Monday, November 24, 2014

Physics 9702 Doubts | Help Page 23

  • Physics 9702 Doubts | Help Page 23


Question 122: [Electric field]
Which diagram shows electric field pattern of an isolated negative point charge?

Reference: Past Exam Paper – November 2002 Paper 1 Q37



Solution 122:
Answer: B.
Electric field lines are drawn with a direction going (away) from positive to negative. So, the electric field lines should be drawn towards a negative charge and away from a positive charge.








Question 123: [Matter > Elastic deformation]
Graph is a load-extension graph for wire undergoing elastic deformation.

How much work is done on wire to increase extension from 10 mm to 20 mm?
A 0.028 J                     B 0.184 J                     C 0.28 J           D 0.37 J

Reference: Past Exam Paper – June 2014 Paper 11 Q21



Solution 123:
Answer: C.
Work done = Force x distance moved in direction of force

Since the load is given in kg, it should be converted in newton (weight) by multiplying by g (= 9.81).
Additionally, the gradient of the graph is not zero [graph is not a horizontal line], so the load is not constant {that is, to increase the extension, a greater should be used}.

Work done required = (area under graph between extension 10mm and 20mm) x g

Area under graph = 0.5 x [sum of parallel sides] x height
Area under graph = 0.5 x [1.9+3.8] x (10x10-3) = 2.85 x 0.01

Work done = (Area under graph) x g = (2.85x0.01) x 9.81 = 0.28J









Question 124: [Waves > Phase difference]
Diagram shows two sinusoidal waveforms.

At time t = 0 waves are in phase. At dotted line, t = 18 s.
At which time is phase difference between the two oscillations ⅛ of a cycle?
A 4.0 s                         B 4.5 s                         C 8.0 s                         D 9.0 s

Reference: Past Exam Paper – November 2011 Paper 12 Q26



Solution 124:
Answer: B.
At time t = 18s, phase difference between the 2 waves = 180o (since at t = 18s, waveform P corresponds to a positive maximum displacement while waveform Q corresponds to a negative maximum displacement. They are in anti-phase {out of phase by 180o})

A phase difference of 360o corresponds to a cycle {Phase difference of 360o means that the waves are in phase – they have the same exact form. A cycle means 1 wavelength / period.}.

1 cycle corresponds to phase difference of 360o
1/8 of a cycle corresponds to phase difference of (360/8 =) 45o

But, a phase difference of 180o between the 2 waves occurs at time t = 18s.
So, a phase difference of 45o occurs at a time t = (18/180) x 45 = 4.5s









Question 125: [Kinematics > Linear motion > Graph]
Student throws ball in the positive direction vertically upwards.
Ball makes elastic collision with ceiling, rebounds and accelerates back to the student’s hand in a time of 1.2 s.
Which graph best represents acceleration of the ball from the moment it leaves the hand to the instant just before it returns to hand?

Reference: Past Exam Paper – November 2010 Paper 12 Q7



Solution 125:
Answer: D.

As defined by the question, the vertical upwards direction is taken as positive.
The force of gravity acts on the ball as it moves upwards. Since the acceleration due to gravity is downwards, its initial value should be negative. [A and B are incorrect]

Acceleration is defined as the rate of change of velocity. {Acceleration = (final velocity – initial velocity) / time}
After collision with the ceiling, the (upward) speed of the ball changes from a positive value to a (downward) speed with a negative value. Thus, the acceleration it undergoes during collision is downwards.

Therefore, in addition to the downward acceleration due to gravity (which also has a negative value as defined in the question), the total acceleration is more negative during collision.









Question 126: [Waves > Polarization]
When liquid crystal display of calculator is observed through a polarising film, the display changes as the film is rotated.
Which property describes radiation from the calculator display?
A unpolarised
B longitudinal wave
C transverse wave
D wave with a 3 cm wavelength

Reference: Past Exam Paper – June 2013 Paper 13 Q25



Solution 126:
Answer: C.
The polarizing film is used to polarize the radiation (wave) from the calculator display. Since the display changes as the film is rotated, it means that the radiation has been polarized. [A is incorrect]

Definition of some terms:
A transverse wave is a wave in which the oscillations of the wave particles (not ‘movement’) are perpendicular (or right-angled) to the direction of the propagation of the wave (direction of energy transfer).
A longitudinal wave is a wave in which the oscillations of the wave particles are parallel to the direction of the propagation of the wave.
Polarization is said to occur when the oscillations are in one direction in a plane, (not just “in one direction”) normal to the direction of propagation.

Transverse waves can be polarized while longitudinal waves cannot (since oscillations and direction of propagation are in the same direction). [B is incorrect and C is correct]









Question 127: [Force > Torque]
Diagram shows two pulley wheels connected by a belt.

Wheel Q is driven by motor and rotates clockwise at constant rate. Wheel Q puts tension in the top portion of belt, which in turn drives wheel P. Lower portion of belt is slack and has no tension. Weight of the belt and frictional forces are negligible.
Diameter of P is 150 mm. Diameter of Q is 100 mm. Torque applied to Q is 3.0 N m.
What is the tension in belt and the torque on wheel P?

Reference: Past Exam Paper – November 2009 Paper 11 Q13 & Paper 12 Q12



Solution 127:
Answer: D.
The tension in the belt between P and Q is the same. (Tension put in belt by wheel Q = Tension in belt at P)

Consider wheel Q: Torque = force x perpendicular distance of force from pivot (radius). The force is actually the tension, T.
Torque applied to Q = 3.0Nm
3 = T x [(100/2) x 10-3]
Tension, T = 3 / [(100/2) x 10-3] = 60N

Note that since the torque is provided by just the top part of the belt, the radius is considered, not the diameter.

Torque on wheel P = 60 x [(150/2) x 10-3] = 4.5Nm









Question 128: [Energy > Work done]
Ball of mass m is thrown up to height h in air with initial velocity v, as shown.

Air resistance is considered negligible. Acceleration of free fall is g.
What is the total work done by gravitational force on ball during its flight from P to Q?
A zero                         B ½mv2                      C mgh                         D 2mgh

Reference: Past Exam Paper – June 2013 Paper 13 Q14



Solution 128:
Answer: A.
Work done = Force x distance moved in direction of force
The gravitational force acts towards the surface.
For the work done, we only need to consider the initial and final position. That is, the total work done depends only on the difference in height.

Since the ball remains on the ground (at both P and Q), the distance moved in the direction of the gravitational force (which acts towards the surface) is zero (difference in height is zero).





5 comments:

  1. 9702/13/m/j/13 q14, 21, 32

    ReplyDelete
    Replies
    1. Q14 is explained as Solution 128 above. Others will be added later

      Delete
    2. 21 available as q 133 at
      http://physics-ref.blogspot.com/2014/11/physics-9702-doubts-help-page-24.html

      Delete
    3. Q32 has been added as solution 141 at
      http://physics-ref.blogspot.com/2014/11/physics-9702-doubts-help-page-25.html

      Delete
    4. it's as solution 142, not 141

      Delete

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