# 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

**negative. So, the electric field lines should be drawn**__to__**a negative charge and away from a positive charge.**__towards__

__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 = 180

^{o}(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 180

^{o}})

A phase difference of 360

^{o}corresponds to a cycle {Phase difference of 360

^{o}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 360

^{o}

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

^{o}

But, a phase difference of 180

^{o}between the 2 waves occurs at time t = 18s.

So, a phase difference of 45

^{o}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).

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

ReplyDeleteQ14 is explained as Solution 128 above. Others will be added later

Delete21 available as q 133 at

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

Q32 has been added as solution 141 at

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

it's as solution 142, not 141

Delete