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 (destru