IJC H1 PHY P1 SOLUTIONS

1   Innova Junior College 2011 Prelims 2 H1 Physics Paper 1 solutions Question Answer Question Answer Question Answer 1 D 11 B 21 D 2 C 12 C 22 B 3 A 13 C 23 C 4 B 14 D 24 B 5 B 15 D 25 D 6 B 16 B 26 A 7 C 17 B 27 B 8 B 18 D 28 C 9 A 19 B 29 B 10 D 20 C 30 C

Qn S o 1    8  He (ro pla 2 G 3. Ap 3 Fo th 4 Ho Ve (ta Ti U s 2. v 5 Dr 3s th Ac 6 Fo m m th T o ½ 7 Ne = olution MM V D     2M M     1 14.89 20  122. 233   ence, density ound off 814 acing as 100 PE = mgh = 5) = 96138 pproximatel or a uniform e accelerat orizontal dis ertical dista aking downw me taken, t sing s = ut + 0 = (0) + ½ = 19 m s1 raw a tange s, then estim is tangent. cceleration or elastic co omentum o asses m) a e system is otal KE befo ½ mv2 = mv2 et force ver 1000 N (up 2 4 2 MM DD L   2 DL DL   0.1 0.2 2.5 5. 35 100  (to y = 800 ± 10 .89 to the sa 0) = (70) x (9.8 J ly 100 kJ mly increasin tion is const stance trave ance travele wards as po t = (12 / v) + ½ a t2, ½ (9.81) (12 ent to the cu mate the gra = gradient ollision, the of the system nd kinetic e s conserved ore collision . rtically = 100 pwards) 2M D L .1 0 o 1 S.F.) 0 ame digit 81) x (4 x ng speed, tant. eled = 12 m d = 2.0 m ositive) / v)2, urve at t = adient of = 1.0 m s-2 total m (2 energy of d. n = ½ mv2 + 000 – 9000 2 Qn S o 11 To = 8 = 4 12 By Ins 13 Mi = 5 = 7 m 14   15 Inf X- Sp sa + 16 Wa P h do Q fro Giv mo 17 Fo A q λ = olution orque of coup 8.0 x (0.60 si 4.2 N m y definition, stantaneous nimum Work 50 x 9.81 x 1 780 J 2x   3. 0 25. 0 fra-red: low f rays: high f, peed of all EM me (speed o ave is movin has to be in t ownwards bu has to be in om P. ven its positi oving upward or X: quarter wave = 4L and hen Q  ple = F x d in60) power = rat s power = F k Done = incr .6 3. 8   rad , large λ small λ M radiation in of light) g from left (Q this quarter s t moving upw this quarter a on, it is displ ds. e is formed in nce fundame te of work d Force x velo rease in GPE n vacuum is Q) to right (P since it is dis wards. as it is 1.25λ laced upwar n the tube su ental frequen P  done. ocity E the P). splaced λ away ds and uch that: cy for X,

3   Net force horizontally = 500 N (right wards) Resultant force = 2 21000 500  = 1100 N fx = 4v L For Y: Half a wave is formed in the tube such that λ = 2L and hence fundamental frequency for Y, f y = 2v L Hence, the ratio will be 1:2 8 Ave F = ∆P/∆t = (P1 – P2) / (t2 – t1) 18 Since the waves came from the same source at O (started in phase) and constructive interference occurred at X, this means that the path difference must be equal to integer times of the wavelength. Wavelength = 28 mm Path OX = 400 mm Path OY must be equal to 400 + n(28) Only possible answer is 456 mm (D) 9 Taking moments about the pivot, W (1.0) = 50 N (0.5) W = 25 N 19 Amplitude of vibration is minimum at nodes (destructive interference always). Amplitude of vibration is maximum at anti-nodes (constructive interference always). 10 The spring constant k of each spring is W/3 = kx k = W / 3x Now that 2W is hung with 2 springs, then each spring will carry a load of W. Let the new extension be y. For each spring, W = (W / 3x) y y = 3x 20 Charge flows = area under the graph = ½ (100 mA + 20 mA) (8) = 480 mC

4   Qn Solution Qn Solution 21 Definition of Ohm’s Law 22 The minimum resistance of variable resistor is zero. Hence the p.d. across it is zero. The maximum resistance of variable resistor is 50 k. Hence p.d. is V = 9.0 V x 50 k / (50 k + 10 k) = 7.5 V 23 At 4.0 V, the resistances of P is 4.0 / 1.0 = 4.0 . At 4.0 V, the resistances of Q is 4.0 / 0.5 = 8.0 . The effective resistance of P and Q is (1/4 + 1/8)1 = 2.67  24 Using right hand grip rule for currents in A and C, both magnetic fields at O are pointing towards B (tangent to circular field pattern). So the resultant field will also point in that same direction. 25 B field due to P is also circular in pattern, which will coincide with current in Q. Since current in Q is parallel (or in the same direction) as the B field that P applies on Q, there is no force induced. 26 Distance is halved means that the field strength experienced by each wire is doubled. Both currents are further doubled means that the force experienced by each wire will be 4 times. In total, the force will be 8 times. 27 Force = PN tt   change i n momentum of each phot on Nh t    2 Hence rate of photon arrival N t   =          34 20 9 6.63 102. 50 10 2 580 10 =   1.1 x 107 s‐1 28 hf = eVs + Φ eVs = hf - Φ Vs = (h/e)f - Φ/e Hence, gradient = h/e h = gradient x e = eV 1 / (f1 – f0) 29 Φ = hf - eVs = (6.63 x 10-34)(3 x 108)/(150 x 10-9) – (1.6 x 10-19) x (1.9) = 1.022 x 10-18 J = 6.4 eV 30 2 2 2 1 2 2 2 h peV mv mm    

5   This means that 2 1 V  and hence 1   