# 9702 November 2013 Paper 21 & 22 Worked Solutions | A-Level Physics

Paper 21 and Paper 22 are similar.

**Question 1****(a)**

__Choose any 2 :__

Kelvin (K), Ampere (amp/A), Mole (mol), Candela (Cd)

**(b)**

(i) Work =
Force x Distance = (Mass x Acceleration) x Distance :

**kg**x**ms**x^{-2}**m**
Work :

**kg m**^{2}s^{-2}__Alternatively:__

Energy = ½ mv

^{2}or mc^{2}:**kg (ms**=^{-1})^{2}**kg m**^{2}s^{-2}
(ii) Density = mass/ volume :

**kgm**^{-3}
Acceleration of free fall, g :

**ms**^{-2}
Cross-sectional area of wire, A :

**m**^{2}
Original length, l

_{0}:**m**
Therefore, C :

**kg m**/ (^{2}s^{-2}**kg m**)^{-3}**(**^{2}**ms**)^{-2}^{2}**m**(^{2}**m**)**=**^{3}**kg**^{-1}ms^{-2}

**Question 2**

**(a)**

Time-base setting = 0.20μscm

^{-1}
That is, 1cm (1 cell) represents 0.20μs

Speed of pulse, v = 3 x 10

Speed of pulse, v = 3 x 10

^{8}ms^{-1}
Number of cells between peaks of the
2 pulses = 4

Speed, v = distance, d / time, t

Therefore, distance travelled by
pulse, d = v x t

t = 4 x 0.20 = 0.80μs

d = 3 x 10

^{8}x 0.80 x 10^{-6}= 240 m
Since the pulse travels from the
source to the reflector, and from the reflector back to the source,

Distance between source and
reflector = 240/2 = 120m

(b)

Speed of sound = 300ms

^{-1}
Speed of light = 3 x 10

^{8}ms^{-1}
Time taken by sound wave = 240 / 300
= 0.8s

^{ }
This time corresponds to 4 cells (4cm)
since the distance 240m corresponds to 4 cells.

Therefore,

4 cells represent 0.8s

1 cell represent 0.8/4 = 0.2s

Time-base setting is 0.20scm

^{-1}[unit required]

__Question 3__**{Detailed explanations for this question is available as Solution 673 at Physics 9702 Doubts | Help Page 135 -**

*http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-135.html*}

__Question 4__**(a)**

The torque of a couple is the product
of

__one of the forces__multiplied by the__perpendicular distance between the forces__.**(b)**

**(i)**

Weight at P – vertically down

__Choose any 1:__

Normal reaction - vertically up

Contact force at P – at point of
contact with the pin

(ii)

Torque = 35 x 0.25 x 2 = 17.5 ≈ 18 Nm

(iii)

The two 35N forces are equal and
opposite and the weight and the upward / contact / reaction force are equal and
opposite.

(iv)

The wheel is not in equilibrium since the (resultant) torque is not zero.

__Question 5__**Detailed explanations for this question is available as Solution 540 at Physics 9702 Doubts | Help Page 106 -**}

*http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-106.html*

__Question 6__

__Question 7__**(a)**

(i)

__Choose any 1:__

The direction of the fields is the
same.

Fields are uniform.

Constant electric field strength

E = V/d where E is the uniform electric
field between the 2 parallel plates, separated by the distance d and V is the
electric potential.

(ii)

Reduce the potential difference
across the

__plates__
Increase the separation

__of the plates__
(iii)

α-particles have opposite charge to β-particles
(deflection in opposite direction)

β-particles have a range of
velocities OR energies (different deflections) while α-particles all have the
same velocity OR energy (constant deflection)

α-particles are more massive (do not
use ‘heavier’) (deflection is less for greater field strength)

**(b)**

W = 234, X = 90, Y = 4 and Z = 2

**(c)**

A = 32, B = 16, C = 0 and D = -1

hi

ReplyDeletehi can i have solution for paper 42 nov 2010

ReplyDeletesolutions have already been provided.

DeleteFor question 6(b)(II) to calculate the current in the wires , the resistance we use is 39.08 ohm, (which is included the resistance of heater)

ReplyDeleteBut however, to calculate the power loss in the wires , in question 6(b)(III) ,the resistance used is only the two wires. Why? And how do I know when should I use which resistance

I have added some for this part.

DeleteHi i didnt understand in question 5(a)(ii) why the distance was divided by 4? If you know please help and explain

ReplyDeleteHi

ReplyDeleteI didnt understand question 5 (a(ii) why the distance was divided by 4? If u can explain more this will be very helpful thanks!

ReplyDeleteThe details have been added

Deletecan i please have the solution to may june 2013 Q6 Paper 21.

ReplyDeleteSee solution 676 at

Deletehttp://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-136.html