal components

**than the magnitude of the diagonal vector itself {this is obvious since the diagonal vector and its components form a right-angled triangle with the diagonal vector as the hypotenuse. The hypotenuse has the longest length in a right-angled triangle}.**

__less__Therefore, in these diagrams, the

__horizontal and vertical__of the diagonal vectors should be equal in length (and opposite in direction) as the

*components*__horizontal and vertical__already given.

*vectors*Diagram A:

The horizontal and vertical components of the diagonal vector are equal in magnitude and opposite in direction to the respective vectors already given. So, in diagram A, the system of forces is in equilibrium.

Diagram B:

The sum of the upward vertical components of the 2 diagonal vectors is greater than the downward vertical vector already present. So, the system of forces is not in equilibrium.

Diagram C:

The sum of the horizontal components (toward the left) of the 2 diagonal vectors is smaller than the horizontal vector (toward the right) already present. So, the system of forces is not in equilibrium.

Diagram D:

The system is not balanced vertically since the downward vertical component of one of the diagonal vectors is greater than the upward vertical component of the other diagonal vector.

__Question 104: [Ohm’s law > Resistance > Resistivity]__Wire PQ is made of three different materials, with resistivities ρ, 2ρ and 3ρ. There is current I in this composite wire, as shown.

Which graph best shows how potential V along the wire varies with distance x from P?

**Reference:**

*Past Exam Paper – November 2010 Paper 12 Q34*

__Solution 104:__**Answer: B.**

**Ohm’s law:**

**Potential difference, V = IR**

The resistance R of a wire
proportional to the resistivity ρ
of the material from which the wire is made [R = ρL / A].

When the resistivity of
the material is greater, its resistance is also greater and thus, the potential

**difference**across it is greater. So, the potential difference is greater across the 3ρ material.
Note that the graph
gives the potential V along the wire on the y-axis while the V from V = IR
represents the potential

**– that is, the change in potential.**__difference__
Therefore, when the resistivity of
the material is greater, the potential difference across it is greater. This means
that the

__reduction__is potential V is greatest for the part where the resistivity is 3ρ.
The change in potential {potential difference, ΔV} as the distance x changes (Δx) is represented by the gradient. Note that for a
specific value of resistivity {along a specific length
of wire, L}, the potential difference (gradient) is constant {and non-zero unlike the graph of D where the gradient is
zero}. [D is incorrect]

{Resistance of the
wire also depends on the length if wire being considered. So, as x increases,
the length of wire being considered increases and so, as x increases, the
difference in potential V at that particular value of x is greater. That is, in
general as x increases, V decreases, even if the resistivity was constant}

As the resistivity of the wire
increases, the potential difference across it also increases. This is
represented by an increase in gradient of the straight lines as distance x
reaches the materials of higher resistivity. [A is
incorrect since the gradient is constant everywhere] [C is incorrect as the gradient of the straight lines decreases
as x increases]

Nov 2010. variant 11

ReplyDeleteQ37

i'm confused here between option A & C

It's explained as solution 120 at

Deletehttp://physics-ref.blogspot.com/2014/11/physics-9702-doubts-help-page-22.html

Nov 2009, variant 11

ReplyDeleteQ9

Q12

Q14

Q13 I'm confused between option B and C

Some of the questions are already solved. Check at

Deletehttp://physics-ref.blogspot.com/2014/10/9702-november-2009-paper-11-worked.html

The others will be added soon

June 2008

ReplyDeleteQ39

Q38

Q37

Q34

Q32

Q28

Q27

Q28

Q25

Q17 - what exactly is meant by linear momentum?

Q16

Q16 is explained as solution 136 at

Deletehttp://physics-ref.blogspot.com/2014/11/physics-9702-doubts-help-page-24.html

Others will be added soon

Q 17 added as solution 139 at

Deletehttp://physics-ref.blogspot.com/2014/11/physics-9702-doubts-help-page-25.html

Q39 added as solution 145 at

Deletehttp://physics-ref.blogspot.com/2014/11/physics-9702-doubts-help-page-25.html

Q25 added as solution 146 at

Deletehttp://physics-ref.blogspot.com/2014/11/physics-9702-doubts-help-page-26.html

Q38 added as solution 154 at

Deletehttp://physics-ref.blogspot.com/2014/11/physics-9702-doubts-help-page-26.html

Q27 explained as solution 161 at

Deletehttp://physics-ref.blogspot.com/2014/11/physics-9702-doubts-help-page-27.html

A28 added as solution 170 at

Deletehttp://physics-ref.blogspot.com/2014/11/physics-9702-doubts-help-page-28.html

why is a voltmeter connected in parallel always?what difference would it make if it was connected in series?

ReplyDelete9702_w09_qp_11

questions 25,28,29, 30

9702_s10_qp_11

questions 25, 26,27, 28,35

by connecting the voltmeter in parallel, it measures the p.d. across the component it is connected to. If it is connected in series, it will measure the p.d. between the points to which it is connected but since it is now NOT connected in parallel to any component, the potential at both points of the voltmeter would be the same, causing the difference between them (called p.d.) to be zero. so, it irrelevant to connect it in such a way.

DeleteFor the questions, check at

http://physics-ref.blogspot.com/2014/10/9702-november-2009-paper-11-worked.html

and

http://physics-ref.blogspot.com/2014/11/9702-june-2010-paper-12-worked.html

more questions will be solved later

more questions have been added

Deleteall of them have been added

DeleteO/N 2002 q10 and q36

ReplyDeleteFor Q10, check at

Deletehttp://physics-ref.blogspot.com/2015/03/physics-9702-doubts-help-page-91.html

Q36 is explained at

Deletehttp://physics-ref.blogspot.com/2015/03/physics-9702-doubts-help-page-92.html

nov 2011 paper 13 Q5

DeleteSee Q7 at

Deletehttp://physics-ref.blogspot.com/2014/10/physics-9702-doubts-help-page-1.html

nov 2010 paper 11

ReplyDeletequestion 34 how to attempt it?

Check solution 554 at

Deletehttp://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-108.html

june 2011 paper 41 question 11 b (ii),(iii),(iv) i have no idea how to solve it! pls help

ReplyDeleteCheck solution 561 at

Deletehttp://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-110.html

can you please help me with 9702/21/O/N/09

ReplyDeleteQ5

Q6

Q7

Please !

Q5 as solution 251 at

Deletehttp://physics-ref.blogspot.com/2014/12/physics-9702-doubts-help-page-41.html

Q6 as solution 612 at

http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-121.html

Q7 as solution 412 at

http://physics-ref.blogspot.com/2015/03/physics-9702-doubts-help-page-77.html

For solution 101, the formula is supposed to be mgh = 1/2 mv^2 - 1/2 mu^2 right? Instead of 1/2 mu^2 - 1/2 mv^2.. or it doesn't matter? But if I use 1/2mv^2-1/2mu^2 then I couldn't get the answer.

ReplyDeleteThanks :)

It's the change in kinetic energy (positive value).

DeleteInitial KE - Final KE

check your calculation again

for solution 100 if we consider potential at point B as 1V and then 1.5-1=0.5V that is also correct way?

ReplyDeleteyeah, as long as the p.d. is 0.5V

DeleteFor Question 102, why is the number of nucleon in new nucleus A-1?

ReplyDeleteShouldn't it be A-4 since alpha particle has 4 nucleons?

For Question 102 why is the nucleon number of the nucleus A-1?

ReplyDeleteit's a proton, not an alpha particle.

Deleteproton = +1

for the total to be A, the nucleon number should be (A-1)

such that (A-1) + 1 = A