2025 TJC H2 Phy Prelim P1 Solution

2025 H2 Physics P1 Solutions 1 A Units of Q is C = A s Units of V is J C-1 = kg m2 s-2 / A s Hence units of C = A2 s4 m-2 kg-1 2 A Average of h A – 6.63 , B – 6.63, C – 6.65, D – 6.62 A, B and D are accurate Range of variation of h A - 0.04, B – 0.02, C – 0.07, D – 0.02 B and D are precise 3 C Δ𝑣 𝑣 × 100% = 1 2 (0.05 4.30 + 20 1450 + 0.05) × 100% = 3.8% ≈ 4% 4 A Initial velocities for both stones, u = 0 s = ut + ½ gt2 = ½ gt2 For the same time interval, Δt, both stones fall down by the same distance, s. Therefore the distance between them will always remain the same as that at the point of release. 5 D 2 6 2(27)(900)(0.4 10 )(3.0) 0.087 NF N Av −= =  = 6 B The weight of the water converted to ice is the same as that of the ice. 7 B Instantaneous power required to drive the escalator friction friction 3 3 sin 3062 20 9.81 0.75 2.5 1060 7.1 10 W 7.1 kW Fv P Mg v P =+ =  + =     +  = =

8 C Drag at 21.0 m s-1: 110 5.24 kN21 Df == Drag at 15.0 m s-1: 2 ' 15.0 ' 2.67 kN21.0 D D D f ff =  = Force of engine at 15.0 m s-1: 110 7.33 kN15 engineF == Hence resultant force = 7.33 – 2.67 = 4.66 kN 9 B Work done = mass x change in potential = (1) (final potential – initial potential) = − 3𝐺𝑀 2𝑅 − (− 𝐺𝑀 𝑅 ) = − 𝐺𝑀 2𝑅 10 C Top: Ftop +mg = mr2 Ftop = 77 [4(2/3.7)2 – 9.81] = 133 N (downwards) Bottom: : Fbottom = mg + mr2 Fbottom = 77 [4(2/3.7)2 + 9.81] = 1640 N (upwards) 11 D Resultant force towards the sun equals to mac. 12 D GPE of orbiting satellite Ep = MmG r− Gravitational force = centripetal force  2 2 Mm vGmrr =  211 22 k MmE mv G r==  2pkEE=− ' ppE E E=+  '22 kkE E E− =− +  ' 2 kk EEE=− 13 A Total number of moles of gas is constant. Since PV = nRT, we take the initial number of moles in the smaller (A) and larger bulb (B) to be n and 8n respectively. Total number of moles = 9n.

At new equilibrium, the pressure will be the same for both. ' ' (80 273.15) (8) (9 ') (9 ') (10 273.15) fA fB PV n RT n R P V n n RT n n R = = + = − = − + Dividing: '(353.15) 8 (9 ')(283.15) 9 ' 8 353.15 ' 283.15 9 8 353.15 1' 283.15 ' 0.8194 0.8194 0.18 Vn V n n nn n n n nn n n n n = − − = =+ =  = − = 14 A 71.0 )4 5sin( )8 52sin( 0 0 0 0 8 5 −= =  = =   x x T Tx x x Tt 15 C ( ) 2 21 22E m f x = ( )( ) ( ) ( ) ( ) ( ) 2 2 2 2 2110.5 2 3.0 0.40 0.72 222 0.72 m f x m f x E  =  = 16 D 2 1 2 1 ' 150 2.0 1200 ' 0.71 PI rd r IA A r A A mm =   = = 17 C Unpolarised light after passing through a polarizer, its intensity is halved (therefore we eliminate options A/B). Amplitude remains at A. (The component of E field perpendicular to the polariser axis is absorbed (e.g. AY is absorbed), leaving the transmitted light having amplitude AX)

Since the polarization angle is 75° (or 105o), by resolving the electric field, ooA' =A cos75 =A sin15 18 A First minima position: b sin  =  0.010 x 10-3 x 0.05 =  5.0 x 10-7 m =  19 D Electric potential energy U = qV Since charge of electron is –ve, and potential at X is +ve while that at Y is –ve,  UY > UX Magnitude of electric force F = qE. Since the E field strength at Y > E field strength at X  FY > FX 20 D Break in +ve wire of cable  voltmeter reading = 0 when connected to X or Y Break in connection within motor  voltmeter reading = 24 V when connected to X and 0 V when connected to Y Break in –ve wire of cable  voltmeter reading = 24 V when connected to X or Y 21 A In dark condition: 3 36 5 10 6 0.00605 10 5 10VV =  = +  In bright condition: 3 33 5 10 6 5.05 10 1 10VV =  = +  22 C 4 22 RIRIP rms== II rms 2= II o 22= 23 A The point charge’s velocity is parallel to resultant magnetic flux density at the centre of the two wires. Thus magnetic force is zero. 24 A 25 A By Fleming’s left hand rule (to find force on an electron in the rod), the electrons in the rod will accumulate at P, causing Q to be of higher electrical potential.

26 C If the magnetic flux linkage is a sine function, the induced e.m.f. will be a cosine function, and vice-versa. Hence, the phase difference between them is /2 rad. 27 D Option A is incorrect because photoelectric effect can occur even for very low intensity but sufficiently high frequency radiation (above the threshold frequency f0). Option B is incorrect because photoelectrons are emitted only if the frequency of radiation is greater than a minimum frequency known as threshold frequency f0; since 𝜆0 = 𝑐 𝑓0 , the wavelength of the radiation must be smaller than the threshold wavelength, which is instead a maximum value. Option C is incorrect because the maximum kinetic energy and thus speed of the photoelectrons is independent of intensity but depends on the frequency of radiation and work function of the metal, as given by ℎ𝑓 = Φ + 𝐸𝑘,𝑚𝑎𝑥. Option D is correct since 𝐼 = 𝑁𝑃 𝑡 ℎ𝑓 𝐴 , at constant intensity and increased frequency, the rate of photons incident on the metal decreases, the lesser the rate of emission of photoelectrons. 28 D proton with the same v but bigger m than electron will have a much larger momentum p (more than 1000 times) Rings will not be observable. 29 A 00 800PQAA += (1) 120 60 120 30 00 11 8022 PQAA     +=       00 11 804 16 PQAA += 004 1280PQAA + = (2) (2) − (1) : 03 480PA = 0 160PA = 30 B Only  decay results in a change in mass number, and each  decay causes the mass number to decrease by 4. Hence the difference in the mass number of the end product and the parent nuclide must be a multiple of 4.