# 9702 June 2010 Paper 21 Worked Solutions | A-Level Physics

__Question 1 (Units)__
Unit is often expressed with prefix.
For example, gram may be written with prefix ‘kilo’ as kilogram. Prefix
represents a power-of-ten. In this case, power-of-ten is 10

^{3}.
Complete Fig to show each prefix with
its symbol and power-of-ten:

Prefix Symbol Power-of-ten

Kilo k 10

^{3}
Nano n

**10**^{-9}
Centi

**c**10^{-2}**Mega**M 10

^{6}

**Tera**T 10

^{12}

__Question 2__**(a)**

Complete Fig to show whether each of
the quantities listed is vector or scalar:

Distance moved: scalar

Speed: scalar

Acceleration: vector

**(b)**

Ball falls vertically in air from
rest. Variation with time t of distance d moved by ball is shown in Fig.

(i)

Reference to Fig, how it can be
deduced that

1.

Ball is initially at rest:

The gradient (of the graph)
represents the speed / velocity (can be scored here or in 2). The initial
gradient {gradient at t = 0} is zero {so, the initial velocity is zero}.

2.

Air resistance is not negligible:

The gradient (of the line / graph)
becomes constant {terminal velocity is reached}

(ii)

Use Fig to determine speed of ball
at time of 0.40 s after it has been released:

{The gradient of the
tangent at t = 0.4s should be calculated (the tangent only touches the point on
the curve at t = 0.4s, and no other points on the curve). This represents the
instantaneous speed.

[By drawing the tangent at
t = 0.4s, I obtained a straight line passing through points (0.21, 0) and (0.6,
1.1) which I used to calculate the gradient. Gradient = (1.1 – 0) / (0.6 –
0.21) = 2.82ms

^{-1}]
Note that calculating the
ratio of 0.55 m / 0.40 s would give the average speed over the first 0.4s,
which is not what is required here.}

Speed of ball at time t = (2.8 ±
0.1) ms

^{-1}^{ }

(iii)

On Fig, sketch graph to show
variation with time t of distance d moved by ball for negligible air
resistance:

The graph is a curved line which is
never below the given line {air resistance opposes
motion, so in the absence of air resistance, the speed is greater} and
starts from zero. It is a continuous curve with increasing gradient {speed increases under gravity} and the line never
becomes vertical {a vertical line would correspond to
an infinite velocity which is not possible} or straight {a straight line corresponds to a constant velocity – if air
resistance was present, this would be the terminal velocity}

{A graph with increasing
gradient is shown. Fit it with the description given. It should look a bit like
the second graph given (but drawn better [with no straight line]– I could not draw a proper graph on
the computer)}

__Question 3__**{Detailed explanations for this question is available as Solution 627 at Physics 9702 Doubts | Help Page 124 -**

*http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-124.html*}

__Question 4__**(a)**

The diffraction of a wave occurs
when a

__wave__(front) passes by / incident on an edge / slit. The wave bends / spreads (into the geometrical shadow)**(b)**

Laser produces narrow beam of
coherent light of wavelength 632 nm. Beam is incident normally on diffraction
grating, as shown.

Spots of light are observed on
screen placed parallel to grating. Distance between grating and screen is 165
cm. Brightest spot is P. Spots formed closest to P and on each side of P are X and
Y. X and Y are separated by distance of 76 cm. Number of lines per metre on
grating:

{Do not confuse the
equation for diffraction grating and that for double slit.

Note that the equation for
diffraction grating (not for the double slit) is dsinθ = nλ where d is the slit separation, θ is the angle made by the
n

^{th}maxima and λ is the wavelength. The diffraction grating formula does not directly contain the separation of the slit and the screen (D, which is 165cm), but to calculate the angle, we can use this D (this D is not the same as the slit separation, d).
For double slit, x = λD /a
where x = fringe separation, λ = wavelength, D = distance between slits and
screen and a = slit separation.}

{distance PX = PY = 76 / 2
= 38cm. Let angle made by first maxima (spots at X and Y) to the horizontal = θ}

tan θ
= 38 / 165 [= 0.2303]

Angle θ = 13

^{o}[= 12.969]
dsinθ
= nλ

{For the first maxima, n =
1. So, n is known here}

(Slit separation,) d (= [632x10

^{-9}] / sin13) = 2.81x10^{-6}m
{One line corresponds to a
separation of 2.81x10

^{-6}m. So, the number of lines in 1m is obtained as follows:}
Number of lines per metre (= 1/d) =
3.6x10

^{5}**(c)**

Grating in (b) is now rotated about
axis parallel to incident laser beam, as shown. State what effect, if any,
rotation will have on positions of spots P, X and Y:

The spot at P remains in the same
position. The spots at X and Y rotate through 90

{Since the grating has been rotated by 90

P is the 0

^{o}.{Since the grating has been rotated by 90

^{o}, the lines are now perpendicular to the positions they were previously. So, diffraction occurs in a plane perpendicular to the previous one.P is the 0

^{th}(central) maxima. It is always found on the same line of incident of the light. So, it remains in the same position even if the grating is rotated}**(d)**

In another experiment using
apparatus in (b), student notices that distances XP and PY, as shown, are not
equal. Reason for difference:

EITHER The screen is not parallel to
the grating OR The grating is not normal to the (incident) light

__Question 5__**(a)**

An electric field is a region / area
where a charge experiences a force.

**(b)**

Electric field between earthed metal
plate and 2 charged metal spheres is illustrated.

(i)

On Fig, label each sphere with (+)
or (-) to show its charge:

{Direction of electric
field is from +ve to –ve}

Sphere on left-hand side is (+) and
sphere on right-hand side is (-)

(ii)

On Fig, mark region where magnitude
of electric field is

1.

Constant (label region C):

A correct region labeled C is within
10mm of the central plate

(otherwise within 5mm of plate)

2.

Decreasing (label region D):

A correct region labeled D is the
area of the field not included for (b)(ii)1

**(c)**

Molecule has its centre P of
positive charge situated distance of 2.8 × 10

^{–10}m from its centre N of negative charge, as illustrated. Molecule is situated in uniform electric field of field strength 5.0 × 104 V m^{–1}. Axis NP of molecule is at angle of 30° to uniform applied electric field. Magnitude of charge at P and at N is 1.6 × 10^{–19}C.
(i)

On Fig, draw arrow at P and arrow at
N to show directions of forces due to applied electric field at each of these
points:

{Electric field is from
+ve to –ve. So, the arrow at P is in the same direction as the applied field
(towards the right) and the arrow at N is in the opposite direction of the
applied field (towards the left)}

Arrows through P (towards the right)
and N (towards the left) in correct directions

(ii)

Torque on molecule produced by
forces in (i):

Torque = Force x

__Perpendicular__distance (between the forces)
Torque = [(1.6x10

^{-19})(5.0x10^{4})] [(2.8x10^{-10})sin30] = 1.1x10^{-24}Nm

__Question 6__
Electric heater is to be made from
nichrome wire. Nichrome has resistivity of 1.0 × 10

^{–6}Ω m at operating temperature of heater. Heater is to have power dissipation of 60 W when potential difference across its terminals is 12 V.**(a)**

For heater operating at its designed
power,

(i)

Current:

Power, P = VI

60 = 12I

Current, I (= 60 / 12) = 5(.0)A

(ii)

Show that resistance of nichrome
wire is 2.4Ω:

EITHER V = IR or P = I

^{2}R or P = V^{2}/ R
EITHER 12 = 5R or 60 = 5

^{2}R or 60 = 12^{2}/ R
Resistance of nichrome wire, R = 2.4Ω

**(b)**

Length of nichrome wire of diameter
0.80 mm required for heater:

Resistance of wire, R = ρL / A

Cross-sectional area, A = π (0.4x10

^{-3})^{2}[= 5.03x10^{-7}]
Length, L = 2.4 (5.03x10

^{-7}) / (1.0x10^{-6}) = 1.2m**(c)**

Second heater, also designed to
operate from 12 V supply, is constructed using same nichrome wire but using
half the length of that calculated in (b). Explain quantitatively effect of
this change in length of wire on power of heater:

The resistance is halved {since R is proportional to L}. EITHER The current is
doubled {since I = V / R} OR The power is
proportional to 1 / R. So, the power is doubled

__Question 7__
One of isotopes of uranium is
uranium-238 (

^{238}_{92}U).**(a)**

Isotopes are nuclei / atoms {of the same element} with the same proton / atomic
number but contain different numbers of neutrons / different atomic mass

**(b)**

For nucleus of Uranium-238,

(i)

Number of protons: 92

(ii)

Number of neutrons: (238 – 92 =) 146

**(c)**

A uranium-238 nucleus has radius of
8.9 × 10

^{–15}m. For uranium-238 nucleus,
(i)

Mass:

{1u = 1.66x10

^{-27}kg}
Mass (= 238u) = 238 (1.66x10

^{-27}) = 3.95x10^{-25}kg
(ii)

Mean density:

{Assuming the nucleus is
of a spherical shape, volume of sphere = (4/3)πr

^{3}}
Volume of nucleus = (4/3) π (8.9x10

^{-15})^{3}[= 2.95x10^{-42}]
Density (= mass/volume) = (3.95x10

^{-25}) / (2.95x10^{-42}) = 1.3x10^{17}kgm^{-3}**(d)**

Density of lump of uranium is 1.9 ×
10

^{4}kg m^{–3}. Using answer to (c)(ii), what can be inferred about structure of atom:
{Since the density of the
nucleus is large} The nucleus
contains

__most__of the mass of the atom. EITHER The nuclear diameter is__very much__less than that of the atom OR The atoms is mostly (empty) space.
Thank you

ReplyDeletethanks for the help

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