# Physics 9702 Doubts | Help Page 137

__Question 678: [Electric field]__
An electric field exists in space
between two charged metal plates.

Which of the following graphs shows
variation of electric field strength E with distance d from X along the line
XY?

**Reference:**

*Past Exam Paper – June 2003 Paper 1 Q35 & November 2006 Paper 1 Q30 & June 2012 Paper 12 Q31*

__Solution 678:__**Answer: C.**

Electric field strength = V / x

where V is the p.d. between the 2
plates and x is the separation of the plates.

Since the separation of the plates and
the p.d. between the plates are constant, the electric field strength is also
constant anywhere between the plates.

__Question 679: [Waves > Double Slit experiment]__
Laser is placed in front of a double
slit, as shown in Fig.1.

Laser emits light of frequency 670
THz. Interference fringes are observed on screen.

**(a)**Explain how interference fringes are formed

**(b)**Show that wavelength of the light is 450nm

**(c)**Separation of the maxima P and Q observed on screen is 12 mm. Distance between the double slit and the screen is 2.8 m.

Calculate separation of the two
slits

**(d)**Laser is replaced by a laser emitting red light. State and explain the effect on the interference fringes seen on screen

**Reference:**

*Past Exam Paper – June 2014 Paper 22 Q7*

__Solution 679:__**(a)**The

__waves__from the double slit are coherent / constant phase difference. The

__waves__(from each slit) overlap / superpose / meet.

Maximum / bright fringes are
observed where the path difference is nÎ»
OR phase difference is n(360

^{o}) / 2Ï€n rad
OR Minimum / dark fringes are observed
where the path difference is (n + ½)Î» OR phase difference is (2n + 1)180

^{o}/ (2n + 1)Ï€ rad**(b)**

v = fÎ»

So, wavelength, Î» = (3x10

^{8}) / (670x10^{12}) = 448 (or 450) nm**(c)**

{12mm represents the

**separation**of 9 bright fringes (from the diagram). P and Q are bright fringes. Note that if we include the fringes at P and Q, there is 10 fringes – but we need to consider the__separation__here.}
Separation of bright fringes = 12 /
9 mm

Separation of 2 slits, a (= DÎ»/w) = (28)(450x10

^{-9}) / ({12/9}x10^{-3}) = 9.5x10^{-4}m**(d)**Red light has a larger / higher / longer wavelength (must be comparison). So, the fringes are further apart / larger separation

__Question 680: [Measurement > CRO]__
A microphone detects musical note of
frequency f. The microphone is connected to cathode-ray oscilloscope (c.r.o.).
Signal from the microphone is observed on the c.r.o. as illustrated in Fig.1.

Time-base setting of the c.r.o. is
0.50 ms cm

^{–1}. Y-plate setting is 2.5 mV cm^{–1}.**(a)**Use Fig.1 to determine

(i) amplitude of the signal,

(ii) frequency f,

(iii) actual uncertainty in f caused
by reading the scale on the c.r.o.

**(b)**State f with its actual uncertainty.

**Reference:**

*Past Exam Paper – November 2014 Paper 23 Q2*

__Solution 680:__**(a)**

(i)

{On the scale, the
amplitude is distance of 2.2cm from the zero line.}

Amplitude scale reading is 2.2 (cm)

{The Y-plate setting is
2.5 mV cm

^{–1}. That is, 1cm corresponds to 2.5 mV.}
Amplitude = 2.2 × 2.5 = 5.5 mV

(ii)

Time period scale reading = 3.8 (cm)

{Time-base setting of the
c.r.o. is 0.50 ms cm

^{–1}.}
Time period = 3.8 × (0.5 × 10

^{–3}) = 0.0019 (s)
{Frequency = 1 / period}

Frequency f = 1 / 0.0019 = 530 (526)
Hz

(iii)

{We need to find the
actual uncertainty in the frequency f.

The uncertainty in the
reading of the scale corresponds to 1 small square. Note that 5 small squares
is 1cm, as indicated on the diagram. So, 1 small square is 0.2cm.}

Time period scale reading
= 3.8 (cm). This corresponds to a period of 0.0019 (s).

EITEHR

If we measure from the
zero position to 1 period, the value read is 3.8cm on the scale and the
uncertainty is 0.2cm. This corresponds to a percentage uncertainty of (0.2 /
3.8) x 100% = 5.3%

OR

If we measure from the
zero position to 2 periods (measuring more periods reduces the uncertainty),
the value read is 7.6cm on the scale and the uncertainty is 0.2cm. This corresponds
to a percentage uncertainty of (0.2 / 7.6) x 100% = 2.6%}

EITHER Uncertainty in reading = ±
0.2 in 3.8 (cm) OR 5.3%

OR Uncertainty in reading = ± 0.2 in
7.6 (cm) OR 2.6%

{Frequency f = 1 / period
T. So, there will be the same percentage uncertainty in the frequency, as the
percentage uncertainty in the period. The frequency f was calculated to be 530
(526) Hz.

EITHER Uncertainty in f = 5.3%
of 526 OR 2.6% of 526}

EITHER Actual uncertainty = 5.3% of
526 = 27.7 or 28 Hz

OR Actual uncertainty = 2.6% of 526
= 13 or 14

**(b)**State f with its actual uncertainty.

{The uncertainty should be
expressed to one significant figure with the frequency quoted to the same
power-of-ten as the uncertainty.}

Frequency = 530 ± 30 Hz or 530 ± 10
Hz

__Question 681: [Electric field]__
Diagram shows an electron, with
charge e, mass m, and velocity v, entering a uniform electric field of strength
E.

Direction of the field and the
electron’s motion are both horizontal and to the right.

Which expression gives the distance
x through which electron travels before it stops momentarily?

A x = mv / E B x = mv / Ee C
x = mv

^{2}/ 2E D x = mv^{2}/ 2Ee**Reference:**

*Past Exam Paper – June 2009 Paper 1 Q29*

__Solution 681:__**Answer: D.**

Electric force on the electron = Ee

This force is the resultant force on
the electron and is equal to ma, where a is the acceleration. The resultant force on the electron causes it
to decelerate until it comes to rest.

So,

Magnitude of acceleration a = Ee / m

(The acceleration of the electron should
be taken as negative as it opposes motion – that is, it is a deceleration.)

Equation for uniformly accelerated
motion: v

^{2}= u^{2}+ 2as
Initial velocity of electron
(corresponding to u in the equation above) is v (as given in the question).
Final velocity of the electron = 0.

0 = v

^{2}+ 2(-a)x
Re-arranging and substituting the
value of a in the equation gives

v

^{2}= 2 (Ee/m) x
Thus, x = mv

^{2}/ 2Ee
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ReplyDeleteSee question 683 at

Deletehttp://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-138.html

Here is the first question from paper 1:

ReplyDelete12/M/J/10 Q.6

Go to

Deletehttp://physics-ref.blogspot.com/2014/11/9702-june-2010-paper-12-worked.html

21/O/N/14 Q.3(d), Q.4(a)(iii), Q.8(b)(i)

ReplyDeleteFor question 8, see solution 726 at

Deletehttp://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-147.html

22/O/N/14 Q.1(c)(i)

ReplyDeleteSee solution 688 at

Deletehttp://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-139.html

May june 2003 p1 q 22

ReplyDeleteSee solution 849 at

Deletehttp://physics-ref.blogspot.com/2015/06/physics-9702-doubts-help-page-169.html

Hi, your blog has been very helpful to me in my teaching. Do you have a compilation of these such that I can download it all? Much appreciated.

ReplyDeleteYour blog has been very helpful to me in my teaching, do you have a link where I can download these for future reference? Much appreciated

ReplyDeleteI'm glad that it's helping people around the world. Unfortunately i have not posted a compilation online. But you can still use the site directly in class. You would need an internet connection.

DeleteAnyway, do you mind sharing with us the way you use the site with the students? It would be really appreciated.