# Physics 9702 Doubts | Help Page 151

__Question 745: [Current of Electricity]__
Graph shows how the electric current
I through a conducting liquid varies with the potential difference V across it.

At which point on graph does the
liquid have the smallest resistance?

**Reference:**

*Past Exam Paper – November 2005 Paper 1 Q34 & November 2014 Paper 11 & 12 Q34*

__Solution 745:__**Answer: C.**

Note that the gradient of an I-V
graph is

**NOT**the reciprocal of the resistance. This is a common misconception among students. The resistance is not related to the gradient of the line. This is true for a straight line graph – Ohm’s law applying.
Ohm’s law: V = IR

Resistance however is defined as the
potential difference per unit current (R = V/I) and therefore the liquid has
the smallest resistance where the ratio I/V is largest.

Or, put another way, a straight line
from the origin to a point on the graph has to have maximum gradient. This will
be a line which is a tangent to the graph at C.

The steepness of the tangent tells
us how large the gradient is (when compared to another line), but to calculate
the VALUE of resistance at a specific value of I or V, we need to take the
ratio of V/I for a SINGLE point. The gradient deals with 2 points, which is
incorrect to evaluate the resistance.

__Question 746: [Vectors]__
Vectors P and Q are drawn to scale.

Which diagram represents vector (P +
Q)?

**Reference:**

*Past Exam Paper – November 2012 Paper 12 Q2*

__Solution 746:__**Answer: A.**

Vector addition of 2 vectors is
performed graphically as follows:

1 The first vector (say P) is drawn.

2 The second (Q) is drawn is its
tail at the head of the first vector (P).

3 The resultant start at the tail of
the first vector (P) and ends at the head of the second vector (Q).

Vector Q is a little longer than vector
P, so the resultant should point a little lower than the horizontal [as in choice A], but not almost downwards [as in choice B] since the vertical components of
the 2 vectors are almost equal.

__Question 747: [Measurement]__
Diagram shows part of a thermometer.

What is the correct reading on
thermometer and the uncertainty in this reading?

reading
/ °C uncertainty in
reading / °C

A 24 ±1

B 24
±0.5

C 24
±0.2

D 24.0
±0.5

**Reference:**

*Past Exam Paper – November 2014 Paper 13 Q4*

__Solution 747:__**Answer: D.**

We need to be careful with
instrument readings from scales. We cannot just read in whole numbers when scales
can usually be read to at least a half a division.

In this question, it was expected (by
the examiners) therefore that the candidates would give 24.0 ± 0.5, making the
last figure in the value correspond to the single figure in the error
estimation.

Note that this is not always the
case. In some cases, the uncertainty is the same as the least division that can
be read. But here, it is possible (and clear enough) to read at least half the
division.

__Question 748: [Matter]__
Which statement applies to the
boiling but

**not**to the evaporation of a liquid?
A All the bonds between molecules in
the liquid are broken.

B At normal atmospheric pressure,
the process occurs at one temperature only.

C Energy must be provided for the
process to happen.

D The separation of the molecules
increases greatly.

**Reference:**

*Past Exam Paper – November 2004 Paper 1 Q19*

__Solution 748:__**Answer: B.**

For both boiling and evaporation,
the bonds between the molecules in the liquids should be broken. [A is incorrect] Energy must be provided for both
processes the happen. [C is incorrect]

In both boiling and evaporation,
when the bonds are broken, the phase of the molecules changes from liquid to
gas. In the gas state, the separation of the molecules increases greatly. This
is true for both boiling and evaporation. [D is
incorrect]

Boiling and evaporating differ in
this aspect: at normal atmospheric pressure, boiling occurs at one temperature
only, while evaporation occurs over a range of temperatures.

__Question 749: [Current of Electricity]__
A researcher has two pieces of
copper of same volume. All of the first piece is made into a cylindrical
resistor P of length x.

All of the second piece is made into
uniform wires each of same length x which he connects between two bars of
negligible resistance to form a resistor Q.

How do the electrical resistances of
P and Q compare?

A P has a larger resistance than Q.

B Q has a larger resistance than P.

C P and Q have equal resistance.

D Q may have a larger or smaller
resistance than P, depending on the number of wires made.

**Reference:**

*Past Exam Paper – June 2007 Paper 1 Q37*

__Solution 749:__**Answer: C.**

Resistance R of a wire = ρL / A

Since both are copper, the
resistivity ρ is the same.

The lengths of both resistors through
which current would flow are the same (x).

Also, since the 2 pieces of copper
are of the same volume, the total cross-sectional area should be the same since
the same length is used. So, both P and Q have equal resistance.

A thick wire is just many thin ones
laid side by side.

A long wire is just many short ones
laid end to end.

__Question 750: [Measurement > C.R.O]__
A signal that repeats periodically
is displayed on screen of a cathode-ray oscilloscope.

The screen has 1 cm squares and the
time base is set at 2.00 ms cm

^{–1}.
What is frequency of this periodic
signal?

A 50 Hz B 100 Hz C
125 Hz D 200 Hz

**Reference:**

*Past Exam Paper – November 2013 Paper 11 & 12 Q4*

__Solution 750:__**Answer: B.**

We need to identify the complete
shape of the signal that repeats itself.

Consider the 4

^{th}block horizontally. There is a straight line at the equilibrium position. Then, there is a very small pulse up, followed by a relative larger pulse down, then one up and finally another pulse down with a portion being a straight line. Including the 4^{th}block, the shape that was just described takes up 5 blocks in total (horizontally).
Notice that the shape on the left of
the screen of the c.r.o. does not have the ‘very small pulse up’ displayed
while the shape on the right does not have the ‘second pulse down’ displayed.
So, we have only consider the central shape for the period.

The period T
corresponds to 5 squares

Period T =
5(2.00) = 10ms = 0.01s

Frequency, f =
1/T = 1 / 0.01 = 100Hz

in q748, how can A apply to evaporation? Evaporation only occurs at the surface, so only those bonds between the molecules at the surface will be broken, not all the bonds between the liquid molecules. so the molecules which arent on the surface would still have attractive forces between them.

ReplyDeleteI believe the questions meant all the bonds between the molecules BEING evaporated.

DeleteBecause even when it's boiling, not 'all the bonds' between the molecules are broken at the same time. If that was the case, the liquid would instantly disappear into a gas.

this is how i did question 749

ReplyDeleteVolume(v)=xArea of P

Ap=v/x

so Rp=resistivity*x/Ap

=resistivity*x*x/Ap

V=7*x*Aq (volume of P=Volume of Q)(lowercase q represents each individual wire in Q)

Aq=v/(7x)

Rq=resistivity*x/Aq (R of one wire in Q)

Rq=7*resistivity*x*x/v

1/RQ=1/Rq+1/Rq+1/Rq+1/Rq+1/Rq+1/Rq+1/Rq (because all the individual q wires are connected in parallel, so dis is how we wd have to obtain their total resistance, ryt)

RQ=Rp

is this method correct?

The working is OK but it's like proving the result I gave above.

DeleteWe don't have time for these in the exam.

Of course, physics can be proved by the calculations involved. But some times, we need to analyze the situation carefully and come up with simple conclusions that are PHYSICAL. - Just like I did above.

Taking several thin wires in parallel is just like taking a single thick wire.

And taking several short wires end to end (in series) is just like taking a single long wire.

11/M/J/11 Q.27,32

ReplyDeleteFor 11/M/J/11 Q.32, see solution 515 at

Deletehttp://physics-ref.blogspot.com/2015/03/physics-9702-doubts-help-page-100.html

For solution 245, first you say gradient does not represent reciprocal of resistance but later on you say we must draw a line to find the maximum gradient at a point... im confused by this please help

ReplyDeletewhen taking about gradient, we are referring to the gradient of the tangent at that point.

DeleteWhat I'm saying here is to draw a line from the origin to that point. this is different from a tangent.

I dont understand question 745. Gradient of IV graph is reverse of R? If not, why this reciprocal concept can be used to expalin IV graph of filament bulb? Thank you.

ReplyDeleteGradient = change in y / change in x

DeleteIn IV graph, gradient = change in I / change in V

But, Ohm's law: V = IR giving R = V/I

Compare the equation for gradient and that for R=V/I. The are reciprocal. So, to have an idea of the resistance from the graph, we should consider the reciprocal of the gradient.

But then again, the reciprocal of the gradient does not give the actual VALUE of resistance. This has already been explained above.