RANK BOOSTER TEST SERIES
12th and 13th STUDENTS
FULL TEST - 4 [PAPER - 1]
TARGET IIT - JEE - 2016
Date : 17-05-2016
Duration :3 Hours
Max. Marks : 264
_________________________________________________________________________________________
INSTRUCTIONS
In each part of the paper, Section-A contains 8 questions, Section - B contains 10 questions & Section-C contains
2 questions. Total number of pages are 24. Please ensure that the Questions paper you have received contains ALL
THE QUESTIONS in each section and PAGES.
SECTION - A
Q.1 to Q.8 are Integer answer type questions (whose answer is 1 digits [0 to 9]) & carry 4 marks each. No Negative
Marking.
SECTION - B
Q.1 to Q.10 are Multiple choice Questions has four choices (A), (B), (C), (D) out of which one or more than one is/are
correct & carry 4 marks each. 2 mark will be deducted for each wrong answer.
SECTION - C
Q.1 & Q. 2 are "Match the Column" Type questions and you will have to match entries in column - I with the entries in
Column - II. One or More entries in Column - I may match with one or more entries of Column - II. For each entry in
Column - I, + 2 for correct answer, 0 if not attempted and – 1 in all other cases.
NOTE : GENERAL INSTRUCTION FOR FILLING THE OMR ARE GIVEN BELOW.
1.
Use only blue/black pen (avoid gel pen) for darkening the bubble.
2.
Indicate the correct answer for each question by filling appropriate bubble in your OMR answer sheet.
3.
The Answer sheet will be checked through computer hence, the answer of the question must be marked by shading the
circles against the question by blue/black pen.
4.
While filling the bubbles please be careful about SECTIONS [i.e. Section-A (include Integer type), Section - B [Multiple type]
& Section-C (Comprehension type)].
394,50 - Rajeev Gandhi Nagar Kota, Ph. No. : 93141-87482, 0744-2209671
IVRS No : 0744-2439051, 52, 53, www. motioniitjee.com , [email protected]
RANK BOOSTER PROGRAM [FULL TEST - 4]
Page # 2
PART - I [MATHEMATICS]
SECTION - A
[ fo' y s
"k. kkRed i z'u i zd kj ]
SECTION - A
[INTEGER ANSWER TYPE]
Q.1 to Q.8 are INTEGER ANSWER TYPE Questions.
(The answer of each of the questions is 1 digits)
x
1.


2
Let f(x)   t  t  1 dt  3, 4, then the
0
i z-1 l si z-8 r d fo' y s"k.kkRed i z'u gS
A¼
i zR; sd i z'u dkmÙkj dsoy
1 va
d ksesnhft
; sA
x
1.
ekuk f(x)    t2  t  1 dt  3, 4, gSr c Qy u
0
difference between the greatest and the least
m
n
dsvf/kdr e r FkkU; w
ur e ekuksadschp vUr j gS gSr c
m
values of the function is
then m - 9n
n
2.
Let
f(x) be a differentiabl e
function
m - 9n Kkr
2.
f ( x )  f ( y)
xy
 x, y
 =
2
 2 
satisfying f 
dhft , A
ekuk f(x) vody uh; Qy u gSt ks
f ( x )  f ( y)
xy
 x, y  R dksl a
rq
"V
f
 =
2
 2 
2
2
 R and f(0) = 0. If
 f (x)  sin x 
2
dx is
0
If f(x) = (2x – 3)5 +
2
dx
0
42).
4
x + cos x and g is 3.
3
the inverse function of f, then g (2) is equal
to P, then find 7P
 f (x)  sin x 
U; w
ur e gS
] r c f(– 42) dk eku Kkr dj ksA
minimised, then find the value of f(–
3.
dj r kgSr Fkkf(0) = 0 gS
A; fn
; fn f(x) = (2x – 3)5 +
4
x + cos x r Fkkg , f
3
dk i zfr y kse Qy u gS
] r c g(2) dk eku P dscj kcj gS
r c 7P Kkr dhft , A
(SPACE FOR ROUGH WORK)
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Just when the caterpillar thought the world was ending, he turned into a butterfly.
RANK BOOSTER PROGRAM [FULL TEST - 4]
Page # 3
1
4.
n 
5.
1
Lim  x n sin (nx ) dx is equal to
4.
n 
0
All the three vertices of an equilateral triangle
lie on the parabola y = x2, and one of its
sides has a slope of 2. The x-coordinates of
the three vertices have a sum equal to
5.
p
q
Let y = f(x ) be t he c onti nuous and
differentiable function which satisfies the



Given that | a | = 1, | b | = 1 and a vector p
is defined by

p =

a


 a

a   

 ........
 ab
2 
2100  299 







, d l eckgqf=kHkq
t dsl Hkhr hu ' kh"kZi j oy ; y = x2 i j
x-funs
Z
' kka
d ksdk; ksx
6.
ekuk y = f(x) l r r ~ r Fkk vody uh; Qy u gSt ks
l EcU/kf(x + f(x)) = x + f(x)  x  R dksl Urq
"V
dj r kgSr c oØ r Fkkj s[kkx = 1 } kj ki fj c) {ks=kQy Kkr
dhft , A
7.

fn; k x; k gSfd | a | = 1, | b | = 1 r Fkk , d l fn' k

d kj i fj Hkkf"kr gSfd
p bl i z

p =
100 times

a


 a

a   

 ........
 ab
2 
2100  299 



 

100 times

If maximum value of | p | is in the form of
1
p
dscj kcj gSt gk¡p r Fkkq i j Li j
q
vHkkT; /kukRed i w
. kkZ
a
d gSr c (q – p) dk eku gS&
relation f(x + f(x)) = x + f(x)  x  R then
find the area bounded by the curves and the
line x = 1
7.
0
fLFkr gS
] r Fkkfdl h, d Hkq
t kdh<ky 2 gS
Ar huks' kh"kksZds
where p and q are relatively prime positive
integers. Then the value of (q – p) is
6.
Lim  x n sin (nx ) dx cj kcj gS%
where ,  I (ranging from 0 to 9)
2
then the value of  +  is equal to
1

; fn | p | d k egÙke eku
,  I (i fj l

2
: i esa gSt gk¡
j 0 l s9 r d ) r c  +  dkeku cj kcj
gksxk%
(SPACE FOR ROUGH WORK)
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:[email protected]
RANK BOOSTER PROGRAM [FULL TEST - 4]
8.
The locus of a point that divide a chord of
slope of 2 of the parabola y2 = 4x internally
in the ratio 1 : 2 is a parabola if the vertex
of parabola is (, ) then the value of
Page # 4
8.
, d fcUnqt ksi j oy; y2 = 4x dh2 <ky oky h , d t hok
dks 1 : 2 esavUr %foHkkft r dj r k gS
; dk fcU
nq
i Fk , d
i j oy ; gS
A ; fn i j oy ; d k ' kh"kZ(, ) gSr c
 729     2 


100


 729     2 

 must be.
100


dk eku gksxk %
SECTION - B
SECTION - B
[ cgqoS
d fYi d
[MULTIPLE OBJECTIVE TYPE]
Q.1 to Q.10 has four choices (A), (B), (C), (D)
i z-1 l si z-10 esapkj fodYi
out of which ONE OR MORE THAN ONE is correct
; k, d l svf/kd^^ l ghgS
A
1.
1.
If an angle  be divided into two parts such
that the tangent of one part is m times the
tangent of the other, then their difference 
is not given by
i z'u i zd kj
]
(A), (B), (C), (D) gS
] ft
uessal s^^, d
; fn dks.k dksnksHkkxksesabl i zd kj foHkkft r fd; kx; kgS
fd , d Hkkx dhLi ' kZj s[kknw
l j sHkkx dhLi ' kZj s[kkdhm
xq
uhgSr c mudkvUr j fdl ds} kj kughafn; kt kr kgS%
(A) cos  
m 1
cos 
m 1
(A) cos  
m 1
cos 
m 1
(B) sin  
m 1
sin 
m 1
(B) sin  
m 1
sin 
m 1
(C) sin  
m 1
cos 
m 1
(C) sin  
m 1
cos 
m 1
(D) cos  
m 1
sin 
m 1
(D) cos  
m 1
sin 
m 1
(SPACE FOR ROUGH WORK)
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RANK BOOSTER PROGRAM [FULL TEST - 4]
Page # 5
2.
If two vertices of an equilateral triangle are
A (– a, 0) and B (a, 0), a > 0 and the
third vertex C lies above x-axis then the
equation of the circumcircle of triangle ABC
is
(A) x2 + y2 – 2ay = a2
(B) x2 + y2 –
2.
(A) x2 + y2 – 2ay = a2
(B) x2 + y2 –
3 ay = a2
log
equal to
(A) 
(C) (3, 5)
1
x
x

 x 1  
 log 2 
  > 0, is
 x 2 

3.
(B) (–5, –2)
(D) (5, )
Which of the following statement is(are)
correct?
(A) The lines
y6
x4
z6
=
=
and
1
3
1
y2
z3
x 1
=
=
are orthogonal.
2
2
1
(B) The planes 3x – 2y – 4z = 3 and x – y – z
= 3 are orthogonal.
(C) The funct ion f (x) = ln(e–2 + ex ) is
monotonic increasing  x  R.
(D) If g is the inverse of the function, f (x)
= ln(e–2 + ex) then g(x) = ln(ex – e–2).
vlfedk
log
1
x
x

 x 1  
 log 2 
  > 0 dks larq"V djus
 x 2 

okys x ds okLrfod ekuksa dk leqPp; gS
(A) 
(B) (–5, –2)
(C) (3, 5)
(D) (5, )
4.
4.
3 ay = 3a2
(D) 3x2 + 3y2 – 2ay = 3a2
The set of real values of x satisfying the
inequality
3 ay = a2
(C) 3x2 + 3y2 – 2
(C) 3x2 + 3y2 – 2 3 ay = 3a2
(D) 3x2 + 3y2 – 2ay = 3a2
3.
;fn ,d leckgq f=Hkqt ds nks 'kh"kZ A (– a, 0) rFkk B
(a, 0), a > 0 rFkk rhljk 'kh"kZ C, x-v{k ds Åij
fLFkr gS, rks f=Hkqt ABC ds ifjo`Ùk dh lehdj.k gksxh%
fuEu esa ls dkSulk@dkSuls dFku lR; gS?
(A)
js[kk,sa
x4
y6
z6
x 1
=
=
rFkk
=
3
1
1
1
y2
z3
=
yEcdks.kh; gSA
2
2
lery 3x – 2y – 4z = 3 rFkk x – y – z = 3
yEcdks.kh; gSA
(C) Qyu f (x) = ln(e–2 + ex) lHkh x  R ds fy;s
,dfn"V o/kZeku gSA
(D) ;fn g, Qyu f (x) = ln(e–2 + ex) dk çfrykse
gS, rks g(x) = ln(ex – e–2) gSA
(B)
(SPACE FOR ROUGH WORK)
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RANK BOOSTER PROGRAM [FULL TEST - 4]
5.
5x  1,
x2

x
Let f (x) =  5 | 1  t | dt , x  2

0
Page # 6
5.
then which of the following statement(s) is
5x  1,
x2

x
ekuk f (x) =  5 | 1  t | dt , x  2

0
gks] rks fuEu esa ls dkSulk@dkSuls dFku vlR; gSa ?
(are) incorrect ?
(A) f (x) is continuous but not differentiable
(A) x = 2
ij f (x) lrr~ gS ijUrq vodyuh; ugha gSA
at x = 2.
(B) x = 2 ij f (x) lrr~
(B) f (x) is not continuous at x = 2.
ugha gSA
lHkh x  R ds fy, f (x) vodyuh; gSA
(C) f (x) is differentiable for all x  R.
(C)
(D) The right hand derivative of f (x) at
(D) x = 3 ij f (x)
dk nk;k¡ vodyt fo|eku ugha gSA
x = 3 does not exist.
6.
Tangents to the parabola at the extremities
6.
oÙ̀k x2 + y2 = 5 r Fkk i j oy ; y2 = 4x dh , d
of a c om mon c hord AB of the ci rc l e
mHk; fUk"V t hokAB dsfl j ksl si j oy; i j [ khphx; hLi ' kZ
x2 + y2 = 5 and the parabola y2 = 4x intersect
j s[kk, safcUnqT i j i zfr PNsn dj r hgS
A , d oxZABCD bl
at the point T. A square ABCD is constructed
t hok i j cuk; k t kr kgSt ksfd i j oy ; dsHkhr j fLFkr gS
on this chord lying inside the parabola then
(A) [(TC)2 + (TD)2]2 = 6400
r c&
(A) [(TC)2 + (TD)2]2 = 6400
(B) TC and TD are equal and irrational
(C) TC > TD
(D) [(TC) + (TD)]2 = 6400
(B) TC
o TD l eku o vi fj es; gS
A
(C) TC > TD
(D) [(TC) + (TD)]2 = 6400
(SPACE FOR ROUGH WORK)
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Just when the caterpillar thought the world was ending, he turned into a butterfly.
RANK BOOSTER PROGRAM [FULL TEST - 4]
Page # 7
7.
If x  1, a  0 and the sum of the series
 sec 1

n 1 


8.
x  cos ec 1
a
1
2
(B) a 
(C) a 
1
2
(D) None of these
Let P(x) =
; fn x  1, a  0 gSr Fkk Js.kh
n
x 
 is finite then


(A) a 
cot2x
7.
 sec 1

n 1 


1
2
 1  tan x  tan 2 x 


 1  cot x  cot 2 x 


8.
x  cos ec 1
a
n
x 



dk; ksx i fj fer gS
]rc
(A) a 
1
2
(B) a 
(C) a 
1
2
(D) bues
al
ekuk P(x) =
cot2x
Then which of the following is(are) incorrect?
(A) Range of P(x) is [1, 2].
2
 cos x  cos 3x  sin 3x  sin x 
 gSr c fuEu esa
+ 
2 (sin 2x  cos 2x )


l sdkS
ul k@l sfodYi l ghughagS
a\
(A) P(x)
(B) The value of P(18°) + P(72°) is 3.
sdksbZugha
 1  tan x  tan 2 x 


 1  cot x  cot 2 x 


2
 cos x  cos 3x  sin 3x  sin x 
 .
2 (sin 2x  cos 2x )


+ 
1
2
dk i fj l j [1, 2] gS
a
.
(B) P(18°) + P(72°) dkeku 3 gS
a
.
(C) Range of P(x) is [0, 1].
(C) P(x)
(D) The value of P(18°) + P(72°) is 5.
(D) P(18°) + P(72°) eku 5 gS
a
.
dki fj l j [0, 1] gS
a
.
(SPACE FOR ROUGH WORK)
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RANK BOOSTER PROGRAM [FULL TEST - 4]
9.
Let f : R  [1, ) be a quadratic surjective
function such that f(2 + x) = f(2 – x) and
f(1) = 2. If g : (– , ln 2]  [1, 5) is given
by g (ln x) = f(x) then which of the following
is(are) correct ?
(A) The value of
(B) g–1(x) = ln
(C) g–1
2 
(x) = ln 2 
Page # 8
9.
f (3) is equal to 2.
(C) g–1

x  1
x 1
dsekuksdk ; ksx t ksfd l ehdj .k f (x) = 5 dks
la
rq
"V dj r sgS
a
] 4 gksxkA
(D) x
equation f (x) = 5 is 4.
From the point P (0, 1) tangent lines PA and
PB are drawn to the hyperbola
x2
y2

 1.
1
8
Which of the following statement(s) is(are)
correct?
(A) The x-intercept of tangent with negative
gradient is equal to 3.
(B) The acute angle between tangents PA
and PB is equal to tan 1

(x) = ln 2 
(B) g–1(x) = ln 2 

x  1
x 1
(D) The sum of values of x satisfying the
10.
ekukf : R  [1, ) f} ?kkr vkPNknd Qy u bl i zd kj
gS
afd f(2 + x) = f(2 – x) r Fkkf(1) = 2 gS
A; fn g
: (– , ln 2]  [1, 5) gSt ksg (ln x) = f(x) ds} kj k
fn; kt kr kgS
a
A r c fuEu esal sdkS
ul k@l sl ghgS
a\
(A) f (3) dk eku 2 dscj kcj gS
a
A.
4
.
3
(C) The area of triangle formed by the
tangents PA and PB and their chord of
contact is equal to 27.
(D) Least distance of mid-point of AB from
x-axis is equal to 8.
10.
vfr i j oy ;
x2
y2

 1 i j fcUnqP (0, 1) l sLi ' kZ
1
8
j s[kk, saPA r FkkPB [ khpht kr hgS
a
A
r c fuEu dFkuksaesal sdkS
ul k@dkS
ul sl ghgS
a\
(A) Li ' kZj s
[ kkdkx
3 dscj kcj
vUr%[ k.Mft l dh<ky _ .kkRed gS
a
gS
a
A
(B) Li ' kZj s
[ kkvksaPA r FkkPB dschp U;w
u dks.k tan 1
(C) Li ' kZj s
[ kkPA r FkkPB r FkkmudhLi ' kZt
4
gS
A
3
hokl scus
f=kHkq
t dk{ks=kQy 27 dscj kcj gS
a
A
(D) AB
dse/; fcUnqdhx-v{kl sU; w
ur e nw
j h8 gS
a
A
(SPACE FOR ROUGH WORK)
Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)
Just when the caterpillar thought the world was ending, he turned into a butterfly.
RANK BOOSTER PROGRAM [FULL TEST - 4]
Page # 9
SECTION - C
[MATCH THE COLUMN TYPE]
Match the entries in Column - I with the entries in
Column- II. One or more entries in Column - I may
match with one or more entries in Column - II.
1.
(A)
Column – I
Column – II
Let f be continuous and
(P) 0
the function F is defined as
x
SECTION - C
[l q
esy u l q
phi zd kj ]
Lr EHk- I dhi zfo"Vh; ksdkLr EHk - II dhi zfo"Vh; ksadsl kFkl q
esy u
dhft ; sALr EHk- I es, d ; k, d l svf/kd i zfo"Vh; kaLr EHk- II ds
, d ; kvf/kd i zfo"Vh; ksadsl kFkl q
esfy r gksl dr hgS
A
1.
(A)
Lr EHk &I
Lr EHk&II
ekuk f l r r ~gSr Fkk Qy u F
(P) 0
bl i zd kj i fj Hkkf"kr gSfd
t

2
F (x) =   t ·  f (u )du  dt


0
1

x
F (x) =
where f (1) = 3, then F'(1) + F''(1)
has the value equal to
(B)
t gk¡f (1) = 3, r c F'(1) + F''(1)
For each value of x a function
(Q) 1
f (x) is defined as min
{2x + 3,
t

 t 2 · f (u )du  dt
 

0
1

dk eku cj kcj gS
A
(B)
x
dsi zR; sd eku dsfy , , d
Qy u bl i zd kj i fj Hkkf"kr gSfd
( x  4) , 3(6 – x)}.
3
min.{2x + 3,
Maximum value of f (x) is
(C)
Lim
x 1
 ln x 
1
( x 1) tan x
Exponent of 2 in the binomial
coefficient
500C
212
is
( x  4) , 3(6 – x)}
3
r c f (x) dkvf/kdr e eku gksxkA
(R) 2
(C)
(D)
(Q) 1
(S) 3
(D)
Lim
x 1
 ln x 
f} i n xq
. kka
d
1
( x 1) tan x
500C
212
es2 dk
?kkr ka
d gksxk
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(R) 2
(S) 3
RANK BOOSTER PROGRAM [FULL TEST - 4]
2.
(A)
Column – I
Column – II
The number of solution(s)
(P) 1
of the equation
z2 = 4z + | z |2 +
(B)
16
| z |3
Page # 10
2.
(A)
Lr EHk &I
Lr EHk&II
l ehdj .k
(P) 1
z2 = 4z + | z |2 +
16
| z |3
dsgy ksdhl a
[ ; kgksxh
is
(where z = x + iy, x, y  R,
(t
i2 = –1 and x  2)
i2 = –1 r Fkkx  2)
Let two non-collinear vectors
(Q) 72


a and b inclined at an angle
(B)
gk¡z = x + iy, x, y  R,
ekuknksvl j s[kh; l fn' k
(Q) 72

2

a r Fkk b dks.k
ij
3
2

be such that | a | 3 and
3

| b | 4 . A point P moves so
bl i zd kj >q
d sgq
, gSfd


| a | 3 r Fkk | b | 4 ., d A fcUnqP
that at any time t the position
bl i zd kj xfr eku gSfd fdl h
vector OP (where O is the origin)
l e; t i j bl dk fLFkfr l fn' k OP
is given as
¼
t gk¡ O ew
y fcUnqgS½


OP = (et+e–t) a +(et – e–t) b .


OP = (et+e–t) a +(et – e–t) b
If the least distance of P from
} kj k fn; k t kr k gS; fn P l sew
y fcUnqdh
origin is
2
a  b where
a, bN then find the value of (a+b).
U; w
ur e nw
jh 2
a  b gS
t gk¡a, bN r c (a+b) dkeku Kkr dj ksA
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RANK BOOSTER PROGRAM [FULL TEST - 4]
Page # 11
(C)


Let a  3î  ĵ  k̂ , b  4î  2ˆj  4 k̂
(R) 288
(C)


ekuk a  3î  ĵ  k̂ , b  4î  2ˆj  4 k̂

r Fkk c  2î  2ˆj gS
A ; fn V1 l ekUr j
and c  2î  2ˆj . If V1 is the
volume of parallelopiped whose
"kV~
Qy d dkvk; r u gSft l dsr hu
three coterminous edges are
     
la
xkehfdukj sa  b, b  c, c  a

(R) 288
    
the vectors a  b, b  c, c  a
gSr Fkk V2 pr q
"Qy d dk vk; r u gS
and V2 is the volume of
ft l dsr hu l xka
ehfdukj s
tetrahedron whose three
     
a  b, b  c, c  a gSr c (V1 + V2)
coterminous edges are
     
the vector a  b, b  c, c  a .
dkeku Kkr dj ksA
Find the value of (V1 + V2).
(D)
(D)
Let the equation of the plane
(S) 11
containing the line x–y–z–4=0
= x + y + 2z – 4 and is parallel
ekukj s[kkx–y–z–4=0
= x + y + 2z – 4 dksl
(S) 11
ekfgr
dj usoky sr Fkkl er y ksa2x + 3y + z = 1
to the line of intersection of the
o x + 3y + 2z = 2 dhi zfr PNsnh
planes 2x + 3y + z = 1 and
j s[kkdsl ekUr j l er y dhl ehdj .k
x + 3y + 2z = 2 be
x + Ay + Bz + C = 0 gSr c
x + Ay + Bz + C = 0.
| A + B + C | Kkr
dhft , A
Find | A + B + C |.
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RANK BOOSTER TEST SERIES_FULL TEST - 4
Page # 12
PHYSICS [PAPER - II]
SECTION - A
[INTEGER ANSWER TYPE]
Q.1 to Q.8 are INTEGER ANSWER TYPE Questions.
(The answer of each of the questions is 1 digits)
1.
Particles of sand are sprinkled on to the
surface of water in a beaker, filled to a depth
SECTION - A
[ fo' y s
"k. kkRed i z'u i zd kj ]
i z-1 l si z-8 r d fo' y s"k.kkRed i z'u gS
A¼
i zR; sd i z'u dkmÙkj dsoy
1 va
d ksesnhft
1.
; sA
, d i k=kesat y dhl r g i j j sr dsd.kksdkfNM+
d ko fd; k
of 45.45 mm to reach the bottom. The sand
x; kgS
] xgj kbZr d i gq
pusdsfy, At y Lr j 45.45 mm
particles are spherical and have a diameter
gS
Aj sr dsd.kxksy h; gSr Fkk0.10 mm. O
; kl dsgS
Ai k=k
of 0.10 mm. The beaker is kept in a lift moving
1.2 m/s2. dsÅ/oZR
oj .kl
with an upward acceleration of 1.2 m/s2 .
sxfr dj r hgq
bZ, d fy ¶V esa
Estimate the minimum time to reach the
j [ kk gS
A d.kks} kj k r y r d i gq
pusdsfy , U; w
ur e l e;
bottom by the particles if viscosity of water
d hx. kuk d hft , ] ; fn t y d h' ; kur k = 1.1 × 10–
 = 1.1 × 10–3 Nsm–2, density of water is
3
1000 kg/m3, density of sand is 2000 kg/m3
Nsm–2,
t y dk ?kuRo 1000 kg/m3, j sr dk ?kuRo
2000 kg/m3
and g is 9.8 ms . (Express your answer in s)
r Fkkg is 9.8 ms–2. (vki dkmÙkj s esa
–2
2.
O
; Dr dhft , )
The temperature T and density  of an ideal
diatomic gas obey the equation T(1 – ) = k,
2.
r ki T r Fkk?kuRo dh, d f} i j ek.kq
d vkn' kZxS
l l ehdj .k
where k is a constant, If V and V' are
T(1 – ) = kdk i ky u
respectively the volumes, then it is found
; fn V r FkkV' Øe' k%vk; r u gS
] r c ; g i k; kt kr kgSfd
 15 
that
– 
 V '
p
 2 n  when    32. Then, what
p'
 V 
 V '
 V   32
 
dj r hgS
] t gk¡k , d fu; r ka
d gs]
15


– 
gS
] p  2 n  A r c n dk eku D; k gS
\
p'
is the value of n?
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RANK BOOSTER TEST SERIES_FULL TEST - 4
Page # 13
3.
Four cylindrical rods of metal are welded as
shown in Figure. The
A
R
rod AB is tapped at
R=
R
2
one- third points. If R 3
R
is
the
thermal D
R=
3
resistance of the rod R3
R
AB, the effe ct i v e C R = 6
R
thermal resistance of 3
the network of rods
B
R
between AB is
. What is the value of
38  n
n?
3.
3
pkj csy ukdkj NM+
ksadksfp=kkuq
l kj osYMfd; kx; kgS
ANM+
AB , d fr gkbZfcU
nq
v ksai j Vsij A
R
R=
gS
A ; fn R NM+AB dkÅ"eh; R
2
3
i zfr j ks/k gS
] r c AB dse/;
R
D
R=
3
NM+
ksadsusVodZdki zHkkohÅ"eh; R
3
2
2
1
4.
Inside a solid sphere of radius R and mass M,
a spherical cavity of
R
R/2
R
radius
is made, such
P
2
C
C
Q R
that the surface of the
cavity passes through
the centre C1 of the sphere. C2 is the centre
of the cavity. A particle of mass m is released
from rest at a point P, a distance R from the
surface of the sphere nearer to the cavity
as shown in the Figure. The speed with which
the particle strikes the point Q is given by u
1
=
R
i zfr j ks/k
38  n
D; k gS
\
4.
R
6
xksy sdsvUnj
1
2
pky l sVdj k, xkog u =
R
2
f=kT; k
nGM
3R
2
} kj knht kr hgS
An dk
eku Kkr dj ksA
nGM
. Find the value of n.
3R
A parallel beam of monochromatic light falls
on a thin biconvex lens with equal radius of
curvature R and of refractive index 1.5. After
two internal reflections, one at each face of
the lens, the beam of light comes out of the
lens and meets the optic axis at a distance
R f=kT; kr FkkM nz
O
; eku dsBksl
R1=
dhxksy h; xq
fgdkbl i zd kj
cuhgSfd xq
fgdk dhl r g
R/2
R
P
xksy s ds dsUnz C1 l s gksd j
C
C
Q R
xq
t j r hgS
AC2 xq
fgdkdkdsUnz
gS
AfcUnqP l sfoj kekoLFkkl s
, d m nzO
; eku dsd.kdksNksM+
kx; kgS
] xq
fgdkdsl ehi
xksy sdhl r g l sR nw
j hi j fp=kkuq
l kj ] d.kfcUnqQ l sft l
5.
5.
gS
An dkeku
3
C
R
3
B
, d , do.khZ
; i zd k' kdhl ekUr j che , d R oØr kf=kT; k
, oa1.5 vi ofr Z
r fxj r h gS
A nksvkUr fj d i j kor Z
uksads
i ' pkr ~
] , d yS
l dhi zR;sd l r g i j ] che y S
a
l l sckgj vkr h
a
gSr Fkky S
l dsdsUnzl s3 cm dhnw
a
j hi j i zd kf' kd v{ki j
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RANK BOOSTER TEST SERIES_FULL TEST - 4
Page # 14
3 cm from the centre of the lens. The focal
length of the lens is given by f = 3x cm. Find
the value of x.
6.
A 30 rpm record plate has a 6 kHz tone cut
in the groove. The number of full waves
recorded in a groove is of the order of 1.2 ×
10x. What is the value of x?
7.
A potential difference of 160 V is applied
between the plates of a parallel plate
capacitor. Midway between the plates, an
e l e ct ron and a prot on are rel ease d
simultaneously, such that electron is released
from rest and proton is projected at right
angles to the negative plates with an initial
speed. If the electron and proton reach the
positive and negative plates at the same
instant of time respectively, then find the
initial speed of the proton in the unit of
1
 106 ms–1.
3
me
[Take m  2000, Me = 9 × 10–31 kg and
p
e = 1.6 × 10–19 C]
8.
A charged particle of mass m and charge q is
shot along a horizontal rough plane with
coefficient of friction . A uniform magnetic
field is applied at right angles to the plane. If
the charged particles enters the magnetic
field region perpendicular to the uniform field
with velocity v0, then the radius fo curvature
of the path after time t 
0
is givey by
8g
7mv0
, where n is an integer. What is the
nqB
value of n?
r
fey r hgS
Ay S
a
l dhQksd l nw
j hf = 3x cm } kj knht kr h
gS
A x dkeku Kkr dhft , &
6.
, d 6 kHz l q
j ksadh30 rpm fj dkMZIysV [ kka
psl sdVhgS
A
[ kka
psesadVhdq
y iw
. kZr j a
xs1.2 × 10x dksfV dhgS
Ax dk
eku D; k gS
\
7.
, d l ekUr j Iy sV l a
/kkfj =k dhIy sVksadse/; 160 V dk
foHkokUr j y xk; k x; k gS
A Iy sVksadse/; e/; fcUnql s] , d
bysDVªkW
u r Fkk, d i zksVksu dksbl i zd kj , d l kFkNksM+
kx; k
gSfd bysDVªkW
u dksfoj kekoLFkkl sr Fkki zksVksu dks_ .kkRed
Iy sV dhvksj y Ecor ~fdl hi zkj fEHkd osx l si z{ksfi r fd; k
t kr k gS
A ; fn by sDVªkW
u r Fkk i zksVksu Øe' k%/kukRed r Fkk
_ .kkRed Iy sVksai j l e; ds, d {k.ki j i gq
¡psr c i zksVksu
dhi zkj fEHkd pky
[
1
 106 ms–1 dsek=kd
3
esaKkr dhft , &
me
 2000, M =9×10–31 kg and e=1.6×10–19 C
e
mp
yhft , ]
8.
, d m nzO
; eku r Fkk q vkos'k  dk ?k"kZ
. k xq
. kka
d ds
dkd.k [ kq
j nj s{kS
fr t r y i j nkxkt kr k gS
A , d l e: i
pq
Ecdh; {kS
=kr y dsy Ecor ~vkj ksfi r fd; kt kr kgS
A; fn
vkosf' kr d.kl e: i {kS
=kdsy Ecor ~{ks=kesav0 osx l si zos'k
dj r kgS
]rc t 
r
7mv0
nqB
0
8g
l e; i ' pkr ~i FkdhoØr kf=kT; k
} kj knht kr hgS
] t gk¡n , d i w
. kkZ
d gS
A n dk
eku Kkr dhft , \
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RANK BOOSTER TEST SERIES_FULL TEST - 4
Page # 15
SECTION - B
SECTION - B
[ cgqoS
d fYi d i z'u i zd kj
[MULTIPLE OBJECTIVE TYPE]
Q.1 to Q.10 has four choices (A), (B), (C), (D)
out of which ONE OR MORE THAN ONE is correct
i z-1 l si z-10 esapkj fodYi
1.
; k, d l svf/kd^^ l ghgS
A
2.
Which of the followin statements is/are false?
(A) The magnitude of momentum of a heavy
object is greater than that of a light object
moving at the same speed
(B) In a perfectly inelastic collision, all the
initial kinetic energy of the colliding bodies is
dissipated
(C) The momentum of a system of colliding
bodies may be conserved even though the
total mechanical energy may not be
(D) The velocity of the center of mass of a
system is the system's net momentum divided
by its total mass
Two satellites S1 and S2 revolve round a planet
in coplanar circular orbits in the same sense.
Their periods of revolution are 1 hr and 8 hr
respectively. The orbital radius of S1 is 104
km when S2 is closest to S1. Then,
(A) the speed of S2 relative to S1 is  × 104
km/hr.
(B) the angular speed of S1 as actually

observed by an astronaut in S2 is rad / hr .
6
(C) the angular speed of S2 as actually
7
rad / hr.
observed by an astronaut in S1 is
4
(D) the speed of S2 as actually observed by
an astronaut in S1, when S1 and S2 are
farthest from each other is 3 × 104 km/hr.
1.
]
(A), (B), (C), (D) gS
] ft
uessal s^^, d
fuEu esal sdkS
ul k@dkS
ul sdFku vl R; gS
\
(A) l eku pky i j , d H
kkj hoLr qdsl a
osx dki fj ek.kgYdh
oLr ql svf/kd gksxk
(B) , d i w
. kZ
r %vi zR; kLFk VDdj esa
] VDdj dj usoky h
oLr q
v ksadhi w
j hi zkj fEHkd xfr t Åt kZmRl ft Z
r gkst kr hgS
A
(C) VDdj
dj usoky hoLr q
v ksadsfudk; dkl a
osx l a
j f{kr
gksl dr kgS
] t cfd i w
. kZ; ka
f=kd Åt kZdkl a
j {k.kughaHkhgksA
(D) , d fudk; dsnz
O
; eku dsUnzdhxfr fudk; dsdq
y
la
osx r Fkkbl dsdq
y nzO
; eku dkvuq
i kr gS
2.
nksS1 r FkkS2 mi xzg , d vFkZesa, d ghxzg dsi fj r %xksy h;
d{kkesa?kw
e j gsgS
Abuds?kw
. kZ
u dsvkor Z
d ky Øe' k%1 hr
r Fkk8 hr gS
At c S2, S1 dsfudVr e gS
] r c S1 dhd{kh;
f=kT; k104 km gS
Ar c]
(A) S2 dhS1 dsl ki s
{kpky  × 104 km/hr gS
A
(B) mi xz
g S2 esal sS1 dhvUr fj {k ; k=kh} kj k okLr fod

i zsf{kr dks.kh; pky rad / hr gS
A
6
(C) mi xz
g S1 esal sS2 dhvUr fj {k ; k=kh} kj k okLr fod
7
rad / hr gS
i zsf{kr dks.kh; pky
A
4
(D) S1 es
al sS2 dhokLr fod i zsf{kr pky ] nkc S1 r FkkS2
, d nw
l j sl svf/kdr e nw
j hgS3 × 104 km/hr gS
A
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RANK BOOSTER TEST SERIES_FULL TEST - 4
3.
4.
A wooden plank OP of length 1m and uniform
cross - section 's' is hinged at one end to
the bottom of a tank. The tank is filled with
water up to a height h m. The specific gravity
s of the plank is 0.5. In the equilibrium
position, the plank makes angle  with the
vertical as shown in Figure. Then, if  = 45º,
(A) the value of
P
h is 0.5 m.
(B) horizontal
component of
reaction at O is
zero.
(C )
ot he r
t h i n g s
remaining the
O
same, centre of buoyancy would get shifted
if s is altered.
(D) l ocati on of centre of buoyancy i s
 s 
proportional to   .
  
A conducting shell, having charge 15mC and
radius a is surrounded by another concentric
conducting shell of charge –15 mC and inner
radius 2a and outer radius 3a as shown in
the Figure. If the inner shell is grounded, then,
(A) the charge on
–Q
inner shell is 6 mC.
+Q = 15 mC
(B) the charge on
outer surface of
a
outer shell is zero
(C) the charge on
inner shell is zero
3a
2a
(D) the charge on
inner surface of
outer shell is – 6 mC
Page # 16
3.
, d 1m yEckbZr Fkkl e: i dkV 's' dkVS
d dsr y ds, d
a
fl j sl sVa
xkgq
v ky dM+
hdkCykW
d OP gS
AVS
a
d t y l shm
Å¡pkbZr d Hkj k gS
A xq
Vdsdk fof' k"V xq
: Ro s, 0.5 gS
A
fp=kkuq
l kj l kE; koLFkkdhfLFkfr esaml dkÅ/ok/Zkj dsl kFk
dks
. kcukr k gS
A r c ; fn  = 45º gS
]
(A) h dk eku
0.5 m gS
P
(B) O i j i z
fr fØ; kdk
{kS
fr t ?kVd ' kw
U; gS
A
(C) ; fn s cny
fy ; k t k, r ksckdh
j kf' k; ksadsl eku j gr s
gq, mRi y kou d k
O
dsUnz foLFkkfi r gks
t kr kgS
A
 
(D) mR
i y kou dsdsUnzdhfLFkfr  s 
  
4.
dsl ekuq
i kr hgS
A
, d 15mC vkos'k r Fkk a f=kT; k dk pky d dks'k vkos'k
–15 mC r FkkvkU
r fj d f=kT; k2a r Fkkckg~
; f=kT; k3a ds
la
d sfUnz; pky d dks'k l si fj c) gS
A ; fn vkUr fj d dks'k
Hkw
&l Ei fdZ
r gS
]rc
–Q
(A) vkU
r fj d dks'ki j
+Q = 15 mC
vkos'k6mC gS
A
a
(B) ckg~
; dks'kdhckg~
;
l r g i j vkos'k' kw
U; gS
A
3a
2a
(C) vkU
r fj d dks'k i j
vkos'k' kw
U; gS
(D ) ckg~
; d ks' k d h v kUr fj d l r g i j
– 6 mC vkos
' kgS
A
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RANK BOOSTER TEST SERIES_FULL TEST - 4
Page # 17
5.
Two charged particles M and N enter a space
of a uniform magnetic field with velocities
perpendicular to the magnetic field. The paths
are as shown in Figure below. Momentum of
the chargd particles are p1 and p2 respectively,
then
(A)
the
charge of
M is
p
p
>
q
q
greater
N
M
than that
M
N
of N
if p1 = p2
(B) the momentum of N is greater than that
of M
(C) the specific charge of M is greater than
that of N, if v is same
(D) the speed of M is less than that of N
5.
p1 = p2 gSr ks
Mij
Mark the correct statements(s):
(A) Direction of wave propagation is along
the normal to wavefront
(B) For a point source of light, the shape of
wavefronts can be considered to be plane at
very large distance from the source
(C) A point source of light is placed at the
focus of a thin spherical lens, then the shape
of the wavefront for emerged light can be
plane
(D) The shape of the wavefront for the light
incident on a thin spherical lens is plane, the
shape of the wavefront corresponding to
emergent light would be always spherical
vkos'kN
p
p
>
l svf/kd gS
A
q
q
N
M
(B) N dk
M
N
la
osx M l s
vf/kd gS
A
(C) M dkfof' k"V vkos
' kN l svf/kd gS
] ; fn v l eku gS
(D) M
6.
6.
, d l e: i pq
Ecdh; {ks=k esapq
Ecdh; {ks=k dsy Ecor ~nks
vkosf' kr d.kM r FkkN i zos'kdj r sgS
Ai Fkfp=kkuq
l kj uhps
fn; sx; sgS
Avkosf' kr d.kksadsl a
osx Øe' k%p1 r Fkkp2 gS
]rc
(A)
; fn
dhpky N l sde gS
l ghdFku@dFkuksadksfpfUgr dhft , %
(A) r j a
x dsxeu dhfn' kkr j a
xnS
/; ZdsvfHkyEc dsvuq
fn' k
gS
A
(B) i z
d k' k, d fcUnqL=kksr dsfy , ] r j a
xkxzdhvkdf̀r L=kksr
l scgq
r vf/kd nw
j hi j l er y ekuht kr hgS
A
(C) , d i z
d k' k dk fcUnqL=kksr , d i r y sxksy h; y S
a
l ds
Qksd l i j j [ kk x; k gS
] r c fuxZ
r i zd k' k dsr j a
xkxzdh
vkdf̀r l er y gksl dr hgS
(D) i r y sxks
y h; y S
l i j vki fr r i zd k' k dsr j a
xkxzdh
vkdf̀r l er y gS
] fuxZ
r fdj .kdsvuq
: i rja
xkxzdhvkdf̀r
ges'skkxksy h; gksxh
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RANK BOOSTER TEST SERIES_FULL TEST - 4
7.
In the circuit given below, if R1 = 240 and
R2 = 120 respectively, L is a lamp of 300 W
then
Page # 18
7.
fn; sx; sfuEu i fj i Fk esa
] ; fn Øe' k%R1 = 240 r Fkk
R2 = 120 gS
( L 300 W dh, d y S
Ei gS
] rc
A
A
120V
60
120V
60
R2
R2
300W L
300W L
(A) R1 es
avi O
; ; ' kfDr 60 W gS
A
(B) R2 es
avi O
; ; ' kfDr 120 W gS
A
(C) i fj i Fkdki z
Hkkohi zfr j ks/k50gS
A
(D) vehVj A 6 A i <+
r k gS
A
(A) power dissipated in R1 is 60 W
(B) power dissipated in R2 is 120 W
(C) effective resistance of the circuit is 50
(D) the ammeter A reads 6 A
8.
R1
R1
A cylinder has 40.0 cm radius and is 50.0 cm
deep. It is filled with air at 20ºC and 1.00
atm (Figure a). A 20.0 kg piston is now slowly
lowered into the cylinder compressing the air
8.
, d csy u 40.0 cm f=kT; kr Fkk50.0 cm xgj kgS
A; g
20ºC
r Fkk 1.00 atm (fp=ka) i j ok; ql sHkj kgS
A vc
, d 20.0 kg dkfi LVu /khjs&/khjsok; qdkscsy u esal EihfM+
r
trapped inside (Figure b). Finally, a 75.0 kg
dj r kgq
v kuhpsvkr kgS
Av{kesa
] bl dsckn , d 75.0 kg
man stands on the piston further compressing
dk, d vknehfi LVu i j ok; qdksl a
i hfMr dj r sgq
, [ kM+
k
the air, without any appreciable change in
gksrkgS
] fcukfdl hr ki dsi ; kZ
Ir i fj or Z
u dsAok; q
e.Myh;
temperature. The atmospheric pressure p0 =
nkc p0 = 1 × 105 N/m2 gS
] rc &
1 × 10 N/m . Then,
5
2
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RANK BOOSTER TEST SERIES_FULL TEST - 4
Page # 19
50cm
hi
Fig.a
50cm
Fig.b
Fig.c
Fig.a
(A) when the man steps on the piston, it
moves down by 0.011m.
(B) the temperature to which the gas be
heated so as to raise the piston and the man
back to hi is 24.2ºC.
(C) when the man stands on the piston the
process involved is isothermal.
(D) when the gas is heated, the process
involved is isobaric.
9.
A particle of charge 32 nC is in a uniform
electric field. In addition to the electric force,
another force acts on the particle so that
when it is released from a point O at rest, it
moves against the electric field. The kinetic
energy of the particle after it has moved a
distance of 8 cm from O found to be 4.8 ×
10–5 J at point A and the additional force has
done a work of 6.4 × 10–5 J. Point out correct
statement/s from the following:
(A) The work done by the electric field is
16J.
(B) The potential at point O is less than that
at A by 500 V.
(C) The magnitude of electric field is 6.25 ×
103 NC–1
(D) The point O is at higher potential than
the point A by 500 V.
hi
Fig.b
Fig.c
t c vknehfi LVu i j p<+
r k gS
] ; g 0.011m uhps
t kr kgS
A
(B) og r ki ft l r d xS
l dksxeZdj usi j fi LVu r Fkk
vknehhi r c i q
u%i gq
¡p t k, 24.2ºC gS
A
(C) t c vknehfi LVu ; g p<+
r k gS
] i zØe l er ki h; gS
A
(D) t c xS
l xeZdht kr hgS
] i zØe l enkch; gS
A
(A)
9.
, d 32 nC vkos'k dk d.k , d l e: i fo| q
r {kS
=k esagS
A
fo| q
r cy dsvfr fj Dr ] , d vU; cy bl i zd kj gSfd t c
bl sfcUnqO l sfoj kekoLFkkl sNksM+
kt kr kgS
] ; g fo| q
r {ks=k
dsfoi j hr xfr dj r k gS
A bl dsO l s8 cm dhnw
j hr ;
d j us d s i ' pkr ~d .k d h xfr t Åt kZfcUnq A i j
4.8 × 10–5 J gSr Fkkvfr fj Dr cy 6.4 × 10–5 J dk; Z
dkpq
d k gS
A fuEu esal sl ghdFku pq
fu, &
(A) fo|
(B)
fcUnqO i j foHko A i j 500 V l sde gS
A
(C) fo|
(D)
q
r {ks=k} kj kfd; kx; kdk; Z16J gS
A
q
r {ks=kdki fj ek.k6.25 × 103 NC–1 gS
A
fcUnqO, A i j 500 V l smPpr j l sfoHko i j gS
A
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RANK BOOSTER TEST SERIES_FULL TEST - 4
10.
A plane spiral made of stiff smooth wire is
rotated with a constant angular velocity  in
a horizontal plane about the fixed vertical
axis O. A small sleeve M slides along that
spiral without friction. Find its velocity v'
relative to the spiral as a function of the
distance  form the rotation axis O if the
initial velocity of the sleeve is equal to v'0.
(A) v '20  22
2
0
Page # 20
10.
, d n`<+fpdusr kj dkl fi Z
y r y fu; r dks.kh; osx l s
fLFkj Å/okZ
/kj v{kO dsi fj r %{kS
fr t r y esa?kw
. kZ
u dj j gk
gS
A, d NksVkvkoj .kM l fi Z
y dsvuq
fn' k?k"kZ
. kl sfQl y r k
gS
Abl dkosx v' ?kw
. kZ
u v{kO l snw
j hdsQy u ds: i esa
l fi Z
y dsl ki s{kKkr dhft , ] ; fn vkoj .kdki zkj fEHkd osx
v'0 gS
A
V'
(A) v '20  22
(B) v '   
(C)
v '20  22
(D)
v '20  22
O
2
0
[MATCH THE COLUMN TYPE]
Match the entries in Column - I with the entries in
Column- II. One or more entries in Column - I may
match with one or more entries in Column - II.
A charged spherical conductor A of radius rA
is surrounded by a hollow conductor B
of radius rB such
q
B
that r B > r A as
A
shown in Figure. A
r
is given a charge
q
qA and that of B,
qB. VA and VB are
r
the potential of A
and B respectively.
Now match the items in Column I with those
in Column II for the conditions indicated in
Column I.
B
A
A
B
2 2
(B) v '   
M
2
0
O
M
2 2
(C)
v'   
(D)
v '20  2 2
SECTION - C
1.
V'
2 2
SECTION - C
[l q
esy u l q
phi zd kj ]
Lr EHk- I dhi zfo"Vh; ksdkLr EHk - II dhi zfo"Vh; ksadsl kFkl q
esy u
dhft ; sALr EHk- I es, d ; k, d l svf/kd i zfo"Vh; kaLr EHk- II ds
, d ; kvf/kd i zfo"Vh; ksadsl kFkl q
esfy r gksl dr hgS
A
1.
rA f=kT; kdkvkos
f' kr pky d
[ kks[ky spky d B l sbl
i zd kj f?kj kgSfd rB > rA
xksy kA , d rB f=kT; kds
qB
B
A
tS
l kfd fp=kesan' kkZ
;k
r
q
gS
A A i j vkos'k qA r Fkk
B i j qB fn; kgS
AA r Fkk
r
B dsfoH
ko Øe' k%
VA r Fkk
VB gS
A vc Lr EHk I dh
Lr EHk l si zfr f"B; ksadk fey ku dhft , ] t ksLr EHk II dh
fLFkfr ; ksadksn' kkZ
r hgS%
A
A
B
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RANK BOOSTER TEST SERIES_FULL TEST - 4
Page # 21
Column I
(A) qA = + m C and qB
= +n C and m < n
Column II
(P) potential
difference
depends on qA only
(Q) VA > VB
(B) qA = – m C and
qB = – n C and m > n
(C) qA = – m C and
(R) VA < VB
qB = – n C and m < n
and A and B are
connected by a
conducting wire
(D) qA = – m C and
(S) charge
qB = – n C and m > n
flows from A to B
and A and B are connected
by a conductiing wire
(T) qA = 0
2.
Match the entries of Column I with that of
column II
Column I
Column II
(A) For a particle moving in (P) The acceleration
a circle
may be perpendicular
to its velocity
(B) For a particle moving in (Q) The acceleration
a straight line velocity
may be in the direction
of
(C) for a particle undergoing (R) The acceleration
projecticle motion with
may be at some angle


angle of projection  :
  0     with the
2


0
velocity
2
(D) For a particle is moving (S) The acceleration
in space
may be opposite to its
velocity
Column I
(A) qA = + m C r Fkk
qB = +n C r Fkkm < n
(B) qA = – m C r Fkk
qB = – n C r Fkkm > n
(C) qA = – m C r Fkk
qB = – n C r Fkkm < n
r Fkk A r Fkk B pky d r kj
Column II
(P) foH
kokUrj dsoy
qA i j fuH
kZ
j dj r kgS
(Q) VA > VB
(R) VA < VB
l st q
M+
kgS
(D) qA = – m C r Fkk
qB = – n C r Fkkm > n
r Fkk A r Fkk B pky d r kj
(S) vkos
' kA l sB dhvksj
cgr kgS
l st q
M+
kgS
(T) qA = 0
Lr EHkI dsl kFk Lr EHkII dksfey kb, &
Lr EHkI
Lr EHk II
(A) , d oR̀
r esaxfr ' khy
(P) R
oj .kbl dsosx ds
, d d.k dsfy ,
y Ecor gksl dr kgS
A
(B) , d l h/khj s
[ kkesaxfr ' khy
(Q) R
oj .kosx dh
, d d.k dsfy ,
fn' kkesagksl dr kgS
A
(C) i z
{ksi.kdks.k l s
(R) R
oj .kosx dsl kFk
2.
0

2
i z{ksI; xfr esa
, d d.k dsfy ,
(D) d.kvU
r fj {kesa
xfr ' khy gksusdsfy ,
:


  0    
2

dks.ki j gksl dr kgS
A
(S) R
oj .kbl dsosx dh
foi j hr gksl dr kgS
A
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RANK BOOSTER TEST SERIES_FULL TEST - 4
Page # 22
CHEMISTRY [PAPER - 1]
SECTION - A
[INTEGER ANSWER TYPE]
SECTION - A
[ fo' y s
"k. kkRed i z'u i zd kj ]
Q.1 to Q.8 are INTEGER ANSWER TYPE Questions.
(The answer of each of the questions is 1 digits)
i z-1 l si z-8 r d fo' y s"k.kkRed i z'u gS
A¼
i zR; sd i z'u dkmÙkj dsoy
1.
If the percentage of water of crystallization
in MgSO4. xH2O is 13%. What is the value of
x?
2.
The graph of compressibility factor (Z) vs. P
for one mole of a real gas is shown in following diagram. The graph is plotted at constant temperature 273 K. If the slope of
 dZ 
 is
graph at very hi gh pre ssure 
 dP 
1 va
d ksesnhft
1.
; fn MgSO4. xH2O esafØLVy hdj .k dst y dk i zfr ' kr
13% gSr ksx dk eku D; k gS
?
2.
, d eksy okLr fod xS
l dsfy , l Ei hM~
; r k xq
. kka
d (Z)
fo: ) P dkvkj s[kuhpsfp=kesafn[ kk; kx; kgS
AxzkQ fu; r
r ki 273 K i j cuk; k x; k gS
A ; fn cgq
r mPp nkc
 dZ 

ij
 dP 
K–1 mole
fn; k gq
v k gS
%NA = 6 × 1023 r FkkR =
22.4
and R =
L atm
273
K–1 mole
22.4
L atm
273
–1
–1
 dZ 
1
atm1


 dP  2.8
Z
 dZ 
1
atm1


 dP  2.8
Z
 1 
xzkQ dh<ky dk eku  2.8  atm–1 gS
] r ks


, d eksy okLr fod xS
l v.kq
v ksadkvk; r u Kkr dj ksA(1L/
mole es
)a
 1 

 atm–1, then calculate volume of one
 2.8 
mole of real gas molecules (in 1L/mole)
Given : NA = 6 × 1023
; sA
P
P
3.
Count the total number at – O bonds which
are having equal length in HSO–4 .
3.
– i j O ca
/kksdhdq
y
la
[ ; kKkr dj kst ksHSO–4 esacj kcj
y EckbZj [ kr sgS
\
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RANK BOOSTER TEST SERIES_FULL TEST - 4
Page # 23
4.
5.
How many geometrical isomers are possible
for given compound.
CH = N – OH
CH3 – CH = C
CH = N – OH
4.
Total no. of orbital taking part in the
5.
CH = N – OH
CH3 – CH = C
CH = N – OH
d j .kesaHkkx y susoky sd{kdksadhdq
yla
[;k
MnO–4 dsl a
gS
\
hybridised of MnO–4 .
6.
6.
fn; sx; s; kS
fxdksdsdq
y fdr usT; kfer h; l eko; oh
l EHko gS
A
CH3 – CH – CH = CH – CH – CH3
OH
OH
OH
Stereoisomers for given compound will be -
7.
In Nido Borane How many hydrogen atom in
Formula (taking n = 1).
8.
When one litre of a saturated solution of PbCl2
(mol. wt. = 278) is evaporated, the residue
is found to weight 2.78 g. If Ksp of PbCl2 is
represented as y × 10–6 then find the value
of y.
CH3 – CH = CH – CH –CH = CH – CH3
fn; sx; s; kS
fxd dsdq
y fdr usf=kfoe l eko; ohl EHko
gksxsA
7.
uk; Mkscksjsu esal w
=k esafdr usgkbMªkst u i j ek.kqgksrsgS
A
(y s
r sgSn = 1).
8.
PbCl2 (mol. wt. = 278) ds, d
yhVj l kUrfj r foy; u
dkokf"i r fd; kx; kr ksvo' ks"kdkHkkx 2.78 g i k; kx; k
gS
A; fn PbCl2 dkKsp, y × 10–6 } kj kfu: fi r gksrkgS
r ksy dk eku Kkr dj ks\
SECTION - B
SECTION - B
[MULTIPLE OBJECTIVE TYPE]
[ cgqoS
d fYi d
Q.1 to Q.10 has four choices (A), (B), (C), (D)
i z-1 l si z-10 esapkj fodYi
out of which ONE OR MORE THAN ONE is correct
; k, d l svf/kd^^ l ghgS
A
1.
For the reacti on A
i z'u i zd kj
]
(A), (B), (C), (D) gS
] ft
uessal s^^, d
 B, the rate l aw
d[A]
1/2
expression is 
d t = k [A] . If initial
concentration of [A] is [A]0, then
1.
d[A]
vfHkfØ; kA  B dsfy , ] nj fu; e O
;a
t d  dt =k
[A]1/2 gS
a
A ; fn [A] dhi zkj fEHkd
l kUnzrk[A]0 gS
a
] r c&
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RANK BOOSTER TEST SERIES_FULL TEST - 4
Page # 24
(A) The integerated rate expression is
(A)
2 1/ 2
1/ 2
k = (A 0  A )
t
l ekdy u nj O
;a
t d k=
(B)
(B) The graph of
2[ A ]10/ 2
(D) vfH
kfØ; kds75% i w
. kZgksusdsfy ,
[ A]0
t3/ 4 =
k
2.
The correct order of increasing X – O – X
bond angle is (X = H, F or Cl) :
(A) H2O > Cl2O > F2O
(B) Cl2O > H2O > F2O
(C) F2O > Cl2O > H2O
(D) F2O > H2O > Cl2O
3.
Which of the following molecules are chiral ?
OH
HO
(B)
[ A]0
k
gS
a
A
X - O - X ca
/kdks.kdkc<r kgq
v kØe gS(X = H, F or
Cl) :
(A) H2O > Cl2O > F2O
(B) Cl2O > H2O > F2O
(C) F2O > Cl2O > H2O
(D) F2O > H2O > Cl2O
3.
fuEu esal sdkS
ul kv.kqfdj S
y gS?
CH3
OH
H
(A)
HO
fy ; kx; kl e;
2.
CH3
OH
gksxkA
K
t
(C) v) Zvk; qdky 1 / 2 = 2[ A ]1 / 2
0
K
(D) The time taken for 75% completion of
(A)
A Vs t dk xzkQ
A Vs t will be
(C) The half life period t1 / 2 =
reaction t 3 / 4 =
2 1/ 2
(A 0  A1 / 2 ) gS
a
A
t
H
CH3
(B)
OH
HO
H
HO
H
CH3
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RANK BOOSTER TEST SERIES_FULL TEST - 4
Page # 25
OH
(C)
OH
CH3
(D)
CH2
(C)
OH
4.
5.
6.
7.
CH3
(D)
CH2
OH
Select correct statement about hydrolysis
of BCl3 and NCl3
(A) NCl3 is hydroysed and gives HOCl but BCl3
is not hydrolysed.
(B) Both NCl3 and BCl3 on hydrolysis gives
HCl
(C) NCl3 on hydrolysis gives HOCl but BCl3
gives HCl.
(D) Both NCl3 and BCl3 on hydrolysis gives
HOCl.
Which statement/s is/are correct
(A) A solution is prepared by addition of
excess of AgNO3 solution in KI solution. The
charge likely to develop on colloidal particle
is positive.
(B) The effects of pressure on physical
adsorption is high if temperature is low.
(C) Ultracentrifugation process is used for
preparation of lyophobic colloids.
(D) Gold number is the index for extent of
gold plating done.
Which of the following species is (are)
isostructural with XeF4 ?
(A) ICl4–
(B) I5–
(C) BrF4–
(D) XeO 4
Which of the following has been arranged in
order of decreasing dipole moment ?
4.
BCl3 r FkkNCl3 dst y h; dj .kdsfy , l ghdFku gS\
(A) NCl3 t y h; Ñr gks
r k gSvkS
j HOCl nsrk gSy sfdu
BCl3 t y h; Ñr ughgks
r kgS
(B) NCl3 r FkkBCl3 nks
uksat
y h; Ñr gksd j HCl nsrsgS
(C) NCl3 t y h; Ñr gks
d j HOCl nsrk gSt cfd BCl3,
HCl ns
r kgS
A
(D) nks
uksaNCl3 r FkkBCl3 t yh; Ñr gksd j HOCl nsrsgS
5.
dkS
ul kdFku l R; gS
a
&
dsvkf/kD; dsl kFk KI foy ; u esavkosf' kr
dksy kbMhd.kksai j /kukos'kgksrkgS
a
A
(A) AgNO3
(B) H
kkS
fr d
vf/k' kks"k.ki j nkc dki zHkko mPp gksxk] ; fn r ki
fuEu gS
A
(C) vYVªkl
sfUVªQ; w
t u i zØe nzoj ks/khdksy kbMhdkscukusesa
mi ; ksx gksrkgS
a
A
(D) Lo.kZl
6.
7.
a
[ ; k] Lo.kZIy sfVa
x dsi zl kj dsfy , bUMsDl gS
A
fuEufy f[ kr esal sfdl Li h'kht dhl a
j pukXeF4 dsl eku
gS
(A) ICl4–
(B) I5–
(C) BrF4–
(D) XeO 4
fuEufy f[ kr dksmudsf} /kzq
o vk/kw
. kZds?kVr sgq
; sØe esa
O
; ofLFkr dhft , ?
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RANK BOOSTER TEST SERIES_FULL TEST - 4
Page # 26
(A) CH3Cl > CH3F > CH3Br > CH3I
(B) CH3F > CH3Cl > CH3Br > CH3I
(C) CH3Cl > CH3Br > CH3I > CH3F
(D) CH3F > CH3Cl > CH3I > CH3Br
Products
Products are:
(A) Ph – CH2–I
(C) Ph – I
9.
(A)
(B)
(C)
(D)
10.
HI
O — CH2
8.
(A) CH3Cl > CH3F > CH3Br > CH3I
(B) CH3F > CH3Cl > CH3Br > CH3I
(C) CH3Cl > CH3Br > CH3I > CH3F
(D) CH3F > CH3Cl > CH3I > CH3Br
HI
O — CH2
8.
mRikn gS:
(B) Ph – CH2–OH
(D) Ph – OH
Pick out the incorrect statement among the
following
Change i n Gibb's function at constant
temperature and pressure in a process
envolving no non PV work is equal to qrev –
qirr
For a reaction with S 0 is positive, as
temperature is increased, Keq for the reaction
increases necessarily
Exothermic reactions have lesser tendency
to g o i n forw ard di re ct i on at hi gher
temperature
Decrease in Gibb's function at constant
temperature and pressure is equal to non PV
work done by system in spontaneous process
Correct statement(s) about fol l owi ng
compounds
9.
(A)
(B)
(C)
(D)
10.
(A) Ph – CH2–I
(B) Ph – CH2–OH
(C) Ph – I
(D) Ph – OH
fuEufy f[ kr esavl R; dFku dkpq
uko dj ks
fxC
l Qyu dkfu; r r ki o nkc i j i fj or Z
u t ksdksbZukW
u
PV dk; Zughans
r k gS
] qrev – qirr dsl eku gS
a
A
fdl hvfHkfØ; kdsfy, S /kukRed gS
at S
l s&2 r ki c<k; k
t kr kgS
aKeq dkeku vfHkfØ; kesavko' ; d : i l sc<r kgS
A
a
Å"ek{ksihvfHkfØ; kdsfy , mPp r ki i j vkxst kusdhi zof̀Ùk
de gksrhgS
a
A
Lor %i zfØ; kesafxCl Qy u esal eku r ki nkc i j dehr a
=k
} kj k non PV dsdk; Zcj kcj gksrkgS
a
A
fn; sx, ; kS
fxdksdsckj sesal ghdFku gS
a
A
OH OH
OH OH
HO
OH
H
H
(P)
(Q)
OH
COOH
COOH
HO
(P)
mRikn
H
OH
(Q)
OH
OH
H
OH
COOH
COOH
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RANK BOOSTER TEST SERIES_FULL TEST - 4
Page # 27
HO
HO
OH
(R)
(R)
(A) Compound (P) have cos
(A)
; kS
fxd (P) l fEer r y gS
A
(B) ; kS
fxd (P) (Q) vkS
j (R) eht ksgS
A
(C) ; kS
fxd (R) ckg~
; i zfr dkj dsdkj.ki zd k'kh; fuf"Ø; gS
A
(D) ; kS
fxd (Q) vkS
j (R) - cU/kdspkj ksavksj ?kw
e l dr s
gS
ay sfdu f=kohe l eko; or kesaoghgksrkgS
A
(B) Compound (P) (Q) & (R) are meso
(C) Compound (R) is optically inactive due to
external compensation
(D) Compound (Q) and (R) can rotate along
-bond but stereochemistry will remain same
SECTION - C
[l q
esy u l q
phi zd kj ]
SECTION - C
[MATCH THE COLUMN TYPE]
Match the entries in Column - I with the entries in
Column- II. One or more entries in Column - I may
match with one or more entries in Column - II.
1.
Match the column
Column–I
(A) Magenetic moment in a paramagnetic
substance
(B) Magnetic moment in a ferromagnetic
(C) Magnetic moment in a antiferromagnetic
(D) Magnetic moment in a ferrimagnetic
Column–II
(P)
(S)
Lr EHk- I dhi zfo"Vh; ksdkLr EHk - II dhi zfo"Vh; ksadsl kFkl q
esy u
dhft ; sALr EHk- I es, d ; k, d l svf/kd i zfo"Vh; kaLr EHk- II ds
, d ; kvf/kd i zfo"Vh; ksadsl kFkl q
esfy r gksl dr hgS
A
1.
fey ku dhft , &
Lr a
Hk–I
(A) vuq
pq
Ecdh; i nkFkZesapq
Ecdh; vk?kw
. kZ
(B) y kS
gpq
Ecdh; i nkFkZesapq
Ecdh; vk?kw
. kZ
(C) i z
fr y kS
g pq
Ecdh; i nkFkZesapq
Ecdh; vk?kw
. kZ
(D) y ?kqy kS
g pq
Ecdh; i nkFkZesapq
Ecdh; vk?kw
. kZ
Lr a
Hk–II
(P)
(Q)
(R)
OH
(Q)
(R)
(S)
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RANK BOOSTER TEST SERIES_FULL TEST - 4
2.
Match the column
Column-I
O
O
S
2.
Match the column
Column-I
O
O
OH
SH – CH – NH2
(B)
OH
SH – CH – NH2
OH
CH=CH–COOH
OH
CH=CH–COOH
(C)
(C)
(D)
S
(A)
(A)
(B)
Page # 28
CH3 – CH – CH2 – OH
Column-II
(P) p – d Resonance
(Q) p – p Resonance
(R) O/P Director
(S)  – p Resonance
(T) M/P Director
(D)
CH3 – CH – CH2 – OH
Column-II
(P) p – d vuq
ukn
(Q) p – p vuq
ukn
(R) O/P fn' kk
(S)  – p vuq
ukn
(T) M/P fn' kk
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