# 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 R

_{1}, R_{2}and R_{3}?
R

A 2 1 5_{1}/ kΩ R_{2}/ kΩ R_{3}/ kΩB 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 R

_{3}= 2 – 0 = 2V
p.d. across R

_{2}= 3 – 2 = 1V
p.d. across R

_{1}= 5 – 3 = 2VThe 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 R

_{1}, current I = 2 / R

_{1}

For R

_{2}, current I = 1 / R

_{2}

For R

_{3}, current I = 2 / R

_{3}

Consider choice A:

For R

_{1}, current I = 2 / 2 = 1A, For R

_{2}, current I = 1 / 1 = 1A, For R

_{3}, current I = 2 / 5 = 0.4A

The currents are not all the same

Consider choice B:

The currents across R

_{1}, R

_{2}and R

_{3}are 2/3A, ½A and 1A respectively.

Consider choice C:

The currents across R

_{1}, R

_{2}and R

_{3}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 V

_{P}= 4.0 V
{Peak power dissipated = V

_{P}^{2 }/ R
Mean power dissipated =
Peak power dissipated / 2}

Mean power dissipated = V

_{P}^{2 }/ 2R = 4^{2}/ (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 p

_{1}at time t_{1}. At time t_{2}the momentum is p_{2}.
What is magnitude of the average
force acting on the ball between times t

_{1}and t_{2}?**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 (t

_{2}– t_{1}), the momentum changes from p_{1}to p_{2}, however the direction of the momentum also changes. So, the momentum change is (p_{1}– p_{2}). Note that the ‘sign’ of the momentum is already included in p_{2}since it is shown below the horizontal line in the graph. For example, p_{2}can be -2, -5, -10, … This makes (p_{1}– p_{2}) positive.
Average force = Δp / Δt = (p

_{1}– p_{2}) / (t_{2}– t_{1})

__Question 549: [Dynamics]__Golf ball of mass m is dropped onto a hard surface from height h

_{1}and rebounds to a height h

_{2}.

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√(2gh

_{1}) – m√(2gh

_{2})

B m√(2gh

_{1}) + m√(2gh

_{2})

C m√[2g(h

_{1}–h

_{2})]

D m√[2g(h

_{1}+h

_{2})]

**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 mv

_{1}^{2}= mgh_{1}
Speed v

_{1}as the ball reached the surface = √(2gh_{1})
Similarly, the kinetic energy of the
ball just as it leaves the surface is converted to potential energy at height h

_{2}.
½ mv

_{2}^{2}= mgh_{2}
Speed v

_{2}just as the ball leaves the surface = √(2gh_{2})
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 (v

_{2}+v_{1})
Total change in momentum = m√(2gh

_{1}) + m√(2gh_{2})
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 E

_{P}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 = Q

_{1}Q_{2}/ 4πε_{0}r
{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.

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?

ReplyDeleteYes. 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.

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

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

DeleteConsider the following example involving number for better understanding.

DeleteLet’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.

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

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

Delete+ 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

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

ReplyDeleteSorry, that was a mistake. I have already correctly it.

DeleteThanks

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?

DeleteNo, 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

DeleteI'm sorry, but I don't get question 551.

ReplyDeleteIsn'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

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