# Physics 9702 Doubts | Help Page 152

__Question 751: [Current of Electricity > Resistance]__
A conductor consists of three wires
connected in series. The wires are all made of same metal but have different cross-sectional
areas. There is current I in the conductor.

Point Y on the conductor is at zero
potential.

Which graph best shows variation of
potential V with distance along the conductor?

**Reference:**

*Past Exam Paper – November 2014 Paper 11 & 12 Q33*

__Solution 751:__**Answer: A.**

Resistance R = ρL / A

Ohm’s law: V = IR

The resistance R of the wire is
inversely proportional to the cross-sectional area. So, the thickest conductor has the least resistance.

Now from Ohm’s law, the potential
difference V is proportional to the resistance R. The thickest conductor has
the least resistance, so the smallest voltage drop occurs across the thickest
conductor.

Consider a part of the wire with constant
cross-sectional area – say, the thickest conductor.

Resistance R = ρL / A

So, along the thickest conductor, as
the distance from X increases, the length L of the conductor also increases. Since
the resistance is directly proportional to the length L, as we go along the
same conductor (of constant cross-sectional area), the resistance R increases
and so, the potential difference also changes. That is, the potential V is not
constant (not a straight line) for any of the conductors of constant
cross-sectional area. The stepped
changes in C and D are unphysical. [C and D are
incorrect]

As explained above, the thickest
conductor has the least resistance, so the smallest voltage drop occurs across
the thickest conductor.

The potential drop with distance is represented
by the gradient of the graph. So, the value of the gradient should be smallest
for the thickest conductor. [B is incorrect] Since
there is a potential DROP, the gradient is negative.

__Question 752: [Measurement > Estimates]__
What is approximate kinetic energy
of an Olympic athlete when running at maximum speed during a 100 m race?

A 400 J B 4000 J C
40 000 J D 400 000 J

**Reference:**

*Past Exam Paper – November 2012 Paper 12 Q3*

__Solution 752:__**Answer: B.**

In this question, we need to
estimate the maximum speed and mass of an Olympic athlete.

A 100m race in an Olympic
competition is usually completed in about 10s by the fastest persons.

Let the maximum speed of the Olympic
athlete be about 10ms

^{-1}.
Let the mass of the Olympic athlete be
about 80kg.

Kinetic energy = ½ mv

^{2}= ½ (80)(10)^{2}= 4000J

__Question 753: [Waves > Diffraction]__
A diffraction grating with 500 lines
per mm is used to observe diffraction of monochromatic light of wavelength 600
nm.

Light is passed through a narrow
slit and grating is placed so that its lines are parallel to the slit. Light
passes through the slit and then the grating.

Observer views the slit through the
grating at different angles, moving his head from X parallel to the grating,
through Y, opposite the slit, to Z parallel to the grating on the opposite
side.

How many images of the slit does he
see?

A 3 B
4 C 6 D
7

**Reference:**

*Past Exam Paper – June 2011 Paper 11 Q27*

__Solution 753:__**Answer: D.**

For diffraction grating: d sinθ = nλ

The largest value of θ is 90° on
both sides of Y. We need to identify the largest order n for which an image can
be seen (constructive interference).

There are 500 lines in 1×10

^{-3}m.
Slit separation, d = (1×10

^{-3}) / 500 m
Value of n = d sinθ / λ = (1×10

^{-3}) (sin90°) / (500) (600×10^{-9}) = 3.33
Since n can only be an integer, the
largest order is n = 3. So, there are 3 images on both sides of Y. An image
will also be seen at Y, which is directly along the source.

Number of images = 3 + 3 + 1 = 7

__Question 754: [Waves > Period]__
Diagram shows a square-wave trace on
the screen of a cathode-ray oscilloscope. A grid of 1 cm squares covers the
screen. Time-base setting is 10 ms cm

^{–1}.
What is approximate frequency of the
square wave?

A 70 Hz B 140 Hz C
280 Hz D 1400 Hz

**Reference:**

*Past Exam Paper – June 2010 Paper 11 Q4 & Paper 12 Q5 & Paper 13 Q3*

__Solution 754:__**Answer: B.**

In the diagram shown, there are
approximately 8 periods shown. This takes 6 squares (= 6cm).

Time-base setting is 10 ms cm

^{–1}.
Time corresponding to 6 cm = 6 x 10
= 60ms

1 period T = 60 /8 = 7.5ms

Frequency = 1 / T = 1/ (7.5x10

^{-3}) = 133.3Hz.
So, answer: B.

__Question 755: [Current of Electricity]__
An extension lead is used to connect
a 240 V electrical supply to heater as shown.

A voltmeter measures potential
difference (p.d.) across the heater as 216 V and an ammeter measures the
current through the heater as 7.7 A.

What is the total resistance of the
extension lead?

A 3.1 Ω B 6.2 Ω C
28 Ω D 31 Ω

**Reference:**

*Past Exam Paper – November 2014 Paper 13 Q36*

__Solution 755:__**Answer: A.**

The extension lead is connected in series
with the heater and the supply.

The same current of 7.7 A flows
through both the heater and the extension lead.

From Kirchhoff’s law, the sum of p.d.
in a circuit is equal the e.m.f. in the circuit here.

Let the p.d across the extension
lead be V.

V + 216 = 240

p.d. V = 240 – 216 = 24 V

Ohm’s law: V = IR

Total resistance R = V / I = 24 /
7.7 = 3.1 Ω

__Question 756: [Matter > Deformation]__
A simple crane consists of rigid
vertical pillar supporting a horizontal beam.

Weight W is lifted by a rope at the
end of the beam.

What are the forces at points X, Y
and Z due to the weight W?

force
at X force at Y force at Z

A tension
compression tension

B tension
tension compression

C compression
tension compression

D compression
compression compression

**Reference:**

*Past Exam Paper – June 2007 Paper 1 Q20*

__Solution 756:__**Answer: B.**

With the weight W placed there, the
horizontal beam would very slightly be curved up (like the upper part of a
semi-circle). This is only slightly since it is a rigid pillar. It may not even
be observed by our eyes.

Thus, the part around point Z would
be compressed. Force at Z: compression

As the horizontal beam is now
slightly curved, the part around Y would be extended. Force at Y: tension.

The same is true for the area around
X, which is on the outside of the pillar. Force at X: tension.

11/M/J/11 Q.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

in question 756, the horizontal prt of the pillar wd curve downwards and the vertical either crve or move toward the right, right?

ReplyDeleteThe right end move a little bit downwards but as said above, the horizontal part itself would be like the upper semi-circle.

DeleteThe vertical part would curve towards the left, just imagine a circle being formed.

but why would the vertical part curve towards the left?

DeleteJust try to imagine the horizontal part as the upper semi-circle and the vertical part being a continuation of that circle

ReplyDeleteQu 755, why are we not multiplying the resistance by 2? Resistances in series add up, right?

ReplyDeleteIt is already accounted for as we have used the total remaining p.d. If we tried to multiply by 2, we need to divide that remaining p.d by 2 first

DeleteThank you!

DeleteFor question 751 I would of assumed that the graph would start of with 0 Voltage as there is no resistance. Then in my mind the voltage would increase as the length increases and resistance is increasing.

ReplyDeleteNote that the potential DIFFERENCE is given by Ohm’s law: V = IR, not the POTENTIAL. Current flows from a point of higher potential to a point of lower potential. So, the point from which current is coming should have relatively the highest potential, compared to the point it is moving towards.

Deletefor solution 751 you said as the length increases the resistance so pd should increase then ?

ReplyDeletesince the potential keeps on changing, we cannot have a horizontal line in the graph (a horizontal line would indicate that the potential is constant)

Deleteyea it wont be a horizontal line but shouldnt it be a straight line showing pd increasing?

DeleteNo, it should decrease. The potential is maximum at first and then decreases to zero. Consider a battery for example. The positive terminal is at high potential while the negative terminal is at zero. So, potential decreases.

Delete9702/12/O/N/12 q6 plz

ReplyDeletesee link at

Deletehttp://physics-ref.blogspot.com/2014/08/9702-november-2012-paper-12-worked.html

Thank youuuu

Delete