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P25-16 Code 0 MAIN PATTERN ONLINE TEST-4 (MONT-4) XI TARGET : JEE (MAIN+ADVANCED) 2017 COURSE : JA**, 01JA, 01EA, 01JB, 02JB, 01EB, 01JR, 05JR, 01ER, 05ER Date (fnuka d ) : 21-03-2017 Time: 3 Hours(l e; : 3 ?k.Vs a) Max. Marks (eg ke v a d ) : 360 Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose. i;k bu funsZ'kksad ks/;ku l si<+saA v kid ks5 feuV fo'ks"k : i l sbl d ke d sfy , fn;sx;sgSaA INSTRUCTIONS / funs Z'k: General : This booklet is your Question Paper. Do not break the seals of this booklet before being instructed to do so by the invigilators. Blank spaces and blank pages are provided in the question paper for your rough work. No additional sheets will be provided for rough work. A. 3. Blank papers, clipboards, log tables, slide rules, calculators, cameras, cellular phones, pagers and electronic gadgets are NOT allowed inside the examination hall. 3. 4. Write you name and roll number in the space provided on the back cover of this booklet. 4. 5. Using a black ball point pen, darken the bubbles on the upper original sheet. 5. 6. DO NOT TAMPER WITH/MUTILATE THE ORS OR THE BOOKLET. 6. 7. On breaking the seals of the booklet check that it contains all the 90 questions and corresponding answer choices are legible. Read carefully the Instructions printed at the beginning of each section. 7. B. Filling the ORS Use only Black ball point pen only for filling the ORS. B. 8. Write your Roll no. in the boxes given at the top left corner of your ORS with black ball point pen. Also, darken the corresponding bubbles with Black ball point pen only. Also fill your roll no on the back side of your ORS in the space provided (if the ORS is both side printed). Fill your Paper Code as mentioned on the Test Paper and darken the corresponding bubble with Black ball point pen. 8. 9. ORS ij If student does not fill his/her roll no. and paper code correctly and properly, then his/her marks will not be displayed and 5 marks will be deducted (paper wise) from the total. 10. ;fn fo|kFkhZ viuk jksy uEcj rFkk isij d ksM lgh vkSj mfpr rjhd s ugha Hkjrk gS rc mld k ifj.kke jksd fy;k t kosxk rFkk iz'u&i=k esaizkIrkad ls5 vad d kV fy, t kosaxsaA 9. 10. 1. ;g iqfLrd k vkid k iz'u&i=k gSA bld h eqgjsarc rd u rksMsat c rd fujh{kd ksad s}kjk bld k funsZ'k u fn;k t k;sA 2. d Pps d ke d s fy;s [kkyh i`"B vkSj [kkyh t xg bl iqfLrd k esagh gSA d Ppsd ke d sfy, d ksbZvfrfjDr d kxt ughafn;k t k;sxkA d ksjs d kxt ] fDyi cksMZ (CLIP BOARD)] ykWx rkfyd k] LykbM: y] d SYd qy sVj] d Sejk] lsy Q ksu] ist j vkSj fd lh izd kj d sbys DV kfud mid j.k ijh{kk d {k esavuqefr ugha gSA bl iqfLrd k d sfiNysi`"B ij fn, x, LFkku esaviuk uke vkSj jksy uEcj fyf[k,A ijh ewy i`"B d s cqy cqy ksa (BUBBLES) d ks d kys ckWy IokbaV d ye lsd kyk d jsaA vks-vkj-,l- (ORS) ;k bl iqfLrd k esagsj&Q sj@fo fr u d jsaA bl iq fLrd k d h eq gjs a rks M+us d s i'pkr~ i;k t k p ys a fd bles alHkh 90 iz 'u vkSj mud sm kj fod Yi Bhd lsi<+ st k ld rsgS A lHkh [ka a Mks ad siz kjaHk es afn;sgq , funs Z'kks ad ks/;ku ls i<+ A a s v ks-vkj-,l (ORS) Hkjuk % ORS d ksHkjusdsfy, ds oy d kysck y iS u dk mi;ks x dhft ,A ORS d slcls ij cka ;sd ksusesafn, x, ck Dl esaviuk jksy uEcj d kysck y ikbUV lsfyf[k, rFkk laxr xksy sHkh d soy d kysisu lsHkfj;sA ORS d sihNsd h rjQ Hkh viuk jksy uEcj fyf[k, (;fn ORS nks uksarjQ Nih gqbZgSA) viuk isij d ksM fyf[k, rFkk laxr xksy ksad ksd kys ck y isu lsd kysd hft ,A (Please read the last page of this booklet for rest of the instructions) i;k 'ks"k funsZ'kksad sfy ;sbl iqfLrd k d sv fUre i`"B d ksi<+sA Resonance Eduventures Ltd. CORPORATE OFFICE : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005 Ph.No. : +91-744-3012222, 6635555 | Toll Free : 258 5555 Reg. Office : J-2, Jawahar Nagar, Main Road, Kota (Raj.)-324005 | Ph. No.: +91-744-3192222 | FAX No. : +91-022-39167222 P25-16 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in | CIN: U80302RJ2007PLC024029 t c rd ifjos"kd funsZ'k ughansarc rd iz'u i=k d h lhy d ksugha[kksay sA 2. l kekU; % DO NOT BREAK THE SEAL WITHOUT BEING INSTRUCTED TO DO SO BY THE INVIGILATOR A. 1. PHYSICS PHYSICS PART-I : PHYSICS SECTION - I Straight Objective Type This section contains 15 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE is correct. [k.M- I lh/ks oLrqfu"B izdkj bl [k.M esa 15 cgq&fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) gSa] ftuesa ls flQZ ,d lgh gSA 1. Two particles moving parallel to the x-axis perform SHM with same amplitude (A) and frequency (f). At a certain instant they are found at distance A 2 from the mean position on opposite sides such that their velocties are in the same direction. Then the phase difference between their SHMs is : x-v{k ds lekUrj xfr dj jgs nks d.k] leku vk;ke (A) rFkk vko`f k (f) ls ljy vkorZ xfr dj jgs gSA fdlh A {k.k ij os] ek/; fLFkfr ls nwjh ij] foijhr vksj bl izdkj ik;s tkrs gS fd buds osx leku fn'kk esa gSA 2 rc buds ljy vkorZ xfr ds e/; dykUrj gksxk: (A) 90 (B) 120 (C) 180 Space for Rough Work (dPps (D) 270 dk;Z ds fy, LFkku) Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PLC024029 PJP/JFMONT4120117C0-1 2. In the figure is shown a spring mass system oscillating in uniform gravity. If we neglect all dissipative force, it will keep on oscillating endlessly with constant amplitude and frequency. Accompanying graph shows how displacement x of the block from the equilibrium position varies with time t. Now at a certain instant t = te when the block reaches its lowest position, gravity is switches off by some unknown mechanism. Which of the following graphs would correctly describes the changes taking place due to switching off the gravity ? fp=kkuqlkj ,d fLiazx nzO;eku fudk; ,d leku xq:Roh; {ks=k esa nksyu xfr dj jgh gSA ;fn lHkh izfrjks/kh cyksa dks ux.; ekuk tk;s rks ;g fu;r vk;ke o vko`f k ls ,d leku :i ls nksyu xfr esa gSA blds lkFk fn;k x;k xzkQ ek/; fLFkfr ls CykWd dh fLFkfr x esa le; t ds vuqlkj ifjorZu dks n'kkZrk gSA ;fn fdlh {k.k t = te tc CykWd viuh fuEure fLFkfr (lowest position) ij gS] vpkud xq:Roh; {ks=k dks 'kwU; dj fn;k tkrk gS] rks fuEu esa ls dkSulk xzkQ d.k dh fLFkfr esa le; t ds vuqlkj ifjorZu dks n'kkZrk gS ? (A) (B) (C) (D) Space for Rough Work (dPps dk;Z ds fy, LFkku) Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PLC024029 PJP/JFMONT4120117C0-2 PHYSICS PHYSICS 3. In the figure shown, the time period and the amplitude respectively when m is released from rest when the spring is relaxed is: (the inclined plane is smooth) (A) 2 , m m g sin k k (B) 2 , m sin 2 m g sin k k (C) 2 , m m g cos k k (D) none of these fn[kk;s x;s fp=k esa] fLizax ruko jfgr ,oa fojkekoLFkk esa gS rks m dks eqDr NksM+us ij vkorZdky rFkk vk;ke e'k% gksaxs (ur ry fpduk gSA) 4. (A) 2 , m m g sin k k (B) 2 , m sin 2 m g sin k k (C) 2 , m m g cos k k (D) buesa ls dksbZ ugh A glass tube bent into a circle of radius R is rigidly fixed in the vertical plane, the top end open to the atmosphere. A liquid is filled in the tube so that liquid column forms a semicircle. If the liquid is disturbed slightly, the time period of oscillation will be : ,d dk p dh V~;wc R f=kT;k ds o` k esa eqM+h gqbZ gS tks n`<+rkiwoZd /okZ/kj ry esa fLFkj gS ftldk ijh fljk ok;qe.My esa [kqyk gqvk gSA V~;wc esa ,d nzo bl dkj Hkjk tkrk gS fd nzo LrEHk v)Zo` k cukrk gSA ;fn nzo gYdk lk foLFkkfir djsa rks nksyu dk vkorZdky gksxk & Space for Rough Work (dPps (A) 2 R g (B) 2 R g (C) 2 Space for Rough Work / (dPps dk;Z ds fy, LFkku) R 2g (D) 2 R 2g dk;Z ds fy, LFkku) Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PLC024029 PJP/JFMONT4120117C0-3 PHYSICS PHYSICS 5. At t = 0, a transverse wave pulse travelling in the positive x direction with a speed of 2 m/s in a long wire is described by the function y = 6 , given that x 0. Transverse velocity of a particle at x2 x = 2m and t = 2 seconds is : (A) 3 m/s t = 0 (B) 3 m/s (C) 8 m/s (D) 8 m/s ij] rkj esa /kukRed x fn'kk esa 2 eh-@lS- ls xfr dj jgh vuqizLFk rjaxksa dks y = 6 x2 ls izznf'kZr djrs gSaA ;g fn;k x;k gS fd x 0 gSA d.k dk vuqizLFk osx x = 2 eh- rFkk t = 2 lSd.M ij gksxk & (A) 3 eh-@lS- 6. (B) 3 eh-@lS- (C) 8 eh-@lS- (D) 8 eh-@lS- Sinusoidal waves 5.00 cm in amplitude are to be transmitted along a string having a linear mass density equal to 4.00 10 2 kg/m. If the source can deliver a average power of 90 W and the string is under a tension of 100 N, then the highest frequency at which the source can operate is (take 2 = 10) : (A) 45.3 Hz (B) 50 Hz (C) 30 Hz (D) 62.3 Hz T;ko h; rjaxs ftudk vk;ke 5.00 lseh- gS dks ml jLlh ds vuqfn'k izlkfjr fd;k tkrk gS ftldk js[kh; nzO;eku ?kuRo 4.00 10 2 fd-xzk-@eh- gSA vxj L=kksr ls vf/kdre 90 okV tkZ iznku dh tk ldrh gS rFkk jLlh esa ruko 100 U;wVu gS rks og mPpre vko`f k ftl ij L=kksr dk;Z dj lds] gksxh &( 2 = 10 ekusa) : (A) 45.3 gV~Zt (B) 50 gV~Zt (C) 30 gV~Zt Space for Rough Work (dPps (D) 62.3 gV~Zt dk;Z ds fy, LFkku) Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PLC024029 PJP/JFMONT4120117C0-4 PHYSICS PHYSICS 7. The figure shows at time t = 0 second, a rectangular and triangular pulse on a uniform wire are approaching each other. The pulse speed is 0.5 cm/s. The resultant pulse at t = 2 second is le; t = 0 lsd.M ij fp=k ij ,d vk;rkdkj rFkk f=kHkqtkdkj rjaxs (pulse) ,d le:i rkj ij ,d nwljs dh vksj vk jgh gSA rjax dh pky 0.5 cm/s gSA t = 2 lsd.M ij ifj.kkeh rjax gksxhA (A) (B) (C) (D) Space for Rough Work (dPps dk;Z ds fy, LFkku) Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PLC024029 PJP/JFMONT4120117C0-5 PHYSICS PHYSICS 8. 1 kg/m. A 20 tuning fork of 50 Hz is found to be in resonance with the horizontal part of wire between pulley and block A. (Assuming nodes at block A and pulley). Now at t = 0, system is released from rest. The ratio of time gap between successive resonance with the same tuning fork starting from t = 0. (take g = 10 m/s2) 1 n'kkZ;s x;s fudk; esa] jsf[k, nzO;eku ?kuRo kg/m dk ,d rkj nks nzO;ekuksa ds e/; ca/kk gSA f?kjuh o CykWd 20 A ds e/; rkj dk {kSfrt Hkkx 50 Hz ds Lofj=k f}Hkqt ds lkFk vuqukn esa ik;k tkrk gSA ekfu, dh f?kjuh o In the system shown, the wire connecting two masses has linear mass density of CykWd A ij fuLiUn curs gS A vc t = 0 ij] fudk; dks fojkekoLFkk ls NksMk tkrk gSA t = 0 ls izkjEHk gksus ds i'pkr ~leku Lofj=k f}Hkqt ds lkFk ekxr vuqukn ds e/; le; vUrjky dk vuqikr gksxk (fyft, g = 10 m/s2) 50Hz 4kg A 60cm B 4kg (A) 2 : 1 9. (B) 1 : 2 (C) 1 : 2 1 (D) 1: 2 If 1 and 2 are the lengths of air column for two air column for two consecitive resonance position when a tuning fork of frequency f is sounded in a resonance tube, then end correction is : ;fn ok;q LrEHk dh yEckbZ 1 rFkk 2 ds fy, nks ekxr vuqukn fLFkfr;kW gS tc vuqukn uyh f vko`f k ds Lofj=k f}Hkqt ds lkFk vuqukfnr gksrh gS] rc fljk la'kks/ku gksxk : (A) ( 2 3 1 ) 2 (B) ( 2 3 1 ) 2 (C) Space for Rough Work (dPps ( 2 1 ) 2 (D) (3 2 1 ) 4 dk;Z ds fy, LFkku) Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PLC024029 PJP/JFMONT4120117C0-6 PHYSICS PHYSICS 10. A student is performing the experiment of Resonance Column. The diameter of the column tube is 4cm. The frequency of the tuning fork is 512 Hz. The air temperature is 38 C in which the speed of sound is 336 m/s. The zero of the meter scale coincides with the top end of the Resonance Column tube. When the first resonance occurs, the reading of the water level in the column is ,d Nk=k vuqukn uyh LrEHk dk iz;ksx dj jgk gSA LrEHk&uyh dk O;kl 4 cm vkSj Lofj=k&f}Hkqt dh vko`f k 512 Hz gSA ok;q d rkieku 38 C gS vkSj mlesa /ofu dh pky 336 m/s gSA ehVj&Ldsy dk 'kwU; LrEHk&uyh ds ijh fljs ds Bhd lkeus gSA izFke vuqukn dh voLFkk esa LrEHk&uyh esa ikuh ds Lrj dk ikB~;kad gSA (A) 14.0 cm 11. (B) 15.2 cm (C) 16.4 cm (D) 17.6 cm Two point sources of sound emitting sound of frequency 340 Hz each move relative to a stationary observer. One source moves away from the observer while the other moves towards him with the same speed. The observer hears beats of frequency 3Hz. The speed of the sound source is approximately equal to : (speed of sound = 340 m/sec) (A) 3m/sec (B) 1.5 m/sec (C) 2 m/sec (D) cannot be calculated nks fcUnq /ofu L=kksr] R;sd dh vko`f k 340 Hz gSA R;sd ,d fLFkj s{kd ds lkis{k xfreku gS rFkk R;sd dh pky leku gSA buesa ls ,d s{kd ls nwj dh vksj tcfd vU; s{kd dh vksj xfreku gSA ;fn s{kd 3Hz vko`f k ds foLiUn lqurk gS rks /ofu L=kksr dh pky yxHkx gksxhA : (/ofu dh pky = 340 m/sec) (A) 3m/sec (B) 1.5 m/sec (C) 2 m/sec (D) x.kuk ugha dj ldrs 12. A uniform rod is kept at smooth horizontal surface, a constant force is applied on the rod in horizontal direction at end A . Find the ratio of energy stored per unit volume at end A to the energy stored per unit volume in the middle of rod. ,d le:i NM+ fpduh {kSfrt lrg ij j[kh gqbZ gS] NM+ ds A fljs ij {kSfrt fn'kk esa ,d fu;r cy vjksfir fd;k tkrk gSA fljs A ij lafpr izfr ,dkad vk;ru dh tkZ rFkk NM+ ds e/; fcUnq esa lafpr izfr ,dkad vk;ru dh tkZ dk vuqikr gksxkA B (A) 2 (B) 4 A (C) 8 Space for Rough Work (dPps F (D) 10 dk;Z ds fy, LFkku) Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PLC024029 PJP/JFMONT4120117C0-7 PHYSICS PHYSICS 13. A uniform rod of length and cross sectional area A rotates about an axis passing through one end of the rod. The extension produced in the rod due to centrifugal force is ( is weight of rod per unit length & is angular speed of rotation of rod, E is young's modulus of elasticity of rod) yEckbZ o A vuq LFk dkV {ks=kQy dh ,d le:i NM+ mlds ,d fljs ls xqtjus okyh v{k ds lkis{k ?kwe jgh gSA vidsUnzh; cy ds dkj.k NM+ esa mRiUu lkj gksxkA (NM+ dh fr bdkbZ yEckbZ dk Hkkj gS rFkk NM+ ds ?kw.kZu dh dks.kh; pky gS] NM+ dk ;ax R;kLFkku xq.kkad E gS) (A) 14. 2 gE (B) 2 3 3gE (C) 2 3 gE (D) 3gE 2 3 A sphere of density falls vertically downward through a fluid of density . At a certain instant its velocity is u. The terminal velocity of the sphere is u0. Assuming that stokes s law for viscous drag is applicable, the instantaneous acceleration of the sphere is ?kuRo dk ,d xksyk ?kuRo ds nzo esa /okZ/kj uhps dh vksj fxjrk gSA fdlh {k.k ij bldk osx u gSA xksys dk lhekUr osx u0 gSA ekfu;s fd ';ku d"kZ.k ds fy, LVkWd fu;e ykxw gS] bl {k.k xksys dk rkR{kf.kd Roj.k gksxkA u (A) g (B) 1 g u (C) 1 1 g u0 Space for Rough Work (dPps u (D) 1 g u0 dk;Z ds fy, LFkku) Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PLC024029 PJP/JFMONT4120117C0-8 PHYSICS PHYSICS 15. A thin horizontal movable plate is separated from two fixed horizontal plates P1 and P2 by two highly viscous liquids of coefficient of viscosity 1 and 2 as shown, where 2 = 4 1. Area of contact of movable plate with each fluid is same. If the distance between two fixed plates is h, then the distance h1 of movable plate from upper fixed plate such that the movable plate can be moved with a constant velocity by applying a minimum constant horizontal force F on movable plate is (assume velocity gradient to be uniform in each liquid). ,d iryh {kSfrt xfr djus ;ksX; IysV nks fLFkj {kSfrt IysVksa P1 o P2 ds e/; Hkjs vR;Ur ';ku nzoksa dks fp=kkuqlkj i`Fkd djrh gS ftuds ';urk xq.kkad 1 o 2 fp=kkuqlkj gS tgk 2 = 4 1 gSA xfr djus ;ksX; IysV dk izR;sd nzo ds lkFk lEidZ {ks=kQy leku gSA ;fn nksuksa fLFkj IysVksa ds e/; nwjh h gS] ij okyh fLFkj IysV dh] xfr dj ldus okyh IysV ls nwjh h1 bl izdkj Kkr dhft, fd xfr djus ;ksX; IysV ij U;wure {kSfrt fu;r cy F vjksfir djus ij IysV fu;r osx ls xfr dj lds (;g ekfu, fd izR;sd nzo esa osx izo.krk leku gS) (A) h 4 (B) h 2 (C) Space for Rough Work (dPps 2h 3 (D) h 3 dk;Z ds fy, LFkku) Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PLC024029 PJP/JFMONT4120117C0-9 PHYSICS PHYSICS SECTION - II Integer value correct Type This section contains 15 questions. The answer to each question is a single digit integer, ranging from 0 to 9 (both inclusive). [k.M - II iw.kk d eku lgh dkj bl [k.M esa 15 'u gSaA R;sd 'u dk m kj 0 ls 9 rd nksuksa 'kkfey ds chp dk ,dy vadh; iw.kk d gSA 16. Figure shows the kinetic energy K of a simple pendulum versus its angle from the vertical. The pendulum bob has mass 0.2 kg. The length of the pendulum is equal to m then find . (g = 10 m/s2). 2 fp=k esa ljy yksyd dh xfrt tkZ K rFkk m/okZ/kj ls dks.k ds chp xzkQ n'kkZ;k x;k gSA yksyd dk nzO;eku 0.2 fdxzk- gSA ;fn ljy yksyd dh yEckbZ 2 eh- ds cjkcj gS rks dk eku Kkr djksA (g = 10 eh-@ls-2). K(mJ) 15 10 5 -100 0 Space for Rough Work (dPps 100 (mrad) dk;Z ds fy, LFkku) Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PLC024029 PJP/JFMONT4120117C0-10 PHYSICS PHYSICS 17. The uniform solid cylinder can roll without sliding on a horizontal surface as shown in the figure. The mass of cylinder is 10 kg and spring constant for both the ideal springs is 300 N/m. The angular frequency of centre of mass of small oscillations is 2 rad/s Find the value of . ,dleku Bksl csyu fp=kkuqlkj {kSfrt lrg ij fcuk fQlys yksVuh xfr dj ldrk gSA csyu dk nzO;eku 10 kg gS rFkk nksauks vkn'kZ fL axksa ds fy, fL ax fu;rkad 300 N/m gSA nzO;eku dsUnz ds vYi nksyu dh dks.kh; vko`f k 2 rad/s gSA dk eku Kkr djksA 18. A thin uniform rod is suspended in vertical plane as a physical pendulum about point A. The time period of oscillation is To. Not counting the point A, the number 'n' of other points of suspension on rod such that the time period of oscillation (in vertical plane) is again To. Then the value of n is : (Since the rod is thin, consider one point for each transverse cross section of rod) ,d iryh ,dleku NM+ dks HkkSfrd isUMqye (physical pendulum) ds :i esa /okZ/kj ry esa fcUnq A ds ifjr% yVdk;k tkrk gSA nksyu dk vkorZdky T0 gSA NM+ ds bl yVdu fcUnq A ds vfrfjDr NM+ ij n ,sls yVdu fcUnq (suspension point) dhydhr fcUnq gS ftuds fy, vkorZdky T0 gh gksxk rks 'n' dk eku gSA (D;ksafd NM+ iryh gS] vr% izR;sd vuqizLFk dkV {ks=k dks ,d fcUnq dh rjg yhft;s) A B Space for Rough Work (dPps dk;Z ds fy, LFkku) Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PLC024029 PJP/JFMONT4120117C0-11 PHYSICS PHYSICS 19. Find the amplitude (in cm) of a particle due to superposition of two SHM along the same line. x1 = 5 sin (t 37 ) ; x2 = 6 cos t. (x1 and x2 are in cm and t in second) ,d dh ljy js[kk ds vuqfn'k ,d d.k ij fuEufyf[kr nks ljy vkorZ xfr ds v/;kjksi.k ls kIr vk;ke (cm esa) dk eku D;k gksxk ; x2 = 6 cos t. (x1 rFkk x2 lseh0 esa ,oa t lSd.M esa gS) x1 = 5 sin (t 37 ) 20. A ring of radius r made of wire of density is rotated about a stationary vertical axis passing though its centre and perpendicular to the plane of the ring as shown in figure. Determine the angular velocity (in rad/s) of ring at which the ring breaks. The wire breaks at tensile stress . Ignore gravity. (Take = 4 and r = 1m) /kuRo ds inkFkZ ds rkj ls cuh r f=kT;k dh ,d oy; blds dsUnz ls xqtjus okyh rFkk oy; ds ry ds yEcor~ fLFkj /okZ/kj v{k ds lkis{k fp=kkuqlkj /kw.kZu dj jghs gSA oy; dk dks.kh; osx (jsfM;u / lSds.M es) Kkr dhft, tc oy; VwV tkrh gSA rkj ruu izfrcy ij VwVrk gSA xq:Ro dks ux.; ekusa ( = 4 rFkk r = 1m ysosaA) r 21. At an instant a point A, lying on a transverse wave in a horizontal string, is moving towards mean position with a velocity 1 m/s and the slope of the string at A is tan(sin-1(1/50)). Find the power transfer in Watt, through the point A at this instant if the tension in the string is 50N. ,d {kSfrt Mksjh esa cuh vuqizLFk izxkeh rjax ds fy;s] fdlh fcUnq A dk fdlh {k.k osx 1 m/s ek/; fLFkfr dh rjQ gSA Mksjh ds fcUnq A ij fn;s x;s {k.k ij izo.krk tan(sin-1(1/50) gSA bl {k.k ij fcUnq A ls fdruh 'kfDr okWV esa LFkkukUrfjr gks jgh gksxh ;fn Mksjh esa ruko 50 N gksxkA Space for Rough Work (dPps dk;Z ds fy, LFkku) Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PLC024029 PJP/JFMONT4120117C0-12 PHYSICS PHYSICS 22. A long wire PQR is made by joining two wires PQ and QR of equal radii. PQ has a length 4.8 m and mass 0.06 kg. QR has length 2.56 m and mass 0.2 kg. The wire PQR is under a tension of 80 N. A sinusoidal wave pulse of amplitude 3.5 cm is sent along the wire PQ from the end P. No power is dissipated during the propagation of wave pulse. If amplitude (in mm) of reflected pulse from jucntion Q is 5A then find A. ,d yEck rkj PQR, leku f=kT;k ds nks rkjksa PQ rFkk QR dks tksM+dj cuk;k x;k gSA PQ dh yEckbZ 4.8 m rFkk nzO;eku 0.06 kg gSA QR dh yEckbZ 2.56 m rFkk nzO;eku 0.2 kg gSA rkj PQR esa ruko 80 N gSA 3.5 cm vk;ke dk ,d T;ko h; LiUn fljs P ls rkj PQ ds vuqfn'k Hkstk tkrk gSA rjax LiUn ds lapj.k ds nkSjku dksbZ 'kfDr O;f;r ugha gksrh gSA lfU/k Q ls ijkofrZr LiUn dk vk;ke (mm esa) 5A gS rks A dk eku Kkr djksA 23. The speed of sound in a mixture of n1 = 2 moles of He, n2 = 2 moles of H2 at temperature T = 972 5 K is 100 m/s. Find . (Take R = T= 972 K 5 Kkr 24. 25 J/mole-K) 3 rki ij He ds n1 = 2 eksy rFkk H2 ds n2 = 2 eksy ds feJ.k esa /ofu dh pky 100 m/s gSA djksA (R = 25 J/mole-K) 3 A siren creates a sound level of 60 dB at a location of 500 m from the speaker. The siren is powered by a battery that can deliver a total energy of 1kJ. Assuming that the efficiency of the siren is 2%. Determine to the nearest integer the total time (in seconds) the siren can sound. ,d lk;ju 60 dB Lrj dh /ofu Lihdj ls 500 m dh fLFkfr ij mRiUu djrk gSA lk;ju dks cSVjh ls 'kfDr nh tkrh gS tks 1kJ dh dqy tkZ ns ldrk gSA ekuk fd lk;ju dh n{krk 2% gSA lk;ju }kjk mRiUu dh tk ldus okyh /ofu dk dqy le; (lSd.M esa) fudVre iw.kkZad esa Kkr djsaA Space for Rough Work (dPps dk;Z ds fy, LFkku) Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PLC024029 PJP/JFMONT4120117C0-13 PHYSICS PHYSICS PHYSICS A sound wave propagating along x-axis, in medium of density 1 = 1.5 kg/m is transmitted to a medium of density 2 = 3kg/m3 as shown. 1 = 1.5 kg/m3 ds ?kuRo ds ek/;e esa x v{k ds vuqfn'k lapfjr /ofu dh rjax] 2 = 3kg/m3 ?kuRo ds ek/;e esa fp=kkuqlkj ikjxfer gksrh gSA ek/; e I ek/; e II > v ki fr r /ofu r j ax The equation of excess pressure developed by wave in medium and that in medium respectively are ek/;e rFkk esa rjax }kjk mRiUu nkc vf/kD; dh lehdj.k e'k% gS x t 400 (in S units) (S bdkbZ p1 = 4 10 2 cos esa) x p2 = 3 10 2 cos t (S bdkbZ eas) (in S units) 1200 If the intensity of transmitted wave is 2 (wave in medium ) and intensity of incident wave is 32 2 1 (wave in medium ), then find the value of . 1 ;fn ikjxfer rjax dh rhozrk 2 (ek/;e esa rjax) gS rFkk vkifrr rjax dh rhozrk 1 gS (ek/;e esa rajx) rks, 32 2 1 dk eku Kkr djksA Space for Rough Work (dPps dk;Z ds fy, LFkku) Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PLC024029 PJP/JFMONT4120117C0-14 PHYSICS 25. 3 26. Two longitudinal sinusoidal pressure waves one having lower frequency of 2Hz & both travelling in same direction through the same medium as shown in figure are superimposed. The difference in frequency of the two waves is f Hz. then find f. nks vuqnS/;Z T;ko h; nkc rjaxksa esa ls ,d U;wure vko`f k dh rjax dh vko`f k 2Hz gS rFkk nksauks leku fn'kk o leku ek/;e esa xfr djrh gqbZ fp=kkuqlkj vk/;kjksfir gksrh gS rks bu nksuksa rjxks dh vko`fr esa vUrj f Hz gS rks f D;k gksxk & 27. A point source of sound emiting sound of frequency 700 Hz and observer starts moving from a point along mutually perpendicular directions with velocity 20 m/s and 15 m/s respectively. If change in observed frequency by observer is 10x Hz then calculate 'x'. [speed of sound in 334 m/sec] /ofu dk ,d fcUnq L=kksr tks 700 Hz vko`f k dh /ofu mRiUu djrk gS rFkk ,d s{kd ,d fcUnq ls fp=kkuqlkj ijLij yEcor~ fn'kkvksa esa e'k% 20 m/s rFkk 15 m/s ls xfr djuk kjEHk djrs gSA ;fn s{kd }kjk sf{kr vko`f k esa ifjorZu 10x Hz gS rc 'x' Kkr djksA [/ofu dh pky 334 m/sec gS] Space for Rough Work (dPps dk;Z ds fy, LFkku) Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PLC024029 PJP/JFMONT4120117C0-15 PHYSICS PHYSICS 28. A capillary tube of radius 0.20 mm is dipped vertically in water. If the height of the water column raised in the tube is 2.5h (in/cm) then find h. (surface tension of water = 0.075 N/m and density of water = 1000 kg/m3. Take g = 10 m/s2 and contact angle 0 ). ,d ds'kuyh ftldh f=kT;k 0.20 mm gSA ty esa /okZ/kj Mwch gSA ;fn uyh esa mBs gq, ty LrEHk dh pkbZ 2.5h lseh esa gS rks h dk eku Kkr djksA (ty dk i`"B ruko = 0.75 N/m vkSj ty dk ?kuRo = 1000 kg/m3 g = 10 m/s2 o lEidZ dks.k 0 ysa) 29. Two opposite forces F1 = 120N and F2 = 80N act on an heavy elastic plank of modulus of elasticity y = 2 1011 N/m2 and length L = 1m placed over a smooth horizontal surface. The cross-sectional area of plank is A = 0.5m2. If the change in the length of plank is x 10 9m, then find x ? nks foifjr cy F1 = 120N rFkk F2 = 80N ,d Hkkjh izR;kLFk IykWad ij dk;Zjr gSA Iykad dk izR;kLFkrk xq.kkad y = 2 1011 N/m2 rFkk yEckbZ L = 1m gS rFkk ;g ?k"kZ.k jfgr {kSfrt lrg ij j[kk gSA IykWad dk vuqizLFk dkV {ks=kQy A = 0.5m gSA ;fn IykWad dh yEckbZ esa ifjorZu x 10 9m gS rks x Kkr djks& 2 F2 30. F1 In the given arrangement shown in figure, a mass M can be hung from a string, that passes over a light pulley. The string connected to a vibrator having constant frequency. When the value of M is either 16 Kg or 25 Kg standing waves are observed, however, no standing waves are observed with any mass between these values. The largest mass for which standing waves could be observed is n 102 Kg. Find the value of n . (string is very light) fp=k esa iznf'kZr O;oLFkk esa M nzO;eku jLlh ls yVdk gqvk gSA rFkk jLlh gYdh f?kjuh ls xqtj jgh gSA jLlh fu;r vko`f k ds nkSfy=k ls tqM+h gqbZ gSA tc M dk eku 16 Kg ;k 25 Kg gS] rc vizxkeh rjax izsf{kr gkrh gSA tcfd nzO;eku ds fn;s x;s ekuksa ds e/; dksbZ vizxkeh rjax izsf{kr ugha gksrh gSA vizxkeh rjax izsf{kr djus ds fy, nzO;eku dk vf/kdre eku n 102 Kg gks rks n dk eku Kkr djksA (jLlh gYdh gS ) Virbrator nksfy=k L L Space for Rough Work (dPps dk;Z ds fy, LFkku) M M Space for Rough Work / (dPps dk;Z ds fy, LFkku) Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)- 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PLC024029 PJP/JFMONT4120117C0-16 PHYSICS PHYSICS CHEMISTRY SECTION I ([k.M- I) Straight Objective Type (lh/ks oLrqfu"B izdkj) This section contains 15 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE is correct. (bl [k.M esa 15 cgq&fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) gSa] ftuesa ls flQZ ,d lgh gSA) 31. Which of the following is not correctly matched [Temperature 25 C] 8 (A) 1.0 10 M Solution of HCl pH < 7 8 (B) 1.0 10 M Solution of NaOH pH > 7 (C) Acidic buffer solutions pH < 7 (D) A solution of salt of weak acid and weak base may have pH < 7 fuEu es ls dkSulk lgh lqesfyr ughs gS\ [rki 25 C ] (A) HCl dk1.0 10 8 M foy;u (B) NaOH dk 1.0 10 M foy;u (C) vEyh; cQj foy;u (D) nqcZy vEy rFkk nqcZy {kkj ds yo.k dk foy;u j[k ldrk gSA 8 32. pH < 7 pH > 7 pH < 7 pH < 7 5 The ionization constant of the HOCl is 2.5 10 . Calculate the pH of 0.08M solutions of hypochlorous acid. [log 1.41=0.15] HOCl dk vk;uu fLFkjkad 2.5 10 5 gSA gkbiksDyksjl vEy ds 0.08M foy;uks dh pH ifjdfyr dhft,A [log 1.41=0.15] (A) 2.85 (B) 3.50 (C) 5.7 Space for Rough Work dPps (D) 3.85 dk;Z ds fy, LFkku Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 |CIN: U80302RJ2007PLC024029 CJP/JFMONT4120117C0-17 CHEMISTRY PART-II : CHEMISTRY Atomic masses : [H = 1, D = 2, Li = 7, C = 12, N = 14, O = 16, F = 19, Na = 23, Mg = 24, Al = 27, Si = 28, P = 31, S = 32, Cl = 35.5, K = 39, Ca = 40, Cr = 52, Mn = 55, Fe = 56, Cu = 63.5, Zn = 65, As = 75, Br = 80, Ag = 108, I = 127, Ba = 137, Hg = 200, Pb = 207] CHEMISTRY The solubility of Sr(OH)2 at 298K is 19.23 g/L of solution. Find the pH of the solution. (given atomic mass of Sr = 87.4) [log 3.1= 0. 5] 298K ij Sr(OH)2 dh (fn;k foys;rk foy;u esa 19.23 g/L gSA foy;u dh pH Kkr dhft,A gS% Sr dk ijek.oh; nzO;eku = 87.4) [log 3.1= 0. 5] (A) 11.50 34. (B) 9.87 (D) 13.5 What is the pH of 0.50 M aqueous NaCN solution pKb of CN is 4.70? [log2 = 0.30] 0.50 M tyh; NaCN foy;u dh pH D;k gSA CN dk pKb 4.70 gS\ [log2 = 0.30] (A) 7.85 35. (C) 7.3 (B) 11.5 (C) 8.7 (D) 10.2 pH of water is 7.0 at 25 C. If water is heated to 70 C, then (A) pH will decrease and solution becomes acidic. (B) pH will increase but solution will be neutral. (C) pH will remain constant as 7. (D) pH will decrease but solution will be neutral. 25 C ij ty dh pH 7.0 gSA ;fn ty dks 70 C rd xeZ djrs gS] rc& (A) pH ?kVsxh rFkk foy;u vEyh; gks tkrk gSSA (B) pH c<+sxh rFkk foy;u mnklhu gksxkA (C) pH 7 fu;r (D) pH ?kVsxh jgrh gSA ysfdu foy;u mnklhu gksxkA Space for Rough Work dPps dk;Z ds fy, LFkku Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 |CIN: U80302RJ2007PLC024029 CJP/JFMONT4120117C0-18 CHEMISTRY 33. Which of the following statement is not correct. (A) An aqueous solution of borax is alkaline. (B) Borax contains tetranuclear units. (C) When borax is heated in a Bunsen burner flame with CoO a green coloured bead is formed. (D) Boric acid has polymeric layered structure. fuEu es ls dkSulk dFku lgh ugh gS\ (A) cksjsDl dk tyh; foy;u {kkjh; gksrk gSA (B) cksjsDl es prq"dukfHkdh; bdkbZ;k gksrh gSA (C) tc cksjsDl dks CoO ds lkFk cqUlu cuZj Tokyk es xeZ fd;k tkrk gS] rks gjs jax dk eudk (bead) curk gSA (D) cksfjd vEy cgqyd ijrh; lajpuk j[krk gSA 37. Select the set representing correct informations about boron hydrides. (i) Diborane is a colourless, highly toxic gas. (ii) Diborane on heating with ammonia gives borazine known as inorganic benzene . (iii) In a molecule of diborane total four atoms lie in one plane. (iv) Bridge [B H] bonds are shorter than terminal [B H] bonds in diborane. (v) Diborane is produced on large scale by the reaction of BF3 with sodium hydride. (A) (i), (ii), (v) (B) (ii), (iii), (iv), (v) (C) (i), (iii), (iv) (D) (i), (ii), (iv), (v) cksjkWu gkbM kbMks ds ckjs es lgh lwpuk iznf'kZr djus okys leqPp; dk p;u dhft,& (i) Mkbcksjsu jaxghu] mPp fo"kSyh xSl gSA (ii) MkbZcksjsu dks veksfu;k ds lkFk xeZ djus ij ;g cksjkthu nsrk gS] tks ^^vdkcZfud csUthu** dgykrk gSA (iii) MkbZcksjsu ds ,d v.kq es ,d ry es dqy pkj ijek.kq fLFkr gksrs gSaA (iv) MkbZcksjsu es lsrq [B H] ca/k] vUrLFk [B H] ca/kks ls NksVs gksrs gSaA (v) MkbZcksjsu lksfM;e gkbM kbM ds lkFk BF3 dh vfHkf ;k }kjk cM+s iSekus ij mRikfnr gksrk gSA (A) (i), (ii), (v) (B) (ii), (iii), (iv), (v) (C) (i), (iii), (iv) Space for Rough Work dPps (D) (i), (ii), (iv), (v) dk;Z ds fy, LFkku Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 |CIN: U80302RJ2007PLC024029 CJP/JFMONT4120117C0-19 CHEMISTRY CHEMISTRY 36. CHEMISTRY 38. Select the incorrect statement about elements of Boron family. CHEMISTRY (A) Aluminium is most abundant element in the earth s crust. (B) Aluminium is least electronegative element in boron family. (C) Metal borides are used in nuclear industry as protective shields and control rods. (D) Orthoboric acid is generally used as a mild antiseptic. cksjkWu ifjokj ds rRoksa ds ckjs es xyr dFku pqfu,& (A) ,Y;qfefu;e HkwiiZVh es lokZf/kd ckgqY; rRo gSA (B) ,Y;qfefu;e cksjkWu ifjokj es U;wure oS|qr_.kh rRo gSA (C) /kkrq cksjkbM ukfHkdh; m|ksx es j{kk dop rFkk fu;fU=kr NM+ks ds :i es iz;qDr gksrk gSA (D) vkWFkksZcksfjd 39. vEy lkekU;r% ,d gYds izfrjks/kh ds :i es iz;qDr gksrk gS The type of hybridization of boron in diborane is MkbZcksjsu es cksjkWu ds ladj.k dk izdkj gS& (B) sp2 (A) sp 40. 3 (C) sp (D) dsp 2 In which of the following molecule, -electron density in benzene ring is minimum ? fuEu esa ls dkSuls v.kq dh csUthu oy; esa -bysDV kWu ?kuRo U;wure gSA OH NO2 (A) NO2 OCH2 CH3 (B) (C) (D) NO2 Space for Rough Work dPps dk;Z ds fy, LFkku Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 |CIN: U80302RJ2007PLC024029 CJP/JFMONT4120117C0-20 CHEMISTRY How many electrons are there in the following species es fdrus bysDV kWu gSa& fuEu Lih'kht (A) 2 42. CHEMISTRY 41. (B) 1 (C) 4 (D) 3 C1 C2 bond is shortest in fuEu esa ls fdlesa C1 C2 cU/k U;wure gSA 1 2 (A) 43. 1 2 H (B) 1 2 (C) 1 2 (D) The kind of delocalization involving sigma bond orbtial is called. (A) Hybridization (B) Electromeric (C) Hyperconjugation (D) Conformation foLFkkuhdj.k dk izdkj ftlesa flXek cU/k d{kd iz;qDr gksrk gS] dgykrk gS& (A) ladj.k (B) bysDV ksesfjd (C) vfrla;qXeu Space for Rough Work dPps (D) la:i.k dk;Z ds fy, LFkku Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 |CIN: U80302RJ2007PLC024029 CJP/JFMONT4120117C0-21 CHEMISTRY Which of the following is the strongest group. fuEu es ls dkSulk izcyre lewg gSA CH3 (A) N H (B) N CH3 H 45. (C) N CH3 H (D) N CH3 CH3 CH3 Which of the following molecule has the maximum SIR ? fuEu es ls fdl v.kq dk SIR eku vf/kdre gS\ CH3 CH3 CH3 (A) H3C CH3 C CH3 H3C (B) C C CH3 CH3 CH3 N CH3 N H3C (C) CH3 (D) N N Space for Rough Work dPps dk;Z ds fy, LFkku Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 |CIN: U80302RJ2007PLC024029 CJP/JFMONT4120117C0-22 CHEMISTRY 44. CHEMISTRY Integer value correct Type (iw.kk d eku lgh dkj) This section contains 15 questions. The answer to each question is a single digit integer, ranging from 0 to 9 (both inclusive). (bl [k.M esa 15 'u gSaA R;sd 'u dk m kj 0 ls 9 rd nksuksa 'kkfey ds chp dk ,dy vadh; iw.kk d gSA) 46. What is the pH of solution made by mixing equal volume of 0.05M H2SO4, 0.1M HNO3).0.1M HCl? 0.05M H2SO4, 0.1M HNO3).0.1M HCl 47. ds leku vk;ru dks fefJr djus ls cus foy;u dh pH D;k gS\ How many of the following mixtures can make a buffer solution ? (a) NH4Cl and NaOH in 2 : 1 mole ratio (b) CH3COONa and HCl in 1 : 1 mole ratio (c) CH3COONa and HCl in 2 : 1 mole ratio (d) CH3COONa and HCl in 1 : 2 mole ratio (e) NaCl and H2SO4 in 2 :1 mole ratio (f) 0.2 M CH3COOH and 0.1 M CH3COONa fuEu es ls fdrus feJ.k cQj foy;u cuk ldrs gSa \ (a) 2 : 1 eksy vuqikr esa NH4Cl rFkk NaOH (b) 1 : 1 eksy vuqikr esa CH3COONa rFkk HCl (c) 2 : 1 eksy vuqikr esa CH3COONa rFkk HCl (d) 1 : 2 eksy vuqikr esa CH3COONa rFkk HCl (e) 2 :1 eksy vuqikr esa NaCl rFkk H2SO4 (f) 0.2 M CH3COOH rFkk 0.1 M CH3COONa Space for Rough Work dPps dk;Z ds fy, LFkku Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 |CIN: U80302RJ2007PLC024029 CJP/JFMONT4120117C0-23 CHEMISTRY SECTION II ([k.M II) CHEMISTRY Consider the aqueous solutions of following salts. NaHSO4, NaCl, NH4NO3, Ba(NO2)2, (CH3NH3)2 SO4, C5H5NHBr if y number of acidic solutions z number of basic solutions Identify y + z KBr, NaCN CHEMISTRY 48. fuEu yo.kks ds tyh; foy;uks dk voyksdu dhft,& ;fn NaHSO4, Ba(NO2)2, y vEyh; NaCl, (CH3NH3)2 SO4, NH4NO3, C5H5NHBr KBr, NaCN foy;uks dh la[;k z {kkjh; foy;uks dh la[;k y + z igpkfu,& 49. 5 A certain weak acid has Ka = 1.0 10 the equilibrium constant for its reaction with a strong base x is 10 . Find the value of x. ,d fuf'pr vEy&yo.k Ka = 1.0 10 5 j[krk gSA izcy {kkj ds lkFk bldh vfHkf ;k dk lkE; fu;rkad 10x gSA x dk eku Kkr dhft,A 1/ x 50. K sp The solubility of A2X3 is y mol dm . It's solubility product is Ksp. If y 108 3 . Identify the value of x. 1/ x A2X3 3 dh foys;rk y mol dm gSA bldk foys;rk xq.kuQy Ksp K gSA ;fn y sp 108 gS] rks x dk eku igpkfu,A Space for Rough Work dPps dk;Z ds fy, LFkku Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 |CIN: U80302RJ2007PLC024029 CJP/JFMONT4120117C0-24 CHEMISTRY The formula of borax is Na 2 [ BwOx (OH ) y ] . zH 2O w, x, y and z are integer ranging from 1 to 9, calculate the value of (y + z ) (w + x ) cksjsDl dk lw=k Na2 [ BwOx (OH ) y ] . zH 2O gSA w, x, y rFkk z 1 ls 9 rd iw.kk d ijkl gSA (y + z ) (w + x ) dk 52. eku ifjdfyr dhft,A Most stable oxidation state of boron is +x while it is +y for thallium. Calculate x + y. cksjkWu dh lokZf/kd LFkk;h vkWDlhdj.k voLFkk +x gS tcfd FkSfy;e ds fy, ;g +y gSA x + y ifjdfyr dhft,A 53. Boron can show maximum covalency x . Find the value of x. cksjkWu vf/kdre la;kstdrk x n'kkZ ldrk gSA x dk eku Kkr dhft,A 54. The number of 3- centre 2- electron bonds in Al2Cl6 is/are : Al2Cl6 es 3-dsUnz 2- bysDV kWu ca/kks dh la[;k gS@gSa& Space for Rough Work dPps dk;Z ds fy, LFkku Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 |CIN: U80302RJ2007PLC024029 CJP/JFMONT4120117C0-25 CHEMISTRY 51. CHEMISTRY AlCl3 dissolves in excess of aqueous sodium hydroxide to form Al(OH)x AlCl3 tyh; 56. lksfM;e gkbM kWDlkbM ds vkf/kD; es ?kqydj Al(OH)x cukrk gSA x dk eku gS& Find out number of +I groups +I lewgks dh la[;k crkb;sA F H C F C H F O O H N H N CH3 H C O O 57. . The value of x is C O H O H H Find out number of M groups M lewgks dh la[;k crkb;sA O C R , O C O R , O N C R , H O H C N R , NO2 , O O , C O H O OH, H Space for Rough Work dPps dk;Z ds fy, LFkku Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 |CIN: U80302RJ2007PLC024029 CJP/JFMONT4120117C0-26 CHEMISTRY 55. CHEMISTRY 58. In how many of the following structures resonance is possible. 59. CHEMISTRY fuEu esa ls fdruh lajpukvksa esa vuqukn laHko gSA In the given structure find out the number of hydrogen involved in the hyperconjugation. nh xbZ lajpuk esa vfrla;qXeu esa iz;qDr gkbM kstu dh la[;k crkb;sA 60. How many of the following species can show resonance fuEu es ls fdruh Lih'kht vuqukn n'kkZrh gSA Space for Rough Work dPps dk;Z ds fy, LFkku Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 |CIN: U80302RJ2007PLC024029 CJP/JFMONT4120117C0-27 MATHEMATICS SECTION I ([k.M- I) Straight Objective Type (lh/ks oLrqfu"B izdkj ) This section contains 15 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE is correct. bl [k.M esa 15 cgq&fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) gSa] ftuesa ls flQZ ,d lgh gSA 61. A five digit number is formed using the digits 5, 6, 7, 8, 9 with no repetition allowed, then the probability that the number is a multiple of 5 is fcuk iqujko`f k ds 5, 6, 7, 8, 9 ls ,d ik p la[;k cukbZ tkrh gS] rc la[;k ds 5 ls foHkkftr gksus dh izkf;drk Kkr dhft,A (A) 62. 1 3 (B) 1 6 (C) 1 5 (D) 1 10 From a pack of cards, all the hearts are taken out. From these cards, cards are drawn out one by one without replacement until the Ace of heart comes. The probability that Ace of heart comes in th 5 draw (A) 1 13 (B) 12 13 (C) 4 13 (D) None of these ,d rk'k dh xM~Mh ls lHkh iku ds i kksa dks fudky fy;k tkrk gSA bu i kks ls ,d ,d i ks fcuk iquZfoLFkkfir ds rc rd fudkyk tkrk gS tc rd fd iku dk bDdk ugha vk tkrk gSA iku ds bDdk ds 5 oh ckj esa vkus dh izkf;drk gS (A) 1 13 (B) 12 13 (C) Space for Rough Work ( dPps 4 13 (D) buesa ls dksbZ ugha dk;Z ds fy, LFkku) Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PLC024029 MJP/JFMONT4120117C0-28 MATHEMATICS PART-III : MATHEMATICS MATHEMATICS A fair die is thrown 20 times. The probability that on the 10 throw, the fourth six appears (A) 20 C10 56 (B) 120 620 57 (C) 84 610 56 610 (D) None of these ,d fu"i{kikrh flDds dks 20 ckj Qsadk tkrk gSA rc 10 oha ckj ikls ds Qsad esa PkkSFkh ckj 6 vkus dh izkf;drk Kkr dhft,A (A) 64. 20 C10 56 (B) 120 620 57 (C) 84 610 56 610 (D) buesa ls dksbZ ugha In a certain town, 40% of the people have brown hair, 25% have brown eyes and 15% have both brown hair and brown eyes. If a person selected at random from the town has brown hair, the probability that he also has brown eyes is fdlh 'kgj esa 40% yksxksa ds Hkwjs cky gS rFkk 25% yksxksa dh Hkwjh vk[ks gS rFkk 15% yksxksa ds Hkwjs cky rFkk Hkwjh vka[ks nksuksa gSA ,d O;fDr dks 'kgj ls ;kn`PN;k pquk tkrk gS rFkk mlds cky Hkwjs ik;s tkrs gSA rc mldh vka[ks Hkh Hkwjh gks bldh izkf;drk Kkr dhft,A (A) 65. 1 5 (B) 3 8 (C) 1 3 (D) 2 3 One mapping is selected at random from all mappings of the set S = {1, 2, 3, ....n} into inself. If the 3 probability that the mapping is one-one is , then the value of n is 32 (A) 2 (B) 4 (C) 3 (D) None of these leqPp; S = {1, 2, 3, ....n} ls S esa ifjHkkf"kr lHkh izfrfp=k.kksa esa ls ,d izfrfp=k.k dks ;kn`PN;k pquk trk gSA ;fn izfrfp=k.k ds ,dSdh gksus dh izkf;drk (A) 2 (B) 4 3 32 gks rc n dk eku gS (C) 3 Space for Rough Work ( dPps (D) buesa ls dksbZ ugha dk;Z ds fy, LFkku) Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PLC024029 MJP/JFMONT4120117C0-29 MATHEMATICS 63. th 66. Three integer are chosen at random from the set of first 20 natural numbers. The chance that their product is a multiple of 3 is izFke 20 izkd`r la[;kvksa ls rhu iw.kkZad ;kn`PN;k pqus tkrs gSA muds xq.kuQy dks 3 dk xq.kt xq.kkad dk gksus dh izkf;drk Kkr dhft,A (A) 67. 194 285 (B) 1 57 13 19 (C) (D) 11 19 If a and b are chosen at randomly from the set consisting of number 1, 2, 3, 4, 5, 6 with 2 a x bx replacement. Then the probability that lim x 0 2 x = 6 is la[;kvksa 1, 2, 3, 4, 5, 6 dks j[kus okys leqPp; ls nks la[;k, a vkSj b dks izfrLFkkiu ds lkFk pquk tkrk gSA 2 rc izkf;drk Kkr dhft, fd (A) 68. 1 3 ax bx lim x 0 2 (B) 1 9 x = 6 gS (C) 1 4 (D) 2 9 Let A be a set of n ( 3) distinct elements. The number of triplets (x, y,z) of the A elements in which at least two coordinates is equal to ekuk A ,d leqPp; gS ftlesa n( 3) fofHkUu vo;o gSA A ds vo;oks ds lHkh f=kdks (x, y, z) dh la[;k ftlesa ds de ls de nks funsZ'kkad cjkcj gS] gksxh n 3 (A) P3 n (B) n P3 Space for Rough Work ( dPps 2 (C) 3n 2n 2 (D) 3n (n 1) dk;Z ds fy, LFkku) Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PLC024029 MJP/JFMONT4120117C0-30 MATHEMATICS MATHEMATICS MATHEMATICS Total number of six-digit in which one and only odd digits appear is (A) 5 (6!) 2 6 vadks (A) 70. (C) 1 (6!) 2 (D) None of these dh dqy la[;k ftuesa dsoy fo"ke vad vkrs gks gksxh - 5 (6!) 2 (B) 6! (C) 1 (6!) 2 (D) buesa ls dksbZ ugha The total number not more that 20 digits that are formed by using the digits 0, 1, 2, 3 and 4 is (A) 5 20 0, 1, 2, 3 (A) 5 71. (B) 6! 20 (B) 5 1 20 (C) 5 +1 (D) None of these vkSj 4 ls cuus okyh lHkh la[;kvksa dh la[;k ftlesa 20 vad ls T;knk ugh gS] gksxh - 20 20 (B) 5 1 20 (C) 5 +1 3 (D) buesa 2 ls dksbZ ugha 2 The number of even divisors of the number N = 12600 = 2 .3 .5 7 is la[;k N = 12600 = 23.32.527 ds lHkh le Hkktdksa dh la[;k gS (A) 72 72. (B) 54 (C) 45 (D) 36 Ten IIT and 2 DCE students seat in a row. The number of ways in which exactly 3 IIT students sit between 2 DCE students is (A) 10 C3 2! 3! 8! (B) 10! 2! 3! 8! (C) 5! 2! 9! 8! (D) None of these 10 IIT rFkk 2 DCE fo|kFkhZ ,d iafDr esa cSBrs gSA rjhdks dh la[;k gksxh ftlesa 3 IIT ds Nk=k 2 DCE ds Nk=kksa ds e/; cSBrs gS (A) 10 C3 2! 3! 8! (B) 10! 2! 3! 8! (C) 5! 2! 9! 8! (D) buesa Space for Rough Work ( dPps ls dksbZ ugha dk;Z ds fy, LFkku) Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PLC024029 MJP/JFMONT4120117C0-31 MATHEMATICS 69. MATHEMATICS The last digit of (1! + 2! + ..... + 2005!) (1! + 2! + ..... + 2005!) 500 (A) 9 74. 500 is dk vfUre vad gS (B) 2 (C) 7 (D) 1 Total number of ways of selecting two number from the set {1, 2, 3, 4 ...... 3n} so that their sum is divisible by 3 is leqPp; {1, 2, 3, 4 ...... 3n} ls 2 vad pqus tkrs gSA mu rjhdksa dh la[;k Kkr dhft, tc mudk ;ksxQy 3 ls foHkkftr gS (A) 75. 2n2 n 2 (B) 3n2 n 2 2 (C) 2n n 2 (D) 3n n The total number of positive integral solutions of 15 < x1 + x2 + x3 20 is equal to (A) 685 (B) 785 (C) 1125 15 < x1 + x2 + x3 20 ds /kukRed iw.kkZad gyksa dh la[;k cjkcj gS - (A) 685 (B) 785 (C) 1125 Space for Rough Work ( dPps (D) None of these (D) buesa ls dksbZ ugha dk;Z ds fy, LFkku) Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PLC024029 MJP/JFMONT4120117C0-32 MATHEMATICS 73. SECTION II ([k.M II) Integer value correct Type (iw.kk d eku lgh dkj) This section contains 15 questions. The answer to each question is a single digit integer, ranging from 0 to 9 (both inclusive). bl [k.M esa 15 'u gSaA R;sd 'u dk m kj 0 ls 9 rd nksuksa 'kkfey ds chp dk ,dy vadh; iw.kk d gSA 76. If the probability of a six-digit number N whose six digits 1, 2, 3, 4, 5, 6 written as random order is divisible by 6 is P, then 6 vad dh la[;k N dh izkf;drk ftlds vad 1, 2, 3, 4, 5, 6 dks ;kn`fPNd e esa fy[kus ij 6 ls foHkkftr gS rc 77. 1 is P 1 P dk eku gS - Seven white balls and three black balls are randomly placed in row. If probability that, no two black balls are placed adjacently is equal to a , where a and b are co-prime to each other, then find the b value of (b a). lkr lQsn xsan vkSj 3 dkyh xsnksa dks ;kn`fPNd ,d iafDr esa j[kk tkrk gSA ;fn izkf;drk fd dksbZ Hkh nks dkyh xsnks dks ikl ikl j[kus dh izkf;drk a b gS tgk a vkSj b izR;sd ,d nqljs dh lg vHkkT; la[;k gS rc (b a) dk eku gSaA Space for Rough Work ( dPps dk;Z ds fy, LFkku) Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PLC024029 MJP/JFMONT4120117C0-33 MATHEMATICS MATHEMATICS 78. Suppose A & B are two events with P(A) = 0.5 and P(A B) = 0.8. Let P(B) = P. If A & B are mutually exclusive and P(B) = q if A and B are independent events then the value of q is p ekuk fd A vkSj B nks ?kVuk, gS ftuesa P(A) = 0.5 rFkk P(A B) = 0.8 ekuk P(B) = P ;fn A vkSj B ijLij viothZ gS rFkk P(B) = q ;fn A vkSj B Lora=k ?kVuk, gS rc 79. q p dk eku gS - A If A and B are two events such that P(A) = 0.6 and P(B) = 0.8, if the greatest value that P can B have is P, then value of 8p is A ;fn A vkSj B nks ?kVuk, bl izdkj gS fd P(A) = 0.6 vkSj P(B) = 0.8 ;fn P dk vf/kdre eku P gS rc B 8p dk 80. eku gS - Two cards are drawn from a well shuffled pack of 52 cards. The probability that one is a heart and and the other is a king is P, then the value of 104P is rk'k ds 52 i kksa dh xM~Mh ls nks i ks ;kn`fPNd fudkys tkrs gS rc muesa ls ,d iku dk i kk rFkk nqljk jktk dk i kk gksus dh izkf;drk P gS rc 104P dk eku gS - Space for Rough Work ( dPps dk;Z ds fy, LFkku) Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PLC024029 MJP/JFMONT4120117C0-34 MATHEMATICS MATHEMATICS 81. A die is thrown 3 times. The chance that highest number shown on the die is 4 is P, then value of 1 P is (where [.] represents greatest integer function) 1 ,d iklk 3 ckj Qsdk tkrk gSA ikls ij 4 ls mPp la[;k vkus dh izkf;drk P gS rc dk eku gS (tgk [.] P eg ke iw.kkZad Qyu gSA) 82. A five digit number of the from x y z y x is chosen, probability that x < y is x y z y x ds 83. , then the value of is 15 :i dh ik p vadks dh la[;k pquh tkrh gS tcfd x < y dh izkf;drk 15 gS rc dk eku gS - Number of numbers that can be formed with the help of the digits 1, 2, 3, 4, 3, 2, 1 so that odd digits always occupy odd places is x. Then, sum of digits of x is 1, 2, 3, 4, 3, 2, 1 dh lgk;rk ls cukbZ tk ldus okyh la[;kvksadh la[;k x gS tcfd fo"ke vad fo"ke LFkku ij vkrs gS rc x ds vadksa dk ;ksxQy gS - Space for Rough Work ( dPps dk;Z ds fy, LFkku) Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PLC024029 MJP/JFMONT4120117C0-35 MATHEMATICS MATHEMATICS MATHEMATICS Number from 1 to 1000 are divisible by 60, but not by 24 is 1 ls 1000 85. rd dh la[;kvksa dh la[;k tks 60 ls foHkkftr gS] ijUrq 24 ls ughA The number of three- digit numbers having only two consecutive digits identical is N, then value of 1 N 2 2 is 1 rhu vad dh la[;kvksa dh la[;k gS ftuesa dsoy nks ekxr vad loZle gS] N gS rc 86. N 2 2 dk eku gS - If N is the number of ways in which a person can walk up a stair ways which has 7 steps if he can take 1 or 2 steps up the stairs at a time, then the value of N is 3 ,d O;fDr 7 lh<+h;ka pydj p<+ ldrk gS ftlesa og 1 ;k 2 dne ,d ckj esa py ldrk gS ;fn blds ep; N gS rc N 3 dk eku gS - Space for Rough Work ( dPps dk;Z ds fy, LFkku) Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PLC024029 MJP/JFMONT4120117C0-36 MATHEMATICS 84. 87. Number of n digit numbers which consist of the digits 1 and 2 only, if each digit is to be used at least once is equal to 510, then n is equal to n vadks dh la[;kvksa dh la[;k ftlesa vad 1 ;k 2 dsoy gS ;fn izR;sd vad ds de ls de ,d ckj vkus ds ep; 510 gS rc n cjkcj gS - 88. From 6 Girls and 6 boys, the number of ways to choose atleast 2 persons is 2 x 1. Then x 6 yMdh 89. rFkk 6 yMdksa ls de ls de 2 O;fDr ds pquus ds ep; 2x x 1 gS rc x 2 x is 2 gS - From Dehradun to Bombay there are two routs rail and road. From Bombay to Goa, there are four routes, air rail, road and Sea. In how many ways a person can go from Dehradun to Goa via Bombay. nsgjknwu ls eqEcbZ ds fy, nks jkLrs jsyekxZ vkSj lM+d ifjogu gSA eqEcbZ ls xksok ds fy, pkj jkLrs jsyekxZ] ok;qekxZ] lM+d ifjogu vkSj leqnzh ekxZ gSA fdrus rjhdks ls ,d O;fDr nsgjknwu ls eqEcbZ gksrs gq, xksok tk ldrk gSA 90. There are 20 books on Algebra and Calculus in one library. For the greatest number of selections each of which consists of 5 books of each topic. If the possible number of Algebra books are N, N then value of is 2 ,d iqLrdky; esa chtxf.kr vkSj dyu dh 20 iqLrds gS izR;sd p;u dh vf/kdre la[;k ds fy, tcfd N izR;sd v/;k; dh izR;sd p;u esa 5 iqLrds gS] ;fn chtxf.kr dh iqLrds lEHkkfor la[;k N esa gS rc dk 2 eku gS - Space for Rough Work ( dPps dk;Z ds fy, LFkku) Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PLC024029 MJP/JFMONT4120117C0-37 MATHEMATICS MATHEMATICS MAIN PATTERN ONLINE TEST-4 (MONT-4) XI TARGET : JEE (MAIN+ADVNACED) 2017 DATE : 21-03-2017 COURSE : JA**, 01JA, 01EA, 01JB, 02JB, 01EB, 01JR, 05JR, 01ER, 05ER HINTS & SOLUTIONS l a d sr ,oagy PART-I PHYSICS 1. y 24 = t (x 2t)3 Phase difference = 2 d y kUrj = 2 6. (C) P = P= 1 2 2 A V 2 using V = T (d k iz ;ksx d jusij) 1 2 2 A T 2 = = 2 45 = 90 ij 24 = 3 m/s. ( 2)3 Vy = 1 1 cos 2 =2 x = 2, t = 2 2P A 2 f= T 1 = 2 2 2P A 2 T using data f = 30 Hz. 3. vk d M+ksd smi;ksx ls f = 30 Hz. At equilibrium : lkE;koLFkk esa 7. mg sin = kA A= mgsin K At t = 2 second, the position of both pulses are separately given by fig.(a) and fig. (b); the superposition of both pulses is given by fig. (c) t = 2 ls d .M ij nksuksaLianksd ksfp=k (a) o fp=k (b) esav/;kjksi.k d s ckn fp=k (c) esacrk;k x;k gSA and time period for spring block system is : fLaizx Cy kWd fud k; d k vkorZd ky : T= 2 m k Hence vr% (A) 4. P = 2x g F = 2x gA T= 5. 2 A R R = 2 . 2 gA 2g (B) y(x, t = 0) = 6 x2 8. then rks y(x,t) = y 24 = t (x 2t)3 6 (x 2t)2 (C) 4a = 4g T 4a = T T = 20 N a=5 50 = at x = 2, t = 2 n 2 0.6 20 1 20 n=3 Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PLC024029 SOLJP/JFMONT4120117-1 1 2 5t 2 0.2 = t= 11. 0.08 t1 = 0 t2 = 0.08 0.16 0.08 t1 t 2 9. t3 = 0.16 0.08 1 12. 2 1 V =f 4( 1 e) 3V =f 4( 2 e) f1 f2 = 3 v v f v vs v vs f=3 2v s f =3 v vs = 1.5 m/sec 1 (stress)2 2 Y Tension Stress = area izfrcy (Stress) = ruko@{ks=kQ y U= UA TA2 F2 = = =4 Umiddle (Tmiddle )2 (F / 2)2 V = 1 + e 4f 13. 3V = 2 + e 4f dF = dm 2x F= 2V = 2 1 4f Ans. (B) = A 2( 2 x2) 2g F A = V = 2f ( 2 1) e= 2f( 2 1 ) 1 4f 2 2 2 1 4 1 4 2( 2 3 1 ) = 4 = E= 10. = 14. = du dx du = 2 3 . 3gE Fd = 6 ru FB = 4 3 4 r g , mg = r 3 g 3 3 mg Fd FB = ma ; u0 = a First resonance occurs at fundamental frequency Fke vuqukn ewy vko`f k ij gksrk gSA f= V 4( e) e = 0.6 2 = 1.2 cm) +e= 336 = 0.164 m 4 512 +e= V 4f (where t gk 15. 2r 2 ( ) g 9 u 1 1 g u0 Let v be the velocity of the movable plate and F is equal to viscous force ekuk xfr d jus;ksX; Iy sV d k osx v gSo F ';ku cy d scjkcj gSA F = 1 v v 2 A h1 h h1 = 16.4 1.2 = 15.2 cm Ans. (B) Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PLC024029 SOLJP/JFMONT4120117-2 dF =0 dh1 16. h1 = h 3 20. 1 mVm2 = 15 10 3 2 Vm = 0.150 m / s A = 0.150 m / s L m. g = L gL = = 0.150 m / s T = r2 2 A T T 0.150 100 10 3 0.150 = 1.5 m 0.1 L= = d m 2r 2 2 2T = A r r 2 2T sin = 2 rad/s = = 1.5 2 21. = 3m T v 3 2 5 2 10k mv + kx = . 4 2 3m 17. E= 18. When the point of suspension is at a distance x from centre of length of rod, the time period of oscillation is Power 'kfDr P = T. v = T v cos (90- ) = Tv sin =50x 1x1/50=1 watt 2 x T = 2 12x where is length of the rod. g 22. The time period of oscillation will be same (TO) if the point of suspension is a distance x= /2 or v2 = x = /6 from centre of the rod. Thus there will be three additional points. ar = t c y Vd u fcUnqNM+d sd sUnzlsx d h nwjh ij gS] rksnksy u d k vkorZd ky gksxkA T = 2 2 x 12x t gk NM+d h y EckbZgSA g 23. nksy u d k vkorZd ky leku (TO) gS] ;fn y Vd u fcUnqNM+d s d sUnzlsx= /2 ;k x = /6 d h nwjh ij gSA vr% ;gk rhu vfrfjDr fcUnqgSA 19. ai 80 2.56 m/s = 32 m/s 0.2 = v 2 v1 32 80 = 3.5 cm = 15mm. v 2 v1 32 80 RT M 4 2 2 2 M= = 3g 4 2 2 (2 2) 3 = 1 = 1 = f 2 2 3 2 5 Ans. 24. T 2 80 4.8 = 80 m/s 0.06 v= v = x1 = 5 sin (t 37 ) x2 = 6 sin ( t + 90 ) T = 1 v1 = 3 25 1000 972 = 900 m/s 2 3 3 5 9 L = 10 log10 0 = 10 6 w/m2 P = 4 r2 = 3.14 w 1000 2 20 = = 6.36 sec. 3.14 100 3.14 To the nearest integer lehiorhZiw .kkZad ksaesa t= Amplitude vk;ke = 32 42 = 5. Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PLC024029 SOLJP/JFMONT4120117-3 t = 6 sec. L dl 2 25. p = v 0 (F1 F2 )L = 1 10 9 m 2Ay 30. 2 p22 1 v1 = 1 p12 2 v 2 = 9 1.5 400 16 3 1200 6 1 6 32 = = 64 2 3 n T As tension increases number of loop decreases 2L ruko c<+usij y wiksad h la[;k ?kVrh gSA f= = 9 3 = 16 3 2 32 n 160 n 1 250 1 mg = = 2L 2L 2L m = maximum mass vf/kd re nz O;eku 4n = 5n 5 26. and rFkkm = 400 kg n=5 PART-II CHEMISTRY 31. (A) it is a solution of acid so, pH will be less than 7. (B) it is a solution of base so, pH will be greater than 7. (C) pH of acidic buffer solution may be greater than or less than 7. f1 v / 2 =3 f2 v / 6 pH = pKa + log [conjugate base] [Acid] f1 = 3f2 = 3 2 = 6 Hz (f2 = 2Hz) the difference in frequency of the two waves is = f1 f2 (D) pH of salt of WAWB depends on pKa and pKb value of weak acid and weak base. = 6 2 = 4 Hz (A) ;g vEy d k ,d foy ;u gS] vr%pH 7 lsd e gksxhA (B) ;g {kkj d k ,d foy ;u gS ] vr%pH 7 lsvf/kd gksxhA (C) vEy h; cQ j foy ;u d hpH 7 lsvf/kd vFkok 7 lsd e gksld rh gSA nksuksarjxksad h vko`fr esavUrj = f1 f2 = 6 2 = 4 Hz 27. pH = pKa + log [l a;qXeh {kkj] [vEy ] (D) WAWB d sy o.k d h pH nq cZy pKb d seku ij fuHkZ j d jrh gSA 32. 325 334 9 700 = 700 = 650 Hz. 350 334 16 f' = 28. HOCl (aq) H 2O (l ) Initial conc. 0.08 0 at equilibrium 0.08 x x x [ H 3O ][ClO ] [ HOCl] Ka 2 0.075N / m 1 = 3 (0.20 10 m) (1000kg / m3 )(10m / s2 ) 2.5 10 5 = = 0.075 m = 7.5 cm Ans. 7.5 cm. x T T T dx dL = A y T = F1 (F1 F2) 0 izkjfEHkd lkUnzrk 2S cos h= r g 29. H 3O (aq ) ClO ( aq ) lkE; ij We have, gekjsikl F2 vEy rFkk nqcZy {kkj d spKa rFkk x L F1 x2 (0.08 x) 0.08 x 0.08 x = 1.41 10 3 [H+] = 1.41 10 3 M pH = log10 (1.41 10 3) 33. Molar mass of Sr(OH)2 = 121.6 g mol 1 Sr(OH)2 d k eks yj nzO;eku = 121.6 g mol 1 solubility of Sr(OH)2 in mol L 1 = 19.23 0.1581M 121.6 1 19 .23 Sr(OH)2 d h foy s ;rk eksy L esa= 0.1581M 121.6 Sr+2 + 2OH Sr(OH)2 [OH ] = 2 0.1581 M = 0.3162 M Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PLC024029 SOLJP/JFMONT4120117-4 POH = log10 0.3162 POH log10 .32 0.5 PH = +13.5 34. pH = 7 + 7+ 39. H 1 [14 4.70 + log 0.5] = 11.5 2 H H 40. Due to NO2 group M and effect. NO2 lew g d sd kj.k M o izHkkoA 41. 2 bond = 4 electrons. 2 cU/k = 4 by s DV kWu 36. (A) 42. Due to resonance. Na2B4O7 + 7H2O 2NaOH + 4H3BO3 (strong base) (A) Na2B4O7 vuqukn d sd kj.kA orthoboric (weak acid) (B) Correct formula of Borax Na2[B4O5(OH)4].8H2O. (C) It is Blue in colour. (D) Boric acid has layered structure in which BO3 units are joined by hydrogen bonds. 43. Hyperconjugation involved -bond orbtial. vfrla;qXeu es -cU/k d {kd iz;qDr gksrsgSA CH3 + 7H2O 2NaOH + 4H3BO3 (iz cy {kkj) vkFkksZcksfjd vEy (nqcZy vEy ) (B) cks jsDl d k lgh lw=kNa2[B4O5(OH)4].8H2O gSA (C) ;g uhy sja x d k gksrk gSA (D) cks fjd vEy cgqy d ijrh; lajpuk j[krk gS] ft lesBO3 bd kbZ;k gkbM kst u ca/kkslst qM h gksrh gSA Heat 3B2H6 6NH3 2B3N3H6 12H2 Borazine (iii) Six atoms 2B & 4 terminal hydrogen atoms lie in one plane in a diborane molecule. (iv) Bridge B H bonds are longer 134 pm while terminal B H bonds are shorter 119 pm. (v) 2BF3 6NaH 6NaF B2H6 (ii) xeZ 3B2H6 6NH3 2B3N3H6 12H2 cksjkt hu 44. N CH3 Strongest group. CH3 CH3 N CH3 izcy re lewg gSA CH3 45. Due to ortho effect vkFkksZizHkko d sd kj.k 46. M of H+ = (0.1 0.1 0.1)V 3V = 0.1 PH = 1 NH 4Cl NaOH NH 4OH NaCl 47. (a) t=0 2 finally vUr ,d MkbZcksjsu v.kqesN& ijek.kq2B rFkk4 vUrLFk gkbM kst u ijek.kq,d ry esfLFkr gksrsgSaA (iv) ls rqB H ca/k cMs134 pm gksrsgSat cfd vUrLFk B H ca/k NksVs 119 pm gks rsgSaA 2BF3 6NaH 6NaF B2H6 0 1 1 0 1 1 CH 3COO Na HCl CH 3 (c) t=0 2 vUr es 2-1 1 Acidic Buffer Solution earth s crust is oxygen which is non metal. (B), (C) & (D) Factual (B), (C) rFkk(D) rF;kRed 0 {kkjh; cQ j foy ;u 38. (A) It is most abundant metal, the most abundant element in vkWDlht u gksrh gS] t ksv/kkrqgSA 1 1-1 Basic Buffer Solution finally ;g lokZf/kd ckgqY; /kkrq gS] Hkw iiZVh es lokZf/kd ckgqY; rRo es 2-1 1 (iii) (A) B H 1 (pKa + log c) 2 t Sl sgh rki Kw y sfd u [H+] = [OH ] (v) H B 35. As Tempeture Kw but [H+] = [OH ] 37. (ii) H 1 0 0 1-1 1 1 0 1 1 vEy h; cQ j foy ;u (f) CH 3COOH CH 3COONa 0.2 M 0.1 M Acidic Buffer Solution vEy h; cQ j foy ;u Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PLC024029 SOLJP/JFMONT4120117-5 48. The aqueous solution of this salt will be acidic due to dissociation of HSO4 HSO4 For weak acid nq cZy H SO4 Na Cl NaCl Further vkxs HA + H2O 2 NH4NO3 Salt of strong acid & weak base acidic solution due to cationic hydrolysis. KBr SASB type salt, no hydrolysis neutral solution NaCN WASB type salt, Basic solution due to anionic hydrolysis. Ba(NO2)2 WASB type salt, Basic solution due to anionic hydrolysis. 50. Ksp z = 2 NaCN, d s fo;kst u d s d kj.k H SO4 Na2 [ B4O5 (OH )4 ]. 8H 2O Na2 [ B4O5 (OH )4 ]. 8H 2O gSA 52. Most stable oxidation state of boron is +3. Most stable oxidation state of thallium is +1(due to inert pair effect) cksjkWu d h lokZf/kd LFkk;h vkWDlhd j.k voLFkk+3 gSA FkSfy ;e d h lokZf/kd LFkk;h vkWDlhd j.k voLFkk +1(vf ; ;qXe izHkko d sd kj.k) vEy h; gksxkA HSO4 Ka 10 5 14 109 Kw 10 2 3 2 3 2 .3 y w=4 y=4 x = 5 z=8 = (y + z) (w + x) = (4 + 8) (4 + 5) = 12 9 =3 C5H5NHBr y+z=4+2=6 finally, it s a common sense that y + z will be equal to total salts given in question. H + OH ; Kw cksjsDl d k lgh lw=k Ba ( NO2 ) 2 , bl y o.k d k t y h; foy ;u HSO4 + A (aq) ; Ka 51. The correct formula of borax is (CH 3 NH 3 ) 2 SO4 , (aq) K eq acid does not undergo hydrolysis, so neutral] y = 4 NaHSO4, NH4NO3, H + H2O [Salt of strong base and strong vEy d sfy , + 2 Na Cl [izcy {kkj rFkk izcy vEy d k NaCl y o.k t y vi?kVu ugh nsrk gS] vr% mnklhu gS] NH4NO3 iz cy vEy rFkk nqcZy {kkj d k y o.k /kuk;fud t y vi?kVu d sd kj.k vEy h; foy ;u KBr SASB iz d kj d k y o.k] t y vi?kVu ugh gksrk gS] mnklhu foy ;u NaCN WASB iz d kj d k y o.k, _ .kk;fud t y vi?kVu d sd kj.k {kkjh; foy ;u Ba(NO2)2 WASB iz d kj d k y o.k] _ .kk;fud t y vi?kVu d sd kj.k {kkjh; foy ;u z = 2 NaCN, fjDr d-d {kd d h vuqifLFkfr d sd kj.k cksjkWu d h vf/kd re la;kst d rk 4 gS A 54. Cl Cl (CH 3 NH 3 ) 2 SO4 , Cl Al Al Cl Cl Cl Cl Al Cl Al bridge bond is not electron deficient. Cl Al Cl Al y = 4 NaHSO4, NH4NO3, 53. Due to absence of vacant d-orbital that the maximum covalence of boron is 4. lsrqca/k by sDV kWu U;wu ugh gSA C5H5NHBr 55. Solution Ba ( NO2 ) 2 , AlCl3 NaOH Al (OH)3 NaCl y+z=4+2=6 White gelatinous vUr es;g O;ogkfjd Kku gSfd y + z iz'u esfn, x;sd qy y o.kksd scjkcj gksxkA Al(OH)3 OH [Al(OH)4 ] (excess) 49. HA + BOH Weak Strong BA + H2O foy ;u AlCl3 NaOH Al (OH)3 NaCl 'osr ft y sfVuh r nqcZy izcy Or ;k + Al(OH)3 HA + B + OH + B + A + H2O A + H2O ; Keq --------------Eq (1) [ A ] K eq [ HA][OH ] [Al(OH)4 ] ( vkf/kD; ) Or ;k HA + OH OH 56. CH3, O , C O are +I groups O Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PLC024029 SOLJP/JFMONT4120117-6 CH3, O , C O +I lew g izFke 9 Q sad esa3 ckj 6 vkSj 6 ckj 6 ughavkrk vkSj nlohaQ sad gSA N% blfy , ;g Q y d ij ughan'kkZ;sxk fd rc vHkh"V izkf;d rk O 57. M group M gksxh lewg H 3 C3 C O R , C N R , NO2 , C O H O O 6 1 84 56 1 5 1 1.........1 = 6 610 6 6 10 times 9 O 64. 58. P(A) = 0.4 P(B) = 0.25 P(A B) = 0.15 B P(A B) 0.15 3 = P(A) = 0.4 = 8 A so P 65. 59. H H Total mapping = nn Number of one-one mapping is n C1 n 1C1 n 2C1.... 1C1 = n! hence, probability is P = n! n = n 3 4! = 32 44 comparing, we get n = 4 Hindi 60. d qy izfrfp=k.k = nn ,d Sd h izfrfp=k.kksad h la[;k n C1 n 1C1 n 2C1.... 1C1 = n! n! 3 4! = = 32 nn 44 n = 4 rq y uk d jusij] ge ikrsgSfd vr% izkf;d rk P = PART-III MATHEMATICS 61. Unit place must be occupied by 5 Total 5 digit numbers = 5! 5 digit number with 5 at unit place = 4! probability = Hindi 66. 4! 1 5! 5 bd kbZd k vad 5 gksuk pfg, d qy 5 vad ksad h la[;k = 5! d qy vad ft uesabd kbZvad 5 gks= 4! iz kf;d rk= 62. 4! 1 5! 5 Hindi 20 C3 = 1140 mud k xq.kuQ y 3 d k xq.kkad gks bld k vFkZ ;g gqv k fd mu rhuksaesalsd e lsd e ,d la[;k 3 lsfoHkkft r gksuh pkfg,A 3 lsfoHkkft r la[;k,s3, 6, 9, 12, 15 & 18. buesalsd e lsd e ,d d kspqust kusd srjhd ksad h la[;k 6C1 ith draw. Required probability = P(E11 E21 E31 E41 E5) Hindi 12 11 10 9 1 1 13 12 11 10 9 13 14 C2 + 6C2 14C1 + 6C3 = 776 vr% izkf;d rk= ekuk Ei, i osackj esabDd k vkusd h ?kVuk d ksfu: fir d jrk gSA 776 194 1140 285 2 vHkh"V izkf;d rk = P(E11 E21 E31 E41 E5) 12 11 10 9 1 1 = 13 12 11 10 9 13 63. 776 194 1140 285 izFke 20 izkd `r la[;kvksalsrhu iw.kkZad pquusd h d qy fof/k;k = hence, probability = Let Ei be the event denotes drawing of Ace in = Total number of ways of selecting 3 integer from 20 natural numbers = 20C3 = 1140 Their product is a multiple of 3 means atleast one number is divisible by 3. The number divisible by 3 are 3, 6, 9, 12, 15 & 18. number of ways selecting atleast one of then is 6 C1 14C2 + 6C2 14C1 + 6C3 = 776 In the first 9 throws, are should have 3 sixes and 6 nonsixes and a six in the 10th throw and thereafter it does not mater whatever face appears and the required probability is 67. a x bx x lim x 0 2 2 a x b x 2 ax bx 2 = lim 1 . x 0 2 a x 1 b x 1 logab lim = e = ab = 6 x 0 x Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PLC024029 SOLJP/JFMONT4120117-7 Total number of possible ways in which a & b take values is 6 6 = 36 Possible ways is (1, 6), (6, 1) , (2, 3) , (3, 2) Probability = 72. Three IIT students who will be between the IIT students can be selected in 10C3 ways. Now, two DCE students having three IIT students between than can be arranged in 2! 3! ways. Finally, a group of above 5 students and the remaining seven students together can be arranged in 8! ways. hence, total number of ways is 10 C3 2! 3! 8! Hindi 3 IIT 4 1 = 36 9 2 Hindi a x bx x lim x 0 2 2 = a x b x 2 ax bx 2 lim 1 . x 0 2 a x 1 b x 1 logab lim = e = ab = 6 x 0 x a vkS j b d ksy susd slHkh lEHko rjhd s6 6 = 36 lEHko rjhd sgS(1, 6), (6, 1) , (2, 3) , (3, 2) izkf;d rk = d sNk=k t ksDCE d sNk=kksad se/; cSBrsgSd h 10C3 rjhd ks lspquk t k ld rk gSA vc 2 DCE d sNk=k d sNk=k rFkk 3 IIT d s Nk=kksad ks2! 3! rjhd kslsO;ofLFkr fd ;k t k ld rk gSA vUrr% 5 Nk=kksa d k lewg rFkk cps gq, Nk=kksa d ks vkil esa 8! rjhd ksa ls O;ofLFkr d j ld rs gSA vr% d qy rjhd ks d h la[;k 10 C3 2! 3! 8! 73. N = 1! + 2! + .... + 2005! N = (1! + 2! + 3! + 4!) + (5! + .......... + 2005!) N = 33 + an integer with 0 at unit's palce 4 1 = 36 9 N = 33 + ,d iw.kkZad ft ld k bd kbZd k vad 0 gS Hence, N500 is an integer having 1 in its units place. 68. Hindi Total number of triplets without restriction is n n n Number of triplets with all coordinates different is nP3. Required triplets is n3 nP3 n3 n(n 1)(n 2) vr% N500 ,d iw.kkZad gSft ld k bd kbZd k vad 1 gSA 74. 1, 4, 7, ........, 3n 2 3 2 2, 5, 8, ........, 3n 1 3 1 3 fcuk 'krZd slHkh f=kd ksad h la[;k= n lHkh f=kd ksad h la[;k ft lesavy x vy x vo;o gSnP3. Required triplets is vHkh"V f=kd n3 nP3 n n n 3, 6, 9, ........, 3n Hindi Clearly, one of the odd digits 1, 3, 5, 7, 9 will be repeated. Number of selections of the sixth digit is 5 C1 = 5 C2 + nC1 nC1 = Hindi 1, 4, 7, ........, 3n 2 3 2 6! Then the required number of numbers is 5 2! Li"Vr% 1, 3, 5, 7, 9 esals,d vad nksckj vk;sxk NBk vad pquusd h la[;k gS5C1 = 5 Each position can be filled in 5 ways. Hence, the total number of numbers is 520. izR;sd LFkku d ks5 rjhd slsHkj ld rsgSA vr% d qy rjhd ks d h la[;k 520. 71. Hindi 2, 5, 8, ........, 3n 1 3 1 3, 6, 9, ........, 3n = (3)(2 + 1)(2 + 1)(1 + 1) = 54 3 bld k vFkZ gqv k] gesa nks la[;k,a vfUre iafDr ls ;k ,d ,d la[;k izR;sd iz/ke vkSj f}rh; iafDr lsy suk pkfg,A vr% d qy rjhd sgSA n C2 + nC1 nC1 = 75. 3n2 n n(n 1) + n2 = 2 2 15 < x1 + x2 + x3 20 x1 + x2 + x3 = 16 + r, where r = 0, 1, 2, 3, 4 Now, the number of positive integral solution of x1 + x2 + x3 = 16 + r is 15 + rC13 + r = 15 + rC2 Thus, total number of solutions is Number of even divisors is equal to number of ways in which one or more '2' zero or more 3, zero or more'5' and zero or more '7' can be selected, and is given by (3)(2 + 1)(2 + 1)(1 + 1) = 54 le Hkkt d ksad h la[;k mrusrjhd ksad scjkcj gSft uesa,d ;k vf/kd '2' 'kwU; ;k vf/kd '3' 'kwU; ;k vf/kd '5' vkSj 'kwU; ;k vf/kd '7' d kspqukk t k ld rk gSA vkSj rjhd sgS 3n2 n n(n 1) + n2 = 2 2 fn;sx;svad bl izd kj O;ofLFkr fd ;st k ld rsgSA n 6! rc vHkh"V la[;kvksad h la[;k 5 2! 70. 3 That means, we must take two number from last row or one number each from first and second rows. Therefore, the total number of ways is n3 n(n 1)(n 2) 69. Given number can be arranged as 4 15 r C2 r 0 = 15C2 + 16C2 + 17C2 + 18C2 + 19C2 = 19C3 15C3 = 685 Hindi 15 < x1 + x2 + x3 20 x1 + x2 + x3 = 16 + r, t gk r = 0, 1, 2, 3, 4 Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PLC024029 SOLJP/JFMONT4120117-8 vc x1 + x2 + x3 = 16 + r d sgy ksad h la[;k15 + rC13 + r = 15 + r event B : King but not heart (3 card) event C : Heart and king (1 card) Required probability P = P(E) C2 gy ksad h d qy la[;k 4 15 r 12 C2 = 3 1 52 r 0 15 16 17 18 19 19 15 = C2 + C2 + C2 + C2 + C2 = C3 C3 = 685 = 76. Hindi Total cases n(s) = 6! Now sum of given digit is 1 + 2 + 3 + 4 + 5 + 6= 21 which is divisible by 3. Now, we have to form the number which is divisible by 6, then we have to ensure that the digit is unit place is even. Favourable cases = n(A) = 3.5! 3 5! 1 Hence, P(A) = 6! 2 d qy fLFkfr;kn(s) = 6! vc fn, x, vad ksd k ;ksx 1 + 2 + 3 + 4 + 5 + 6 = 21 t ks3 lsfoHkkft r gSA vc la[;k 6 lsfoHkkft r gSrc bd kbZLFkku d k vad le gSA vuq d wy fLFkfr;k= n(A) = 3.5! vr% , P(A) = Hindi 12 C1 C1 3 C1 C1 1 C1 C1 2 52 ekuk C2 104P = 4 ?kVukA : Card is of heart but not king (12 i ks) ?kVukB : jkt k ;k iku d k i kk ugh (3 i ks) ?kVukC : jkt ;k iku d k i kk(1 i ks) vHkh"V izkf;d rk 12 P = P(E) = = 81. 2 52 3 1 12 C1 C1 3 C1 C1 1 C1 C1 52 C2 104P = 4 Highest number in three throws is 4, which means at least one of the throws must be equal to 4. Number of ways when three blocks are filled from {1, 2, 3, 4} = 43 3 5! 1 6! 2 Number of ways when three blocks are filled from {1, 2, 3} = 33 Required number of ways = 43 33 77. Hindi Number of ways of placing 3 black balls at 10 places is 10 C3. Number of ways in which two black balls not together is equal to the number of ways of choosing 3 places marked with X out of eight places. W W W W W W W W Probability, P = Hindi 43 33 6 3 = 37 216 rhu Q sd esamPp la[;k 4 gSA t ksfd ,d Q sd esad e lsd e la[;k 4 d scjkcj gSA ep;ksad h la[;k t cfd rhu fMCCksHkjs j[kusd s ep; 10C3 gSA nksd ky h xsan d slkFk lkFk ugh gksusd s ep; 8 LFkkuksaesalsX y xs fpUg ij esals3 LFkku pquusd s ep; d scjkcj gSA 10 LFkkuks aij 3 d ky h xsan t krsgSA {1, 2, 3, 4} = 43 W W W W W W W W ep;ksad h la[;k t c rhu fMCcsHkjst krsgS {1, 2, 3} = 33 78. When A and B are mutually exclusive P(A B) = 0 t c A vkSj B ijLij vkiot hZr P(A B) = 0 iz kf;d rk, P = ep;ksad h la[;k = 43 33 P(A B) = P(A) + P(B) 0.8 = 0.5 + P or P = 0.3 Now vc, P(A B) = P(A) + P(B) P(A B) 43 33 6 3 79. 80. q =2 p 3 A P(A B) 0.6 P = = = B 4 P(B) 0.8 360 2 = = 6 9 10 10 5 Probability iz kf;d rk= 83. 4 odd digits 1, 3, 3, 1 can be arranged in 4 odd places in 4! = 6 ways that, ever digits 2, 4, 2 can be arranged in 2! 2! ( maximum value of P(A B) = P(A) = 0.6) the three even places in ( P(A B) d k vf/kd re = 6 3 = 18 Let card) eku= P(A) = 0.6) 37 216 82. 0.8 = 0.5 + q (0.5)q q = 0.6 = 3! = 3 ways hence, required ways 2! event A : Card is of heart but not king (12 Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PLC024029 SOLJP/JFMONT4120117-9 Hindi 4 fo"ke vad d ks1, 3, 3, 1 4 fo"ke LFkkuksaij O;ofLFkr d jus d srjhd s 4! = 6 t cfd le va d 2, 4, 2 d ksrhu le 2! 2! LFkku ij 3! = 3 rjhd kslsO;ofLFkr fd ;k t k ld rk gS 2! 86. If x denotes the number of times he can take unit step and y denotes the number of times he can take 2 steps, then 3x + 2y = 7. We must have x = 1, 3, 5 If x = 1, the steps will be 1 2 2 2. Number of ways = vr% vHkh"V ep; If x = 3, the steps will be 1 1 1 2 2 = 6 3 = 18 ways = 84. n(A) = Number divisible by 60 = (60, 120, .......960) = 16 n(B) = Number divisible by 24 = (24, 48, ....... 984) = 41 n(A B) = Divisible by both = 120 + 240 + ........ + 960 = 8 Let ekuk n(A) = 60 lsfoHkkft r Hindi la[;k 4! =4 3! ;fn x = 3 d ne gksxs1 1 1 2 2 la[;k ep; = = (24, 48, ....... 984) = 41 n(A B) = nks uksalsfoHkkft r = 120 + 240 + ........ + 960 = 8 n(A B') = n(A) n(A B) = 16 8 = 8 85. x x ways = when two consecutive digits are 0 d qy ep; = N = 21. when two consecutive digits are 00 = 9 x x x x 2n 2 = 510 2n = 512 n=9 88. Number of ways to select atleast two persons Total ways Number selection 1 person selected 212 1 12 x = 12 Hindi d e lsd e nksO;fDr d spquusd s ep; d q y rjhd s vHkh"V p;u 1 O;fDr pquk t krk gSA t c nks ekxr vad 11, 22 gS= 9 9 = 81 0 0 t c nks ekxr vad 00 gS= 9 x x t c nks ekxr vad 11, 22, 33, ...... N =7 3 87. when two consecutive digits are 11, 22, 33, ...... = 9 8 = 72 Total number of numbers are N = 162 Hindi 5! = 10 2! 3! If x = 5, steps will be 1 1 1 1 1 2 ways = 6C1 = 6 If x = 7, steps will be 1 1 1 1 1 1 7C0 = 1 11, 22 etc = 9 9 = 81 0 N =7 3 ;fn x ep;ksad h la[;k d ksO;Dr d jrk gSft lesaog ,d d ne py rk gSA rFkk y mu ep;kad d h la[;k gSft lesaog 2 d en py rk gSrc 3x + 2y = 7. ;gk x = 1, 3, 5 ;fn x = 1 d ne gksxsa1 2 2 2. = (60, 120, .......960) = 16 n(B) = 24 lsfoHkkft r 5! = 10 2! 3! If x = 5, steps will be 1 1 1 1 1 2 ways = 6C1 = 6 If x = 7, steps will be 1 1 1 1 1 1 7C0 = 1 hence, total number of ways = N = 21. n(A B') = n(A) n(A B) = 16 8 = 8 Hindi 4! =4 3! 212 1 12 x = 12 89. Deheradun Bombay Goa = 9 8 = 72 gS A la[;kvksad h la[;k N = 162 Clearly Li"Vr;k%, 3 4 = 12 ways rjhd s 90. Let r number of books of algebra and (20 r) of calculus Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PLC024029 SOLJP/JFMONT4120117-10 Number of selections = rC5 20 rC5 which has maximum when r = 10 N = 10 ekuk cht xf.kr iqLrd ksla[;k r vkSj d y u d h iqLrd ksd h la[;k (20 r) p;uksd h la[;k = rC5 20 rC5 vf/kd re gksxk t c r = 10 N = 10 Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PLC024029 SOLJP/JFMONT4120117-11 MAIN PATTERN ONLINE TEST-4 (MONT-4) XI TARGET : JEE (MAIN+ADVNACED) 2017 DATE : 21-03-2017 COURSE : JA**, 01JA, 01EA, 01JB, 02JB, 01EB, 01JR, 05JR, 01ER, 05ER ANSWER KEY CODE-0 PART-I PHYSICS 1. (A) 2. (D) 3. (A) 4. (D) 5. (B) 6. (C) 7. (D) 8. (C) 9. (A) 10. (B) 11. (B) 12. (B) 13. (B) 14. (C) 15. (D) 16. (3) 17. (5) 18. (3) 19. (5) 20. (2) 21. (1) 22. (3) 23. (9) 24. (6) 25. (3) 26. (4) 27. (5) 28. (3) 29. (1) 30. (4) PART-II CHEMISTRY 31. (C) 32. (A) 33. (D) 34. (B) 35. (D) 36. (C) 37. (A) 38. (A) 39. (C) 40. (D) 41. (C) 42. (D) 43. (C) 44. (D) 45. (C) 46. 1 47. 3 48. 6 49. 9 50. 5 51. 3 52. 4 53. 4 54. 0 55. 4 56. 3 57. 4 58. 4 59. 2 60. 3 PART- III MATHEMATICS 61. (C) 62. (A) 63. (C) 64. (B) 65. (B) 66. (A) 67. (B) 68. (B) 69. (A) 70. (A) 71. (B) 72. (A) 73. (D) 74. (B) 75. (A) 76. (2) 77. (8) 78. (2) 79. (6) 80. (4) 81. (5) 82. (6) 83. (9) 84. (8) 85. (9) 86. (7) 87. (9) 88. (6) 89. (8) 90. (5) Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PLC024029 SOLJP/JFMONT4120117-12 P25-16 MAIN PATTERN ONLINE TEST-4 (MONT-4) XI Code 0 TARGET : JEE (MAIN+ADVANCED) 2017 DATE : 21-03-2017 | COURSE : JA**, 01JA, 01EA, 01JB, 02JB, 01EB, 01JR, 05JR, 01ER, 05ER 11. Since it is not possible to erase and correct pen filled 11. pa wfd isu lsHkjsx, xksy sfeVkuk vkSj lq/kkjuk laHko ughagS bubble, you are advised to be extremely careful while blfy, vki lko/kkuh iw oZd viusm kj d sxksy ksad ksHkjsaA darken the bubble corresponding to your answer. 12. Neither try to erase / rub / scratch the option nor 12. make the Cross (X) mark on the option once filled. fod Yi d ksu feVk,a@ u L s p d jsavkSj u gh xyr (X) fpUg d ksHkjsaA ORS d ksd kVsu gh Q kMsu gh xUnk ughad jsarFkk Do not scribble, smudge, cut, tear, or wrinkle the d ksbZHkh fu'kku ;k lQ snh ORS ij ughayxk;sA ORS. Do not put any stray marks or whitener anywhere on the ORS. 13. If there is any discrepancy between the written data 13. and the bubbled data in your ORS, the bubbled data fd , vka d Mksaesafojks/kkHkkl gS] rksxksy sfd , vkad Mksad ksgh will be taken as final. C. ;fn ORS esafd lh izd kj d h fy[ksx, vkad MksarFkk xksy s lgh ekuk t kosxkA Question Paper Format C. This question paper consists of three part. Each part iz'u&i=k d k iz k: i bl iz'u&i=k esarhu Hkkx gSaA izR;sd Hkkx esanks[kaM gSaA consists are two section. 14. Section 1 contains 15 multiple choice questions. 14. [ka M 1 esa15 cgqfod Yi iz'u gaSA gj iz'u esapkj fod Yi (A), Each question has Four choices (A), (B), (C) and (D) (B), (C) vkS j (D) gSaft uesalsd soy ,d l gh gSA out of which only ONE is correct. 15. Section 2 contains 15 questions. The answer to 15. [ka M 2 esa15 iz'u gSaA izR;sd iz'u d k m kj 0 l s9 rd nksuksa each question is a single-digit integer, ranging from 'kkfey d schp ,d ,d y v ad h; iw.kkZad gSA 0 to 9 (both inclusive). D. Marking Scheme 16. For each question in Section 1, you will be awarded 4 marks if you darken the bubble corresponding to only the correct answer and zero mark if no bubbles are darkened. In all other cases, minus one ( 1) mark will be awarded. 17. For each question in Section 2, you will be awarded 4 marks if you darken the bubble corresponding to only the correct answer and zero mark if no bubbles are darkened. In all other cases, zero will be awarded. Name of the Candidate (ijh{kkFkhZd k uke) : va d u ;ks t uk D. 16. [ka M 1 es agj iz 'u es ad s oy lgh m kj okyscq y cq y s(BUBBLES) d ksd kyk d jusij 4 va d vkS j d ks bZHkh cq y cw y k d kyk ughad jusij (0) va d iz nku fd , t k;s axs A vU; lHkh fLFkfr;ks aes a_ .kkRed ,d ( 1) va d iz nku fd ;k t k;s xkA 17. [kaM 2 esagj iz'u esad soy lgh m kj oky scqy cqy s(BUBBLES) d ksd ky k d jusij 4 v ad vkSj d ksbZHkh cqy cwy k d ky k ughad jus ij (0) vad iznku fd , t k;saxsA vU; lHkh fLFkfr;ks aes a(0) vad iz nku fd ;k t k;sxkA Roll Number (jks y uEcj) : I have read all the instructions and shall abide by them I have verified all the information filled by the candidate. eSauslHkh funsZ'kksad ksi<+fy ;k gSvkS j eSamud k vo'; iky u d : xk@d : xhA ijh{kkFkhZ}kjk Hkjh xbZlkjh t kud kjh d kseSusa t k p fy ;k gSA ...................................... Signature of the Candidate ...................................... Signature of the Invigilator ijh{kkFkhZd sgLrk{kj ijh{kd d sgLrk{kj Resonance Eduventures Ltd. P25-16 CORPORATE OFFICE : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005 Ph.No. : +91-744-3012222, 6635555 | Toll Free : 1800 258 5555 Reg. Office : J-2, Jawahar Nagar, Main Road, Kota (Raj.)-324005 | Ph. No.: +91-744-3192222 | FAX No. : +91-022-39167222 Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in | CIN: U80302RJ2007PLC024029

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