CLASS XII PHYSICS. (a) 30 cm, 60 cm (b) 20 cm, 30 cm (c) 15 cm, 20 cm (d) 12 cm, 15 cm. where

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1 PHYSICS combintion o two thin lenses with ocl lengths n respectively orms n imge o istnt object t istnce cm when lenses re in contct. The position o this imge shits by cm towrs the combintion when two lenses re seprte by cm. The corresponing vlues o n re () cm, cm (b) cm, cm (c) 5 cm, cm () cm, 5 cm Solution: (b) Initilly cm (ocl length o combintion) Hence by using inlly by using...(ii) where rom equtions (i) n (ii). cm n = cm rom eqution (i)..(iii) lso, ierence o ocl lengths cn written s ( ) 4 5..(iv) rom (iii) (iv) n. thin ouble convex lens hs rii o curvture ech o mgnitue 4 cm n is me o glss with rerctive inex.5. Its ocl length is nerly () cm (b) cm (c) 5 cm () 5 cm Solution: (b) By using R cm cm.. sphericl surce o rius o curvture R seprtes ir (rerctive inex.) rom glss (rerctive inex.5). The centre o curvture is in the glss. point object P plce in ir is oun to hve rel imge in the glss. The line P cuts the surce t point O n The istnce PO is equl to () 5 R (b) R (c) R ().5 R PO O. Solution: () By using v u Where,.5, u = OP, v = O Hence.5 O OP R.5 R.5.5 OP OP R P O OP = 5 R EDUDIGM South - 47/C Hzr Ro, olkt - 79 EDUDIGM North BE 9, Sector, Slt Lke, olkt 74 Web: Emil: contct@euigm.in North: 897/ South: 898/

2 4. The istnce between n object n the screen is cm. lens prouces n imge on the screen when plce t either o the positions 4 cm prt. The power o the lens is () D (b) 5 D (c) 7 D () 9 D Solution: (b) By using Hence power D x 4 cm 4 D 4 P D cm 5 5. Shown in igure here is convergent lens plce insie cell ille with liqui. The lens hs ocl length + cm when in ir n its mteril hs rerctive inex.5. I the liqui hs rerctive inex., the ocl length o the system is () + 8 cm (c) 4 cm (b) 8 cm () cm Liqui Lens Solution: () Here....(i).5...(ii) (iii) By using cm. concve lens o ocl length cm plce in contct with plne mirror cts s () Convex mirror o ocl length cm (c) Concve mirror o ocl length cm (b) Concve mirror o ocl length 4 cm () Concve mirror o ocl length cm Solution: () By using l m Since l m cm + (ter silvering concve lens behve s convex mirror) e m 7. cnle plce 5 cm rom lens, orms n imge on screen plce 75 cm on the other en o the lens. The ocl length n type o the lens shoul be () cm n convex lens (c) +.5 cm n convex lens (b) 8.75 cm n concve lens ().5 cm n concve lens EDUDIGM South - 47/C Hzr Ro, olkt - 79 EDUDIGM North BE 9, Sector, Slt Lke, olkt 74 Web: Emil: contct@euigm.in North: 897/ South: 898/

3 Solution: () In concve lens, imge is lwys orme on the sme sie o the object. Hence the given lens is convex lens or which u = 5 cm, v = 75 cm. By using v u 75 5 = cm. 8. convex lens orms rel imge o n object or its two ierent positions on screen. I height o the imge in both the cses be 8 cm n cm, then height o the object is () cm (b) 8 cm (c) 4 cm () cm Solution: (c) By using O I I O 8 4 cm 9. convex lens prouces rel imge m times the size o the object. Wht will be the istnce o the object rom the lens () m m m (b) ( m ) (c) m () m Solution: () By using m here u u m u m u m u. m. n ir bubble in glss sphere hving 4 cm imeter ppers cm rom surce nerest to eye when looke long imeter. I. 5, the istnce o bubble rom rercting surce is g (). cm (b). cm (c).8 cm (). cm Solution: () By using v u R where u =?, v = cm,. 5,, R = cm. = =.5 C u v =cm R = cm. lsh light is covere with ilter tht trnsmits re light. The electric iel o the emerging 7 5 bem is represente by sinusoil plne wve E x sin(. z. t) V / m. The verge intensity o the bem will be ().8 W / m (b).7 W / m (c).44 W / m ().88 W / m Solution : (b) 8 5 c E 8.85 I v.7 W / m EDUDIGM South - 47/C Hzr Ro, olkt - 79 EDUDIGM North BE 9, Sector, Slt Lke, olkt 74 Web: Emil: contct@euigm.in North: 897/ South: 898/

4 . Wht shoul be the height o trnsmitting ntenn i the T.V. telecst is to cover rius o 8 m () 5 m (b) 8 m (c) 5 m () 79 m Solution : (b) Height o trnsmitting ntenn (8 ) h 8 m Re.4. T.V. tower hs height o m. How much popultion is covere by T.V. brocst, i the verge popultion ensity roun the tower is / m () 9.5 (b) 9.5 (c) 9.5 () 9 Solution : () Rius o the re covere by T.V. telecst Totl popultion covere hr e popultion ensity hr Popultion ensity n electromgnetic rition hs n energy 4.4 ev. To which region o electromgnetic spectrum oes it belong () Inr re region (b) Visible region (c) X-rys region () -ry region e 5 Solution : (c) 4 8 hc..8 m.8 Å. This wvelength belongs to X-ry E region. 5. The minimum wvelength o X-rys prouce in coolige tube operte t potentil ierence o 4 kv is ().Å (b).å (c) Å () Å Solution : () min Å. Å. The X-ry wvelength o L line o pltinum (Z = 78) is. Å. The X ry wvelength o L line o Molybenum (Z = 4) is () 5.4Å (b) 4.Å (c).7å ().5 Å Solution : () The wve length o L line is given by R( z 7.4) ( z 7.4) ( z ( z 7.4) 7.4). (4 7.4) (78 7.4) 5.4 Å. EDUDIGM South - 47/C Hzr Ro, olkt - 79 EDUDIGM North BE 9, Sector, Slt Lke, olkt 74 Web: Emil: contct@euigm.in North: 897/ South: 898/

5 7. The cut o wvelength o continuous X-ry rom two coolige tubes operting t kv but using ierent trget mterils (molybenum Z= 4 n tungsten Z = 74) re () Å, Å (b). Å,. Å (c).44 Å,.8 Å ().44 Å,.44 Å Solution : () Cut o wvelength o continuous X-rys epens solely on the voltge pplie n oes not epen on the mteril o the trget. Hence the two tubes will hve the sme cut o wvelength. Ve 4 hc hc.7 h or m 44 m.44 Å. Ve Three cpcitors o, n re joine in series n the combintion is chrge by mens o 4 volt bttery. The potentil ierence between the pltes o the cpcitor is () 4 volts (b) volts (c) 8 volts () volts Solution: () Equivlent cpcitnce o the network is C eq Chrge supplie by bttery = Ceq.V 4 = 4 C C eq Hence potentil ierence cross cpcitor 4 4 volt. + 4V 9. Two cpcitors ech o cpcitnce re connecte in prllel n re then chrge by V D.C. supply. The totl energy o their chrges in joules is (). (b). (c).4 (). Solution: (c) By using ormul U C V Here C eq U () eq + V EDUDIGM South - 47/C Hzr Ro, olkt - 79 EDUDIGM North BE 9, Sector, Slt Lke, olkt 74 Web: Emil: contct@euigm.in North: 897/ South: 898/

6 . prllel plte cpcitor o re, plte seprtion n cpcitnce C is ille with three ierent ielectric mterils hving ielectric constnts, n s shown in ig. I single ielectric mteril is to be use to hve the sme cpcitnce C in this cpcitor, then its ielectric constnt is given by / / / = re o pltes () (b) (c) () Solution: (b) The eective cpcitnce is given by. C eq ( The cpcitnce o cpcitor with single ielectric o ielectric constnt is ccoring to question C eq C i.e., ) C. chrge +q is revolving roun sttionry + in circle o rius r. I the orce between chrges is then the work one o this motion will be () r (b) r (c) r () Solution: () Since +q chrge is moving on n equipotentil surce so work one is zero. + + q EDUDIGM South - 47/C Hzr Ro, olkt - 79 EDUDIGM North BE 9, Sector, Slt Lke, olkt 74 Web: Emil: contct@euigm.in North: 897/ South: 898/

7 4 cm CLSS XII. our equl chrge re plce t the our corners o boy o sie ech. Work one in removing chrge rom its centre to ininity is () (b) 4 (c) () Solution: (c) We know tht work one in moving chrge is W = V Here W V V ) V W = V ( B lso V 4. 4 / 4 4 O So, W D C. Two point chrge C n 5 C re plce t point n B respectively with B = 4 cm. The work one by externl orce in isplcing the chrge 5 C rom B to C, where BC = cm, 9 ngle BC n 9 Nm / C 4 8 () 9 J (b) J (c) 9 5 J () 9 4 J Solution: () Potentil t B ue to + C chrge is 9 9 V B volt Potentil t C ue to + C chrge is 9 9 V C volt Hence work one in moving chrge +5C rom B to C W 5 ( V C V B ) W 5 J C 5 cm / B cm C + 5 C 4. There is n electric iel E in x-irection. I the work one in moving chrge. C through istnce o metres long line mking n ngle o with the x-xis is 4J, wht is the vlue o E () 4 N/C (b) 8 N/C (c) N/ C () N/C Solution: () By using W q V n V Er cos So, W qe r cos.c m W 4 j. E cos E = N/C O o x EDUDIGM South - 47/C Hzr Ro, olkt - 79 EDUDIGM North BE 9, Sector, Slt Lke, olkt 74 Web: Emil: contct@euigm.in North: 897/ South: 898/

8 5. n electric chrge o C is situte t the origin o X-Y co-orinte system. The potentil ierence between the points. (5, ) n (, 4) will be Solution: (c) () (b) (c) Zero () k V 5 n V V B k V B 5 B (, 4). Two insulte metllic spheres o n 5 cpcitnces re chrge to V n 5 V respectively. The energy loss, when they re connecte by wire, is 5 5 (5, ) (). J (b).8 J (c).75 J ().75 J CC Solution: (c) By using Δ U ( V V ) ; Δ U. 75 J ( C C ) 7. 4 smll rops o mercury, ech o rius r n chrge q colesce to orm big rop. The rtio o the surce ensity o chrge o ech smll rop with tht o the big rop is () : 4 (b) 4 : (c) 4 : () : 4 Smll q / 4r q R Solution: () ; / 4R r Smll So / n Big Big Smll Big 4 since R = n / r n = nq 8 light bulb, cpcitor n bttery re connecte together s shown here, with switch S initilly open. When the switch S is close, which one o the ollowing is true s + () The bulb will light up or n instnt when the cpcitor strts chrging (b) The bulb will light up when the cpcitor is ully chrge (c) The bulb will not light up t ll () The bulb will light up n go o t regulr intervls Solution: () Current through the circuit cn low only or the smll time o chrging, once cpcitor get s chrge it blocks the current through the circuit n bulb will go o. EDUDIGM South - 47/C Hzr Ro, olkt - 79 EDUDIGM North BE 9, Sector, Slt Lke, olkt 74 Web: Emil: contct@euigm.in North: 897/ South: 898/

9 9. Cpcity o prllel plte conenser is when the istnce between its pltes is 8 cm. I the istnce between the pltes is reuce to 4cm, its cpcity will be () (b) 5 (c) () 4 Solution: (c) C C or C C C 4. Wht is the re o the pltes o prllel plte cpcitor, i the seprtion between the pltes is 5 mm ().94 m (b) 4.59 m (c) 9 () 9.8 m.98 m 9 Solution: () By using the reltion C C m EDUDIGM South - 47/C Hzr Ro, olkt - 79 EDUDIGM North BE 9, Sector, Slt Lke, olkt 74 Web: Emil: contct@euigm.in North: 897/ South: 898/