# Physics 9702 Doubts | Help Page 29

__Question 172: [Current of Electricity > Resistance]__
In each arrangement of resistors,
ammeter has a resistance of 2 Ω.

Which arrangement gives largest
reading on ammeter when the same potential difference is applied between points
P and Q?

**Reference:**

*Past Exam Paper – June 2010 Paper 12 Q32*

__Solution 172:__**Answer: D.**

We need to find the combined
resistance of circuit, R.

Current in a loop does not change.

**Ohm’s law: V = IR**. So, I = V / R.

For

**A**, R = 5. I =**V/5**.
For

**B**, R = [1/1 + 1/2]^{-1}+ 2 = 8/3. I =**3V/8**.
For

**C**, I = V / (2+1) =**V/3**(since potential difference is the same across any loop, I = V/ (sum of resistances in a loop)).
For

**D**, I =**V/2**.

__Question 173: [Dynamics > Elastic Collision]__Ball drops onto horizontal surface and bounces elastically.

What happens to kinetic energy of ball during the very short time that it is in contact with the surface?

A Most of kinetic energy is lost as heat and sound energy.

B Kinetic energy decreases to zero and then returns to its original value.

C Kinetic energy remains constant because it is an elastic collision.

D Kinetic energy remains constant in magnitude but changes direction.

**Reference:**

*Past Exam Paper – June 2014 Paper 13 Q18*

__Solution 173:__**Answer: B.**

For elastic collision, the kinetic
energy is not lost [A is incorrect].

However, the kinetic energy does not
remain constant since the speed of the ball changes during this very short
time. At some instant, the speed of the ball is zero (ball
stops). During that instant, all the energy becomes elastic potential
energy [C is incorrect].

Therefore, the kinetic energy
decreases to zero at some point (where the ball stops)
and then returns to its original value (as the ball is
about to leave the surface) [B is correct].

[D cannot be correct
since kinetic energy is a scalar quantity, so it does not have a direction.]

__Question 174: [Current of Electricity > Kirchhoff’s laws]__
Cell X has e.m.f. of 2.0 V and
internal resistance of 2.0 Ω. Cell Y has e.m.f. of 1.6 V and internal
resistance of 1.2 Ω. These two cells are connected to resistor of resistance
0.8 Ω, as shown.

What is current in cell X?

A 0.10 A B 0.50 A C
0.90 A D 1.0 A

**Reference:**

*Past Exam Paper – June 2014 Paper 12 Q33*

__Solution 174:__**Answer: A.**

The e.m.f. of cell X and Y opposes
each other (current flows from the positive terminal.
So, the currents from these 2 cells are in opposite directions).

Overall emf = 2 – 1.6 = 0.4V

**(the total e.m.f in a loop is equal to the sum of potential differences [across the components in series] in the loop)**

__Using Kirchhoff’s second law,__
2.0 – 1.6 = I (2+1.2+0.8)

So, current I = V / R = 0.4 /
(2+1.2+0.8) = 0.10A.

__Question 175: [Dynamics > Collisions]__
The diagram shows two identical spheres
X and Y.

Initially, X moves with speed v
directly towards Y. Y is stationary. The spheres collide elastically.

What happens?

**Reference:**

*Past Exam Paper – November 2006 Paper 1 Q11 & June 2010 Paper 11 Q12 & June 2010 Paper 12 Q10*

__Solution 175:__**Answer: D.**

For elastic collision, both the momentum
and kinetic energy are conserved.

Let mass of 1 sphere = m

__Before collision,__
Momentum = mv

Kinetic energy = ½ mv

^{2}

__After collision,__
Momentum should be conserved.

For A, momentum is conserved.

B is an impossible case since it
would imply that sphere Y does not exist (even though
momentum is conserved, this is physically impossible).

For C, sum of (magnitude of) momentum after collision is mv (but the spheres moves in opposite directions).

For D, momentum is conserved

If the speeds of both spheres become
½ v [as suggested by A and C],

Sum of Kinetic energies after
collision = ½ m(v/2)

^{2}+ ½ m(v/2)^{2}= mv^{2}/4
Kinetic energy is not
conserved in these 2 cases (A and C).

Only choice D will have both the
momentum and kinetic energy conserved.

__Question 176: [Forces > Torque]__
Spanner is used to tighten nut as
shown.

Force F is applied at right-angles
to spanner at a distance of 0.25 m from centre of the nut. When nut is fully tightened,
the applied force is 200 N.

What is resistive torque, in an
anticlockwise direction, preventing further tightening?

A 8 N m B 42 N m C
50 N m D 1250 N m

**Reference:**

*Past Exam Paper – November 2010 Paper 12 Q12*

__Solution 176:__**Answer: C.**

The 200N force acts at a distance of
0.25m from the centre of the nut.

To prevent further tightening, the
resistive torque should be equal to the torque by the 200N force.

Resistive torque = Torque by the
200N force = 200 x 0.25 = 50Nm

__Question 177: [Dynamics > Momentum]__
Object travelling with velocity v
strikes wall and rebounds as shown.

Which property of object is not
conserved?

A kinetic energy

B mass

C momentum

D speed

**Reference:**

*Past Exam Paper – November 2012 Paper 11 Q11*

__Solution 177:__**Answer: C.**

The question asks which

**is**__property of the object__**not**conserved.
Kinetic energy = ½ mv

^{2}. So, it is conserved.
Mass is also conserved as it remains
the same.

Speed (which
is a scalar – the choice is not ‘velocity’) is also conserved as it
remains v.

The property which is not conserved
here is the

**momentum****. Momentum is a vector quantity. Initial momentum = mv. Final momentum = - mv.**__of the object__
Note that even if the momentum of
the object is not conserved, the momentum of the

__whole system__should be conserved. So, the key word in the question is ‘property of the object’, not the whole system.

__Question 178: [Kinematics > Linear motion]__
Diagram shows barrel suspended from
frictionless pulley on building. Rope supporting the barrel goes over pulley and
is secured to a stake at bottom of the building.

A man stands close to stake. Bottom
of the barrel is 18 m above man’s head. The mass of barrel is 120 kg and mass
of the man is 80 kg.

Man keeps hold of rope after untying
it from the stake and is lifted upwards as barrel falls.

What is man’s upward speed when his
head is level with the bottom of barrel? (Use g = 10 m s

^{–2}.)
A 6 m s

^{–1}B 8 m s^{–1}C 13 m s^{–1}D 19 m s^{–1}**Reference:**

*Past Exam Paper – June 2012 Paper 11 Q12*

__Solution 178:__**Answer: A.**

There is a downward force
on each sides of the rope due to the weights of the barrel and the man. First,
the resultant force in the system should be calculated.

(Downward) force (due to weight) of
barrel = mg = 120 x 10 = 1200N

(Downward) force (due to weight) of
man = mg = 80 x 10 = 800N

**Net (resultant) force in the system = 1200 – 800 = 400N**

This force acts downwards at the
side of the rope where the barrel is located. So, the barrel moves downwards
while the man moves upwards.

The force causes a resultant
acceleration in the system. Since it is the resultant force in the SYSTEM, the
masses of both the man and the barrel should be taken into account.

Resultant force = ma = 400

(120 + 80) a = 400

**Acceleration, a = 400 / 200 = 2ms**

^{-2}
The same acceleration acts on both
the man and the barrel.

For the man’s head to be
level with the bottom of the barrel, both of them should travel a distance of (18
/ 2 =) 9m.

Consider the equation of uniformly
accelerated motion:

**v**^{2}= u^{2}+ 2as
Initial velocity, u = 0

Acceleration, a = 2ms

^{-2}
Distance travelled, s = 9m

v

^{2}= 02 + 2(2)(9)
Final speed = √(2x2x9) = 6ms

^{-1}

__Question 179: [Waves]__
Diagram shows two waves X and Y.

Wave X has amplitude 8 cm and
frequency 100 Hz.

What are the amplitude and frequency
of wave Y?

**Reference:**

*Past Exam Paper – November 2008 Paper 1 Q24 & June 2013 Paper 12 Q25*

__Solution 179:__**Answer: D.**

Amplitude is the maximum
displacement. From the diagram, the amplitude of Y is half that of X. So,
amplitude of Y = 8 / 2 = 4cm.

Consider 1 period of X
(starting a time = 0) on the diagram. It can be seen that for this amount of
time, there are 3 periods of wave Y. So, the period of wave Y is a third of
that of wave X.

Period of wave Y = (1/3) x Period of
wave X

**Frequency = 1 / Period**

Since frequency is
inversely proportional to the period, the frequency of wave Y is 3 times that
of wave X.

Frequency of wave Y = 3 (100) =
300Hz

OR

For X, period = 1 / 100 = 0.01s.

As seen from diagram, period of Y is
one third that of X (in wave Y, 3 oscillations occurs for the same amount of
time), so period of Y = 0.01 /3.

Frequency of Y = 1 / [0.01/3] =
300Hz.

(or you could directly obtain the
frequency to be 3 times that of X if you understand the concept of frequency
well)

__Question 180: [Electric Field > Electric field Strength]__Diagram shows two parallel metal plates connected to a d.c. power supply through resistor.

There is a uniform electric field in region between plates.

Which change would cause a decrease in strength of the electric field?

A small increase in the distance between the plates

B small increase in the potential difference between the plates

C small increase in the value of the resistor

D small increase to the area of both plates

**Reference:**

*Past Exam Paper – June 2011 Paper 12 Q30*

__Solution 180:__**Answer: A.**

**Electric field strength, E = V / d**

where V is the potential difference
between the plates and d is the separation of the plates

Since E is inversely proportional to
d, increasing the distance between the plates causes a decrease in the strength
of the electric field.

Note that the plates are
not contributing any resistance in the circuit. So, increasing the value of the
resistor will NOT cause the potential difference across the plates to decrease.

__Question 181: [Current of Electricity > Resistance]__In circuit below, reading V

_{T}on voltmeter changes from high to low as temperature of the thermistor changes. Reading V

_{L}on voltmeter changes from high to low as level of light on the light-dependent resistor (LDR) changes.

Readings V

_{T}and V

_{L}are both high.

What are the conditions of temperature and light level?

**Reference:**

*Past Exam Paper – June 2013 Paper 12 Q38*

__Solution 181:__**Answer: C.**

**Ohm’s law: V = IR.**Therefore, for the voltmeter to read a high value, the resistance of the component to which it is connected must be high.

For both reading V

_{T}and V_{L}to be high, the resistance of the thermistor should be low (so that the resistance of the fixed resistor, to which the voltmeter is connected, is much higher than the resistance of the thermistor) and the resistance of LDR should be high (since the voltmeter is connected across it).
A

**is a type of resistor whose resistance varies significantly with temperature, more so than in standard resistors. ΔR = k ΔT. Depending on k, resistance may increase or decrease with temperature (For A-level, k is normally taken to be only negative. That is, resistance decreases with an increase in temperature).**__thermistor__
A

**or light-dependent resistor (LDR) or photocell is a light-controlled variable resistor. The resistance of a photoresistor decreases with increasing incident light intensity; in other words, it exhibits photoconductivity.**__photoresistor__

__Question 182: [Current of Electricity > Resistance]__Four identical resistors are connected in three networks below.

Which arrangement has highest total resistance and which has lowest?

**Reference:**

*Past Exam Paper – June 2012 Paper 12 Q38*

__Solution 182:__**Answer: C.**

**Hint: Try to simplify each part of the circuit.**

Let the resistance of a resistor = R

For diagram 1:

1/R

_{1}= (1/R + 1/(2R) + 1/R) = 5/2R
So, total resistance

**R**_{1}= 2R/5
For diagram 2:

Resistance of upper part = R + (1/R
+ 1/R)

^{-1}= R + R/2 = 3R/2
1/R

_{2}= 2/3R + 1/R = 5/3R.
So, total resistance

**R**_{2}= 3R/5
For diagram 3:

1/R

_{3}= 1/3R + 1/R = 4/3R
So, total resistance

**R**_{3 }= 3R/4
Awesome... Can you make Topic/ Chapter wise Ansers???

ReplyDeleteThanks. I intend to, but not righ now. It will be after I finish the yearly past papers.

DeleteIt's basically a classification of the questions in these papers. SO, if you know the reference of a particular question, you can just look up for it in the particular solved past paper for that year here.

Many variants of the yearly past papers are not available yet?

ReplyDeleteMany questions from these variants have already been solved but the link has not yet been added there. I plan to do it later.

DeleteFor the moment, the can state the specific doubts from these variants and I'll try to help if possible.

Just referent to the variant, year question number ,.... do NOT include links to them

Cab you solve q 11 o/n8 paper1

ReplyDeleteCan you solve q11 o/n 8 paper1

ReplyDeleteSee solution 425 at

Deletehttp://physics-ref.blogspot.com/2015/03/physics-9702-doubts-help-page-80.html

solution 9702/11/m/j/12 please??????

ReplyDeletewhich questions?

Deletecan u do june 2008 paper 4 q1 (b)

ReplyDeleteCheck solution 1095 at

Deletehttp://physics-ref.blogspot.com/2016/01/physics-9702-doubts-help-page-233.html

thank you very much for your explanations and solutions. this blog has truly helped me and my friends a lot for physics.

ReplyDeleteCan you solve 9709/11/mj/12 question no 15

ReplyDeletesee solution 777 at

Deletehttp://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-157.html

For 9702/11/M/J/12 Q12, there's an alternative method right? using conservation of energy.

ReplyDeletelost GPE of barrel = gain GPE of man + gain KE man + gain KE barrel

then calculate chg in velocity, since u=0m/s, thus v=6 m/s.

Of course there are other ways.

Deletedo you mind writing the complete solution?

You are a hero! I swear I respect you and your kindness so much! Keep up the amazing work and thanks for helping me ace my Physics!

ReplyDeleteLiterlly GOD Bless You thank you for your elaborated explanations on the MCQ's mainly.... :)

ReplyDeleteThis is really good

ReplyDelete