® P1-21

MAIN PATTERN CUMULATIVE TEST-4 (MCT-4)

TARGET : JEE (MAIN + ADVANCED)
CODE

1
COURSE : VIKAAS (JA) | BATCH : 01JA, IA, JAZA
Time (le;) : 3 Hours (?k.Vs) Maximum Marks (vf/kdre vad) : 300

Ñi;k bu funsZ'kksa dks /;ku ls i<+saA vkidks 5 feuV fo'ks"k :i ls bl dke ds fy, fn;s x;s gSaA
(Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose).

INSTRUCTIONS TO CANDIDATES

Question paper has three (03) parts: Physics, Chemistry ç'u&i=k eas rhu (03) Hkkx gS % (HkkSfrdh ] jlk;u foKku ,oa
and Mathematics.
xf.kr)
Each part has a total 30 questions divided into two (02) çR;sd Hkkx esa dqy 30 iz'u gS tks nks (02) [kaMks esa foHkkftr gS
sections (Section-1 and Section-2). ¼[kaM 1 vkSj [kaM 2)
Total number of questions in Question paper are 90 and iz'u&i=k esa ç'uksa dh dqy la[;k % 90 ,oa vf/kdre vad % rhu
Maximum Marks are Three Hundred only (300). lkS (300) gSaA
ç'uksa ds çdkj vkSj ewY;kadu ;kstuk,¡ (Type of Questions and Marking Schemes)
[kaM–1 ¼vf/kdre vad : 80½ | SECTION-1 (Maximum Marks : 80)
This section contains TWENTY (20) questions bl [kaM esa chl (20) iz'u gSaA
Each question has FOUR options (1), (2), (3) and (4) izR;sd iz'u esa pkj fodYi (1), (2), (3) rFkk (4) gSaA bu pkj fodYiksa
ONLY ONE of these four option is correct) esa ls ds oy ,d fodYi lgh gSaA
 Marking scheme:  vadu ;kstuk
 Full Marks : +4 If the corresponding to the
answer is darkened
 iw.kZ vad % +4 ;fn flQZ lgh fodYi gh pquk x;k gSA
 Zero Marks : 0 If none of the options is chosen  'kwU; vad % 0 ;fn fdlh Hkh fodYi dks ugha pquk x;k gS
(i.e. the question is unanswered). ¼vFkkZr~ iz'u vuqÙkfjr gS½A
 Negative Marks : –1 In all other cases.  _.k vad % –1 vU; lHkh ifjfLFkfr;ksa esaA

[kaM–2 ¼vf/kdre vad : 20½ | SECTION – 2 : (Maximum Marks : 20)

This section contains TEN (10) questions. The answer to bl [kaM esa nl (10) iz'u gSA izR;sd iz'u dk mÙkj la[;kRed eku
each question is NUMERICAL VALUE with two digit integer (NUMERICAL VALUE) gSa]tks f}&vadh; iw.kkZad rFkk n'keyo
and decimal upto two digit.
f)&vadu eas gSA
If the numerical value has more than two decimal places ;fn laa[;kRed eku esa nks ls vf/kd n’'keyo LFkku gS] rks la[;kRed eku dks
truncate/round-off the value to TWO decimal placed. n'keyo ds f)&LFkku rd VªadsV@jkmaM vkWQ (truncate/round-off) djsaA
There are 10 Questions & you have attempt any 5
Questions. If a student attempts more than 5 questions, bl [ka M esa 10 iz'u gaS ftuesa ls vkidks dsoy fdUgh 5 iz'uksa dk
then only first 5 questions which he has attempted will mÙkj nsuk gS ;fn vki 5 ls vf/kd iz'uksa dk mÙkj nsrs gS] rks mÙkj
be checked. fn;s x;s izFke 5 iz'uksa dh gh tk¡p dh tk;sxhA
 Marking scheme  vadu ;kstuk
 Full Marks : +4 If the corresponding to the  iw.kZ vad % +4 ;fn flQZ lgh fodYi gh pquk x;k gSA
answer is darkened
 Zero Marks : 0 In all other cases  'kwU; vad % 0 vU; lHkh ifjfLFkfr;ksa esaA
ijh{kkFkhZ dk uke : ………………………………………………………………jksy uEcj : ………………………..…………………
eSaus lHkh funsZ'kksa dks i<+ fy;k gS vkSj eSa mudk eSusa ijh{kkFkhZ dk ifjp;] uke vkS j jksy uEcj
vo'; ikyu d:¡xk@d:¡xhA dks iwjh rjg tk¡p fy;k gSA
------------------------ ------------------------
ijh{kkFkhZ ds gLrk{kj ijh{kd ds gLrk{kj
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 vkWfIVdy fjLikal 'khV (ORS) Hkjus ds funsZ’'k (INSTRUCTIONS FOR OPTICAL RESPONSE SHEET (ORS)
Darken the appropriate bubbles on the original by applying Åijh ewy i`"B ds vuq:i cqycq yksa (BUBBLES) dks i;kZIr ncko
sufficient pressure. Mkydj dkyk djsaA
The original is machine-gradable and will be collected by the ewy i`"B e'khu&tk¡ p gS rFkk ;g ijh{kk ds lekiu ij fujh{kd ds
invigilator at the end of the examination. }kjk ,d=k dj fy;k tk;sxkA
Do not tamper with or mutilate the ORS. vks-vkj-,l- dks gsj&Qsj@fod`fr u djsaA
Write your name, roll number and the name of  the viuk uke] jksy ua- vkSj ijh{kk dsanz dk uke ewy i`"B esa fn, x,
examination centre and sign with pen in the space provided [kkuksa esa dye ls Hkjsa vkSj vius gLRkk{kj djsaA buesa ls dksbZ Hkh
for this purpose on the original. Do not write any of these
details anywhere else. Darken the appropriate bubble under tkudkjh dgha vkSj u fy[ksaA jksy uEcj ds gj vad ds uhps vuq:i
each digit of your roll number. cq ycqys dks dkyk djsaA
ORS ij cqycqyksa dks dkyk djus dh fof/k % (DARKENING THE BUBBLES ON THE ORS) :
Use a BLACK BALL POINT to darken the bubbles in the
Åijh ewy i`"B ds cqycq yksa dks dkys ckWy ikbUV dye ls dkyk djsaA
upper sheet.
Darken the bubble COMPLETELY. cq ycqys dks iw.kZ :i ls dkyk djsaA
Darken the bubble ONLY if you are sure of the answer. cq ycqyksa dks rHkh dkyk djsa tc vkidk mÙkj fuf'pr gksA
cqycqyksa dks dkyk djus dk mi;qDr rjhdk ;gk¡ n'kkZ;k x;k gS %
The correct way of darkening a bubble is as shown here :

There is NO way to erase or "un-darkened bubble. dkys fd;s gq;s cqycq ys dks feVkus dk dksbZ rjhdk ugha gSA
The marking scheme given at the beginning of each section
 gj [k.M ds izkjEHk esa nh x;h vadu ;kstuk esa dkys fd;s x;s rFkk dkys
gives details of how darkened and not darkened bubbles are
evaluated.
u fd;s x;s cqycqyksa dks ewY;kafdr djus dk rjhdk fn;k x;k gSA.
For example, if answer ‘SINGLE DIGIT’ integer type below : mnkgj.k ds fy, , ;fn mÙkj ‘,dy vadh;’ iw.kkZad gS rc :
0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9
For example, if answer ‘DOUBLEDIGIT’ integer type below : mnkgj.k ds fy, , ;fn mÙkj ‘f)&vadh;’ iw.kkZad gS rc :
0 0 0 0
1 1 1 1
2 2 2 2
3 3 3 3
4 4 4 4
5 5 5 5
6 6 6 6
7 7 7 7
8 8 8 8
9 9 9 9
FOR DECIMAL TYPE QUESTIONS OMR LOOKS LIKE : n'keyo iw .kkZad@la[;kRed vadksa ds fy, ORS fuEu çdkj gS :
COLUMN COLUMN
1 2 . 3 4 1 2 . 3 4
0 0 0 0 0 0 0 0
1 1 1 1 1 1 1 1
2 2 2 2 2 2 2 2
3 3 3 3 3 3 3 3
4 4 4 4 4 4 4 4
5 5 5 5 5 5 5 5
6 6 6 6 6 6 6 6
7 7 7 7 7 7 7 7
8 8 8 8 8 8 8 8
9 9 9 9 9 9 9 9
st
If answer is 3.7, then fill 3 in either 1st or 2nd column and 7 in ;fn mÙkj 3.7 gS rc 3 dks 1 ;k 2nd dkWye esa Hkjsa rFkk 7 dks 3rd ;k
rd th
3 or 4 column. 4th dkWye esa HkjsaA

If answer is 3.07 then fill 3 in 1st or 2nd column ‘0’ in 3rd column ;fn mÙkj 3.07 gS rks 3 dks 1st dkWye ;k 2nd dkWye esa Hkjsa rFkk ‘0’
and 7 in 4th column. dks 3rd dkWye esa rFkk 7 dks 4th dkWye esa HkjsaA
If answer is, 23 then fill 2 & 3 in 1st and 2nd column ;fn mÙkj 23 gS rc 2 dks 1st dkWye esa ] 3 dks 2nd dkWye esa tcfd
respectively, while you can either leave column 3 & 4 or fill rd
3 vkSj 4th dkWye dks [kkyh NksM+ nsa ;k ‘'kwU; Hkj nsaA
‘0’ in either of them.

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 ®

TEST TYPE : MAIN PATTERN CUMULATIVE TEST-4 (MCT-4)

TARGET : JEE (MAIN+ADVANCED) 2021
COURSE NAME : VIKAAS (01JA)

PART-A (Hkkx– A)

SECTION – 1 : (Maximum Marks : 80)

 This section contains TWENTY (20) questions.
 Each question has FOUR options (1), (2), (3) and (4) ONLY ONE of these four option is correct
 Marking scheme :
 Full Marks : +4 If ONLY the correct option is chosen.
 Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered).
 Negative Marks : –1 In all other cases

[kaM 1 : (vf/kdre vad : 80)

 bl [kaM esa chl (20) iz'u gSaA
 izR;sd iz'u esa pkj fodYi (1), (2), (3) rFkk (4) gSaA bu pkjksa fodYiksa esa ls dsoy ,d fodYi lgh gSaA
 vadu ;kstuk :
 iw.kZ vad % +4 ;fn flQZ lgh fodYi gh pquk x;k gSA
 'kwU; vad % 0 ;fn dksbZ Hkh fodYi ugha pquk x;k gS ¼vFkkZr~ iz'u vuqÙkfjr gS½A
 _.k vad % –1 vU; lHkh ifjfLFkfr;ksa esaA

1. The bob of a simple pendulum executes simple harmonic motion in water with a period t, while the
period of oscillation of the bob is t0 in air. Neglecting frictional force of water and given that the density
of the bob is (4/3) × 1000 kg/m3. What relationship between t and t0 is true?
,d ljy vkorhZ nksyd ty esa vkorZ dky t ds lkFk ljy vkorZ xfr dj jgk gS] tcfd gok esa bldk vkorZ dky
t0 gSA ty dk ?k"kZ.k ux.; ekfu;s rFkk ;g fn;k x;k gS fd nksyd dk ?kuRo (4/3) × 1000 kg/m 3 gS rks t rFkk t0 esa
lgh lEcU/k gksxkA
(1) t = t0 (2) t = t0/2 (3*) t = 2t0 (4) t = 4t0
Sol. The time period of simple pendulum in air
ok;q esa ljy yksyd dk vkorZ dky
 
T = t0 = 2    ........... (i)
 g 

, being the length of simple pendulum.

, ljy yksyd dh yEckbZ gS
In water, effective weight of bob
ty esa xksyd dk izHkkfo Hkkj
w’ = weight of bob in air – upthrust
w’ = ok;q esa xksyd dk Hkkj – mRIykou cy
  Vgeff = mg – m’g
=Vg – ‘ Vg = (–‘ )Vg
where  = density of bob,
;gk  = xksyd dk ?kuRo
' = density of water
' = ty dk ?kuRo

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   – '  '
  geff =   g = 1 –  g
     

 ' 
  t = 2 1 –  g  (ii)
   

 
 1 
t 1  
Thus vr% = =
t0   '   1000 
1–
1 –    4 / 3   1 0 0 0 
    

= 2 t = 2 t0

2. A man can swim certain distance in still water up and down in time t1. If he swims to some distance
down stream the river and returns back to same point he takes time t2. Then :
(1) t1 = t2 (2*) t1 < t2 (3) t1 < t2 (4) t1 & t2 can't be compared
,d O;fDr fLFkj ikuh esa rSjrs gq, fuf'pr nwjh rd tkdj okil vkus esa t1 le; ysrk gSA ;fn og ,d unh esa ikuh
ds cgko dh fn'kk esa leku nwjh rd tkdj okil izkjfEHkd fcUnq rd vkus esa t2 le; ysrk gks rksA
(1) t1 = t2 (2*) t1 < t2 (3) t1 < t2
(4) t1 rFkk t2 dh rqyuk ugh dh tk ldrh gSA
2d d d 2vd 2d
Sol. t1 = t2 =  = = 2
v v u v u v
2
u
2
2 u
v 
v
2d
t2 =
 u 
2

v 1  
2
 v 

t2 > t1.

3. A cart of mass M has a block of mass m attached to it as shown in the figure. Co-efficient of friction
between the block and cart is . What is the minimum acceleration of the cart so that the block m does
not fall ?
nzO;eku M dh ,d xkM+h ij m nzO;eku dk ,d xqVdk fp=kkuqlkj tqM+k gSA xkM+h o xqVds ds e/; ?k"kZ.k xq.kkad  gSA
xkM+h dks fdrus U;wure Roj.k ls pyk;k tk;s rkfd xqVdk fxjs ughaA

(1)  g (2) /g (3*) g/ (4) none dksbZ ugha

Sol.

Solving from the frame of cart, we get

xkM+h ds funsZ'k ra=k ls gy djus ij gesa izkIr gksrk gS

N = ma, mg = N
g
 mg = ma  a=


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4. The system is in equilibrium and at rest. Now mass m is removed from m . Find the time period and
1 2
amplitude of resultant motion. Spring constant is K.
fudk; lkE;koLFkk esa ,oa fLFkj gSA vc nzO;eku m1 dks m2 ls gVk ysrs gSA ifj.kkeh xfr dk vkorZdky ,oa vk;ke
Kkr djksA fLçax fu;rkad K gSA
/ // // // // // // // // // / // // //

m1
m2

m2 m 1g m2 m 2g
(1*) T = 2 ;A= (2) T = 2 ;A=
K K K K

m1 m 2g m1 m 2g
(3) T = 2 ;A= (4) T =  ;A=
K K K K
Sol. Initial extension in the spring
fLçax esa izkjfEHkd f[kapko
(m 1  m 2 ) g
x=
K
m 2g
Now, if we remove m , equilibrium position(E.P.) of m will be below natural length of spring.
1 2 K

m 2g
vc , ;fn m1 dks gVk ysrs gS rks m2 dh lkE;koLFkk fLçax dh lkekU; yEckbZ ls uhps gksxhA
K

/ // // // // // // / // // // // // // / // // // // // // / // // // // // //

N .L
m 2g
(m 1+ m 2)g
K
K E .P
m 1g
K

// // // // // // // // // // // // // // // // // // // // // // // // // //

N .L
m 2g
(m 1+ m 2)g
K
K l kE; ko L Fkk
m 1g
K

At the initial position, since velocity is zero i.e. it is the extreme position.

çkjfEHkd fLFkfr ij pwafd osx 'kwU; gS blfy, ;g lhekUr fLFkfr ij gSA

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5. Two men of equal masses stand at opposite ends of the diameter of a turntable disc of a certain mass,
moving with some angular velocity. The two men start moving towards the center of the turntable at
equal rates. Then choose the correct option:
dqN dks.kh; osx ls ?kweus okyh ,d ?kw.khZ pdrh ftldk dqN fuf'pr nzO;eku gS] ds O;kl ds foijhr fljksa ij leku
nzO;eku ds nks O;fDr [kM+s gSA nksuksa O;fDr leku nj ls ?kw.kZu pdrh ds dsUnz dh vksj pyuk izkjEHk djrs gS rks lR;
lgh fodYi dk p;u dhft,A
(1*) kinetic energy of the system will increase.
?kw.kZu dh xfrt ÅtkZ c<+rh gSA
(2) kinetic energy of system will decrease.
?kw.kZu dh xfrt ÅtkZ ?kVrh gSA
(3) Angular momentum of the system will increase.
fudk; dk dks.kh; laosx c<+rk gSA
(4) Angular momentum of the system will decrease.
fudk; dk dks.kh; laosx ?kVrk gSA
Sol. 11 = 22
Since, men move towards middle of turn table  decreases hence 2 increases.
pwafd, O;fDr dsUnz dh vksj xfr djrs gS blfy,  ?kVsxk QyLo:i 2 c<sxkA
2
L
 KE = ; L = constant and  decreases so KE increases vr% xfrt ÅtkZ c<sxhA
2

6. A uniform disc of mass m and radius R is resting an a smooth horizontal surface and it is free to rotate
about its own axis which is vertical. A small particle of mass m collides its periphery with a velocity of V
in tangential direction. If the collision is perfectly in-elastic and the particle sticks to the periphery, the
angular velocity of the system just after the collision will be :
m nzO;eku vkSj R f=kT;k dh ,d le:i pdrh fpduh {kSfrt lrg ij j[kh gS vkSj vius Loa; ds Å/okZ/kj v{k ds
lkis{k ?kweus ds fy, Lora=k gSA m nzO;eku dk ,d d.k V pky ls pdrh dh ifj/kh ls Li'kZ js[kh; fn'kk esa Vdjkrk
gSA ;fn VDdj iw.kZ vizR;kLFk gS vkSj VDdj ds ckn d.k pdrh dh ifj/kh ls fpid tk, rks VDdj ds Bhd ckn
fudk; dk dks.kh; osx gksxk%

V 2V 3V V
(1) (2*) (3) (4)
3R 3R 2R R

Sol. Applying angular momentum conservation about the rotating axis

?kw.kZu v{k ds ifjr% dks.kh; losxa lja{k.k ls

Li = Lf

 mR2 2

O + mvR =   mR f
 2 
 

2V
 
3R

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7. A uniform metre stick is held vertically with one end on the floor and is allowed to fall. The speed of the
other end when it hits the floor assuming that the end at the floor does not slip :
,d ehVj dh ,d le:i NM+ Å/okZ/kj j[kh tkrh gSA bldk ,d fljk Q'kZ ij gS vkSj ;g fxjus nh tkrh gS ;g
ekurs gq,s fd Q'kZ okyk fljk fQlyrk ugha gSA tc ;g Q'kZ ij Vdjkrh gS rc nwljs fljs dh pky Kkr djks A
(1) 4g (2*) 3g (3) 5g (4) g
Sol.

using energy conservation ÅtkZ lja{k.k ls

1
mg = 2
2 2
2
1 m
mg = . 2
2 2 3
3g
 = 1m =

VA =  = 3g =  3g 
8. A ball ‘A’ of mass M collides elastically with another identical ball ‘B’ at rest as shown in figure. Initially
velocity of ball ‘A’ is u m/s. After collision :
fp=kkuqlkj M nzO;eku dh ,d xsan 'A', leku xsan ‘B’ tksfd fLFkj gS] ls çR;kLFk VDdj djrh gSA ;fn çkjEHk esa ‘A’
dk osx u m/s gS rks VDdj ds ckn :

(1) speed of ball A is u cos. (2) speed of ball A is u

(3*) speed of ball A is u sin (4) None of these
(1) A dk osx u cos gksxkA (2) B dk osx u gksxkA
(3*) A dh pky usin gksxh (4) buesa ls dksbZ ugha
Sol.

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 vB = ucos
VA = usin

9. A particle of mass 1 kg is thrown vertically upwards with speed 100 m/s. After 5s it explodes into two
parts. One part of mass 400g comes back with speed 25 m/s, what is the speed of other part just after
explosion?
(1*) 100 m/s upwards (2) 600 m/s upwards (3) 100 m/s downward (4) 300 m/s upward
1 kg ds ,d fi.M dks 100 m/s ds osx ls Å/okZ/kj Åij dh vksj Qsadk tkrk gSA 5s i'pkr~ ;g nks [k.Mksa esa
foLQksfVr gks tkrk gSA 400g dk ,d fi.M 25 m/s dh pky ls okfil ykSVrk gS] foLQksV ds rqjUr ckn nwljs [k.M
dk osx gksxk ?
(1*) 100 m/s Åij dh vksj (2) 600 m/s Åij dh vksj (3) 100 m/s uhps dh vksj (4) 300 m/s Åij dh vksj
Sol. Velocity of particle after 5 s
v = u – gt
v = 100 – 10 × 5
= 100 – 50 = 50 m/s (upwards)
Conservation of linear momentum gives
Mv = m1v1 + m2v2 ......(i)
Taking upward direction positive, the velocity v1 will be negative.
 v1 = – 25 m/s, v = 50 m/s
Also M = 1 kg, m1 = 400 g = 0.4 kg
and m2 = (M – m1) = 1 – 0.4 = 0.6 kg
Thus, Eq. (i) becomes,
1 – 50 = 0.4 × (–25) + 0.6 v2
or 50 = – 10 + 0.65 v2
or 0.6 v2 = 60
60
or v2 = = 100 m/s
0 .6
As v2 is positive, therefore the other part will move upwards with a velocity 100 m/s.
Sol. 5 lsd.M ckn d.k dk osx]
v = u – gt
v = 100 – 10 × 5
= 100 – 50 = 50 eh@ls (Åij dh vksj)
laosx laj{k.k ds fl)kUr ls
Mv = m1v1 + m2v2 ......(i)
Åij dh fn'kk /kukRed ysus ij] osx v1 _.kkRed gksxk
 v1 = – 25 eh@ls, v = 50 m/s eh@ls
rFkk m = 1 kg, m1 = 400 g = 0.4 kg
m2 = (m – m1) = 1 – 0.4 = 0.6 kg
leh- (i) ls
1 × 50 = 0.4 × (–25) + 0.6 v2
or 50 = – 10 + 0.65 v2
or 0.6 v2 = 60
60
or v2 = = 100 eh@ls
0 .6
  v2 /kukRed gS vr% nwljs Hkkx dk osx 100 eh@ls gksxk rFkk bldh fn'kk Åij dh vksj gksxhA

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10. A thin rod having mass m and length 4 is free to rotate about horizontal axis passing through a point
distant  from one of its end, as shown in figure. It is released, from the horizontal position as shown :
m æO;eku o 4 yEckbZ dh ,d leku iryh NM+] blds ,d fljs ls  nwjh ls fp=kkuqlkj xqtjus okyh {kSfrt v{k ds
lkis{k ?kweus ds fy, Lora=k gSA bls n'kkZ;h xbZ {kSfrt fLFkfr ls NksM+k tkrk gS :
x
What will be acceleration of centre of mass at this instant
bl {k.k ij æO;eku dsUæ dk Roj.k D;k gksxk \
3g 2g 3g 2g
(1*) (2) (3) (4)
7 7 5 5
Sol. Torque equation
cyk?kw.kZ lehdj.k
Hinge = Hinge 
 m (4 )2 2

mg =   m  
 12 
 
3g
=
7
3g
Tangential acceleration =  =
7
3g
Li'kZ js[kh; Roj.k =  =
7
Radial acceleration = 2  = 0
f=kT;h; Roj.k = 2  = 0
3g
Ans.
7

11. When a person throws a meter stick it is found that the centre of the stick is moving with speed 10 m/s
and left end of stick with speed 20 m/s. Both points move vertically upwards at that moment. Then
angular speed of the stick is:
tc ,d vkneh us ,d ehVj NM+ dks QSadk rks ;g ns[kk x;k fd mlds dsUnz dh pky 10 m/s gS vkSj ckb± Nksj dh
pky 20 m/s gSA nksuksa fcUnq ml {k.k Åij dh vksj tk jgs Fks] rks NM+ dh dks.kh; pky gksxh :
(1*) 20 rad/ sec    (2) 10 rad/sec
(3) 30 rad/sec (4) none of these buesa ls dksbZ ugha
20  10
Sol. Angular velocity (dks.kh; osx) w = = 20 rad/sec.
0 .5

12. A particle moving on a circular path travels first one third part of circumference in 2 sec & next
n
one third part in 1 sec. Average angular velocity of the particle is (in rad/sec). Then value
9
of n.
o`Rrkdkj ekxZ ij xfr djrk gqvk ,d d.k ifjf/k dk izFke ,d frgkbZ Hkkx 2 sec esa rFkk vxyk ,d frgkbZ
n
Hkkx 1 sec esa r; djrk gSA d.k dk vkSlr dks.kh; osx gks (jsfM;u@lsd.M esa) rks n dk eku gksxkA
9
(1*) 4 (2) 0 (3) 3 (4) 2

to ta l  d qy 
Sol. arg = <> =
to ta l tim e d qy le;

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 2 / 3  2 / 3 4
= = rad/sec.
2 1 9

13. A mass M, attached to a horizontal spring, executes SHM with a amplitude A 1. When the mass M
passes through its mean position then a smaller mass m is placed over it and both of them move
 A1 
together with amplitude A2. The ratio of  
is :
 A2

,d {kSfrt dekuh ls c¡/kk ,d nzO;eku M vk;ke A1 ls ljy vkorZ xfr dj jgk gSA tc nzO;eku M viuh ek/;
voLFkk ls xqtj jgk gS] rc ,d NksVk nzO;eku m blds Åij j[k fn;k tkrk gS vkSj vc nksukas vk;ke A2 ls xfr
 A1 
djrs gSA  
dk vuqikr gS :
 A2
1/ 2 1/ 2
M M  m  M  M  m
(1) (2) (3)   (4*)  
M  m M M  m  M 
Ans. (4)
Sol. C.O.L.M.
js[kh; laosx laj{k.k ls (C.O.L.M.)
MVmax = (m + M)Vnew , Vmax = A1 1
M Vm ax
Vnew =
(m  M )
vc, Vnew = A2.2
M .A 1 K K
= A 2
(m  M ) M (m  M )
1/ 2
M A1 m  M
A2 = A1   
(m  M ) A 2
 M 

14. A parachutist drops freely from an aeroplane for 10 s before the parachute opens out. Then he
descends with a net retardation of 2.5 ms–2. If he bails out of the plane at a height of 2495 m and
g = 10 ms–2, his velocity on reaching the ground will be
(1) 2.5 ms–1 (2) 7.5 ms–1 (3*) 5 ms–1 (4) 10 ms–1
,d isjkW'kwV/kkjh] isjkW'kqV [kksyus ls 10 s iwoZ rd ok;q;ku ls LorU=krkiwoZd fxjrk gSA blds ckn og 2.5 ms–2 ds
ifj.kkeh eanu ls uhps vkrk gSA ;fn og 2495 m dh Åpk¡bZ ij isjkW'kwV dks [kksyrk gS rks mldk tehu ij igq¡pus ds
le; osx gksxk g = 10 ms–2 :
(1) 2.5 ms–1 (2) 7.5 ms–1 (3*) 5 ms–1 (4) 10 ms–1
Sol. Suppose the man drops at A, from A to B he is falling freely & than at B parachute opens out & he falls
with a retardation of 2.5 m/s2.
A 2
a 1 = – 1 0 m /s

B
2495 m

2
a 2 = 2 . 5 m /s
C
 AB = 1/2 × 10 × 102 = 500 m
 BC = AC – AB = 2495 – 500 = 1995 m.
Velocity at B,
VB = gt = 10 × 10 = 100 m/s 
Velocity at C,
2 2
VC = VB  2 a y = 100  2  2 .5  (  1 9 9 5 ) = 25 = 5 m/s.

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Sol. ekuk fd O;fDr A ij dwn tkrk gS rFkk xq:Ro ds v/khu LorU=k :i ls fxj jgk gSA tc ;g B ij igq¡p tkrk gS rc
isjk'kwV [kqyrk gS rFkk 2.5 m/s2 dk eUnu mRiUu gksrk gSA.
A 2
a 1 = – 1 0 m /s

B
2495 m

2
a 2 = 2 . 5 m /s
C
 AB = 1/2 × 10 × 102 = 500 m
 BC = AC – AB = 2495 – 500 = 1995 m.
B ij osx,
VB = gt = 10 × 10 = 100 m/s 
C ij osx,
2 2
VC = VB  2 a y = 100  2  2 .5  (  1 9 9 5 )

= 25 = 5 m/s  .

15. A rigid body rotates about a fixed axis with variable angular velocity equal to  – t, at the time t, where
, are constants. The angle through which it rotates before its stops ?
,d n`<+ fi.M fLFkj v{k ds lkis{k ?kwe jgk gS vkSj bldk dks.kh; osx le; ds lkFk  – t dh rjg ls ifjofrZr gks
jgk gS tgka , fLFkjkad gSaA :dus ls igys ;g fi.M fdruk dks.k ?kwesxk\
2 2 2 2 2
  –   –  ( –  ) 
(1*) (2) (3) (4)
2 2 2 2

16. For the equilibrium condition shown, the cords are strong enough to withstand a maximum tension
100 N. What is the largest value of W (in newton) that can be suspended :
çnf'kZr lkE;koLFkk dh fLFkfr esa Mksjh vf/kdre 100 N dk ruko lgu dj ldrh gSA W dk vf/kdre eku (U;wVu
esa) gksxk tks vkyfEcr fd;k tk ldrk gS :

53°

A
53°

W
(1) 100 N (2*) 35 N (3) 80N (4) 55 N
Sol.
100N

53°

53°
T
w

T sin 53° – 100 cos 53° = 0  T = 75 N

Also rFkk, 100 sin 53° – T cos 53° – W = 0

 W = 35 N

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17. STATEMENT–1 : For a particle performing SHM, its speed decreases as it goes away from the mean
position.
STATEMENT–2 : In SHM, the acceleration is always opposite to the velocity of the particle.
oDrO;–1 : ljy vkorZ xfr dj jgs ,d d.k dh pky de gksrh gS tc ;g ek/; fLFkfr ls nwj tkrk gSA
oDrO;–2 : ljy vkorZ xfr esa Roj.k lnSo d.k ds osx dh foijhr fn'kk esa gksrk gSA
(1) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
(2) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
(3*) Statement-1 is True, Statement-2 is False
(4) Statement-1 is False, Statement-2 is True.
(1) oDrO;-1 lR; gS, oDrO;-2 lR; gS; oDrO;-2 oDrO; -1 dk lgh Li"Vhdj.k gSA
(2) oDrO;-1 lR; gS, oDrO;-2 lR; gS ; oDrO;-2 oDrO;-1 dk lgh Li"Vhdj.k ugha gSA
(3*) oDrO; -1 lR; gS, oDrO;-2 vlR; gSA
(4) oDrO; -1 vlR; gS , oDrO;-2 lR; gSA
Sol. Acce is opposite to displacement therefore when it moves away from mean its velocity decreases.
Since velocity and acceleration are opposite in direction.
Roj.k foLFkkiu ds foifjr gS vr% tc d.k ek/; fLFkfr ls nwj tkrk gS] rks osx ?kVrk gS] vkSj osx rFkk Roj.k foifjr
fn'kk esa gksrs gSA

18. A metre stick swinging about its one end oscillates with frequency f 0. If the bottom half of the stick was
cut off, then its new oscillation frequency will be :
,d fljs ds lkis{k >wy jgh ,d ehVj NM+ f0 vko`fÙk ls nksyu djrh gSA vxj NM+ dk uhps dk vk/kk Hkkx dkV fn;k
tk, rks nksyuksa dh u;h vko`fÙk D;k gksxh &
(1) f0 (2*) 2 f0 (3) 2f0 (4) 2 2 f0
1 mg
Sol. f0 =
2 
where,  is distance between point of suspension and centre of mass of the body.
Thus, for the stick of length L and mass m :
L
m .g.
1 2 1 3g
f0 = =
2 2
(m L / 3 ) 2 2L

when bottom half of the stick is cut of

m L
.g.
1 2 4 1 3g
f 0’ = =  = 2 f0 Ans.
2 m (L / 2 )
2
2 L
2 3

1 mg
Sol. f0 =
2 

;gk¡,  fuyEcu fcUnq o oLrq ds nzO;eku dsUnz ds chp dh nqjh gSA

vr% nzO;eku m rFkk L yEckbZ dh NM+ ds fy,
L
m .g.
1 2 1 3g
f0 = =
2 2
(m L / 3 ) 2 2L

tc NM+ dks vk/kk dkV ysa rks

m L
.g.
1 2 4 1 3g
f 0’ = =  = 2 f0 Ans
2 m (L / 2 )
2
2 L
2 3

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19. The total work done on a particle is equal to the change in its kinetic energy.
,d d.k ij fd;k x;k dqy dk;Z bldh xfrt ÅtkZ esa ifjorZu ds cjkcj gksrk gS
(1*) always
(2) only if the forces acting on the body are conservative
(3) only if the forces acting on the body are gravitational
(4) only if the forces acting on the body are elastic.
(1*) lnSo
(2) ;fn d.k ij dsoy laj{kh cy yxrs gaSA
(3) ;fn d.k ij dsoy xq:Rokd"kZ.k cy yxrs gaSA
(4) ;fn d.k ij dsoy çR;kLFk cy yxrs gSaA
Sol. By work energy theorem
W all = K

20. Figure shows a plot of potential energy function U(x) = kx 2 where x = displacement and k = constant.
Identify the correct conservative force function F(x)
fn;k x;k fp=k fLFkfrt ÅtkZ dk Qyu U(x) = kx2 tgk¡ x = foLFkkiu rFkk k = fu;rkad gSA laj{kh cy F(x) dk lgh
Qyu gSA

(1*) (2) (3) (4)

dU
Sol. F= –
dx
 U(x) = kx2

F = –2kx

SECTION – 2 : (Maximum Marks : 20)

 This section contains TEN (10) questions. The answer to each question is NUMERICAL VALUE with two
digit integer and decimal upto two digit.
 If the numerical value has more than two decimal places truncate/round-off the value to TWO decimal
placed.
 There are 10 Questions & you have attempt any 5 Questions. If a student attempts more than 5
questions, then only first 5 questions which he has attempted will be checked.
 Marking scheme :
 Full Marks : +4 If ONLY the correct option is chosen.
 Zero Marks : 0 In all other cases
[kaM 2 ¼vf/kdre vad% 20)
 bl [kaM esa nl (10) iz'u gSA izR;sd iz'u dk mÙkj la[;kRed eku (NUMERICAL VALUE) gSa] tks f}&vadh; iw.kkZad
rFkk n'keyo f)&vadu eas gSA
 ;fn la[;kRed eku esa nks ls vf/kd n’'keyo LFkku gS ] rks la[;kRed eku dks n'keyo ds nks LFkkuksa rd VªadsV@jkmaM
vkWQ (truncate/round-off) djsaA
 bl [kaM esa 10 iz'u gaS ftuesa ls vkidks dsoy fdUgh 5 iz'uksa dk mÙkj nsuk gS ;fn vki 5 ls vf/kd iz'uksa dk mÙkj nsrs gS] rks mÙkj
fn;s x;s izFke 5 iz'uksa dh gh tk¡p dh tk;sxhA
 vadu ;kstuk :

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  iw.kZ vad % +4 ;fn flQZ lgh fodYi gh pquk x;k gSA
 'kwU; vad % 0 vU; lHkh ifjfLFkfr;ksa esaA

21. Block A of mass m after sliding from an inclined plane, strikes elastically another block B of same mass
at rest. Find the minimum height H so that ball B just completes the circular motion (R = 5m).
m nzO;eku dk ,d CykWd ur ry ls fQlydj] fojkekoLFkk esa fLFkr leku nzO;eku ds nwljs CykWd B ls izR;kLFk
VDdj djrk gSA U;wure Å¡pkbZ H dk eku Kkr djks] ftlls fd CykWd B, o`Ùkh; xfr dks iwjk dj ldsA(R = 5m)

Ans. H = 12.50 m

Sol.

For the just completing the circular motion, minimum velocity at bottom in
vB = 5 g R
Energy conservation b/w point A and B
1
MgH + 0 = 0 + mvB2
2
1
MgH = m (5gR)
2
5R
H=
2

gy
o`Ùkh; xfr dks iwjk djus ds fy;s ry ij U;wure osx
vB = 5gR

fcUnq A rFkk B ds e/; ÅtkZ laj{k.k ls

1
MgH + 0 = 0 + mvB2
2
1
MgH = m (5gR)
2
5R
H=
2

22. A force F = (10 + 0.50 x) acts on a particle in the xdirection. Find the work done by this force during a
displacement from x = 0 to x = 2.0 m.
,d d.k ij xfn'kk esa ,d cy F = (10 + 0.50 x) yxrk gSA bl cy ds }kjk x = 0 ls x = 2.0 m rd foLFkkiu esa
fd;k x;k dk;Z Kkr djksA
Ans. 21.00
2

Sol. work done fd;k x;k dk;Z =  f .d x

0

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 2
d x x
23. If a simple harmonic motion is represented by 2
+ = 0, Find out its time period :
dt 100
2
d x x
;fn fdlh ljy vkorZ xfr dks 2
+ =0 ds }kjk fu:fir fd;k tkrk gS] rks bldk vkorZdky Kkr djksA :
dt 100
Ans. 62.80 sec
2
d x x
Sol. 2
= ........ (i)
dt 100
2
d x
We know ge tkurs gS fd a = 2
= – 2x ........ (ii)
dt
From Eq. (i) and (ii), we have
lehdj.k (i) o (ii) ls]
1
2 = 
100
1
=
10
2 1
or ;k = T = 20= 62.8
T 10

24. A particle moves along a straight line. A force acts on the particle which produces a constant power.
It starts with initial velocity 3 m/s and after moving a distance 252 m its velocity is 6 m/s. Find the time
taken (approximately).
,d d.k ljy js[kk ij xfr djrk gSA d.k ij ,d cy yxrk gS] tks fu;r 'kfDr iznku djrk gSA ;g 3 m/s ds
izkjfEHkd osx ls xfr vkjaHk djrk gS rFkk 252 m nwjh pyus ds i'pkr~ bldk osx 6 m/s gks tkrk gSA bl nkSjku
fy;k x;k yxHkx le; Kkr djksA
Ans. 96.00 s
1
Sol. ma .252 = m (36–9)
2
2  252
where a = 2
t
 t = 96 s

25. The acceleration of a certain simple harmonic motion is given by a(in m/s 2) = 50 sin 5t. Find the
amplitude (in meter) of simple harmonic motion.
fdlh ljy vkorZ xfr dk Roj.k a (in m/s2) = 50 sin 5t }kjk O;Dr fd;k tkrk gSA bl ljy vkorZ xfr dk vk;ke
(ehVj ek=kd esa) Kkr dfj,A
Ans. 02.00
Sol. 2A = 50
52A = 50
25 × A = 50
A = 2m

26. A uniform block A of mass 25 kg and length 6m is hinged at C and is supported by a small block B as
shown in the Figure. A constant force 'F' of magnitude 400N is applied to block B horizontally. What is
the speed of B after it moves 1.5 m? The mass of block B is 2.5 kg & the coeffcient of friction for all
contact surfaces is 0.3. [ use  n (3/2) 0.41 and g = 10 ms2 ]
25 kg nzO;eku vkSj 6m yEckbZ dk ,d leku CykWd A dks C ij dhyfdr (hinged) fd;k x;k gS rFkk bls ,d NksVs
CykWd B }kjk fp=kkuqlkj lgkjk iznku fd;k x;k gSA 400N ifjek.k dk ,d cy 'F' dks CykWd B ij {kSfrt :i ls
vkjksfir fd;k tkrk gSA blds 1.5 m xfr ds i'pkr~ B dh pky D;k gS\ CykWd B dk nzO;eku 2.5 kg gS vkSj lHkh
lEidZ lrgksa ds fy, ?k"kZ.k xq.kkad 0.3 gSA [ fn;k gqvk gS  n (3/2) 0.41 vkSj g = 10 ms2 ]

Ans. 323  18.00

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27. A 70 kg man stands in contact against the inner wall of a hollow cylindrical drum of radius 1m rotating
about its vertical axis. The co-efficient of friction between the wall and his clothing is 0.1. What is the
minimum rotational speed of the cylinder to enable the man to remain stuck to the wall (without falling)
when the floor is suddenly removed ? (g = 10 m/s2)
70 kg lagfr dk dksbZ O;fDr 1 m f=kT;k dh fdlh csyukdkj nhokj dh vkUrfjd lrg ds lkFk mlds laidZ eas [kMk
gS tks viuh Å/okZ/kj v{k ds lkis{k ?kwe jgk gSA nhokj rFkk mlds diMksa ds chp xq.kkad 0.15 gSA nhokj dh og
U;wure ?kw.kZ.k pky Kkr dhft,] ftlls Q'kZ dk ;dk;d gVk ysus ij Hkh og O;fDr fcuk fxjs nhokj ls fpidk jg
ldsA (g = 10 m/s2)
Ans. 10.00 sec–1
Sol. The horizontal force N by the wall on the man provides the needed centripetal force : N = m R 2. The
frictional force f (vertically upwards) opposes the weight mg. The man remains stuck to the wall after
the floor is removed if mg = f < N i.e. mg <  m R 2. The minimum angular speed of rotation of the
cylinder is min = g / R = 10 s–1.
O;fDr ij nhokj }kjk yxk;k x;k {kSfrt cy N vko';d vfHkdsUnzh; cy çnku djrk gS: N = m R 2 gSA Hkkj mg ds
foijhr ?k"kZ.kcy f (Å/okZ/kj Åij dh vksj gSA Q'kZ ds gVkus ds ckn Hkh fcuk fxjs nhokj ls fpidk jg ldrk gS ;fn
mg = f < N vFkkZr~ mg <  m R 2 gSA csyu ds ?kweus dh U;wure dks.kh; pky min = g /  R = 10 s–1 gSA

28. A ball collides elastically with a massive wall moving towards it with a velocity of v as shown. The
collision occurs at a height of h above ground level and the velocity of the ball just before collision is 2v
in horizontal direction. Then the distance between the foot of the wall and the point on the ground
2h
where the ball lands (at the instant the ball lands) is n v , where n is :
g

fp=k esa n'kkZ;s vuqlkj ,d xsan ] ,d Hkkjh nhokj tks xsan dh rjQ v osx ls xfr'khy gS ] ls çR;kLFk VDdj djrh gSA
VDdj i`Foh ry esa h Å¡pkbZ ij gksrh gS rFkk VDdj ls Bhd iwoZ xsan dk {kSfrt fn'kk esa osx 2v gSA rc nhokj ds
vk/kkj (foot) ,oa /kjkry ij fLFkr ij ml fcUnq ds e/; dh nwjh ¼/kjkry ij xsan ds fxjus ds {k.k ij½ tgk¡ xsan
2h
fxjrh gS] n v gS] tgkW n gksxk &
g

v 2v

Ans. 03.00
Sol. Solve in the reference frame fixed to the wall.
Before collision, velocity of ball = 3v towards it.
 After elastic collision of ball = 3v away from it (here we have used the result that a light mass is
reversed by a heavy body at rest).
2h
Time of flight =
g

2h
 Distance between wall and ball = 3 v. .
g

Sol. ;gk¡ ge nhokj ds funsZ'k ra=k ds lkis{k bls gy djsaxsA

VDdj ds igys] ckWy dk osx, = 3v ikl vkus okyk
 çR;kLFk VDdj ds ckn] ckWy dk osx = 3v nwj tkus okyk (fLFkj Hkkjh oLrq ls Vdjkus ds ckn ckWy dk osx
iyV tkrk gSA)
2h
tehu ij fxjus esa yxk le; =
g

2h
 fxjrs oDr nhokj vkSj ckWy ds chp nwjh = 3 v. .
g

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29. An old record player of 10 cm radius turns at 10 rad/s while mounted on a 30° incline as shown in the
figure. A particle of mass m can be placed any where on the rotating record. If the least possible
coefficient of friction  that must exist for no slipping to occur is , find 2 3 .
,d iqjkus fjdkWMZ Iys;j dh f=kT;k 10 cm gS tks fp=kkuqlkj 30° urry ij 10 rad/s ls ?kwe jgh gSA m nzO;eku ds
,d d.k dks ?kweus okys fjdkWMZ ij dgh Hkh j[kk tk ldrk gSA d.k o fjdkWMZ esa fQlyu ugha gksus ds fy, lEHko
U;wure ?k"kZ.k xq.kkad dk eku gS, rc 2 3 Kkr dhft,A

30°
Ans. 06.00
Sol. f  mg sin  + mr2
µ mg cos  mg sin  + mr2
 r
2

or ;k µ  tan  +
g cos 

mg sin+ mr 2


30. In the system shown blocks A and B have masses 5m and m respectively. The pulley has moment of
inertia  and there is no slip between the string and the pulley. If the inner and outer radius of the pulley
g
are r and 2r respectively then the angular acceleration of the pulley is , find N. [Take  = 2mr2]
Nr
iznf'kZr fp=k esa CykWd A rFkk B ds nzO;eku Øe'k% 5m rFkk m gSA f?kjuh dk tM+Rok?kw.kZ  gS rFkk f?kjuh ,oa jLlh ds
e/; dksbZ fQlyu ugh gSA ;fn f?kjuh dh vkUrfjd rFkk ckgjh f=kT;k,W Øe'k% r rFkk 2r gks rks bldk dks.kh; Roj.k
g
gSA N dk eku Kkr djksA [ = 2mr2 ekusA]
Nr

Ans. 11.00
Sol. For A : 5m(r) = 5 mg sin37º – T1
5mr = 3mg – T1 .............(i)
For B : m2r = T2 – mg .............(ii)
For pulley,
 = T1r – T22r
 = (3mg – 5mr)r – (mg + mg 2r)2r
 = 3mgr – 5mr2 – 2mgr – 4mr2
 = mgr – 9mr2
( + 9mr2) = mgr
m gr m gr
= 2
= 2
  9m r 1 1m r
g
= .
11 r
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 MAIN PATTERN CUMULATIVE
TEST-4 (MCT-4)
TARGET : JEE (MAIN+ADVANCED) 2022
COURSE : VIKAAS| BATCH : 01JA
QUESTIONS, HINTS & SOLUTIONS
PART : II CHEMISTRY
PAPER
SECTION – 1 : (Maximum Marks : 80)
 This section contains TWENTY (20) questions.
 Each question has FOUR options (1), (2), (3) and (4) ONLY ONE of these four option is correct
 Marking scheme :
 Full Marks : +4 If ONLY the correct option is chosen.
 Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered).
 Negative Marks : –1 In all other cases
[kaM 1 : (vf/kdre vad : 80)
 bl [kaM esa chl (20) iz'u gSaA
 izR;sd iz'u esa pkj fodYi (1), (2), (3) rFkk (4) gSaA bu pkjksa fodYiksa esa ls dsoy ,d fodYi lgh gSaA
 vadu ;kstuk :
 iw.kZ vad % +4 ;fn flQZ lgh fodYi gh pquk x;k gSA
 'kwU; vad % 0 ;fn dksbZ Hkh fodYi ugha pquk x;k gS ¼vFkkZr~ iz'u vuqÙkfjr gS½A
 _.k vad % –1 vU; lHkh ifjfLFkfr;ksa esaA

31. The number of moles of Cr2O72 needed to oxidize 0.03 moles of N2H5 by the reaction

H
N2H5 + Cr2O72   N2 + Cr3+ + H2O is :
fuEu vfHkfØ;k }kjk 0.03 eksy N2H5 dks vkWDlhÑr djus ds fy, fdrus eksy Cr2O72 dh vko';drk gksxh \

H
N2H5 + Cr2O72   N2 + Cr3+ + H2O
(1) 0.13 mol (2) 0.27 mol (3) 0.04 mol (4*) 0.02 mol
Sol. 3 N2H5 + 2 Cr2O72– + 13H+  3N2 + 4Cr3+ + 14H2O

32. Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following
endothermic reaction.
CH4(g) + H2O (g) CO(g) + 3H2(g)

(i) increasing the pressure (a) No shift

(ii) increasing the temperature (b) Back ward shift
(iii) use of catalyst (c) Forward shift
The correct codes for the following changes at equilibrium are :
(1) (i) – a (ii) – b (iii) – c (2*) (i) – b (ii) – c (iii) – a
(3) (i) – c (ii) – b (iii) – a (4) (i) – b (ii) – a (iii) – c
fuEu Å"ek'kks"kh vfHkfØ;k ds vuqlkj Hkki ds lkFk vkaf'kd vkWDlhdj.k }kjk izkd`frd xSl ls MkbZ gkbMªkstu xSl izkIr
gksrh gS &
CH4(g) + H2O (g) CO(g) + 3H2(g)

(i) nkc c<+kus ij (a) dksbZ foLFkkiu ugha

(ii) rki c<+kus ij (b) i'p foLFkkiu
(iii) mRizsjd dk mi;ksx (c) vxz foLFkkiu
lkE; ij fuEu ifjorZuksa ds fy, lgh dksM gSa &
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 (1) (i) – a (ii) – b (iii) – c (2*) (i) – b (ii) – c (iii) – a
(3) (i) – c (ii) – b (iii) – a (4) (i) – b (ii) – a (iii) – c

33. One mol of an ideal gas was taken from A  B as shown in given figure. Magnitude of work involved in
25 J
process is (R = ):
3 mol K
,d vkn'kZ xSl dk ,d eksy fn, x;s fp=k esa n'kkZ, vuqlkj A  B rd fy;k x;k A izØe es fufgr dk;Z dk
25 J
ifjek.k gksxk (R = ):
3 mol K
V

10L B
A
5L

T
300K 600K
(1) 5 KJ (2) 7.5 KJ
(3*) 2.5 KJ (4) None of these buesa ls dksbZ ugha
Sol. w = – nRT
25
=–1× × 300
3
= – 2500 J

34. Chlorine is prepared in the laboratory by treating manganese dioxide with aqueous hydrochloric acid
according to the reaction
HCl + MnO2  MnCl2 + Cl2 + H2O
Amount of HCl which reacts with 17.4 g manganese dioxide is:
fuEu vfHkfØ;k ds vuqlkj tyh; gkbMªksDyksfjd vEy ds lkFk eSaxuht MkbZvkWDlkbM dks mipkfjr djus ls
iz;ksx'kkyk esa DYkksjhu curh gSA
HCl + MnO2  MnCl2 + Cl2 + H2O
17.4 g eSxuht MkbZvkWDlkbM ds lkFk HCl dh fdruh ek=kk fØ;k djrh gS\
(1) 14.6 g (2) 7.3 g (3*) 29.2 g (4) 10.95 g
Sol. 4HCl + MnO2  2H2O + MnCl2 + Cl2
Moles of MnO2 = 0.2
Moles of HCl required = 0.8
Weight of HCl = 0.8 × 36.5 = 29.2 g
gy% 4HCl + MnO2  2H2O + MnCl2 + Cl2
MnO2 ds eksy = 0.2
HCl ds vko';d eksy = 0.8
HCl dk Hkkj = 0.8 × 36.5 = 29.2 g

1
35. Number of electrons which will be present in the sub-shells having m = 0 & ms = – for n = 4 are:
2
1
n = 4 ds fy, m = 0 rFkk ms = – eku j[kus okys midks'kksa esa mifLFkr bysDVªkWuksa dh la[;k gksxh &
2
(1) 8 (2*) 4 (3) 16 (4) 6
Sol. 4s, 4p, 4d & 4f each subshell will have one orbital with m = 0.
4s, 4p, 4d rFkk 4f izR;sd midks'k m = 0 ds lkFk ,d d{kd j[ksxkA

36. pq r s t u v
2 5
4s 3d
If the spin quantum number of 'p' and 'u' is same. The group of electrons (Given in the same bracket in
options) having at least three quantum numbers same are :
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 ;fn 'p' rFkk 'u' dh pØ.k Dok.Ve la[;k leku gS rks de ls de rhu Dok.Ve la[;k j[kus okys bysDVªkWuks dk lewg
(fodYiksa esa leku dks"Bd es fn;s x;s) gaS&
(1) (pq) (rst) (tup) (2) (pr) (qs) (stu) (3*) (pq) (rst) (tuv) (4) None of these buesa ls dksbZ ugha

Sol. pq r s t u v
2 5
4s 3d
n 4 3
 0 2
m 0 –2, –1, 0, 1, 2
pØ.k spin

37. A gas consists of 3 molecules with velocity 3 m/s, 5 molecules with a velocity 5 m/s and 8 molecules
with velocity 8 m/s. The ratio of rms velocity and average velocity of the molecule is :
,d xSl esa 3 m/s osx ;qDr 3 v.kq] 5 m/s osx ;qDr 5 v.kq rFkk 8 m/s osx ;qDr 8 v.kq mifLFkr gSA v.kq ds rms osx
rFkk vkSlr osx dk vuqikr gS&
(1) 1.75 (2) 1.55 (3) 1.35 (4*) 1.05
33  55  88 n v  n2v 2  n3 v 3
Sol. vavg = = 6.125 vavg = 1 1
358 n1  n2  n3
n1v12  n2 v 22  n3 v 32 (3  9)  (5  25)  (8  64) 664
vrms = = = = 6.44
n1  n2  n3 16 16
v rms 6.44
= = 1.05
v avg 6.125

38. In which of the following entropy decreases?

(1) 2NaHCO3(s)  Na2CO3(s) + CO2(g) + H2O(g)
(2) H2(g)  2H(g)
(3) Temperature of crystalline solid increased from 2 K to 50 K.
(4*) Liquid crystallizes into a solid.
fuEu esa ls fdlesa ,UVªkWih de gksrh gS \
(1) 2NaHCO3(s)  Na2CO3(s) + CO2(g) + H2O(g)
(2) H2(g)  2H(g)
(3) fØLVyhd`r Bksl dk rki 2 K ls 50 K rd c<+rk gSA
(4*) nzo Bksl esa fØLVyhd`r gksrk gSA
Sol. Randomness decreases.
;kn`fPNdrk ?kVrh gSA

39. In the following cyclic process 1-2-3-4-1

(1*) work is being done on the gas (2) work is being done by the gas
(3) net work done is zero (4) data insufficient

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 fuEu pØh; izØe 1-2-3-4-1 esa

(1*) xSl ij dk;Z fd;k x;k gSA (2) xSl }kjk dk;Z fd;k x;k gSA
(3) fd;k x;k dqy dk;Z 'kwU; gSA (4) vkadM+s vi;kZIr gSA

Sol. (1)

Anticlockwise cyclic process, thus work is being done on the gas.

okekorZ pØh; izØe gS] vr% xSl ij dk;Z fd;k x;k gSA

40. At moderate pressure, the compressibility factor for a particular gas is given by:
170P
Z = 1 + 0.34 P – ( P in bar and T in Kelvin)
T
The Boyle's temperature of this gas is :
e/;orhZ nkc ij fdlh fo'ks"k xSl ds fy, laihM~;rk xq.kkad fuEukuqlkj gS&
170P
Z = 1 + 0.34 P – (P ckj es rFkk T dsfYou esa)
T
rks bl xSl dk ckW;y rki gS&
(1) 298 K (2) 170 K (3*) 500 K (4) 340 K
Sol. At Boyle's temperature, z = 1
ckW;y rki ij, z = 1
170
T= = 500 K
0.34

41. Which of the following set of compound follow Lewis octel rules.
fuEu esa ls dkSulk ;kSfxdks dk leqPp; v"Vd fu;e dh ikyuk djrk gSA
(1*) SiF4 , NH3, BF4– (2) ICl3 , BF3, PCl5 (3) BCl3, SF6, SF4 (4) BCl3, BeCl2, PCl5
Sol. SiF4 , NH3, BF4–

42. In which of the following species, each atom carries same number of lone pair of electrons on it?
fuEu esa ls fdl iztkfr esa izR;sd ijek.kq ij ,dkdh bysDVªku ;qXeksa dh la[;k leku gSA
(1) XeO3 (2*) XeF2 (3) CO32– (4) O3
Xe
Sol. (1)
O O
O
(2) all three atoms carry 3 lp each (blesa izR;sd ijek.kq ij 3 ,dkdh ;qXe gS)

O
(3) C O
O

(4)
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43. The s and p characters in the bond formed by the central atom is equal in :
fuEu esa ls fdlesa dsUnzh; ijek.kq }kjk fufeZr cU/k esa s o p vfHky{k.k leku gS &
(1) CH3 (2) CH3 (3*) CH  CH (4) CH4
Sol. sp hybridisation has 25% s–character and 75% p– character. sp hybridisation has 33% s – character
3 2

and 67% p–character. While sp–hybridisation has 50% s–character and 50% p–character.
sp3 ladj.k esa 25% s–vfHky{k.k o 75% p–vfHky{k.k gksrs gSA sp2 ladj.k esa 33% s–vfHky{k.k o 67% p– vfHky{k.k
gksrs gSA tcfd sp–ladj.k esa 50% s– vfHky{k.k o 50% p– vfHky{k.k gksrs gSA

44. A nonpolar molecule AX4 have all bond angles equal then which of the following conclusion is
incorrect?
,d v/kzqoh; v.kq AX4 esa lHkh cU/k dks.k leku gS rc fuEu esa ls dkSulk fu"d"kZ lgh ugha gSa\
(1) Molecule may be tetrahedral.
(2) Molecule may be square planar.
(3*) Central atom 'A' must have at least six valence electrons.
(4) Central atom 'A' has either zero lone pair or two lone pairs.
(1) v.kq prq"Qydh; gks ldrk gSA
(2) v.kq oxZ leryh; gks ldrk gSA
(3*) dsfUnz; ijek.kq 'A' ij de ls de N% l;ksth bysDVªkWu gksus pkfg,A
(4) dsfUnz; ijek.kq 'A' ij 'kwU; vFkok nks ,dkdh ;qXe gksus pkfg,A
Sol. For AX4 type molecule
If  = 0, it is non-polar. It must be tetrahedral (or) square planar.
Ex : CH4, SiH4  tetrahedral (Zero lone pairs)
XeF4, ICl4  square planar (Two lone pairs)
Sol. AX4 izdkj ds v.kq ds fy,
;fn  = 0 ;g v/kzoh; gSA ;g prq"Qydh; vFkok oxZ leryh; gksuk pkfg,A
mnk% : CH4, SiH4  prq"Qydh; ('kwU; ,dkdh ;qXe)
XeF4, ICl4  oxZ leryh; (nks ,dkdh ;qXe)

45. Which of the following molecule contains shortest N–O bond ?

fuEu esa ls dkSulk v.kq lcls NksVk ¼y?kqÙke½ N–O cU/k j[krk gS ?
(1*) NOF (2) NO2– (3) NO3– (4) NH2OH
N N
O F O O
Sol. (bond order = 2) (bond order = 1.5)
O
 N
N H
O (bond order = 1.33) O H
O H (bond order = 1)
N N
O F O O
Sol. (ca/k Øe = 2) (ca/k Øe = 1.5)
O
 N
N H
O (ca/k O H
O Øe = 1.33) H (ca/k Øe = 1)

46. How many of following is/are non linear molecule.

fuEu esa ls dkSulk@dksuls vjs[kh; v.kq gS@gSaA
(1) N2O (2*) 3 (3) Cl2– (4) XeF2
Sol. 3 is bent shaped.
3 dh eqM+h gqbZ vkÑfr gksrh gSA

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47. The correct decreasing order of bond order of Si–O, P–O, S–O and Cl–O in SiO44–, PO43–, SO42– and
ClO4– is :
SiO44–, PO43–, SO42– rFkk ClO4– esa Si–O, P–O, S–O rFkk Cl–O ds cU/k Øe dk ?kVrk gqvk lgh Øe gS &
(1*) ClO4– > SO42– > PO43– > SiO44– (2) PO43– > SiO44– > SO42– > ClO4–
(3) SiO44– > PO43– > SO42– > ClO4– (4) SO42– > SiO44– > PO43– > ClO4–
Sol. This is due to increasing extent of back bonding.
,slk i'p cU/ku ds izlkj esa o`f) ds dkj.k gksrk gSA
O O
O Si O O P O
O O
Bond order = 1.0 Bond order = 1.25
cU/k Øe = 1.0 cU/k Øe = 1.25

O O
O S O O Cl O
O O
Bond order = 1.5 Bond order = 1.75
cU/k Øe = 1.5 cU/k Øe = 1.75

48. Total number of bond pair of electrons and lone pair of electrons in CO2 are-
CO2 esa cU/kh bysDVªkWu ;qXe o ,dkdh bysDVªkWu ;qXe dh dqy la[;k gSa &
(1) 2, 8 (2*) 4, 4 (3) 4, 7 (4) 3, 6
Sol. O C O
Total bond pairs = 4
dqy cU/k ;qXe = 4
Total lone pairs = 8
dqy cU/k ;qXe = 8

49. Which of the following statement is correct ?

(1) Octet rule is followed by N in NO2.
(2) BF3 is hypervalent species and PF5 is hypovalent species.
(3*) SO3 does not follow octet rule.
(4) BCl3 has lone pair of electrons on boron.
fuEu esa ls dkSulk dFku lgh gS \
(1) NO2 esa N }kjk v"Vd fu;e dk ikyu fd;k tkrk gSA
(2) BF3 vfrla;ksth rFkk PF5 U;wula;ksth Lih'kht gSA
(3*) SO3 v"Vd fu;e dk ikyu ugha djrk gSA
(4) BCl3 esa cksjsu ij ,dkdh bysDVªkWu ;qXe gksrk gSA

50. Among the followig, the species with identical bond order are
fuEu ;qXeksa esa ls fdl ;qXe esa ,dleku vkca/k Øe ¼ bond order½ ik;k tkrk gS&
(1) CO and N2– (2) O2– and CO (3*) O22– and B2 (4) CO and N2+
(1) CO vkSj N2– (2) O2– vkSj CO (3*) O22– vkSj B2 (4) CO vkSj N2+
Sol.
Number of electron Bond order
O22– 18 10  8
BO = =1
2
B2 10 64
BO = =1
2

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Sol.
bysDVªkWu dh la[;k ca/k Øe
O22– 18 10  8
BO = =1
2
B2 10 64
BO = =1
2

SECTION – 2 : (Maximum Marks : 20)

 This section contains TEN (10) questions. The answer to each question is NUMERICAL VALUE with two
digit integer and decimal upto two digit.
 If the numerical value has more than two decimal places truncate/round-off the value to TWO decimal
placed.
 There are 10 Questions & you have attempt any 5 Questions. If a student attempts more than 5
questions, then only first 5 questions which he has attempted will be checked.
 Marking scheme :
 Full Marks : +4 If ONLY the correct option is chosen.
 Zero Marks : 0 In all other cases

[kaM 2 ¼vf/kdre vad% 20)

 bl [kaM esa nl (10) iz'u gSA izR;sd iz'u dk mÙkj la[;kRed eku (NUMERICAL VALUE) gSa] tks f}&vadh; iw.kkZad
rFkk n'keyo f)&vadu eas gSA
 ;fn la[;kRed eku esa nks ls vf/kd n’'keyo LFkku gS ] rks la[;kRed eku dks n'keyo ds nks LFkkuksa rd VªadsV@jkmaM
vkWQ (truncate/round-off) djsaA
 bl [kaM esa 10 iz'u gaS ftuesa ls vkidks dsoy fdUgh 5 iz'uksa dk mÙkj nsuk gS ;fn vki 5 ls vf/kd iz'uksa dk mÙkj nsrs
gS] rks mÙkj fn;s x;s izFke 5 iz'uksa dh gh tk¡p dh tk;sxhA
 vadu ;kstuk :
 iw.kZ vad % +4 ;fn flQZ lgh fodYi gh pquk x;k gSA
 'kwU; vad % 0 vU; lHkh ifjfLFkfr;ksa esaA

51. If the percentage yield of given reaction is 40%, how many total moles of the gases will be produced, if
8 moles of NaNO3 are taken initially

NaNO3(s)   Na2O(s) + N2 + O2
;fn fn x;h vfHkfØ;k dh izfr'kr yfC/k 40% gS ;fn izkjEHk esa NaNO3 ds 8 eksy fy, tk, rc eqDr ;k mRiUu xSlksa
ds dqy eksyksa dh la[;k D;k gksxh \

NaNO3(s)   Na2O(s) + N2 + O2
Ans: 05.60

52. The ionization energy of He+ is x times that of H. The ionization energy of Li2+ is y times that of H. Find
|y–x|.
He+ dh vk;uu ÅtkZ H dh vk;uu ÅtkZ dh x xquk gSA Li2+ dh vk;uu ÅtkZ H dh vk;uu ÅtkZ dh y xquk gSA
|y–x| Kkr dhft,A
Ans. 05.00
Sol. x = 4, y = 9 I.E. = 13.6 Z2

53. Ammonium hydrogen sulphide dissociates according to the following equation :

NH4HS(s) NH3(g) + H2S (g)
The total pressure at equilibrium at 400 K is found to be 1 atm.
1
The equilibrium constant Kp of the reaction is atm2 then value of x is:
x

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 veksfu;e gkbMªkstu lYQkbM fuEu lehdj.k ds vuqlkj fo;ksftr gksrk gS &
NH4HS(s) NH3(g) + H2S (g)
lkE; ij 400 K ij dqy nkc 1 atm izkIr gksrk gSA
vfHkfØ;k dk lkE; fu;rkad Kp 1 atm2 gS rc x dk eku gS&
x
Ans. (04.00)
Sol. NH4HS(s) NH3(g) + H2S(g)
0 0
P P
Total pressure = P + P = 1 atm
2P = 1 atm
1
P = atm
2
1 1 1
Kp = (pNH3 )(pH2S )    atm2
2 2 4
Hence x = 4
Sol. NH4HS(s) NH3(g) + H2S(g)
0 0
P P
dqy nkc = P + P = 1 atm
2P = 1 atm
1
P = atm
2
1 1 1
Kp = (pNH3 )(pH2S )    atm2
2 2 4
blfy, x = 4

54. The number of completely filled orbitals in 29Cu which have total number of nodes equal to two is :
29Cu esa dqy nks uksM j[kus okys iw.kZiwfjr d{kdksa dh la[;k gSa&
Ans. 09.00
29Cu  1s
Sol. 2 2s2 2p6 3s2 3p6 3d10 4s1

Total nodes 0 1 1 2 2 2 3
(n–1)
Orbitals 1 3 5

So, 9.

Sol. 29Cu  1s2 2s2 2p6 3s2 3p6 3d10 4s1

dqy uksM 0 1 1 2 2 2 3
(n–1)
d{kd 1 3 5

blfy,, 9.

3
55. A monoatomic gas (CV = R) is allowed to expand adiabatically and reversibly from initial volume of 8
2
L at 300 K to a volume of V2 at 240 K. The value of V2 is : [(5)1/2 = 2.236]
3
,d ,dyijekf.od xSl (CV = R) dks izkjfEHkd vk;ru 8 L o rki 300 K ls vafre vk;ru V2 o rki 240 K
2
rd :)ks"e rFkk mRØe.kh; :i ls izlkfjr fd;k tkrk gSA V2 dk eku D;k gS\ [(5)1/2 = 2.236]
Ans. 11.18

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Sol. TV–1 = constant fu;rakd
5 2
=  –1=
3 3
 300 × (8)2/3 = 240 × (V2)2/3  (V2)2/3 = 5
 V2 = (5)3/2 = 5 × 2.236 = 11.18 L

56. By assuming that there are 9 periods in the periodic table, maximum how many elements would be
present in 9th period.
;g ekurs gq, fd vkorZ lkj.kh esa 9 vkorZ gS] fdrus vf/kdre rRo 9th vkorZ esa mifLFkr gksxsa \
Ans. 50.00
Sol. In 9th period, the following subshells should be filled.
9th vkorZ esa] fuEu midks'k Hkjs gksus pkfg,A
9s, 6g, 7f, 8d, 9p
(2e–), (18), (14), (10, (6)

57. Consider the following overlapping (z is inter molecular axis).

The number of positive overlap = ‘x’
The number of negative overlap = ‘y’
The number of zero overlap = ‘z’.
Then x – y + z =
fuEu vfrO;kiu dk voyksdu dfj;s (z vUrjvkf.od v{k gS)
/kukRed vfrO;kiu dh la[;k = ‘x’
_.kkRed vfrO;kiu dh la[;k = ‘y’
'kwU; vfrO;kiu dh la[;k = ‘z’.
rc x – y + z =

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Ans. 02.00
Sol. The number of positive overlap = 4(a,b,c,d)
The number of negative overlap = 4(e,f,g,h)
The number of zero overlap = 2(I,j)
/kukRed vfrO;kiu dh la[;k = 4(a,b,c,d)
_.kkRed vfrO;kiu dh la[;k = 4(e,f,g,h)
'kwU; vfrO;kiu dh la[;k = 2(I,j)

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58. How many lone pair of electrons are present in XeF2.
XeF2 esa fdrus ,dkdh bysDVªkWu ;qXe mifLFkr gSaA
Ans. 09.00

Sol. F—Xe—F

59. In PCl5 90º bond angles are X and maximum number of atoms in same plane are Y. Report your
answer as X – Y.
PCl5 esa 90º ca/k dks.k X gS rFkk leku ry es ijek.kqvks dh vf/kdre la[;k Y gSA viuk mÙkj X – Y ds :i es
nhft,A
Ans. 02.00
Sol. X = 6, Y = 4

60. In how many of the following triatomic molecules/ions, the linear molecules(s)/ion(s) are, where the
hybridization of the central atom does not have contribution from the d-orbital(s) :
BeCl2 , N3– , I3– , XeF2 , CO2 , SO2, ICl2–, SnCl2 and BeF2
fuEu esa ls fdrus f=kijekf.od v.kq@vk;u esa js[kh; v.kq@vk;u gSa tgk¡ dsUnzh; ijek.kq ds ladj.k esa d-d{kd dk
;ksxnku ugh gksrk gSA
BeCl2 , N3– , I3– , XeF2 , CO2 , SO2, ICl2–, SnCl2 and (o) BeF2
Ans. 04.00
Sol. Cl– Be–Cl , , BeF2 , CO2
sp sp

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 MAIN PATTERN CUMULATIVE
TEST-3 (MCT-3)
TARGET : JEE (MAIN+ADVANCED) 2022
COURSE : VIKAAS |
BATCH : 01JA, 01,02,03-iJA

PART : I (MATHEMATICS)
SECTION – 1 : (Maximum Marks : 80)
 This section contains TWENTY (20) questions.
 Each question has FOUR options (1), (2), (3) and (4) ONLY ONE of these four option is correct
 Marking scheme :
 Full Marks : +4 If ONLY the correct option is chosen.
 Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered).
 Negative Marks : –1 In all other cases
[kaM 1 : (vf/kdre vad : 80)
 bl [kaM esa chl (20) iz'u gSaA
 izR;sd iz'u esa pkj fodYi (1), (2), (3) rFkk (4) gSaA bu pkjksa fodYiksa esa ls dsoy ,d fodYi lgh gSaA
 vadu ;kstuk :
 iw.kZ vad % +4 ;fn flQZ lgh fodYi gh pquk x;k gSA
 'kwU; vad % 0 ;fn dksbZ Hkh fodYi ugha pquk x;k gS ¼vFkkZr~ iz'u vuqÙkfjr gS½A
 _.k vad % –1 vU; lHkh ifjfLFkfr;ksa esaA

61. The value of “  ” for which region x + y + 0 and circle x2 + y2 – 6x + 8y + 23 = 0 have exactly one
point common is
“  ” dk eku gksxk ftlds fy, x + y +   0 vkSj o`Ùk x2 + y2 – 6x + 8y + 23 = 0 ds {ks=kQy esa Bhd ,d
mHk;fu"B fcUnq gS -
(1) 3 (2) – 3 (3) 1 (4*) – 1
Sol. This is possible only when line is tangent
;g laHko gksxk dsoy tc js[kk Li'kZ js[kk gSA
perpendicular dist. From center = Radius
dsUnz ls yEcor~ nqjh = f=kT;k
3 4  
 2
2
  1  2
   3 or   1
But when   3 then infinite points are common
jUrq tc   3 gks vuUr mHk;fu"B fcUnq gSA
So blfy,,   1

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62. The area of triangle formed by positive x-axis and normal and tangent to circle x2 + y2 = 4 at (1, 3 ) is
o`Ùk x2 + y2 = 4 ds fcUnq (1, 3 ) ij Li'kZ js[kk vkSj vfHkyEc js[kk rFkk /kukRed x-v{k ls cuk;s x;s f=kHkqt dk
{ks=kQy gS -
1
(1) 3 (2*) 2 3 (3) (4) 4 3
3
Sol. Equation of Tangent is x  3y  4
Equation of Normal is y  3x
1
Hence area of ABC  (4)( 3)
2
Hindi Li'kZ js[kk dh lehdj.k x  3y  4
vfHkyEc dh lehdj.k y  3x
1
vr% ABC = (4)( 3)
2
 2 3

63. If from origin a chord is drawn to the circle x 2 + y2 – 2x = 0, then the locus of the mid point of the chord
is
;fn ewy fcUnq ls o`Ùk x2 + y2 – 2x = 0 ij ,d thok [khph tkrh gS rc thok ds e/; fcUnq dk fcUnqiFk gS -
(1) x2 + y2 + x + y = 0 (2) x2 + y2 + 2x + y = 0 (3*) x2 + y2 – x = 0 (4) x2 + y2 – 2x + y = 0
Sol. T = S1
hx + ky – (x + h) = h2 + k2 – 2h
(0, 0)
– h = h2 + k2 – 2h
x 2 + y2 – x = 0

64. Line 3x + 4y = 12 meets coordinate axes at A and B. A square ABCD is constructed on the side away
from origin. The center of square is (a,b), then a + b is
js[kk 3x + 4y = 12 funsZ'kkad v{kks dks A rFkk B ij feyrh gSA ,d oxZ ABCD Hkqtk ij] ewy fcUnq ls nwj] nwljh rjQ
cuk;k tkrk gS oxZ dk dsUnz (a, b) gS rc a + b dk eku gS - 303
(1) 5 (2) 6 (3*) 7 (4) 8
Sol.
y C

D(4+5cos, 0+5sin)
B(0,3) (a,b)
5
x
O A(4,0)
4 3
Slope of AD =  tan  cos  
3 5
4
AD dh izo.krk =  tan 
3
3 4
x  4  5 cos  = 4  5   7 sin  
5 5
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 y=0+ 5sin   4
70 7 43 7
a  b 
2 2 2 2
7 7
ab  7
2 2

65. The angle between the lines y2sin2 – xy sin2 + x2(cos2 – 1) = 1 is

js[kkvksa y2sin2 – xy sin2 + x2(cos2 – 1) = 1 ds e/; dks.k gS -
  2 
(1) (2) (3) (4*)
3 4 3 2
2 2
Sol. coeff. of x + coeff of y = 0
x2 dk xq.kkad + y2 dk xq.kkad = 0

2
66. Let ABC be a triangle with BAC  & AB = x such that (AB) (AC) = 1. If x varies then the longest
3
possible length of the angle bisector AD is:
2
ekuk ABC ,d f=kHkqt gS ftlesa BAC  rc AB = x, (AB) (AC) = 1 ;fn x pj gS rc dks.k v)Zd AD dh
3
vf/kdre laHkkfor yEckbZ gS -
3
(1) 2 (2) 1 (3*) 1/2 (4)
2
2bc A bx
Sol. AD  cos  pawfd (as c  x )
bc 2 bx
1
ijUrq bx = 1 b
x
x 1
 AD  2

1 x 1
x
x
1
 AD(max) = since min value of denominator is 2 if x > 0
2
1
AD(max) = pwfd gj dk U;wure eku 2 gS ;fn x > 0
2
67. Triangle ABC is right angled at A. The points P and Q are on the hypotenuse BC such that
BP = PQ = QC. If AP = 3 and AQ = 4 then which of the following is incorrect
f=kHkqt ABC, A ij ledks.k gSA fod.kZ BC ij fLFkr fcUnq P rFkk Q bl izdkj gS fd BP = PQ = QC. ;fn
AP = 3 rFkk AQ = 4 rc fuEu esa ls dkSulk fodYi vlR; gS &

(1*) BC = 45 (2*) CA = 33 (3*) AB = 12 (4) AB = 33

Sol. In  ABP esa

c
9 = c2 + x2 – 2cx cosB; but ijUrq cosB =
3x

c
9 = c2 + x2 – 2cx
3x

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 c2 b2
9 = x2 + ....(1) ; ||| ly 16 = x2 + ....(2)
3 3
Equation lehdj.k (1) + (2)

1 2
25 = 2x2 + (b + c2) = 2x2 + 3x2
3

 5x2 = 25  x= 5 ;  BC = 3 5 = 45

c2
From equation lehdj.k (1) 9  5   c= 12
3

b2
From equation lehdj.k (2) 16  5   b= 33
3
68. Number of proper divisors of 32.53 which are odd, is p then find the value of (p-5)
32.53 ds Loa; ds vykok Hkktdksa dh la[;k p gks rks (p-5) dk eku gS &
(1*) 6 (2) 5 (3) 7 (4) 8
Sol. obvious

69. The sum of all 4 digit numbers formed by using 2, 3, 4 and 7 if repetition of digits is not allowed, belongs
to the interval:
vadks 2, 3, 4 rFkk 7 ls cuus okys lHkh pkj vadks okyh la[;vksa dk ;ksx fdl leqPp; esa gS ¼vadksa dh iqukjko`fÙk u gks½
(1) (106000, 106300) (2) (106300, 106600)
(3*) (106600, 106900) (4) (106900, )
4!
Sol. (2  3  4  7)(100  101  10 2  103 )
4
24
= (16 )(1111) = 106656
4
’70. If n is the number of positive integral solutions of x 1 x2 x3 x4 = 210, then which of the following is incorrect
?
(1) n must be a perfect square (2) n must be a perfect 4th power
(3) n must be a perfect 8th power (4*) n must be divisible by an odd prime number
;fn x1 x2 x3 x4 = 210 ds lHkh /kukRed iw.kk±d gyksa dh la[;k n gks rks fuEu esa ls dkSulk fodYi vlR; gS &
(1) n ,d iw.kZ oxZ gS (2) n ,d prqFkZ ?kkr gS

(3) n ,d 8th ?kkr gS (4*) n ,d fo"ke vHkkT; la[;k ls foHkkftr gS

Sol. x1 x2 x3 x4 = 210 = 2 × 3 × 5 × 7
 no. of positive integral solutions
=4×4×4×4
= 44
–5
 1 
71. The coefficient of x20 in the expression of (1 + x2)40  x 2  2  2  is
 x 
–5
 1 
(1 + x2)40  x 2  2  2  ds foLrkj esa x20 dk xq.kkad gS -
 x 

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 (1) 30C10 (2*) 30C25 (3) 1 (4) 0
–10
 1
Sol. Expression = (1 + x2)40  x   
 x
= (1 + x2)30x10
coefficient of x20 in x10(1 + x2)30
= coefficient of x10 in (1 + x2)30
= 30C5
–10
1
Hindi O;atd = (1 + x2)40  x  
 x

= (1 + x2)30x10
x10(1 + x2)30 esa x20 dk xq.kkad

= (1 + x2)30 esa x10 dk xq.kkad

= 30C5

72. The value of k which satisfies the equation 1! + 2! + 3! + ...... + (x – 1)! + x! = k2 and k   is
k dk eku tks lehdj.k 1! + 2! + 3! + ...... + (x – 1)! + x! = k2, tgk¡ k  dks larq"V djrk gS &
(1) 0 (2*) 1 (3) 2 (4*) 3

2
73. The values of x between 0 and 2 which satisfy the equation sin x 8 cos x  1 are in A.P. The
common difference of the A.P. is
2
0 ,oa 2 ds e/; fLFkr x ds ,sls eku tks lehdj.k sin x 8 cos x  1 dks larq"B djrs gS] l-Js- esa gks rks bl l-
Js- dk lkoZvUrj gksxk -
  3 5
(1) (2*) (3) (4)
8 4 8 8
Sol. From the given equation we have
2sin x | cos x | 1/ 2  sin2x  1/ 2 if cos x > 0

And sin2x  1 2 if cos x  0

 when cos x  0, sin2x  1/ 2  x   / 8,3 / 8
When cos x  0, sin2x  1/ 2  x  5  / 8,7 / 8.
So the required values of x are
 / 8, 3 / 8, 5 / 8, 7  / 8
Which form an A.P. with common difference  / 4.
Hindi nh xbZ lehdj.k ds fy;s
2sin x | cos x | 1/ 2  sin2x  1/ 2 ;fn cos x > 0

vkSj sin2x  1 2 if cos x  0

 tc cos x  0, sin2x  1/ 2  x   / 8,3 / 8
tc cos x  0, sin2x  1/ 2  x  5  / 8,7 / 8.
vr% x ds vHkh"V eku
 / 8, 3 / 8, 5 / 8, 7  / 8

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 tks l-Js- cukrs gS ftldk lkoZvUrj  / 4 gS

sin 4 x – cos 4 x  sin2 x cos 2 x  

74. y= , x   0 ,  then
sin 4 x  cos 4 x  sin 2 x cos 2 x  2

sin 4 x – cos 4 x  sin2 x cos 2 x  

y= 4 4 2 2
, x   0 ,  rc -
sin x  cos x  sin x cos x  2

3 1 1 5
(1) – y (2) y1 (3) – y1 (4*) –1 < y < 1
2 2 2 3
1
– cos 2x  sin 2 2x
4 1 – 4 cos 2x – cos 2 2x
Sol. y= =
1 3  cos 2 2x
1 – sin 2 2x
4
(1 + y)cos22x + 4cos 2x + 3y – 1 = 0
D0
5
– y1
3
but when y = 1
ijUrq tc y = 1

x= which is not permissible
2

x= tks fd laHko ugh gSA
2
4n k(k 1)
75. Let Sn = 
k 1
(–1) 2 k2 . Then Sn can take value(s)

(1*) 1056 (2) 1088 (3) 1120 (4*) 1332

4n k(k 1)
ekuk fd Sn =  (–1) 2 k2 , rc Sn fuEu eku ys ldrk gS%&
k 1
(1) 1056 (2) 1088 (3) 1120 (4) 1332
4n k(k 1)
Sol. Sn =  ( 1) 2 k 2 = –12 – 22 + 32 + 42 – 52 – 62 + 72 + 82 +........
k 1
= (32 – 1) + (42 – 22) + (72 – 52) + (82 – 62)...... = 2 [4  6  12  
 14
 20  22  ......]

2n terms

[4  6  12  14  20  22  .....]
  
2n in
= 2[(4 + 12 + 20 ......) + (6 + 14 + 22 ........)]
n terms n terms
n in n in
n n 
= 2  (4  2  (n  1)8) (2  6  (n  1)8)  = 2[n(4 + 4n – 4) + n(6 + 4n – 4)]
 2 2 
= 2(4n2 + (4n + 2)n) = 2(8n2 + 2n) = 4n(4n + 1)
(1) 1056 = 32 × 33 n=8
(2) 1088 = 32 × 34
(3) 1120 = 32 × 35
(4) 1332 = 36 × 37 n=9

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76. If a and b are respectively the first and the last terms of an A.P. as well as of an H.P. both having n
terms. Product of the rth terms of the first series and (n – r + 1)th term of the second series, is
(1*) independent of both n and r and equals ab
(2) independent of both n and r and equals a2b
(3) independent of both n and r and equals ab2
(4) dependent on n and r.
;fn a rFkk b Øe'k% ,d lekUrj Js.kh vkSj ,d gjkRed Js.kh nksuksa ds izFke rFkk vfUre in gS ftuesa inksa dh la[;k
n gSA lekUrj Js.kh ds rosa in rFkk gjkRed Js.kh ds (n – r + 1)osa in dk xq.kuQy gS &

(1*) n rFkk r nksuksa ls Loar=k gS vkSj ab ds cjkcj gS

(2) n rFkk r nksuksa ls Loar=k gS vkSj a2b ds cjkcj gS

(3) n rFkk r nksuksa ls Loar=k gS vkSj ab- ds cjkcj gS

(4) n rFkk r ij fuHkZj gS

Sol. a,(a + d),(a + 2d),...... b

nth term

b–a
 b = a + (n – 1)d  d =
n–1
Tr of A.P. = a + (r – 1)d
(r – 1)(b – a) an – a  (r – 1)(b – a) an  r (b – a) – b
=a+ = =
n –1 (n – 1) n –1

1 1 1
H.P. . ..........
A A D A  (n – 1)D

1 1
where = a and =b
A A  (n – 1)D

1 1 1
 A + (n – 1) D =  + (n – 1) D =
b a b
a–b
 D=
ab(n – 1)

1 1 ab(n – 1)
 Tn – r + 1 = = = ………(2)
A  (n – r)D 1 (n – r)(a – b) an  r(b – a) – b

a ab(n – 1)

 product = ab

77. If one root of the equation x2 + px + q = 0 is 2  3 where p, q  I and roots of the equation rx2 + x + q =

0 are tan 268 and cot 88 then value of p + q + r is:

;fn lehdj.k x2 + px + q = 0 dk ,d ewy 2  3 gS tgk¡ p, q  I rFkk lehdj.k rx2 + x + q = 0 ds ewy tan

268 vkSj cot 88 gS rc p + q + r dk eku gS -

(1) –1 (2*) –2 (3) –3 (4) –4

Sol. x2 + px + q = 0 has one root 2  3 then other root = 2  3

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 x2 + px + q = 0 dk ,d ewy 2  3 gS rc vU; ewy = 2  3 gSA

P
sum of roots ewyksa dk ;ksx =  = 2 3 2 3
1
 P = –4 …(i)
q
product of roots ewyksa dk xq.ku =
1

= 2 3 2 3  
 q=1 …(ii)
now vc rx2 + x + q = 0

roots are ewy gS tan (268) = tan (180 + 88) = tan 88 and vkSj cot 88

q
 Product of roots ewyksa dk xq.ku = tan 88 . cot 88 =
r

q
 1=
r
 q=r …(iii)
 p + q + r = –4 + 1 + 1 = – 2

78. If the equations x 2 + ax + b = 0 and x 2 + bx + a = 0 hav e a common root, then the v alue
of a + b is
;fn x 2 + ax + b = 0 rFkk x 2 + bx + a = 0 dk ,d mHk;fu"B ewy gS] rks a + b dk eku gS&

(1) 1 (2) 0 (3*) – 1

(4) none of these buesa ls dksbZ ughsa
Sol. Let  be a common root of x 2 + ax + b = 0 and x 2 + bx + a = 0. Then,
ekuk x 2 + ax + b = 0 vkSj x 2 + bx + a = 0 dk mHk;fu"B ewy  gS] rks
 2 + a + b = 0 and vkSj  2 + b + a = 0   = 1
Putting  = 1, either of the two, we get a + b = – 1.
 = 1 j[kus ij, a + b = – 1.

a b c 125a3  8b3 – 64c3

79. If = = , then =
2 3 4 abc
a b c 125a3  8b3 – 64c3
;fn = = gks, rc =
2 3 4 abc
(1) 9 (2*)–120 (3) 99 (4) 3
a b c
Sol. = = =
2 3 4
 a = 2
b = 3
c = 4
  5a + 2b – 4c = 10 + 6 – 16 = 0
 (5a)3 + (2b)3 + (– 4c)3 = 3(5a)(2b)(–4c)
 125a3 + 8b3 – 64c3 = – 120abc


80. If 2 5 7 6 a 4 5 6 b is divisible by 15 then
(1) a = 6 and b = 5 (2*) a = 4 and b = 0 (3) a = 4 and b = 5 (4) a = 3 and b = 0
;fn 2 5 7 6 a 4 5 6 b, 15 ls foHkkftr gks] rks
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 (1) a = 6 rFkk b = 5 (2*) a = 4 rFkk b = 0 (3) a = 4 rFkk b = 5 (4) a = 3 rFkk b = 0
Sol. Sum of the digits vadksa dk ;ksxQy = a + b + 35
it is divisible by 15 if a = 4, b = 0 (from options) to be divisible by 15, it must be divisible by 3 & 5
To be divisible by 5, b = 0 or 5 and to be divisible by 3, a + b = 4  a = 4, b = 0
a + b = 7  a = 7, b = 0 or a = 2, b = 5
which are not available in options

SECTION – 2 : (Maximum Marks : 20)

 This section contains TEN (10) questions. The answer to each question is NUMERICAL VALUE with two
digit integer and decimal upto two digit.
 If the numerical value has more than two decimal places truncate/round-off the value to TWO decimal
placed.
 There are 10 Questions & you have attempt any 5 Questions. If a student attempts more than 5
questions, then only first 5 questions which he has attempted will be checked.
 Marking scheme :
 Full Marks : +4 If ONLY the correct option is chosen.
 Zero Marks : 0 In all other cases
[kaM 2 ¼vf/kdre vad% 20)
 bl [kaM esa nl (10) iz'u gSA izR;sd iz'u dk mÙkj la[;kRed eku (NUMERICAL VALUE) gSa] tks f}&vadh; iw.kkZad
rFkk n'keyo f)&vadu eas gSA
 ;fn la[;kRed eku esa nks ls vf/kd n’'keyo LFkku gS ] rks la[;kRed eku dks n'keyo ds nks LFkkuksa rd VªadsV@jkmaM
vkWQ (truncate/round-off) djsaA
 bl [kaM esa 10 iz'u gaS ftuesa ls vkidks dsoy fdUgh 5 iz'uksa dk mÙkj nsuk gS ;fn vki 5 ls vf/kd iz'uksa dk mÙkj nsrs gS] rks mÙkj
fn;s x;s izFke 5 iz'uksa dh gh tk¡ p dh tk;sxhA
 vadu ;kstuk :
 iw.kZ vad % +4 ;fn flQZ lgh fodYi gh pquk x;k gSA
 'kwU; vad % 0 vU; lHkh ifjfLFkfr;ksa esaA

81. If a vertex of a triangle is (1, 1) and the mid points of two sides through this vertex are (–1, 2) & (3, 2).
 
Then centroid of triangle is  1,  . then value of is.
 3
;fn f=kHkqt dk ,d 'kh"kZ (1, 1) gS rFkk bl 'kh"kZ ls tkus okyh nks Hkqtkvksa ds e/; fcUnq (–1, 2) vkSj (3, 2) gSA rc

f=kHkqt dk dsUnz  1,  gS rc dk eku Kkr dhft,A
 3 
Ans. 07.00
(1, 1)

Sol. (–1, 2) (3, 2)

(–3, 3) (5, 3)

 x y   7 
Centroid =  i , i  =  1,   =7
 3 3   3

82. If the straight lines joining the origin and the points of intersection of the curve

5x2 + 12xy  6y2 + 4x  2y + 3 = 0 and x + ky  1 = 0 are equally inclined to the x-axis, then find the
value of | k |.
oØ 5x2 + 12xy  6y2 + 4x  2y + 3 = 0 ,oa js[kk x + ky  1 = 0 ds izfrPNsn fcUnqvksa dks ewy fcUnq ls feykus
okyh ljy js[kk,¡ x-v{k ls leku dks.k ij >qdh gqbZ gks] rks | k | dk eku gS&
Ans. 01.00
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Sol. Homogenize 5x2 + 12xy – 6y2 + 4x – 2y + 3 = 0 by x + ky = 1
5x2 + 12xy – 6y2 + 4x(x + ky) – 2y (x + ky) + 3(x + ky)2 = 0
it is equally indined with x-axes hence coeff. xy = 0 12 + 4k – 2 + 6k = 0 k = – 1
Hindi. 5x2 + 12xy – 6y2 + 4x – 2y + 3 = 0 dks x + ky = 1 dh lgk;rk ls le?kkr cukus ij
5x2 + 12xy – 6y2 + 4x(x + ky) – 2y (x + ky) + 3(x + ky)2 = 0
;s js[kk,sa x-v{k ls leku dks.k ij >qdh gS vr% xy dk xq.kkad = 0 12 + 4k – 2 + 6k = 0 k = 1

3 sin A ab 3
83. In ABC, sin(A – B) = , = 2 and = then value of is
5 sinB c 
3 sin A ab 3
ABC esa sin(A – B) = , = 2 vkSj = rc  dk eku cjkcj gS -
5 sinB c 
Ans. 05.00
 A B 
2 tan  
Sol. sin (A – B) =  2  = 3
 A B 5
1  tan2  
 2 
10x = 3 + 3x2 3x2 – 10x + 3 = 0
10  8 1
x = = 3,
6 3
 A B a b C
tan   = a  b cot 2
 2 
1 2b  b C
= cot  C = 90°
3 2b  b 2
b2 + 4b2 = c2
c =  5 b
3b ab
=
5b c

84. There are 15 seats in a row numbered as 1 to 15. If number of ways in which 4 persons can be seated
such that seat number 6 is always occupied and no two persons sit on the adjacent seats is 384, then
determine .

,d iafä esa 1 ls 15 rd vafdr 15 LFkku gSA ;fn 4 O;fDr bl izdkj cSBsa rkfd LFkku Øekad lnSo Hkjk gks vkSj
mlds nk;sa rFkk ck;sa LFkku ij dksbZ O;fDr uk cSBs rks blds dqy rjhds 384 gks rks  dk eku gS &
Ans. 08.00
Sol. Case-I : 2 left to 6 and 1 right to 6 two can sit in the ways (1, 3), (1, 4), (2, 4) and one who sit right to 6
can sit in 8 ways
 total number of ways = 24
Case-II : 1 left to 6 and 2 right to 6 the person left to 6 in 4 ways
Two right to 6 can be done in following way
x1 + x2 + x3 = 7
x1  1, x2  1, x3  0
Number of integral solution = 21
total ways 21 × 4 = 84 ways
Case-III : All 3 right to 6
x1 + x2 + x3 + x4 = 6(x1  1, x2  1, x3  1, x4  1)
coefficient t3 in (1 – t)–4  6C3 = 20
Case-I+ Case-II + Case-III = 128
Total arrangements = 128 × 4! = 384
Hence = 8

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85. If the tangents at the point P on the circle x2 + y2 + 6x + 6y = 2 meets the straight line 5x – 2y + 6 = 0 at
a point Q on the y axis, then the length of PQ is
;fn o`Ùk x2 + y2 + 6x + 6y = 2 ds fcUnq P ij Li'kZ js[kk,a] ljy js[kk 5x – 2y + 6 = 0 dks y v{k ij fcUnq Q ij
feyrh gS] rc PQ dh yEckbZ gS&
Ans. 05.00

Sol.

At point Q, x = 0
fcUnq Q ij, x = 0
 in line 5x – 2y + 6 = 0
js[kk 5x – 2y + 6 = 0
put x = 0 j[kus ij  y=3
 Q (0,3)
 PQ = 02  32  6(0)  6(3) – 2  5

86. The remainder when 487 + 687 + 32 is divided by 25, is

'ks"kQy gksxk tc 487 + 687 + 32 dks 25 ls foHkkftr fd;k tkrk gSA
Ans. 02.00
Sol. 487 + 687 + 32
(5 – 1)87 + (5 + 1)87 + 32
= (1 + 5)87 – (1 – 5)87 = 2[87C1 · 5 + 87C3 · 53 + ....... + 87C87 · 587] + 32
= 10 · 87 + 2[87C3 · 53 + ....... + 87C87 · 587] + 32 = 902 + an expression divisible by 25
= 10 · 87 + 2[87C3 · 53 + ....... + 87C87 · 587] + 32 = 902 + izlkj djus ij 25 ls foHkkftr fd;k tkrk gSA
902 2
 = 36 +
25 25
 remainder is 'ks"kQy 2 Ans.
sin x sin 3 x sin 9 x  
87. Number of solutions of equation + + = 0 in  0, 
cos 3 x cos 9 x cos 27 x  4

  sin x sin 3 x sin 9 x

 0,  esa lehdj.k + + = 0 ds gyksa dh la[;k Kkr dhft,A
 4 cos 3 x cos 9 x cos 27 x

Ans. 06.00
sin x sin 3 x sin 9 x
Sol. + + =0
cos 3 x cos 9 x cos 27 x
2 sin x cos x 2 sin 3 x cos 3 x 2 sin 9 x cos 9x
 + + =0
2 cos 3 x cos x 2 cos 9 x cos 3 x 2 cos 27 x cos 9x
sin(3 x – x ) sin(9 x – 3 x ) sin(27 x – 9 x)
 + + =0
2 cos 3 x cos x 2 cos 9 x cos 3 x 2 cos 27 x cos 9x
 (tan 3x – tanx) + (tan 9x – tan3x) +(tan27x – tan9x) = 0

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 n
 tan27x = tanx 27x = n+ x x= n 
26
 2 3 4 5 6
 x= , , , ,
26 26 26 26 26 26
1 1 1 1
88. If Sn = 1 + + 2 + ...... + n  1 , n  N, then least value of n such that 2 – Sn < is
2 2 2 100
1 1 1 1
;fn Sn = 1 + + 2 + ...... + n  1 , n  N rc n dk U;wure eku gksxk tks 2 – Sn < dks larq"B djrk gS -
2 2 2 100
Ans. 08.00
  1 n 
1    
 2  n 1 n 1
 1  1 1
Sol. Sn =   ; Sn = 2 –      <
1 2 2 100
1
2
 n – 17  n 8
1 1
89. If  and  are roots of equation x2 – 7x + 1 = 0, then the value of 2
+ is
( – 7) ( – 7)2

1 1
;fn  vkSj  lehdj.k x2 – 7x + 1 = 0 ds ewy gSa] rks 2
+ dk eku gS&
( – 7) ( – 7)2

Ans. 47.00
Sol. 2 – 7 + 1  ( – 7) = – 1
1
  – 7 = –

1 1
 2
+ = 2 + 2
( – 7) ( – 7)2

= ( + )2 – 2 = 47

90. The number of solutions of log2(x – 1) = 2 log2(x – 3) is

log2(x – 1) = 2 log2(x – 3) ds gyksa dh la[;k gS&
Ans. 01.00
Sol. Given fn;k gS log2 (x – 1) = 2 log2 (x – 3)
 log2 (x – 1) = log2 (x – 3)2
 (x – 1) = (x – 3)2
 x2 – 7x + 10 = 0
 x = 2, 5
Now, x = 2 does not satisfy equation as x > 3. vc x = 2 lehdj.k dks larq"B ugha djrk gS D;ksfd x > 3
Hence, x = 5 is only solution. vr% x = 5 dsoy ,d gy gSA

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