2019 GCE A Level 9814 H3 P1 SS [HCI]

Hwa Chong Institution (College) 1 2019 A level paper (9814) suggested solutions 1 (a) Draw the elliptical shape. Correct labelling of the maximum distance between the centre of ellipse to the point on the orbit. 2 (a) Gauss’s Law for magnetic fields states that the total magnetic flux through a closed surface is zero, 0B dA = (b) A magnetic field line that does not form an open loop will imply the presence of magnetic monopoles at the ends of the field line. As magnetic monopoles do not exist, magnetic field lines will always form a closed loop. (c) If a magnetic monopole exists, the magnetic flux calculated based on a closed surface that enclose the magnetic monopole will be non -zero which is inconsistent with Gauss’s Law for magnetic fields. 3 (a) 0B ds = I Ampere’s Law states that where B is the magnetic field strength, ds is a line element along the integration path, is the permeability of free space and I is the current passing through the area enclosed by the integration path. (b) (i) Apply Ampere’s Law along a circular path of radius d centered at the axis of the wire that encloses the wire, when d is larger than the radius of the wire, ( ) 0 0 0 2 2 B ds Bd B d     = = =  I I I (ii) When d is shorter than the radius of the wire, assuming that the current is uniform across the cross section of the wire, Sun Planet semi-major axis, a

Hwa Chong Institution (College) 2 ( ) 0 2 0 2 0 2 2 2 B ds dBd r Bd r      = =   =  I I I Hence, we have the following graph (iii) A magnetic field exists inside the wire as the current flows through all parts of the cross section of the wire, hence when we apply Ampere’s Law over a circular path inside the wire, the path will enclose a fraction of the current that flows in the wire resulting in a non-zero magnetic field in the wire. 4 (a) The centre of mass must lie somewhere along the line of symmetry (blue dashed line). It should also lie to the left of the green dashed line. COMMENT: The C.M. of the completed square is exactly at r. The C.M. of the missing square is exactly at s, on the right of r. So the C.M. of the L-shape must be left of r. X r s

Hwa Chong Institution (College) 3 (b) (i) cosx R t = siny R t =− R is the distance between A and X. (ii) COMMENT: Sketch should show increasing “pitch”. (iii) sin cos yy R t v R t  =−  =− Maximum upward velocity due to rotation 1 (0.50)(25) 12.5 m s R − = = = ()v u at=+ Velocity of centre of mass (9.81)t= Solve for 12.5 9.81 1.274 s t t = = X t=0 t=T t=2T t=3T t=4T

Hwa Chong Institution (College) 4 (iv) 22 0.251325T  === s 1.274 s 1.274 0.2513 5.07tT= =  = After the 5th rotation, the vertical velocity is always downward. So the trajectory stops intersecting itself. The last intersection occurs during the 4th rotation. From the sketch, it looks like the intersection occurs close to the left extreme horizontal position. Estimated time of intersection 4.5 4.5 0.2513 1.13 sT= =  = 5 Define U as the energy stored in the inductor at any time. The potential difference across the inductor is dVL dt= I 00 2 2 TfE f T dU d VLdt dt dU L d dU L d LE == = = =  I III II II I 6 (a) Volume of oil in test-tube, ( ) 2 4 6 3.0 10 10V r h −−= =  ( ) 10 7 22 3.0 10 9.55 10 m 10 h  − − − = =  (b) At air-to-oil boundary, the reflected blue light had a phase change of  rad. At oil-to-water boundary, the reflected blue light had no phase change. Since the blue light interferes constructively at the surface of oil film, the path difference for the two rays of reflected blue light is an odd number of half wavelengths. in air in oil 7 in air 11Path difference = 2 22 where 1.000, 1.390, 9.55 10 m and 486 nm air oil air oil nh m m n n n h  −    = + = +       = = =  = Substituting the values in, it is found that m = 5 (m must be an integer). Rewriting the path difference to find a more accurate value of h, ( ) ( ) 9 7 15 486 102 9.62 10 m2 1.390h − −  += = 

Hwa Chong Institution (College) 5 7 (a) 0 0 0 0 0 0 ln ln ln Nt N tt dN Ndt dN dtN NN N t t N N e N N eN     −− =− =− − =− → =− = → =  (b) (i) 1 1 1/2 ln2 ln2 0.0105 h65.9t −= = = 1 2 1/2 ln2 ln2 0.115 h6.01t −= = = Using ( ) 121 20 21 ttN N e e   −−=− − 2 59N = and 3 1000 900 59 41N = − − = (ii) ( ) 121 20 21 ttN N e e   −−=− − To find maximum N2: ( ) 1221 0 1 2 21 0ttdN N e edt   −−= − + =− 21 21 ttee −− = ( ) 2 12 1 1 2 t t t e ee      − − −== Solving for t, t = 22.8 h.

Hwa Chong Institution (College) 6 (iii) (c) (i) Beta particles are ionizing radiation which can cause tissue damage. (ii) The half life of 99Tc is 211000 years. Number of 99Tc nuclide at 150 h is 773. Therefore, the activity of 99Tc at 150 h, (iii) The activity is very small. It corresponds to approximately one decay every 400 years, which is much longer than a human life.

Hwa Chong Institution (College) 7 8 (a) (i) CE (ii) (b) (i) 2 2 CE (ii) 2CE (iii) RCE V V dq qER dt C =+ =+ (iv) 00 ln q t t RC t RC dq qREdt C dq E q dt R RC dq dt q CE RC q CE t q CE eCE RC CE q CE CEe − − += =− =−− −− =− → = −− =−  It is given that /tq A Be −=+

Hwa Chong Institution (College) 8 A CE B CE RC = =− = (v) tt RC RCdq CE E eedt RC R −− = =− = −I (vi) 2 22 2 2 tt RC RCEEP I R e R eRR −−= = =  Total energy dissipated in the resistor 222 0 Energy 2 t RCE CEe dtR  − ==  (vii) Yes, answer to (b)(vi) (the energy dissipated in the resistor while fully charging the capacitor) is the same as (b)(ii) (work done by the battery fully charging the capacitor) minus (b)(i) (energy stored in the capacitor) (c) (i) Same Q on each capacitor. 66 12 6 9 2.5 10 1.1 10 6.88 10 C Q Q Q Q CC Q −− − = + = +  = (ii) 6 1 6 1 6.875 10 2.75 V2.5 10 QV C − − = = =  2 9 2.75 6.25 VV = − = (iii) ( )9 58 6.25 58 0.10775862 A 25.5 R R =+ = = = I I I

Hwa Chong Institution (College) 9 9 (a) Roll A will land first. Since the drop height is the same for both rolls, the loss in GPE is the same. For roll B, loss in GPE is converted to both rotational and translational KE. (b) Consider the mass of a thin hollow cylinder  is the density per unit area. ( ) 22 out in M R R L   = − Moment of inertia of a thin hollow cylinder about axis through its centre of mass ( ) ( ) 2 2322 r Mr rdr Lr r Ldr      = == I Moment of inertia of a solid hollow cylinder about an axis through its centre of mass, ( ) ( )( ) ( ) 4 3 4 4 2 2 2 2 22 22 4 2 42 1 2 outout in in RR cm R R out in out in out in out in rL r dr L LLR R R R R R M R R     == = − = − + =+ I (c) (i) For solid cylinder, Rin is approximately zero, Rout is R. ( ) 2 21 022 cm out MRMR= + =I (ii) For thin cylindrical shell, Rout is approximately Rin. ( ) 2 2 21 2 cm M R R MR= + =I

Hwa Chong Institution (College) 10 (d) (i) (ii) ( ) ( ) ( ) 2 2 2 , 22 22 1 2 1 32 2 3 out roll left out in out out out in out out in Torque MgR M R R MR MgR M R R gR RR    = = = + +  =+ = + I (iii) By using PCOE, 0rotational translationalGPE KE KE + + = As the roll unrolls, its height changes. Let the change in height be h and outhR = , where  is the angular displacement. ( ) ( ) ( ) ( ) ( ) ( ) ( ) 22 2 2 2 2 2 2 2 2 22 22 11 22 1 1 1 (pure roll)2 2 2 11 322 11 3 2 (angular kinematics)22 2 3 cm out out in out out out in out out in out out in Mgh Mv Mg R M R R MR Mg R M R R Mg R M R R gR RR         =+ = + + =+  =+  = + I (e) For free falling roll A, it travels in a straight vertical track with uniform acceleration. 210 2H gt=+ For un