Tuesday, April 7, 2015

Physics 9702 Doubts | Help Page 107

  • Physics 9702 Doubts | Help Page 107



Question 545: [Current of Electricity]
Potential divider is used to give outputs of 2 V and 3V from a 5 V source, as shown.

What are possible values for resistances R1, R2 and R3?

R1/ kΩ             R2/ kΩ             R3/ kΩ
A         2                      1                      5
B         3                      2                      2
C         4                      2                      4
D         4                      6                      10

Reference: Past Exam Paper – June 2002 Paper 1 Q35



Solution 545:
Answer: C.
The total p.d. across the network of resistors is 5V. The potential at the bottom is 0V and that at the upper terminal is 5V.

The p.d. across a component is the difference in potential between the 2 terminals of that component. From the diagram, it can be concluded that
p.d. across R3 = 2 – 0 = 2V
p.d. across R2 = 3 – 2 = 1V
p.d. across R1 = 5 – 3 = 2V

The sum of p.d. is 2 + 1 + 2 = 5V.
From Ohm’s law: Current I in the network = V / R   for each resistor

Since the resistors are connected in series, the current flowing through them is the same.
For R1, current I = 2 / R1
For R2, current I = 1 / R2
For R3, current I = 2 / R3

Consider choice A:
For R1, current I = 2 / 2 = 1A, For R2, current I = 1 / 1 = 1A, For R3, current I = 2 / 5 = 0.4A
The currents are not all the same

Consider choice B:
The currents across R1, R2 and R3 are 2/3A, ½A and 1A respectively.

Consider choice C:
The currents across R1, R2 and R3 are 0.5A, 0.5A and 0.5A respectively.
The currents are the same.









Question 546: [Current of Electricity > A.C. circuits]
Bridge rectifier consists of four ideal diodes A, B, C and D, connected as shown in Fig.


An alternating supply is applied between terminals X and Y.
(a)
(i) On Fig, label the positive (+) connection to load resistor R.
(ii) State which diodes are conducting when terminal Y of supply is positive.

(b) The variation with time t of potential difference V across the load resistor R is shown in Fig.

Load resistor R has resistance 2700 Ω.
(i) Use Fig to determine the mean power dissipated in the resistor R.
(ii) On Fig, draw symbol for a capacitor, connected so as to increase the mean power dissipated in the resistor R.

(c) Capacitor in (b)(ii) is now removed from the circuit.
Diode A in Fig stops functioning, so that it now has infinite resistance.
On Fig, draw variation with time t of the new potential difference across the resistor R.

Reference: Past Exam Paper – November 2012 Paper 43 Q6



Solution 546:
(a)
(i) The connection to the ‘top’ of the resistor should be labelled as positive


(ii) Diode B and diode D

(b)
(i)
Peak voltage VP = 4.0 V
{Peak power dissipated = VP2 / R
Mean power dissipated = Peak power dissipated / 2}
Mean power dissipated = VP2 / 2R = 42 / (2 × 2700) = 2.96 × 10–3 W

(ii) The capacitor, correct symbol, should be connected in parallel with R

(c) The graph is that of half-wave rectifications. It has the same period and the same peak value.
{Current cannot flow through A. So, there will only be a p.d. across R when terminal Y is positive. When terminal X becomes positive again (since this is an a.c. supply), current in the circuit is zero and hence the p.d. across R is also zero.}











Question 547: [First law of Thermodynamics]
(a) First law of thermodynamics may be expressed in the form ΔU = q + w.
Explain symbols in this expression
+ ΔU:
+ q:
+ w:

(b)
(i) State what is meant by specific latent heat
(ii) Use first law of thermodynamics to explain why specific latent heat of vaporisation is greater than specific latent heat of fusion for a particular substance

Reference: Past Exam Paper – June 2011 Paper 42 & 43 Q4



Solution 547:
(a)
+ ΔU: increase in internal energy
+ q: thermal energy / heat supplied to the system
+w: work done on the system

(b)
(i) Specific latent heat is the (thermal) energy required to change the state of a substance per unit mass without any change of temperature.

(ii)
{The gas is not considered to be an ideal gas here}
When evaporating, there is a greater change in the separation of atoms / molecules.
{This results in a greater increase in internal energy. Internal energy here depends only on the potential energy of the molecules. For specific latent heats, the temperature does not change, and so the kinetic energy also remain constant.}
Additionally, when evaporating, there is a greater change in volume.
{so the gas has to do more work against atmospheric pressure to escape since work done w = pΔV. In this case, w is negative since work is done by the system and so, the latent heat of evaporisation, q is greater.}
Identifies each difference correctly with ΔU and w.
{ΔU = q + w
The specific latent heat of fusion or vaporization are given by q = ΔU – w
For vaporization, ΔU is greater and w (which has a negative value) is also greater. These cause q to be greater.}











Question 548: [Dynamics]
Graph shows variation with time of the momentum of a ball as it is kicked in a straight line.

Initially, momentum is p1 at time t1. At time t2 the momentum is p2.
What is magnitude of the average force acting on the ball between times t1 and t2?
 


Reference: Past Exam Paper – June 2007 Paper 1 Q10



Solution 548:
Answer: B.
Force is defined as the rate of change of momentum.

In a time interval of (t2 – t1), the momentum changes from p1 to p2, however the direction of the momentum also changes. So, the momentum change is (p1 – p2). Note that the ‘sign’ of the momentum is already included in p2 since it is shown below the horizontal line in the graph. For example, p2 can be -2, -5, -10, … This makes (p1 – p2) positive.

Average force = Δp / Δt = (p1 – p2) / (t2 – t1)










Question 549: [Dynamics]
Golf ball of mass m is dropped onto a hard surface from height h1 and rebounds to a height h2.
Momentum of the golf ball just as it reaches the surface is different from its momentum just as it leaves the surface.
What is total change in the momentum of the golf ball between these two instants? (Ignore air resistance.)
A m√(2gh1) – m√(2gh2)
B m√(2gh1) + m√(2gh2)
C m√[2g(h1–h2)]
D m√[2g(h1+h2)]

Reference: Past Exam Paper – November 2014 Paper 13 Q11



Solution 549:
Answer: B.
Momentum = mv

We need to find the velocities at the 2 instants mentioned.

As the golf ball reaches the surfaces after being dropped, all its potential energy is converted to kinetic energy.
½ mv12 = mgh1
Speed v1 as the ball reached the surface = √(2gh1)

Similarly, the kinetic energy of the ball just as it leaves the surface is converted to potential energy at height h2.
½ mv22 = mgh2
Speed v2 just as the ball leaves the surface = √(2gh2)

The direction of motion of the golf ball changes after hitting the hard surface, so the initial momentum should be added.
Total change in momentum = m (v2+v1)
Total change in momentum = m√(2gh1) + m√(2gh2)

The two terms must be well separated and not enclosed by a single square root sign.










Question 550: [Current of Electricity]
Diagram shows an incorrectly connected circuit. Ammeter has a resistance of 0.1 Ω and the voltmeter has a resistance of 1 MΩ.

Which statement is correct?
A The ammeter reads 2 mA.
B The ammeter reads 20 A.
C The voltmeter reads zero.
D The voltmeter reads 2 V.

Reference: Past Exam Paper – June 2013 Paper 11 Q36



Solution 550:
Answer: D.
The resistance of the voltmeter is very much larger than that of the ammeter and the resistor.

Ohm’s law: V = IR


The greater the resistance, the greater is the p.d. across a component. So, the voltmeter reading would be 2V because the combined resistance of the resistor and ammeter is negligible compared to that of the voltmeter. The resistance of the ammeter and resistor in parallel is close to 0.1 Ω, so the potential difference of 2V must be mostly across the much larger resistance of the voltmeter.


All the other statements are incorrect.











Question 551: [Electric field]
(a) Define electric potential at a point.

(b) Two small spherical charged particles P and Q may be assumed to be point charges located at their centres. The particles are in vacuum.
Particle P is fixed in position. Particle Q is moved along line joining the two charges, as illustrated in Fig.

Variation with separation x of electric potential energy EP of particle Q is shown in Fig.

(i) State how magnitude of the electric field strength is related to potential gradient.
(ii) Use answer in (i) to show that the force on particle Q is proportional to the gradient of the curve of Fig.

(c) Magnitude of the charge on each of the particles P and Q is 1.6 × 10–19 C.
Calculate separation of particles at the point where particle Q has electric potential energy equal to –5.1 eV.

(d) By reference to Fig, state and explain
(i) whether the two charges have the same, or opposite, sign,
(ii) effect, if any, on the shape of the graph of doubling the charge on particle P.

Reference: Past Exam Paper – June 2011 Paper 41 Q4



Solution 551:
(a) The electric potential at a point is defined as the work done in bringing unit positive charge from infinity (to that point).

(b)
(i) The magnitude of the electric field strength is the potential gradient
(ii) The electric field strength is proportional to the force (on particle Q). The electric potential gradient itself is proportional to the gradient of the (potential energy) graph given (as stated in part (i)). So the force is proportional to the gradient of the graph.

(c)
An energy of 5.1 eV = 5.1 × (1.6 × 10–19) (J)
Electric potential energy = Q1Q2 / 4πε0r
{When dealing with magnitudes, we may neglect the sign of the electric potential energy.}
5.1 × (1.6 × 10–19) = (1.6 × 10–19)2 / (4π × (8.85 × 10–12) × r)
Separation r = 2.8 × 10–10 m

(d)
(i) Work is got out when the charge separation x decreases, so the charges are of opposite sign

(ii) The energy would be doubled. So, the gradient would be increased.



12 comments:

  1. in question 548, we assume that the force that causes the change in momentum of the ball is in the same direction as that of its initial momentum, but why do we do so?

    ReplyDelete
    Replies
    1. Yes. But the speed changes and so does the momentum. Both its magnitude and direction changes at different stages. So, the direction of the force (both magnitude and direction) also changes accordingly at different stages.

      This IS Newton's 2nd law. When momentum changes with time, there is a force.

      Delete
    2. bt how is the momentum change P1-P2? can you please explain ASAP with the directions of the changes etc?

      Delete
    3. Consider the following example involving number for better understanding.
      Let’s say momentum changes from +8 to -6. On a graph of momentum against time, (we will ignore the time here because it is obvious), the ‘+8’ would be above the time-axis (x-axis) and the ‘-6’ would be below the time-axis. The time-axis would be drawn at ‘momentum = 0’.

      The MAGNITUDE of the change is as follow:
      From +8 to 0, we have a magnitude of 8
      From 0 to -6, we have a magnitude of 6
      The total magnitude of the change is 8 + 6 = 14

      Now, let’s try to write it in a single equaltion:
      Magnitude of change = 8 – (-6) = 14

      If we compare with the question, p1 = 8 and p2 = -6
      So, magnitude of change = p1 – p2

      Here the question asks for magnitude, so we consider only the magnitude.

      Delete
    4. but shouldn't the change be final minus initial, p2-p1?

      Delete
    5. Yeah, but here we are given a graph - so we should consider the momentum as displayed in the graph.

      + we want the magnitude.
      With the direction, the change would be (consider the example using numbers) -6 - 8 = -14
      same value but with a negative sign

      Delete
  2. In 550, it says 'the voltmeter reading will be zero' I am confused

    ReplyDelete
    Replies
    1. Sorry, that was a mistake. I have already correctly it.
      Thanks

      Delete
    2. Shouldn't it be zero for voltmeter reading? As the voltmeter is incorrectly installed, it should be connected in parallel to the part of the circuit you wish to measure. It's true that resistance of the voltmeter is so high that it 'takes up' all the p.d but it shouldn't be giving any reading isn't it so?

      Delete
    3. No, most of the resistance is due to the voltmeter. From V = IR, it will have a p.d. of 2V across it as the other resistance is negligible compared to that of the voltmeter

      Delete
  3. I'm sorry, but I don't get question 551.
    Isn't the formula for electric potential energy: V = k · q1 / r? So, why do we have to multiply both charges: q1 and q2? In any case, what I would do it would be to add both potentials (scalar quantity). Any help will be very appreciated

    ReplyDelete
    Replies
    1. That's the formula for electric potential, not electric potential energy. For formula in the explanation above is the correct one - it's the product of the electric potential with the charge.

      Delete

If it's a past exam question, do not include links to the paper. Only the reference.
Comments will only be published after moderation

Currently Viewing: Physics Reference | Physics 9702 Doubts | Help Page 107