9702 November 2013 Paper 23 Worked Solutions | A-Level Physics
Question 1
Diameter of cylindrical disc = 28mm
Thickness of cylindrical disc = 12mm
Density of material of disc = 6.8 x
103 kgm-3
Density = mass, m /volume
Volume = π(14 x 10-3) 2
x 12 x 10-3 = 7.389 x 10-6 m3
Mass, m = density x volume = 6.8 x
103 x 7.389 x 10-6 = 0.0502kg
Weight = mg = 0.0502 x 9.81 = 0.49N
Question
2
(a)
SI
units for T : s, R : m and M : kg
K =
T2M/R3 units : s2kgm-3
(b)
Uncertainty
in T : 0.5%, R : 1% and M : 2%
K =
T2M/R3
{(ΔK / K) x 100% = [2(ΔT / T)
+ (ΔM / M) + 3(ΔR / R)] x 100%} % uncertainty in K = 2(0.5)% {for T} + 2% {for M} +3(1)% {for R} = 6%
K =
[864002 x 6 x 1024] / (4.23 x 107) = 5.918 x
1011
{(ΔK / K) x 100% = 6%. So, ΔK = (6/100) x K}
(Uncertainty in K =) 6 % of K = 0.355 x 1011
(Uncertainty in K =) 6 % of K = 0.355 x 1011
K =
(5.9 ± 0.4) x 1011 s2kgm-3
Question
3
(a)
(i)
Velocity
is the rate of change of displacement OR displacement change per
unit time taken.
(ii)
Acceleration
is the rate of change of velocity OR velocity change per unit
time taken
(b)
(i)
Initially , the car moves with constant velocity (gradient constant).
Then the velocity decreases in the middle section (gradient
decreases). Finally, in the last section, the car moves with a
constant velocity, which is smaller than the initial constant velocity
(gradient is constant but smaller).
(ii)
Velocity
= 45/1.5 = 30ms-1
(iii)
At t
= 6s, x = 122m and at t = 4s, x = 98m
Velocity
at 4s = (122-98) / 2 = 12ms-1
at t = 1.5s, velocity = 30ms-1 and at t = 4s, velocity = 12ms-1
Acceleration
= (12-30) / (4.0-1.5) = (12-30) / 2.5 = -7.2ms-2
(iv)
F = ma
= 1500 x (-7.2) = -10800N ≈ -11000N
Question
4
{Detailed explanations for this question is
available as Solution 390 at Physics 9702 Doubts | Help Page 72 - http://physics-ref.blogspot.com/2015/02/physics-9702-doubts-help-page-72.html}
Question
5
{Detailed explanations for this question is available as Solution 660 at Physics 9702 Doubts | Help Page 132 - http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-132.html}
Question
6
(a)
e.m.f
is the total energy available per unit charge. The potential difference across
the battery is not equal to the e.m.f of the battery because some of the
available energy is used / lost / wasted / given out in the internal resistance
of the battery. Hence the potential difference available is less than the
e.m.f.
(b)
(i)
V
=IR
Current
in circuit, I = V/R = 6.9 / 5.0 = 1.38A ≈ 1.4A
(ii)
r
= lost volts / current = (9 - 6.9) / 1.38 = 1.5(2)Ω
(c)
(i)
P
= EI = 9 x 1.38 = 12 (12.4) W
(ii)
Efficiency
of battery = output power / total power = VI /EI
= 6.9 / 9 or (9.52) / (12.4) = 0.767 (76.7%)
Question
7
(a)
(i)
The
6 vertical lines from one plate to the other should be equally spaced with the
arrow downwards
(ii)
Electric
field strength E = V /d = 1200 / 40 x 10-3 = 3.0 x 104Vm-1 (allow 1 sf)
(b)
(i)
Force
acting on A, F = Ee = 3.0 x 104 x 1.6 x 10-19 = 4.8 x 10-15
N
(ii)
Torque
of the couple acting on the rod = F x separation of charges
= 4.8 x 10-15 x 15 x 10-3
= 7.2 x 10-17 Nm
(iii)
The
rod will be vertically aligned, parallel to the electric field, with A will be
at the top, close to the +ve plate and B will be at the bottom, close the –ve plate.
The
forces are equal and opposite in the same line. So, there is no resultant force
and no resultant torque.
