H2 PHYSICS. SET B PAPER 1 Answer THE PHYSICS CAFE

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1 H2 PHYSIS Exam papers wih worked soluions (Seleced from Top J) SET PPER 1 nswer ompiled by THE PHYSIS FE 1 P a g e

2 1 Penelope measures he mass and speed of a glider. The percenage uncerainy in her measuremen of he mass is 3% and in he measuremen of he speed is 10%. Her calculaed value of he kineic energy of he glider will have an uncerainy of ns: KE KE 10% 13% 23% 30% M M v (0.10) 0.23 v 2 The size of an aom is of he order of magniude of ns: hus percenage uncerainy = 23% 10-6 m 10-8 m m m The nucleus of he aom has a diameer of abou m, whereas he aomic diameer is abou m. 3 ball was projeced from he ground wih a velociy v a an angle θ o he horizonal. I races a parabolic pah as shown below, reaching a maximum heigh H afer a ime T and hen falls back down o he ground. ir resisance may be assumed o be negligible. Which graph bes shows he variaion of kineic energy E k of he ball wih ime? E k 0 T v θ E k H E k The Physics afe 0 T 0 T E k 0 T ns: Noe ha a maximum heigh, he ball sill has a speed of v sin θ. Hence E k E is no equals o zero a = T. Opions and are ou. KE due o horizonal velociy = consan horizonal line (1) 2 P a g e

3 Verically, v = u + a KE due o verical velociy = ½ m (u + a) 2 quadraic curve (2) Summing up (1) and (2), our answer is. 4 saionary 1800 kg ruck is hi from behind by a 900 kg car. The wo become enangled and move off a he same speed ogeher. If he smaller car was moving a a speed of 20 m s -1 iniially, wha is he final speed of he wo vehicles afer he collision? ns: v 6.67ms 4.20 m s m s m s m s -1 (900)(20) ( )( v) 1 5 Two saellies collide in space inelasically. Wha happens o he kineic energy and momenum of he sysem of he wo saellies? onserved onserved Reduced Reduced Kineic energy onserved Reduced onserved Reduced Momenum ns: 6 sone was hrown verically upwards and subsequenly falls ino a pond. Which graph bes represens he variaion of he sone s acceleraion a wih ime? ssume air resisance o be negligible. a a P a g e

4 a a ns: efore impac wih waer, downward acceleraion a = g = consan poin of impac wih waer, upward acceleraion is maximum. s sone slows down, drag force decreases, hence magniude of acceleraion decreases. When drag force + uphrus = weigh, a = 0 (i.e. erminal velociy reached). Here, vecors downward aken as posiive. 7 The graph below shows he force acing on a body of 2 kg over a period of 4 s. ns: 0 F/N 10 0 If he body was iniially moving a 1 m s -1, wha would be is velociy afer 4 s? 10 m s m s m s m s -1 Toal change in momenum = area under graph = ½ (4)(10) = 20 kg m s -1 p f = p i + p = (2)(1) + 20 = 22 kg m s -1 v f = 22 / 2 = 11 m s The Physics afe 8 paricle of mass m slides down an inclined plane wih an average acceleraion a. The inclined plane makes an angle wih he horizonal. The average fricional force exered by he plane on he mass is m(g - a) m(g + a) sin m(g - a) sin m(g sin - a) 0 /s 4 P a g e

5 ns: mg sin f = ma f = m(g sin a) 9 n objec resing on a horizonal fricionless surface is acceleraed from res by a consan force from a moor. Which of he following graphs shows he variaion of he moor power P wih ime? P P 0 0 P P 0 0 ns: P = Fv = F(u + a) = F(0 + a) = Fa Since F and a are consans, P varies linearly wih. 10 n elecric moor is required o haul a cage of mass 400 kg up a mine shaf hrough a verical heigh of 1000 m in 3 minues. Wha will be is elecrical power required if is overall efficiency is 87%? ns: 2.55 kw 19.0 kw 21.8 kw 25.1 kw Power oupu = mgh / = (400 x 9.81 x 1000) / (3 x 60) = W = 21.8 kw Power oupu/power inpu = 0.87 Power inpu = Power oupu / 0.87 = 21.8 / 0.87 = 25.1 kw 11 suden performing an experimen whirls a rubber bung aached o one end of a sring which passes hrough a glass ube wih smooh openings and has a weigh W hanging a is oher end. The weigh of he rubber bung is much smaller han W. The rubber bung is se ino a horizonal uniform circular moion wih angular speed ω while he lengh of he sring beyond he upper opening of he glass ube is L and his porion of he sring makes an angle θ wih he verical as shown. 5 P a g e

6 Which of he following saemen(s) is/are correc? (1) If L is kep consan, θ will decrease as ω is increased. (2) If θ is kep consan, L will increase as ω is increased. (3) When W is increased, θ will increase. ns: (1) only (2) only (3) only (2) and (3) only Verically, T cos θ = mg --- (i) Horizonally, enripeal force = T sinθ = m(l sinθ)ω 2 T = ml ω (ii) From (ii): When ω is increased and L is kep consan T increased From (i): T increased cos θ mus decrease, since mg unchanged θ mus increase. Hence (1) is false. From (i): θ kep consan T kep consan, since mg unchanged. From (ii): If ω is increased and θ is kep consan (and hence T is kep consan), L mus decrease. Hence (2) is false. The Physics afe When W is increased, he ension in he sring increases and hence a smaller of is verical componen is needed o balance he weigh of he rubber bung. Hence θ will increase. i.e. T cos θ = mg When T increases, since mg unchanged, θ mus increase. Hence (3) is rue. 12 marble of mass 20 g is released from res from he rim of a semi-circular bowl of radius R. eermine, in newons, he normal reacion force exered by he bowl on he marble when he marble is a he boom of he bowl. R 6 P a g e

ns: 0.2 0.4 0.6 0.8 y.o.e., mgr = ½ mv 2 mv 2 = 2mgR he boom of he bowl, N mg = mv 2 / R =2mgR / R = 2mg N = 3mg = 0.

7 ns: y.o.e., mgr = ½ mv 2 mv 2 = 2mgR he boom of he bowl, N mg = mv 2 / R =2mgR / R = 2mg N = 3mg = 0.6 N 13 Two sars of equal mass M move wih consan speed v in a circular orbi of radius R abou heir common cenre of mass as shown in he diagram below. Wha is he ne force on each sar? GM 2 2 Mv zero 2Mv 2 4R 2 2R R ns: Jus apply Newon s Law of graviaion o he wo sars. 14 The diagram below shows how he graviaional poenial varies beween he moon and Earh. which posiion will a paricle experience zero ne force? ns: Since g = -dф/dr., gradien = 0 g = 0 F G = mg = 0. 7 P a g e

8 This is he poin where he graviaional field due o Earh cancels ou he graviaional field due o he Moon so ha he resulan graviaional field srengh g is zero. 15 n objec placed on a horizonal plaform is vibraing verically in simple harmonic moion wih a frequency of 2.0 Hz. The maximum ampliude of oscillaion ha will allow he objec o remain in conac wih he plaform hroughou he moion is ns: 0.78 cm 1.6 cm 2.5 cm 6.2 cm 16 paricle of mass 4 kg moves wih simple harmonic moion and is poenial energy U varies wih posiion x as shown in he diagram. ns: N mg a max Wha is he period of oscillaion of he mass? 0.56 s 1.78 s 1.00 s 2.50 s U = ½ kx 2 = ½ mω 2 x 2 (1.0) = ½ (4)(ω 2 )(0.2) 2 ω 2 = 12.5 (2π/T) 2 = 12.5 T = 1.78 s Take vecors upwards as posiive, N mg = m(-a max ) To find max. amplilude, se N=0 a x = x o Hence a max = g For SHM, a max = ω 2 x o Hence x o = g / ω 2 = g / (2πf) 2 = 9.81 / (4π) 2 = 6.2 cm U /J x / m The Physics afe 8 P a g e

9 17 meal block X, of mass m, a 0 o comes ino conac wih anoher meal block Y, of mass 2m, a 100 o. Hea conducion akes place wih no loss o he surroundings. The final equilibrium emperaure of he blocks is 20 o. If he specific hea capaciies of he wo meals are c x and c y respecively, hen ns: c x = 8 c y c x = 4 c y c x = 2 c y c x = ½ c y Hea loss by Y = Hea gain by X (2m)c Y (100-20) = mc X (20 0) 160 c Y = 20 c x 8 c Y = c x 18 n ideal gas has an iniial volume of m 3 and a a pressure of Pa. When 80 J of energy is supplied o he gas, is volume increases o m 3 while is pressure remains consan. s a resul, he inernal energy of he gas is ns: increased by 110 J increases by 80 J increases by 50 J decreases by 50 J W = -p V = - ( )( )(10-3 ) = - 30 J U = Q + W = (+80) + (- 30) = + 50 J 19 Three polaroid shees P, Q and R are placed along a sraigh line wih a lamp and a deecor as shown. Iniially he direcions of polarizaion of P and Q are parallel bu are boh normal o ha of R. Wha happens o he inensiy I recordered by he deecor when P is being roaed slowly hrough 90 unil is direcion of polarizaion is parallel o ha of R? I remains unchanged. I increases hroughou. I increases and hen decreases. I decreases and hen increases. ns: Since he polarizing axis of Q and R remains perpendicular o each oher hroughou, inensiy I remains consan a zero hroughou. 9 P a g e

10 20 Two waves P and Q are moving owards he righ as shown in he figure below. isplacemen Q irecion of wave Which of he following correcly describes he phase difference beween he wo waves? P lags Q by 4 radians. P leads Q by 4 radians. 3 P lags Q by radians. 2 3 P leads Q by radians. 2 ns: 3 Q leads P by rad. Equivalenly, P leads Q by rad beam of whie ligh was projeced ono a diffracion graing wih 500 lines per mm. How many orders of he enire visible specrum (400 nm 700 nm) can i produce? ns: P Posiion The Physics afe dsinθ = mλ (0.001/500)sinθ = m(700 x 10-9 ) Noe: Use λ = 700 nm no 400 nm since he longer wavelengh ligh will be diffraced more. Hence on he screen, we will observe more orders of brigh fringes corresponding o he shorer wavelengh ligh. Hence he number of OMPLETE specrums we will observe is limied by he longer wavelengh ligh. To find he max. order, sinθ = 1 m = 2.9 max order = 2 Including he zeroh order, oal number of orders of enire visible specrum seen = = P a g e

11 22 suden prepares a double-sli se-up as shown below. However, no inerference fringe can be observed on he screen. ns: suiable improvemen may be using a mercury lamp o provide radiaion of a shorer wavelengh. reducing x. reducing d. increasing a. For observable diffracion, λ >> d and λ >> a. 23 The diagram below shows elecric field lines wih poins PQR on one of he field lines. The disance PQ=QR. If he poenial a P is 0V and he poenial a Q is -200V, wha is a possible value of he poenial a R? ns: -450 V -400 V -350 V +200 V Field lines are narrowing, meaning increasing field srengh. Since E = - dv/dx, poenial gradien is seeper beween Q and R han beween P and Q. Hence he poenial drop from Q o R is greaer han ha from P o Q. Only possible answer is -450 V. 24 eermine he elecric field srengh a he mid-poin beween he wo charges below. 11 P a g e

12 ns: 0 Vm x 10 4 Vm x 10 4 Vm x 10 5 Vm -1 E 4 ( Vm Vm ) (0.5) 4 ( ) (0.5) 25 baery, during is usable lifeime, supplies a consan curren of 40 µ o a clock for 800 days. How much charge does he baery supply during his ime? ns: 20 µ µ Q = I = 40 x 10-6 x 800 x 24 x 60 x 60 = ~ generaor produces 100 kw of power a a p.d. of 10 kv. The power ransmied hrough cables of oal resisance 5. Wha is he power loss in he cables? ns: 50 W 250 W 500 W 1000 W urren passes hrough he circui is I=P/V = 100 kw/10 kv = 10 Power loss in he cables is = I 2 R = 10 x 10 x 5 = 500 W The Physics afe 27 The maerial of a wire has resisiviy 1.3 x 10-8 m. The wire has diameer 0.50 mm and is lengh is jus enough o enable i o be wound ighly round an insulaing rod 30 imes. The rod has diameer 1.5 cm. Wha is he resisance of he wire? 1.1 x x x x P a g e

13 ns: The lengh of he wire is (2 /2)(N) = N The cross-secional area is d 2 /4 Resisance = L/ = ( N) / ( d 2 /4) = 4 N/d 2 = 4 [(1.3 x 10-8 m)(1.5/100 m)(30)] / [(0.5/1000)(0.5/1000)] = 9.4 x Three similar ligh bulbs are conneced o a consan-volage d.c. supply as shown in he diagram. Each bulb operaes a normal brighness and he ammeer (of negligible resisance) regisers a seady curren. ns:. The filamen of one of he bulbs breaks. Wha happens o he ammeer reading and o he brighness of he remaining bulbs? mmeer reading ulb brighness Increases Increases Increases Unchanged Unchanged Unchanged ecreases Unchanged When one filamen breaks, oal circui resisance increases from R/3 o R/2. Hence ammeer reading decreases. Since bulbs are conneced in parallel, volage across each remains he same when here is one less bulb. Power dissipaed by each bulb, given by P = V 2 /R, remains unchanged. 13 P a g e

14 29 The figure below shows a recangular meal frame enering a region of uniform magneic field. Given ha he plane of he meal frame is always normal o he magneic field, and ha he meal frame moves a a consan speed, which graph below bes shows he variaion wih ime of he curren induced in he frame. ns: When he coil is fully in he magneic field, here is no change in flux and i here is no induced emf/curren. When he coil is enering / exiing he magneic field, he rae of change of flux linkage is consan, hence he induced emf should be consan, albei in an opposie direcion. 30 horizonal sraigh wire of lengh 2 m is raised verically hrough a heigh of 3.0 m in 0.20 s. The horizonal componen of he Earh s magneic field srengh a his locaion is T, while he verical componen a his locaion is 8.0 x 10-7 T. Wha is he average e.m.f. induced across he ends of he wire? ns: zero mv 0.12 mv 0.60 mv Induced EMF = lv = (2.0 x 10-5 )(2)(3 / 0.20) = 0.60 mv The Physics afe 14 P a g e

15 31 The diagram shows hree long sraigh wires P, Q and R normal o he plane of he paper. Wires P and R carry currens direced ino he plane of he paper, and wire Q carries a curren direced ou of he paper. ll hree currens have he same magniude. ns: Which arrow bes shows he direcion of he resulan force on wire P? irecion of resulan field due o curren R and Q a P is in direcion of. Using Fleming s Lef Hand Rule, he force is in direcion. 32 charge moves in a circular orbi in a uniform magneic field. Which one of he following saemens is correc? The force on he charge is parallel o he field. The period in he orbi is independen of he speed of he charge. The momenum of he charge is independen of he circle radius. The radius of he circle is direcly proporionae o he charge. ns: Manipulaing qv = mv 2 / r gives qr = mv hence is wrong. Manipulaing qr = mv = mrω = mr(2 /T) gives T = 2 m / q hence is correc. 33 seady curren dissipaes a cerain power in a variable resisor. The resisance has o be halved o obain he same power when a sinusoidal alernaing curren is used. Wha is he r.m.s. value of he alernaing curren? ½ ½ 15 P a g e

16 ns: 2 R = 2 (R/2) 2 = 2 2 = 34 The power dissipaed in a ligh bulb conneced across an a.c. source of peak volage 180 V is 50 W. If wo such ligh bulbs are conneced in series o he elecrical mains of 230 V r.m.s., wha would be he oal power dissipaed in boh he lamps? ns: 50.6 W 81.6 W 100 W 61.6 W Since V o = 180 V, V rms = 180/ 2. R = V rms 2 / P = (180/ 2) 2 / 50 = 324. Power dissipaed in one ligh bulb is = V 2 /R = (230 / 2) 2 / 324 = 40.8 W Power dissipaed in wo ligh bulbs is = 40.8 x 2 = 81.6 W. 35 specromeer is se up o observe he specrum produced by he sodium lamp as shown in Figure The energy levels of he lowes saes of sodium are given in Figure Figure 36.1 Figure 36.2 How many specra lines of unique wavelenghs corresponding o hese energy levels can be observed? ns: because 5 2 = 10 The Physics afe 16 P a g e

17 36 The graph below shows a ypical X-ray specrum produced when elecrons are acceleraed hrough a poenial difference and are hen sopped in a meal arge. inensiy 0 0 wavelengh Which feaure of he graph enables his poenial difference o be deermined? The wavelengh of he peaks on he graph. The maximum wavelengh of he X-rays produced. The maximum inensiy of he X-rays produced. The minimum wavelengh of he X-rays produced. ns: X-rays wih he shores wavelengh have he highes energy. y conservaion of energy, his energy comes from he siuaion when he elecron has a single head-on collision wih he arge aoms, losing all is energy. Wih his we can deermine he energy ha i gains hrough he poenial difference and hence he poenial difference. 37 Why is laser ligh coheren? The excied elecrons are in a measable sae. The sysem is in a sae of populaion inversion. Simulaed emission causes he emied phoon and he inciden phoon o be of he same phase. Phoons of he same energy as ha of he inciden phoons are emied when he elecrons ransi down from a higher energy level. ns: The inciden and he emied phoon are in phase and of he same frequency. Hence is he answer. 38 Which saemen abou semiconducors is correc? Their resisiviy is normally somewha greaer han ha of mos meals. hole in a semiconducor is due o he removal of a proon. Elecrical conducion in an n-ype semiconducor is due o he ransfer of neurons. p-ype semiconducor is produced by doping p-ype impuriies called donors. ns:. 17 P a g e

18 is wrong. hole is due o he removal of an elecron from he valence band. is wrong. I is due o he ransfer of boh elecrons (in he conducion band) and holes (in he valence band), bu wih elecrons as he majoriy charge carriers. is wrong. I is called accepors. 39 In he Ruherford scaering experimen, mos of he alpha paricles passed sraigh hrough he gold foil undefleced. Which one of he following is a correc conclusion from his resul? ns: Mos of he mass of he aom is wihin he aom. The diameer of he nucleus is much less han he diameer of he aom. The nucleus has a posiive charge The aom is overall neural Radioacive nucleus X decays by releasing a β-paricle. The daugher nucleus is 48 d. Wha is X? In In g 111 g ns: 48 The Physics afe 18 P a g e

0 time. 2 Which graph represents the motion of a car that is travelling along a straight road with a uniformly increasing speed?

0 time. 2 Which graph represents the motion of a car that is travelling along a straight road with a uniformly increasing speed? 1 1 The graph relaes o he moion of a falling body. y Which is a correc descripion of he graph? y is disance and air resisance is negligible y is disance and air resisance is no negligible y is speed and