Thursday, November 27, 2014

Physics 9702 Doubts | Help Page 29

  • 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

Using Kirchhoff’s second law, (the total e.m.f in a loop is equal to the sum of potential differences [across the components in series] in the loop)
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 mv2

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 = mv2/4
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 property of the object is not conserved.
Kinetic energy = ½ mv2. 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 of the object. Momentum is a vector quantity. Initial momentum = mv. Final momentum = - mv.

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: v2 = u2 + 2as
Initial velocity, u = 0
Acceleration, a = 2ms-2
Distance travelled, s = 9m

v2 = 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 VT on voltmeter changes from high to low as temperature of the thermistor changes. Reading VL on voltmeter changes from high to low as level of light on the light-dependent resistor (LDR) changes.

Readings VT and VL 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 VT and VL 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 thermistor 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).

A photoresistor 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.








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/R1 = (1/R + 1/(2R) + 1/R) = 5/2R
So, total resistance R1 = 2R/5

For diagram 2:
Resistance of upper part = R + (1/R + 1/R)-1 = R + R/2 = 3R/2
1/R2 = 2/3R + 1/R = 5/3R.
So, total resistance R2 = 3R/5

For diagram 3:
1/R3 = 1/3R + 1/R = 4/3R
So, total resistance R3 = 3R/4




18 comments:

  1. Awesome... Can you make Topic/ Chapter wise Ansers???

    ReplyDelete
    Replies
    1. Thanks. I intend to, but not righ now. It will be after I finish the yearly past papers.

      It'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.

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

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

      For 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

      Delete
  3. Cab you solve q 11 o/n8 paper1

    ReplyDelete
  4. Can you solve q11 o/n 8 paper1

    ReplyDelete
    Replies
    1. See solution 425 at
      http://physics-ref.blogspot.com/2015/03/physics-9702-doubts-help-page-80.html

      Delete
  5. solution 9702/11/m/j/12 please??????

    ReplyDelete
  6. can u do june 2008 paper 4 q1 (b)

    ReplyDelete
    Replies
    1. Check solution 1095 at
      http://physics-ref.blogspot.com/2016/01/physics-9702-doubts-help-page-233.html

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

    ReplyDelete
  8. Can you solve 9709/11/mj/12 question no 15

    ReplyDelete
    Replies
    1. see solution 777 at
      http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-157.html

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

    ReplyDelete
    Replies
    1. Of course there are other ways.
      do you mind writing the complete solution?

      Delete
  10. 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!

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

    ReplyDelete

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