• Physics 9702 Doubts | Help Page 26

Question 146: [Waves > Graph]
Graph shows how height of a water surface at a point in a harbour varies with time t as waves pass the point.

What are p and q?

For June 2008 Paper 1 Q25:

For November 2013 Paper 13 Q25:

p                      q

A         displacement   period
B         displacement   wavelength
C         amplitude        period
D         amplitude        wavelength


Reference: Past Exam Paper – June 2008 Paper 1 Q25 & November 2013 Paper 13 Q25



Solution 146:


June 2008 Paper 1 Q25 - Answer: B.
November 2013 Paper 13 Q25 – Answer: A.

The x-axis indicates the time t. So, q gives the period (which is in units of time) and not the wavelength (which is in units of length).

The y-axis indicates the height of the water surface. The term ‘amplitude’ is used for the maximum displacement. As p does not show the maximum displacement, it is not the amplitude. p only shows the displacement.









Question 147: [Electric Field > Equipotential surface]
Diagram shows two points P and Q which lie 90° apart on circle of radius r.
Positive point charge at centre of the circle creates an electric field of magnitude E at both P and Q.

Which expression gives work done in moving a unit positive charge from P to Q?
A 0                  B E × r                        C E × (πr / 2)               D E × (πr)

Reference: Past Exam Paper – June 2014 Paper 11 Q29 & June 2010 Paper 12 Q28 & June 2010 Paper 13 Q29



Solution 147:

Answer: A.

Work done = Force x Distance moved in direction of force

The force on the charge moved from P to Q does not change if it is moved along the circle.

The force is always at right-angles to the motion, so no work is done.

The problem can also be considered in terms of the electrical potential. The charges P and Q are at the same potential (from the +ve charge at the centre) and so zero work done is the correct answer.

It is incorrect to use W = Ex which would result in answer C.

Question 148: [Waves > Diffraction]
Light passes through diffraction grating ruled at 1000 lines per cm and same wavelength of light also passes through two narrow slits 0.5 mm apart. Both situations produce intensity maxima and minima on screen.
Which statement about separation of the maxima on the screen and the sharpness of the maxima is correct?
A Diffraction grating maxima are less widely spaced and are less sharp than the two-slit maxima.
B Diffraction grating maxima are less widely spaced and are sharper than the two-slit maxima.
C Diffraction grating maxima are more widely spaced and are less sharp than the two-slit maxima.
D Diffraction grating maxima are more widely spaced and are sharper than the two-slit maxima.

Reference: Past Exam Paper – June 2014 Paper 12 Q26



Solution 148:

Answer: D.

The diffraction grating maxima are more widely spaced (since the length of a line in the diffraction grating {= 1/1000 = 0.001cm = 0.01mm} is less than the separation of the 2 narrow slits {= 0.05mm}).

{Consider the spots formed to be black and white. A higher sharpness would cause the spots formed to be ‘whiter’ or ‘blacker’. That is, they can be more distinguished from each other.}

Additionally, the diffraction grating maxima are sharper than the two-slit maxima (since more interference occurs from the waves that get diffracted from more slits in the diffraction grating, - the amplitudes of more diffracted waves contribute to every single maxima).

Question 149: [Current of Electricity > Internal Resistance]
Cell has electromotive force (e.m.f.) of 6 V and internal resistance R. External resistor, also of resistance R, is connected across cell, as shown.

Power P is dissipated by external resistor.
Cell is replaced by a different cell that has e.m.f. of 6 V and negligible internal resistance.
What is the new power that is dissipated in external resistor?
A 0.5P                         B P                  C 2P                D 4P

Reference: Past Exam Paper – June 2014 Paper 13 Q34



Solution 149:

Answer: D.

For the initial cell, total resistance on circuit = R + R = 2R

Current in circuit, I (= V / 2R) = 6 / 2R = 3/R

Power dissipated by the external resistor, P = I²R.

When a different cell with negligible internal resistance is used, the total resistance in the circuit is halved (total resistance in circuit = R) and thus, the current is doubled (from Ohm’s law: V = IR giving current, I = V / R = 6/R).

When the current is doubled, power dissipated (P = I²R) increases by a factor of 4 (since it depends on I²).

Question 150: [Dynamics > Laws of Motion]
Which statement about Newton’s laws of motion is correct?
A First law follows from the second law.
B Third law follows from the second law.
C Conservation of energy is a consequence of the third law.
D Conservation of linear momentum is a consequence of the first law.

Reference: Past Exam Paper – June 2009 Paper 1 Q7



Solution 150:

Answer: A.

It is important to note that the conservation laws (of energy and linear momentum) are NOT consequences of the third or first (Newton’s) law of motion but the conservation laws are fundamental laws to which the other laws are subject to. [C and D incorrect]

Newton’s first law of motion states that a body continues at rest or at constant velocity unless a resultant (external) force acts on it.

Newton’s second law of motion states that the resultant force is equal to the rate of change of momentum.

Newton’s first law does follow from the second law as follows. By using the second law for a body on which the resultant force is zero, the acceleration will also be zero, and so the first law is thus proved from the second law.

Question 151: [Measurements > Estimates]
Which statement is incorrect by factor of 100 or more?
A Atmospheric pressure is about 1 × 10⁵Pa.
B Light takes 5 × 10²s to reach us from Sun.
C Frequency of ultra-violet light is 3 × 10¹²Hz.
D Life-span of a man is about 2 × 10⁹s.

Reference: Past Exam Paper – June 2013 Paper 13 Q3



Solution 151:

Answer: C.

“The frequency of ultra-violet light is 3 × 10¹²Hz” is incorrect. The correct frequency is from about 10¹⁴Hz to 10¹⁷Hz.









Question 152: [Forces > Equilibrium]
Diagrams show two ways of hanging same picture.

In both cases, string is attached to same points on the picture and looped symmetrically over a nail in a wall. Forces shown are those that act on the nail.
In diagram 1, string loop is shorter than in diagram 2.
Which information about magnitude of the forces is correct?
A R₁ = R₂        T₁ = T₂
B R₁ = R₂        T₁ > T₂
C R₁ > R₂        T₁ < T₂
D R₁ < R₂        T₁ = T₂

Reference: Past Exam Paper – June 2014 Paper 11 Q11 & November 2009 Paper 11 Q12 & November 2009 Paper 12 Q11



Solution 152:

Answer: B.

The weight of the picture is the same, so, the upward tension is the same. i.e. R₁ = R₂. [C and D incorrect]

{From this point itself, it is clear that T₁ cannot be equal to T₂. The solution may be chosen as B from here itself. Explanations are available below to show that B is actually the correct answer}

Consider the acute angles (angles between 0^o and 90^o) between the T’s and the vertical. Call these θ. The acute angle formed in diagram 1 (θ₁) is clearly larger than in diagram 2 (θ₂).

For each diagram, the sum of the vertical downward components of the tensions should be equal to the weight, W which is equal to R (R₁ = R₂).

2T₁cosθ₁ = Weight, W            and Weight, W= 2T₂cosθ₂.

Therefore, T₁cosθ₁ = T₂cosθ₂

From the graph of cosine, it can be seen that as acute angle θ increases (angle between 0^o and 90^o), the value of cosθ decreases. Since angle θ₁ > angle θ₂, the value of cosθ₁ < value of cosθ₂. So, for the equation (T₁cosθ₁ = T₂cosθ₂) to hold, tension T₁ > tension T₂.

Question 153: [Current of Electricity > Electrons]
Electric current is passed from thick copper wire through a section of thinner copper wire before entering second thick copper wire as shown.

Which statement about current and speed of electrons in wires is correct?
A Current and speed of the electrons in the thinner wire are both less than in the thicker copper wires.
B Current and speed of the electrons is the same in all the wires.
C Current is the same in all the wires but speed of the electrons in the thinner wire is greater than in the thicker wires.
D Current is the same in all the wires but speed of the electrons in the thinner wire is less than in the thicker wire.

Reference: Past Exam Paper – November 2013 Paper 13 Q33



Solution 153:

Answer: C.

In the situation described, the thinner copper wire acts as a ‘resistor’ since a smaller cross-sectional area is now available for the electrons to flow.

It is a common misconception among many learners of physics to think that electrons slow down in a resistor. But they do not. The electrons speed up to release energy (as it is known that thermal energy is released).

 (A similar situation to this is that water speeds up when it moves from a river to a waterfall.)

Since the wires are connected in series, the current is the same in each of them.

Question 154: [Current of Electricity > Potentiometer]
Unknown e.m.f. E of cell is to be determined using a potentiometer circuit. Balance length is to be measured when the galvanometer records a null reading.
What is the correct circuit to use?

Reference: Past Exam Paper – June 2008 Paper 1 Q38



Solution 154:

Answer: B.

For the galvanometer to give a null reading, the potential difference across it should be zero. This can only occur when the position at which it is connected on the wire gives a potential difference (on the wire) equal to the cell of e.m.f. E. The cell should be connected such that its potential opposes the potential on the wire.  When the potential difference across the galvanometer is zero, no current flows through it and the galvanometer records a null reading. Only circuit B provides such a situation.

Current flows from the positive terminal to the negative terminal. So, the positive terminal of the cell cannot be connected directly to the galvanometer because this will always cause a current to flow through it. [A is incorrect] {Circuit A may be used to calibrate the potentiometer}

Similarly, the galvanometer cannot be connected directly to the negative terminal of the main supply because all the current would return to the negative terminal of the main supply, causing the galvanometer to always give a non-zero reading. [C and D are incorrect]

Question 155: [Electric Field > Moving charge]
Diagram shows path of a charged particle through a uniform electric field, having vertical field lines.

What could give a path of this shape?
A positive charge travelling left to right in a field directed downwards
B positive charge travelling right to left in a field directed downwards
C negative charge travelling right to left in a field directed upwards
D negative charge travelling left to right in a field directed downwards

Reference: Past Exam Paper – June 2013 Paper 13 Q30



Solution 155:

Answer: D.

The direction of the electric field lines is (away) from positive towards negative.
So, if the field is directed downwards, a positive change would experience a force downwards while a negative charge would experience a force upwards. Similarly, if the field is directed upwards, a positive charge would experience a force upwards and a negative charge would experience a force downwards.

One important deduction that should be made from the diagram is that the charge is actually entering the field lines at an angle downward. So, for any of the possible choices for answers, it should be accounted that initially, the path would be along the path of motion of the charge (that is, initially at an angle downward).

Consider A:
If it’s entering from the left, the right part of its path should be downwards since the field is directed downwards. [A is incorrect]

Consider B:
If it’s entering from the right, the left part of its path should be downwards since the field is directed downwards. [B is incorrect]

Consider C:
If it’s entering from the right, the left part of its path should be downwards since the field is directed upwards. [C is incorrect]

Consider D:
By entering from the left (at an angle downwards), the right part of the path should be upwards since the field is directed downwards. [D is correct]