# Physics 9702 Doubts | Help Page 159

__Question 787: [Waves > Diffraction]__
Interference patterns from a
diffraction grating and a double slit are compared.

Using diffraction grating, yellow
light of the first order is seen at 30° to the normal to the grating.

The same light produces interference
fringes on screen 1.0 m from the double slit. The slit separation is 500 times
greater than the line spacing of the grating.

What is the fringe separation on the
screen?

A 2.5 × 10

^{–7}m
B 1.0 × 10

^{–5}m
C 1.0 × 10

^{–3}m
D 1.0 × 10

^{–1}m**Reference:**

*Past Exam Paper – November 2006 Paper 1 Q27*

__Solution 787:__**Answer: C.**

For double slits: Separation of
slits, a = Dλ / w

where D is the distance of the slit
form the screen, w is fringe separation and λ
is the wavelength

For diffraction grating: d sinθ = n λ

Since the same light is used, the
wavelength λ is the same in both cases.

For the diffraction grating, the 1

^{st}order (n = 1) is seen at 30° to the normal (θ = 30°)
d sin30° = 1 λ

Wavelength λ = 0.5d

For double slits,

Separation of slits, a = Dλ / w

Fringe separation, w = Dλ / a

In the question, it is stated that
the slit separation (a) for the double slit is 500 times greater than the line
spacing of the grating (d).

a = 500d

Wavelength λ = 0.5d

Fringe separation, w = Dλ / a = D (0.5d)
/ 500d = 0.5 (1.0) / 500 = 1.0 × 10

^{–3}m

__Question 788: [Matter > Young modulus]__
Behaviour of a wire under tensile
stress may be described in terms of the Young modulus E of the material of the
wire and of the force per unit extension k of the wire.

For wire of length L and
cross-sectional area A, what is the relation between E and k?

A E = A / kL B E = kA / L C
E = kL / A D E = L/ kA

**Reference:**

*Past Exam Paper – June 2011 Paper 12 Q23*

__Solution 788:__**Answer: C.**

Young modulus E = stress / strain

Stress = Force / Area = F / A

Strain = extension / original length
= e / L

Young modulus E = (F/A) / (e/L) = FL
/ Ae

But Hooke’s law: F = ke giving k = F/e.

Young modulus E = kL / A

__Question 789: [Current of Electricity]__
Diagram shows the circuit for a
signal to display a green or a red light. It is controlled by the switch S.

The signal is some way from S to
which it is connected by cable with green, red and black wires. At the signal,
green and red wires are connected to the corresponding lamp and the black wire
is connected to a terminal x to provide a common return. The arrangement is
shown correctly connected and with the switch set to illuminate the red lamp.

During maintenance, wires at the
signal are disconnected and, when reconnected, the black wire is connected in
error to the green lamp (terminal g) instead of terminal x. Red wire is connected
correctly to its lamp and connections at S remain as in the diagram.

When the system is tested with
switch connection to the red wire, what does the signal show?

A the green lamp illuminated
normally

B the red lamp illuminated normally

C the red and green lamps both
illuminated normally

D the red and green lamps both
illuminated dimly

**Reference:**

*Past Exam Paper – June 2012 Paper 12 Q37*

__Solution 789:__**Answer: D.**

The lamps are lit dimly since they
are connected in series.

The same power supply is now providing
current to both lights at the same time.

Ohm’s law: V = IR

The total resistance in the circuit is
higher and thus the current flowing in the circuit is less.

Each of the lights receive a voltage
lower than 12V such that the total voltage across the two of them is 12V. Previously,
all the 12V was being provided to a single lamp.

__Question 790: [Current of Electricity]__
A cell of e.m.f. E and internal
resistance r is connected in series with switch S and an external resistor of
resistance R.

The p.d. between P and Q is V.

When S is closed,

A V decreases because there is a
p.d. across R.

B V decreases because there is a
p.d. across r.

C V remains the same because the
decrease of p.d. across r is balanced by the increase of p.d. across R.

D V remains the same because the sum
of the p.d.s across r and R is still equal to E.

**Reference:**

*Past Exam Paper – November 2011 Paper 11 Q36 & Paper 13 Q38*

__Solution 790:__**Answer: B.**

When the switch S is open, no
current flows in the circuit. So, the p.d. between P and Q is equal to the
e.m.f. E of the cell.

The cell has an internal resistance,
so as the switch is closed (the circuit is now complete for a current to flow) there
is some loss volt (V

_{L}= Ir) across the internal resistance r of the cell. That is, the p.d. between P and Q is now less than the e.m.f. of the cell (V = E – V_{L}= E – Ir).

__Question 791: [Vectors]__
Diagram shows a resultant force and
its horizontal and vertical components.

The horizontal component is 20.0 N
and θ = 30°. What is the vertical component?

A 8.7 N B 10.0 N C
11.5 N D 17.3 N

**Reference:**

*Past Exam Paper – June 2009 Paper 1 Q2*

__Solution 791:__**Answer: C.**

Let the resultant vector be R.

Horizontal component = R cosθ

Vertical component = R sinθ

Horizontal component = R cosθ = 20.0 N

Resultant vector R = 20 / cosθ

Vertical component = R sinθ = (20/cosθ)
sinθ = 20 tanθ = 20 tan30° = 11.5 N

__Question 792: [Waves > Diffraction grating]__
Monochromatic light is incident on
diffraction grating and a diffraction pattern is observed.

Which line of table gives the effect
of replacing the grating with one that has more lines per metre?

**Reference:**

*Past Exam Paper – June 2014 Paper 13 Q30*

__Solution 792:__**Answer: B.**

Diffraction grating: dsinθ = nλ

Replacing the grating with one that
has more lines per metre causes the slit separation, d to decrease.

So, the number of orders of
diffraction visible, n (= dsinθ / λ) decreases [since
d is decreasing] while the angle between first and second orders of
diffraction, θ (= sin

^{-1}(λ/d) and can only be between 0 and 90) increases [between 0 and 90, the greater the argument of sin^{-1}, the closer the angle is to 90 {as d decreases, the argument increases}].
i would need an explanation in june 2006 q28 paper 1

ReplyDeleteSee solution 920 at

Deletehttp://physics-ref.blogspot.com/2015/08/physics-9702-doubts-help-page-189.html

q25 june 2011 variant 11

ReplyDeleteCheck solution 129 at

Deletehttp://physics-ref.blogspot.com/2014/11/physics-9702-doubts-help-page-24.html

In Q790 your reasoning shows A to be correct not B

ReplyDeleteThe explanation has been updated.

Deletesolution 792,

ReplyDeletewhat is argument of sine?

it's already in the explanation notes:

Deletesin^-1(λ/d) and can only be between 0 and 90

if you draw a sine curve for angles between 0 and 90, you would see that it increases