• 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

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.