GENERAL INSTRUCTIONS The Test Booklet consists of 120 questions. There are Two parts in the question paper. The distribution of marks subjectwise in each part is as under for each correct response. MARKING SCHEME : PART-I : MATHEMATICS Question No. 1 to 20 consist of ONE (1) mark for each correct response. PHYSICS Question No. 21 to 40 consist of ONE (1) mark for each correct response. CHEMISTRY Question No. 41 to 60 consist of ONE (1) mark for each correct response. BIOLOGY Question No. 61 to 80 consist of ONE (1) mark for each correct response. PART-II : MATHEMATICS Question No. 81 to 90 consist of TWO (2) marks for each correct response. PHYSICS Question No. 91 to 100 consist of TWO (2) marks for each correct response. CHEMISTRY Question No. 101 to 110 consist of TWO (2) marks for each correct response. BIOLOGY Question No. 111 to 120 consist of TWO (2) marks for each correct response. Date : 04-11-2012 Duration : 3 Hours Max. Marks : 160 KISHORE VAIGYANIK PROTSAHAN YOJANA - 2012 STREAM - SB/SX
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
GENERAL INSTRUCTIONS
� The Test Booklet consists of 120 questions.
� There are Two parts in the question paper. The distribution of marks subjectwise in each
part is as under for each correct response.
MARKING SCHEME :PART-I :
MATHEMATICSQuestion No. 1 to 20 consist of ONE (1) mark for each correct response.
PHYSICSQuestion No. 21 to 40 consist of ONE (1) mark for each correct response.
CHEMISTRYQuestion No. 41 to 60 consist of ONE (1) mark for each correct response.
BIOLOGYQuestion No. 61 to 80 consist of ONE (1) mark for each correct response.
PART-II :MATHEMATICSQuestion No. 81 to 90 consist of TWO (2) marks for each correct response.
PHYSICSQuestion No. 91 to 100 consist of TWO (2) marks for each correct response.
CHEMISTRYQuestion No. 101 to 110 consist of TWO (2) marks for each correct response.
BIOLOGYQuestion No. 111 to 120 consist of TWO (2) marks for each correct response.
Date : 04-11-2012 Duration : 3 Hours Max. Marks : 160
KISHORE VAIGYANIK PROTSAHAN YOJANA - 2012
STREAM - SB/SX
RESONANCE PAGE - 2
KVPY QUESTION PAPER - STREAM (SB / SX)
PART-IOne Mark Questions
MATHEMATICS1. Three children, each accompanied by a guardian, seek admission in a school. The principal wants to
interview all the 6 persons one after the other subject to the condition that no child is interviewed before itsguardian. In how many ways can this be done?rhu cPps] ftuesa izR;sd vius vfHkHkkod ds lkFk gS] ,d fo|ky; esa izos'k ikuk pkgrs gSA fo|ky; ds izkpk;Z ,d ds ckn ,dlHkh 6 O;fDr;ksa ls bl izdkj lk{kkRdkj djuk pkgrs gS fd fdlh Hkh cPPks dk lk{kkRdkj mlds vfHkHkkod ds igys u gksA ;gfdrus rjhds ls fd;k tk ldrk gS ?(A) 60 (B) 90 (C) 120 (D) 180
Sol. Number ways are rjhdksas dh la[;k = !2!2!2
!6 =
8720 = 90 ways
2. In the real number system, the equation 1�x4�3x + 1x68x = 1 has
(A) no solution (B) exactly two distinct solutions(C) exactly four distinct solutions (D) infinitely many solutions
3. The maximum value M of 3x + 5x � 9x + 15x � 25x, as x varies over reals, satisfiesO;atd 3x + 5x � 9x + 15x � 25x, tgk¡ x ,d okLrfod la[;k gS] dk vf/kdre eku (M) fufEufyf[kr dks larq"V djrk gS&(A) 3 < M < 5 (B) 0 < M < 2 (C) 9 < M < 25 (D) 5 < M < 9
Sol. Let ekuk a = 3x and rFkk b = 5x
Therefore blfy, ,3x + 5x � 9x + 15x � 25x = a + b � a2 + ab � b2
= 21
[2 � (a � 1)2 � (b � 1)2 � (a � b)2]
21
(2)
Thus vr%, M = 1 (happens when tc x = 0)Answer (B)
4. Suppose two perpendicular tangents can be drawn from the origin to the circle x2 + y2 � 6x � 2py + 17 = 0,
for some real p. Then |p| is equal to;fn fdlh okLrfod p ds fy, x2 + y2 � 6x � 2py + 17 = 0 ij ewyfcUnq ls nks ijLij yEcor~ Li'kZthok [khaph tk ldrhgS] rc |p| gksxk&(A) 0 (B) 3 (C) 5 (D) 17
RESONANCE PAGE - 3
KVPY QUESTION PAPER - STREAM (SB / SX)
Sol.
Equation of chord of contact T = 0 w.r.t. origin is� 3(x + 0) � p (y + 0) + 17 = 0
3x + py � 17 = 0
By Homoginization
x2 + y2 � (6x + 2py)
17pyx3
+ 17
2
17pyx3
= 0
For perpendicular coeff of x2 + coeff of y2 = 0
1 � 17
3.6 +
.917
+ 1 � 17p2 2
+ 17p2
= 0
34 � 18 + 9 � p2 = 0 p2 = 25 |p| = 5
5. Let a, b, c, d be numbers in the set {1, 2, 3, 4, 5, 6} such that the curves y = 2x3 + ax + b and y = 2x3 + cx+ d have no point in common. The maximum possible value of (a � c)2 + b � d is
;fn fdlh leqPp; {1, 2, 3, 4, 5, 6} ds in a, b, c, d bl izdkj gS fd y = 2x3 + ax + b rFkk y = 2x3 + cx + d ijdksbZ fcUnq mHk;fu"B ugha gS] rc (a � c)2 + b � d dk vf/kdre laHkkfor eku gksxk&(A) 0 (B) 5 (C) 30 (D) 36
Sol. 2x3 + ax+ b 2x3 + cx + d(a � c)x d � b
If a � c = 0 & d � b 0 Max. (a � c)2 + (b � d)
= 0 + 5 Ans.
6. Consider the conic ex2 + y2 � 2e2x � 22y + e3 + 3 = e. Suppose P is any point on the conic and S1, S
2
are the foci of the conic, then the maximum value of (PS1 + PS
2) is
'kkado (conic) ex2 + y2 � 2e2x � 22y + e3 + 3 = e ij ;fn P dksbZ fcUnq gks rFkk S1, S
2 'kkado ds ukfHkd gks] rc
(PS1 + PS
2) dk vf/kdre eku gksxk&
(A) e (B) e (C) 2 (D) e2Sol. e(x2 � 2 ex + e2) + (y2 � 2y + 2) = e
e�ye�x 22
= 1 (a > b)
Ellipse a = , b = e
Req. PS1 + PS
2 = 2
7. Let f(x) = )axcos(�)axcos()axsin()axsin(
, then
(A) f(x + 2) = f(x) but f(x + ) f(x) for any 0 < < 2 (B) f is a strictly increasing function(C) f is strictly decreasing function (D) f is a constant function
10. Let n be a natural number and let a be a real number. The number of zeros of x2n+1 � (2n + 1) x + a = 0 in the
interval [�1, 1] is
(A) 2 if a > 0(B) 2 if a < 0(C) at most one for every vale of a(D) at least three for every value of aeku ysa fd n ,d izkd r la[;k gS rFkk a ,d okLrfod la[;k gSA lehdj.k x2n+1 � (2n + 1) x + a = 0 ds 'kwU;ksa dh la[;kvUrjky [�1, 1] esa gS&(A) 2 ;fn a > 0
(B) 2 ;fn a < 0
(C) izR;sd a ds eku ds fy, vf/kd ls vf/kd ,d(D) izR;sd a ds eku ds fy, de ls de rhu
Sol. f(x) = x2n +1 � (2n + 1) x + a
f (x) = (2n + 1)(x2n � 1)
x = �1 point of local maximax = �1 point of local minima
RESONANCE PAGE - 5
KVPY QUESTION PAPER - STREAM (SB / SX)
graph of y = f(x) when a = 0f(x) is decreasing on [�1, 1]. Thus, f(x) will have atmost one root on [�1,1] that too will happen if f(�1) 0 andf(1) 0 i.e. for a [�2n, 2n]
Answer (C)
11. Let f : R R be the function f(x) = (x � a1) (x � a
2) + (x � a
2) (x � a
3)+ (x � x
3) (x �x
1) with a
1, a
2, a
3 R. Then
f(x) 0 if and only if(A) at least two of a
1, a
2, a
3 are equal (B) a
1 = a
2 = a
3
(C) a1, a
2, a
3 are all distinct (D) a
1, a
2, a
3 are all positive and distinct
;fn f : R R ,d Qyu f(x) = (x � a1) (x � a
2) + (x � a
2) (x � a
3)+ (x � x
3) (x �x
1) tgk¡ a
1, a
2, a
3 R rc
f(x) 0 ;fn vkSj dsoy ;fn(A) a
1, a
2, a
3 esa de ls de nks cjkcj gSA (B) a
1 = a
2 = a
3
(C) a1, a
2, a
3 rhuksa fofHkUu gSA (D) a
1, a
2, a
3 rhuksa /kukRed vkSj fofHkUu gSA
Sol. f(x) = 3x2 � 2(a1 + a
2 + a
3) x + a
1a
2 + a
2a
3 + a
3a
1 0
D 04(a
1 + a
2 + a
3)2 � 4.3.(a
1a
2 + a
2a
3 + a
3a
1) 0
21a + 2
2a + 23a � a
1a
2 � a
2a
3 � a
3a
1 0
21
[(a1 � a
2)2 + (a
2 � a
3)2 + (a
3 � a
1)2] 0
a1 = a
2 = a
3
Ans. (B)
12. The value
/ 22 1
0/ 2
2 1
0
(sinx) dx
(sin x) dx
is
/ 22 1
0/ 2
2 1
0
(sinx) dx
(sin x) dx
dk eku gS&
(A) 1�2
12 (B)
12
12
(C)
2
12 (D) 2�2
Sol. I1 =
2/
0
2)x(sin (sinx)�1 dx
I1 =
2/
0
2)x)(sinx(cos
+ 2
122/
0
2 )x(sin)x(cos
dx
I1 = 2
dxxsindx)x(sin2/
0
122/
0
12
RESONANCE PAGE - 6
KVPY QUESTION PAPER - STREAM (SB / SX)
I1 = 2 (I
2 � I
1) Here I
1 =
2/
0
12)x(sin dx
2
1
=
)12(
2
×
)12(
12
= 22 Ans.
13. The value
2012
2012�
53 dx)1x)x(sin( is
2012
2012�
53 dx)1x)x(sin( dk eku gS&
(A) 2012 (B) 2013 (C) 0 (D) 4024
Sol.
2012
2012�
53 dx)xx(sin + 2012
2012�
dx.1
= 0 + 2012 � (� 2012)
= 4024 Ans.
14. Let [x] and {x} be the integer part and fractional part of a real number x respectively. The value of the integral
16. A purse contains 4 copper coins and 3 silver coins. A second purse contains 6 coins and 4 silver coins. Apurse is chosen randomly and a coin is taken out of it. What is the probability that it is a copper coin?,d FkSyh esa 4 rkacs vkSj 3 pkanh ds flDds gSA ,d nwljh FkSyh esa 6 rkacs vkSj 4 pkanh ds flDds gSA bu nks FkSfy;ksa esa ls ,dFkSyh ;knfPNd fudkyh tkrh gS vkSj mlesa ls ,d flDdk fudkyk tkrk gSA bl ckr dh izkf;drk D;k gS fd flDdk rkacsdk gS ?
(A) 7041
(B) 7031
(C) 7027
(D) 31
Ans. (A)
Sol. req. prob =
106
74
21
=7041
704240
21
Ans.
17. Let H be the orthocentre of an acute�angled triangle ABC and O be its circumcenter. Then HCHBHA
(A) is equal to HO (B) is equal to HO3
(C) is equal to HO2 (D) is not a scalar multiple of HO in general
;fn H ,d fuEudks.k f=kHkqt ABC dk yEc dsUnz gS ,oa O bldk ifjdsUnz gSA rc HCHBHA
18. The number of ordered pairs (m,n) where m, n {1, 2, 3, ......, 50}, such that 6m + 9n is a multiple of 5 isØfer ;qXeksa (m,n) tgk¡ m, n {1, 2, 3, ......, 50} bl izdkj gS fd 6m + 9n, 5 dk xq.kt gSA bl izdkj ds Øfer ;qXeksadh la[;k gksxh&(A) 1250 (B) 2500 (C) 625 (D) 500
Sol. 6m + 9n
Unit digit of 6m is = 6Unit digit of 9n will be = 9 or 1For multiple of 5 unit digit of 9n must be = 9It occur when n = oddTotal number of ordered pair = 50 × 25 = 1250
19. Suppose a1, a
2, a
3, ....., a
2012 are integers arranged on a circle. Each number is equal to the average of its
two adjacent numbers. If the sum of all even indeced numbers is 3018, what is the sum of all numbers?eku ysa fd iw.kk�±d a
20. Let S = 1, 2, 3, ...., n} and A = {(a, b) | 1 a, b n} = S S . A subset B of A is said to be a good subset if(x, x) B for every x S. Then the number of good subsets of A is;fn S = 1, 2, 3, ...., n} rFkk A = {(a, b) | 1 a, b n} = S S . A ds mileqPp; B dks vPNk mileqPp; rc dgktk,xk tc izR;sd x S ds fy, (x, x) B gksA rc A ds vPNs mileqPp;ksa dh la[;k gksxh&
(A) 1 (B) 2n (C) 2n(n�1) (D) 2n2
Sol. Number of element in B = n × n = n2
Number of elements of type (x, x) = n after choosing n elements.We can choose any number of elements from remaining n2 � n elements.
Number of subsets = n�n
n�n2
n�n1
n�n0
n�n2
2222C......CCC
= n�n22
RESONANCE PAGE - 9
KVPY QUESTION PAPER - STREAM (SB / SX)
PHYSICS21. An ideal monatomic gas expands to twice its volume. If the process is isothermal, the magnitude of work
done by the gas is W i . If the process is adiabatic the magnitude of work done by the gas is Wa . Which ofthe following is true
x;k dk;Z W i gSA ;fn izfØ;k :)ks"e gS rks xSl }kjk fd;k x;k dk;Z Wa gSA fuEufyf[kr esa dkSu lgh gS \(A) W i = Wa > 0 (B) W i > Wa > 0 (C) W i > Wa = 0 (D) Wa > W i = 0
Sol.
Since area of PV graph under isothermal curve is greater than area under adiabatic curveSo, w
i > w
a > 0
Ans. (B)
22. The capacitor of capacitance C in the circuit shown is fully charged initially, Resistance is R.
ifjiFk esa n'kkZ, C /kkfjrk dk ,d la/kkfj=k izkjEHk ls iw.kZ :i ls vkosf'kr gSA R çfrjks/k gSA
After the switch S is closed, the time taken to reduce the stored energy in the capacitor to half its initial valueis :
fLop (S) ds can gksus ds ckn la/kkfj=k esa lafpr Å tkZ fdrus le; esa izkjfEHkd eku dh vk/kh gks tk;sxh\
(A) 2
RC(B) RC ln 2 (C) 2RC ln 2 (D)
22lnRC
Sol. Q = Q0e�t/RC
2
Q0 = Q
0e�t/RC
2
1 = e�t/RC Ans. (D)
RCt
2
1n
2nRC
t
t = RC 22n
RESONANCE PAGE - 10
KVPY QUESTION PAPER - STREAM (SB / SX)
23. A liquid drop placed on a horizontal plane has a near spherical shape (slightly flattened due to gravity). Let Rbe the radius of its largest horizontal section. A small disturbance causes the drop to vibrate with frequency
about its equilibrium shape. By dimensional analysis the ratio
3R
c
can be (Here is surface tension,
is density, g is acceleration due to gravity, and k is an arbitrary dimensionless constant)
g xq:Roh; Roj.k vkSj k ,d ;knPN foek jfgr fLFkjkad gSA½
(A)
2gRk(B)
gRk 3
(C)
gRk 2
(D)
gk
Sol. 3R/
=
3R
=
2/1
2
331
MT
LMLT
= T�1 T
= 1
KR/ 3
23
2 KR
K2 = 23
TR
= 22
RTR
K = gR2
Ans. (A)
24. Seven identical coins are rigidly arranged on a flat table in the pattern shown below so that each coin touchesits neighbours. Each coin is a thin disc of mass m and radius r. Note that the moment of inertia of an
individual coin about an axis passing through center and perpendicular to the plane of the coin is 2
mr 2
.
,d {kSfrt est ij leku vkdkj ds lkr flDds n<+rkiwoZd bl rjg ls O;ofLFkr gS fd çR;sd flDdk vius utnhdh flDdksa
dks Nwrk gSA çR;sd flDds dk nzO;eku m vkSj f=kT;k r gSA /;ku nsa fd gj flDds dk ¼flDds ds dsUnz ls xqtjrh gqbZ vkSj flDds
ds ry ds yEcor v{k ds lkis{k½ tM+Ro vk?kw.kZ 2
mr 2
gSA
RESONANCE PAGE - 11
KVPY QUESTION PAPER - STREAM (SB / SX)
The moment of inertia of the system of seven coins about an axis that passes through the point P (the centreof the coin positioned directly to the right of the central coin) and perpendicular to the plane of the coins is
n'kkZ, x,s fp=k esa bu lkrksa flDdksa ds lewg dk fcUnq P ls xqtjrs gq, ,oa flDds ds yacor v{k ds lkis{k tM+Ro vk?kw.kZ D;k
25. A planet orbits in an elliptical path of eccentricity e around a massive star considered fixed at one of the foci.The point in space where it is closest to the star is denoted by P and the point where it is farthest is denotedby A. Let vP and vA be the respective speeds at P and A. Then
oÙk ds ,d ukfHk ij fLFkj gSA rkjs ds lcls utnhd ds fcUnq dks P ,oa lcls nwj ds fcUnq dks A. ls n'kkZ;k x;k gSA ;fn vP
,oa vA Øe'k% P vkSj A ij xzg dk osx gS rks
(A) e1e1
vv
A
P
(B)
A
P
vv
= 1 (C) e1e1
vv 2
A
P
(D) 2
2
A
P
e1
e1vv
RESONANCE PAGE - 12
KVPY QUESTION PAPER - STREAM (SB / SX)
Sol.
Angular momentum conservation about Sv
p (a � ae) = v
A(a + ae)
aeaaea
v
v
A
p
e1e1
v
v
A
p
Ans. (A)
26. In a Young's double slit experiment the intensity of light at each slit is 0. Interference pattern is observedalong a direction parallel to the line S1 S2 on screen S.
;ax f}&f>jh (Young's double slit) iz;ksx esa izR;sad f>jh ij izdk'k fdj.k dh rhozrk 0. gSA O;frdj.k izfr:i dks insZ S
ij S1 S2 js[kk ds lekukUrj ns[kk tkrk gSA
The minimum, maximum, and the intensity averaged over the entire screen are respectively.
27. A loop carrying current has the shape of a regular polygon of n sides. If R is the distance from the centre to
any vertex, then the magnitude of the magnetic induction vector B
at the centre of the loop is
n Hkqtkvksa okys ,d lecgqHkqtkdj can ifjiFk esa /kkjk izokfgr gks jgh gSA ;fn cgqHkqt ds dsUnz ls fdlh 'kh"kZ dh nqjh R gS
rks ifjiFk ds dsUnz ij ;qXedh; izsj.k B
dk ifjek.k gksxkA&
(A) n
tanR2
n 0
(B)
n2
tanR2
n 0
(C)
R20
(D) n
tanR0
RESONANCE PAGE - 13
KVPY QUESTION PAPER - STREAM (SB / SX)
Sol.
B0 =
nsin
nsin
ncosR
i4
0
B0 =
ntan
Ri
20
Ans. (A)
28. A conducting rod of mass m and length is free to move without friction on two parallel long conducting rails,
as shown below. There is a resistance R acorss the the rails. In the entire space around, there is a uniformmagnetic filed B normal to the plane of the rod and rails. The rod is given an impulsive velocity v0.
n'kkZ, x, fp=k esa ,d m nzO;eku vkSj L yackbZ dh pkyd NM nks yach vkSj lekUrj pkyd iVfj;ksa ij fcuk ?k"kZ.k ds xfr
dj ldrh gSA iVfj;ksa ds chp dk izfrjks/k R gSA fp=k dh lrg ds yacor~] gj txg ,d leku pqEcdh; {ks=k B gSA NM dks
(A) will be converted fully into heat energy in the resistor(B) will enable rod to continue to move eith velocity v0 since the rails are frictionless(C) will be converted fully into magnetic energy due to induced current(D) will be converted into the work done against the magnetic field
vUrr% izkjafHkd Å tkZ 21
mv02
(A) iw.kZ :i ls izfrjks/k R esa Å "eh; Å tkZ esa :ikUrfjr gks tk,xhA
(C) mRizsfjr fo|qr /kkjk dh otg ls] iw.kZr;k pqEcdh; Å tkZ esa :ikUrfjr gks tk,xhA
(D) pqEcdh; {ks=k ds f[kykQ dk;Z djus esa O;; gks tk,xhASol. Due to the negative work on rod K.E. will decrease and finally become zero.
Ans. (A)
29. A steady current flows through a wire of radius r, length L and resistivity . The current produces heat in thewire. The rate of heat loss in a wire is proportional its surface area. The steady temperature of the wire isindependent of
,d r f=kT;k] L yackbZ ,oa izfrjks/kdrk ds rkj esa LFkk;h fo|qr /kkjk I izokfgr gksrh gSA /kkjk dh otg ls rkj esa Å "ek dk
fuekZ.k gksrk gSA rkj esa Å "ek ds {k; dh nj rkj ds lrgh {ks=kQy ds vuqikrh gSA rkj dk fLFkj rkieku fuEu esa ls fdlds
Å ij fuHkZj ugha gS\(A) L (B) r (C) (D)
RESONANCE PAGE - 14
KVPY QUESTION PAPER - STREAM (SB / SX)
Sol. R = A
R = 2r
L
2R = K2rLdtdT
2
2
r
L
= K2rL
dtdT
dtdT
= 32
2
r2K
L
Ans. (A)
30. The ratio of the speed of sound to the average speed of an air molecule at 300K and 1 atmospheric pressureis close to
¼300K rkieku ,oa 1 atm nkc ½ ij /ofu dh xfr vkSj ok;q ds ,d v.kq dh vkSlr xfr dk vuqikr fuEu esa ls fdlds fudV
gS \
(A) 1 (B) 300 (C) 300
1(D) 300
Sol.
mRT8mRT
= 85
7 = 0.73
So, closest ans is 1.Ans. (A)
31. In one model of the election, tho electron of mass me is thought to be a uniformly charged shell of radius Rand total charge e, whose electrostatic energy E is equivalent to its mass me via Einstein's mass energyrelation E = mec2 . In this model, R is approximately (me = 9.1 × 10�31 kg, c = 3 × 108 m.s�1,
041
= 9 × 109 Farads m�1, magnitude of the electron charge = 1.6 × 10�19 C)
bysDVªkWu ds ,d fu:i.k esa me nzO;eku ds bysDVªkWu dks R f=kT;k ds leku :i ls vkosf'kr ,d xksyk] ftldk dqy vkos'k
e, gS] le>k tkrk gSA bl vkosf'kr xksys dh fLFkj oS|qr Å tkZ E ,oa nzO;eku me vkbaLVhu ds E = mec2 . laca/k ls nh tkrh
gSA bl fu: i.k esa R d k eku fd l d s d jhc gS\ (me = 9.1 × 10�31 kg, c = 3 × 108 m.s�1,
(A) 1.4 × 10�15 m (B) 2 × 10�13 m (C) 5.3 × 10�11 m (D) 2.8 × 10�35 mSol. U = mc2
22
mcR2
KQ
R2)106.1(109 2199
= 9.1 × 10�31 × (3 × 108)2
R = 2831
2199
)103(101.92
)106.1(109
= 1.4 × 10�15
RESONANCE PAGE - 15
KVPY QUESTION PAPER - STREAM (SB / SX)Ans. (A)
32. A body is executing simple harmonic motion of amplitude a and period T about the equilibrium positionx = 0. Large numbers of snapshots are taken at random of this body in motion. The probability of the bodybeing found in a very small interval x to x + |dx| is highest at
vius lkE; voLFkk x = 0. ds lkis{k ,d fiM+ a vk;ke vkSj T vkorZ dky ls ljy vkorZ xfr (simple harmonic
Sol. Probability of being found is maximum where speed is minimum.So, x = ±a
Ans. (A)
33. Two identical bodies are made of a material for which the heat capacity increases with temperature One ofthese is held at a temperature of 100°C while the other one is kept at 0ºC. If the two are brought into contact,
then, assuming no heat loss to the environment, the final temperature that they will reach is
ij j[kk tkrk gS vkSj nqljs dks 0ºC.ijA ;fn nksuks fiaM ,d nqljs ls laidZ esa vkrs gS vkSj ;fn eku fy;k tk, fd ifjos'k
esa Å "ek dk {k; ugha gksrk] rks fiaMksa dk vafre rki D;k gksxk\(A) 50°C (B) more than 50°C (C) less than 50°C (D) 0°C
(A) 50°C (B) 50°C ls vf/kd (C) 50°C ls de (D) 0°C
Sol. Since heat capacity at high temperature is high. So for same amount of heat transfer T is more at lowertemperature then at higher temperature.So, final temperature is more than 50ºC .
34. A particle is acted upon by a force given by F = �x3 � x4 where and are positive constants. At the pointx = 0, the particle is
,d cy F = �x3 � x4 ¼tgka vkSj /kukRed fLFkjkad gS ½ ,d d.k ij yxk;k tkrk gSA fdlh fcUnq x = 0, ij d.k(A) in stable equilibrium (B) in unstable equiibrium(C) in neutral equilibrium (D) not in equilibrium
(A) fLFkj lkE;koLFkk esa gSA (B) vLFkk;h lkE;oLFkk esa gSA
(C) mnklhu lkE;oLFkk esa gSA (D) lkE;koLFkk esa ugha gSASol. As F = �x3 � x4
At x = 0, F = 0Hence particle is in equilibrium.
35. The potential energy of a point particle is given by the expression V(x) = x + sin (x / ). A dimensionlesscombination of the constants , and is
,d fcUnq d.k dh fLFkfrt Å tkZ dks V(x) = x + sin (x / ).ls fu:fir fd;k tkrk gSA , vkSj ds chp ,d foekghu
(dimensionless) laca/k gSA
(A)
(B)
2
(C)
(D)
Sol. It is clear thatdimension x = ML2T�2 = MLT�2
= ML2T�2
and x = L
So,
will be dimensionless.
RESONANCE PAGE - 16
KVPY QUESTION PAPER - STREAM (SB / SX)
36. A ball of mass m suspended from a rigid support by an inextensible massless string is released from aheight h above its lowest point. At its lowest point it colIides elastically with a block of mass 2m at rest on africtionless surface. Neglect the dimensions of the ball and the block. After the collision the ball rises to amaximum height of
,d << lgkjs ls tqM+h ,d Hkkjghu ,oa f[kpkojfgr rkj ls yVdh m nzO;eku dh ,d xsan dks fuEure fcUnq ls h Å apkbZ
ij mBkdj NksM fn;k tkrk gSA (fp=k nsf[k,) vius fuEure fcUnq ij xsan ,d ?k"kZ.k jfgr lrg ij j[ks 2m nzO;eku ds ,d
37. A particle released from rest is falling through a thick fluid under gravity. The fluid exerts a resistive force onthe particle proportional to the square of its speed. Which one of the following graphs best depicts thevariation of is speed v with time t?
fLFkj voLFkk ls xq:Roh; Roj.k esa fxjrk gqvk ,d d.k ,d xk<sa rjy inkFkZ ls xqtjrk gSA rjy nzO;d.k ij izfrjks/kd cy
Sol. Let ball is moving with speedv at anytime t hence
Hence, mdtdv
= mg � Kv2
On the basis of equation we can rays that speed first increase and become constant when mg = kv2
Hence, Ans. (A)
38. A cylindrical steel rod of length 0.10m and thermal conductivity 50 W.m�1.K�1 is welded end to end to copperrod of thermal conductivity 400 W.m�1.K�1 and of the same area of cross section but 0.20 m long. The freeend of the steel rod is maintained at 100°C and that of the copper rod at 0°C. Assuming that the rods are
perfectly insulated from the surrounding the temperature at the junction of the two rods is
LVhy dh ,d csyukdkj NM+ ftldh yackbZ 0.10m rFkk Å "eh; pkydrk 50 W.m�1.K�1 gS] ,d Nksj ij leku vuqizLFk
Both are connected in series hence heat current in both will be same. So
A5001
T100 =
A20001
0T T = 20ºC
Ans. (A)
39. A parent nucleus X is decaying into daughter nucleus Y which in turn decays to Z. The half lives of X and Yare 4000 years and 20 years respectively. In a certain sample, it is found that the number of Y nuclei hardlychanges with time. If the number of X nuclei in the sample is 4 × 1020 , the number of Y nuclei present in itis
,d ewy ukfHkd X nwljs ukfHkd Y esa {kf;r gksrk gS] tks fQj Z. esa {kf;r gksrk gSA X vkSj Y dh v/kZ vk;q Øe'k% 4000 o"kZ
,oa 20 o"kZ gSA ,d uewus esa ;g ns[kk x;k gS fd Y ukfHkdksa dh la[;k esa le; ds lkFk dksbZ [kkl ifjorZu ugha gks jgk gSA
;fn bl uewus esa X ukfHkdksa dh la[;k 4 × 1020 rks ml le; Y ukfHkdksa dh la[;k D;k gksxh\(A) 2 × 1017 (B) 2 × 1020 (C) 4 × 1023 (D) 4 × 1020
Sol. FromN
1
1 = N
2
2
RESONANCE PAGE - 18
KVPY QUESTION PAPER - STREAM (SB / SX)
4 × 1020 = 40000
)2(n =
20)2(n
N2
N2 = 2 × 107
Ans. (A)
40. An unpolarized beam of light of intensity 0 passes through two linear polarizers making an angle of 30° with
respect to each other. The emergent beam will have an intensity.
Ans. Option �D� Closest, if (�) is change to fodYi �D� lgh gS] D;ksafd (�) dks esa ifjofrZr fd;k tk ldrk gSA
Sol.
45. The most stable conformation of 2, 3-dibromobutane is :2, 3-MkbczkseksC;wVsu dk vf/kdre LFkk;h la:i.k D;k gS \
(A) (B) (C) (D)
Ans. (Bonus)46. Typical electronic energy gaps in molecules are about 1.0 eV. In terms of temperature, the gap is closest to:
v.kqvksa ds bysDVªkWfud ÅtkZ&varjky (energy gaps) dk eku yxHkx 1.0 eV gSA rkieku ds lanHkZ esa ;g varj buesa ls fdldsfudV gS \(A) 102 K (B) 104 K (C) 103 K (D) 105 K
Ans. (B)Sol. KE = | T.E. |
(KE) = (T.E.)
23
× 8.3 × (T) = 1.6 × 10�19 × 6 × 1023
(T) = 3.8106.9 4
× 32
(T) = 7.6 × 103 K i.e. close to 104 K.(T) = 7.6 × 103 K vFkkZr 104 K ds fudV gSA
47. The major final product in the following reaction is :fuEu vfHkfØ;k dk eq[; vafre mRikn D;k gS \
CH3CH
2CN
OH)2
MgBrCH)1
3
3
(A) (B)
(C) (D)
RESONANCE PAGE - 20
KVPY QUESTION PAPER - STREAM (SB / SX)Ans. (C)
Sol.
48. A zero-order reaction, A Product, with an initial concentration [A]0 has a half-life of 0.2 s. If one starts with
the concentration 2[A]0, then the half-life is :
,d 'kwU; dksfV vfHkfØ;k] A mRikn (product), dh v)Z&vk;q 0.2 s gS] tc izkFkfed lkanzrk [A]0 gSA ;fn vfHkfØ;k
(A) 0.1 s (B) 0.4 s (C) 0.2 s (D) 0.8 sAns. (B)Sol. t
1/2 (a)1 � n
II
I
)t()t(
2/1
2/1 =
01
2
1aa
II)t(2.0
2/1 =
0
0A2
A
(t1/2
)II = 0.4 s
49. The isoelectronic pair of ions is :fuEu vk;fud ;qXe lebysDVªkWfud (isoelectronic) gS %(A) Sc2+ and V3+ (B) Mn3+ and Fe2+ (C) Mn2+ and Fe3+ (D) Ni3+ and Fe2+
Ans. (C)Sol.
25Mn2+ and
26Fe3+ both has 23 electrons.
25Mn2+ o
26Fe3+ nksuksa esa 23 bysDVªkWu gSA
50. The major product in the following reaction is :fuEu vfHkfØ;k dk eq[; mRikn D;k gS \
2NaNH
(A) (B) (C) (D)
Ans. (A)
Sol.
51. The major product of the following reaction is :fuEu vfHkfØ;k dk eq[; mRikn D;k gS \
HBr.Conc
(A) (B) (C) (D)
Ans. (B)
Sol.
RESONANCE PAGE - 21
KVPY QUESTION PAPER - STREAM (SB / SX)52. The oxidation state of cobalt in the following molecule is :
moment) 3.83 BM gSA rRo M D;k gS \(A) Co (B) Cu (C) Mn (D) Fe
Ans. (A)Sol. It is high spin complex as Cl� is weak field effect ligand. In [CoCl
4]2� oxidation state of Co is +2 in which 3
unpaired electrons are present which gives the spin-only magnetic moment equal to 3.83 BM.;g mPp pØ.k ladqy gS D;ksafd Cl� ,d nqcZy {ks=k fyxs.M gSA [CoCl
55. Among the following graphs showing variation of rate (k) with temperature (T) for a reaction, the one thatexhibits Arrhenius behavior over the entire temperature range is :fuEu vkjs[k esa vfHkfØ;k nj ds ifjorZu (k) dks rkieku (T) ds lkFk n'kkZ;k x;k gSA iwjs rki ifjlj esa vkgsZfu;l (Arrhenius)
65. Peptic ulcers are caused by(A) a fungus, Candida albicans(B) a virus, cytomegalo virus(C) a parasite, Trypanosoma brucei(D) a bacterium, Helicobacter pylori
66. Transfer RNA (tRNA)(A) is present in the ribosomes and provides structural integrity(B) usually has clover leaf-like structure(C) carries genetic information from DNA to ribosomes(D) codes for proteins
Ans (B)VªkUlQj (tRNA)
(A) jkbckslkse esa fLFkr gS vkSj lajpukRed v[kaMrk nsrk gSA(B) lkekU;r% frifr;k (clover leaf-like) vkdkj dh gksrh gSA(C) DNA ls jkbckslkse rFkk vkuqoaf'kd lwpuk igqapkrk gSA(D) izksVhu cukrk gS
Ans (B)
67. Some animals excrete uric acid in urine (uricotelic) as it requires very little water. This is an adaptation toconserve water loss. Which animals among the following are most likely to be uricotelic?(A) fishes (B) amphibians (C) birds (D) mammals
69. Human chromosomes undergo structural changes during the cell cycle. Chromosomal structure can be bestvisualized if a chromosome is isolated from a cell at(A) G1 phase (B) S phase (C) G2 phase (D) M phase
Ans (D)euq"; ds dksf'kdk pØ ds nkSjku Øksekslkse dh lajpuk esa cnyko gksrk gSA bl cnyko dks Li"V :i ls ns[kus ds fy, pØdh fdl izkoLFkk esa Øksekslkse dk ijh{k.k fd;k tkuk pkfg,\(A) G1 izkoLFkk (B) S izkoLFkk (C) G2 izkoLFkk (D) M izkoLFkk
Ans (D)
70. By which of the following mechanisms is glucose reabsorbed from the glomerular filtrate by the kidney tubule(A) osmosis (B) diffusion (C) active transport (D) passiver transport
71. In mammals, the hormones secreted by the pituitary, the master gland, is itself regulated by(A) Hypothalamus (B) median cortex (C) pineal gland (D) cerebrum
72. Which of the following is true for TCA cycle in eukaryotes(A) takes place in mitochondrion(B) produces no ATP(C) takes place in Golgi complex(D) independent of electron transport chain
Ans (A);wdSfj;ksV ds TCA pØ ds ckjs esa fuEufyf[kr dkSu lk dFku lgh gS\(A) ekbVksdkfUMª;k esa gksrk gSA (B) ATP ugha cukrkA(C) xkWYth dkEIysDl esa gksrk gSA (D) izefLr"d (cerebrum)
Ans (A)
73. A hormone molecule binds to a specific protein on the plasma membrane inducing a signal. The protein itbinds to it called(A) ligand (B) antibody (C) receptor (D) histone
Ans (C)dksf'kdk f>Yyh ij fLFkr ml izksVhu dks D;k dgrs gS tks gkeksZu ls tqMrk gS vkSj ladsr izsfjr djrk gS\(A) layXuh (ligand) (B) xzkgh (receptor) (C) izfrj{kh (antibody) (D) fgLVksu (histone)
Ans (C)
RESONANCE PAGE - 26
KVPY QUESTION PAPER - STREAM (SB / SX)74. DNA mutations that do not cause my functional change in the protein product are known as
75. Plant roots are usually devoid of chlorophyll and cannot perform photosynthesis. However, three are excep-tions. Which of the following plant root can perform photosynthesis(A) Arabidopsis (B) Tinospora (C) Rice (D) Hibiscus
Ans (B)ikS/kksa dh tM+ esa DyksjksfQy ugha gksrk gS] ftlds dkj.k os izdk'k la'ys"k.k (photosynthesis) ugha dj ikrsA buesa ls dkSulk vlk/kkj.k foijhr mnkgj.k gS\(A) vjschMkIlhl (Arabidopsis) (B) VhuksLiksjk (Tinospora)
(C) pkoy (Rice) (D) fgfcLdl (Hibiscus)Ans (B)
76. Vitamin A deficiency leads to night-blindness. Which of the following is the reason for the disease ?(A) rod cells are not converted to cone cells(B) rhodopsin pigment of rod cells is defective(C) melanin pigment is not synthesized in cone cells(D) cornea of eye gets dried
77. In Dengue virus infection, patients often develop haemorrhagic fever due to internal bleeding. This happensdue to the reduction of(A) platelets (B) RBCs (C) WBCs (D) lymphocytes
Ans (A)MsUX;q ok;jl ds laØe.k ls dbZ jksfx;ksa esa vkUrfjd jDrlzko gksrk gSA buesa ls fdl dksf'kdk ds de gksus ls ,slk gksrk gS\(A) jDr foEck.kq (platelets) (B) yky :f/kj df.kdk (RBCs)
78. If the sequence of bases in sense strand of DNA is 5'-GTTCATCG-3, then the sequence ofd bases in its RNAtranscript would be(A) 5'-GTTCATCG-3' (B) 5'GUUCAUCG-3 (C) 5'CAAGTAGC-3' (D) 5'CAAGUAGC -3
Ans (B)vxj DNA ds sence strand dk vuqØe, 5'-GTTCATCG-3, 5'-GTTCATCG-3 5'-GTTCATCG-3 gS] rks mlds RNA
79. A refilex aciton is a quick involuntary response to stimulus. Which of the following is an example of BOTH,unconditioned and conditioned reflex(A) knee jerk reflex(B) secretion of saliva in response to the aroma of food(C) sneezing reflex(D) contration of the pupil in response to bright light
80. In a food chain such as grass deer lion, the energy cost of respiration as a proportion of total assimi-lated energy at each level would be(A) 60%- 30%-20% (B) 20%- 30%-60% (C) 20%- 60%-30% (D) 30%- 30%-30%
Ans (B)?kkl & fgj.k & 'ksj] fd vkgkj Ja[kyk esa dqy laxzghr Å tkZ ds vuqikr esa izR;sd Lrj ij 'olu Å tkZ dh ykxr fdruhgksaxh\(A) 60%- 30%-20% (B) 20%- 30%-60% (C) 20%- 60%-30% (D) 30%- 30%-30%
Ans (B)
PART-IITwo Mark Questions
MATHEMATICS81. Suppose a, b, c are real numbers, and each of the equations x2 + 2ax + b2 = 0 and x2 + 2bx + c2 = 0 has two
distinct real roots. Then the equation x2 + 2cx + a2 = 0 has(A) two distinct positive real roots (B) two equal roots(C) one positive and one negative root (D) no real rootseku ysa fd a, b, c okLrfod la[;k,a gS rFkk izR;sd lehdj.k x2 + 2ax + b2 = 0 ,oa x2 + 2bx + c2 = 0 ds nks fofHkUuokLrfod ewy gSA rc lehdj.k x2 + 2cx + a2 = 0 ds(A) nks fofHkUu /kukRed okLrfod ewy gksxsaA (B) nks leku ewy gksxsaA(C) ,d /kukRed o ,d _ .kkRed ewy gksxkA (D) dksbZ okLrfod ewy ugha gksxkA
gSA eku ysa fd de ls de K vad izkIr djus dh izkf;drk P(K) gks] rc P(K) > 21ds fy, K dk vf/kdre eku D;k gksxk?
(A) 14 (B) 15 (C) 16 (D) 17Sol. P(K) = P (at least K points)
x1 + x
2 + x
3 + ....... x
10 = K
Coeff xK in (x1 + x2)10
= x10(1 + x)10
coeff xk�10 in (1 + x)10
= 10CK�10
K 10
Now P(K) > 21
1010K
102
101
100
10
2
C......CCC
21
10C0 + 10C
1 + ............. + 10C
K�10 > 29
1 + 10 + 45 + 120 + 210 + 252 + ...... > 51210C
0 + 10C
1 + 10C
2 + 10C
3 + 10C
4 + 10C
5 > 512
K � 10 = 5
K = 15Ans. (B)
90. Let f(x) = 1x1x
for all x 1. Let f1(x) = f(x), f2(x) = f (f(x)) and generally fn(x) = f(fn�1(x)) for n > 1. Let P = f1(2)
f2(3) f3(4) f4(5) which of the following is a multiple of P
eku ysa fd lHkh x 1 ds fy, f(x) = 1x1x
gSA ;fn f1(x) = f(x), f2(x) = f (f(x)) vkSj n > 1 ds fy, lkekU;r%
fn(x) = f(fn�1(x)) gS vkSj P = f1(2) f2(3) f3(4) f4(5). buesa ls P dk xq.kt D;k gksxk?(A) 125 (B) 375 (C) 250 (D) 147
Sol. P = f(2) . f(f(3)) f(f(f(4)) ff(f(f(5)
=
13
24
f f
35
f
46
fff
= (3)
13
1
1f
3535
f
1
1f
2323
= (3) (3) (f(4)) f(f(5))
RESONANCE PAGE - 32
KVPY QUESTION PAPER - STREAM (SB / SX)
= 9
35
46
f
= (15)
123
123
= (15) (5) = 75375 is multiple of 75.Ans. (B)
PHYSICS
91. The total energy of a black body radiation source is collected for five minutes and used to heat water. Thetemperature of the water increases from 10.0°C to 11.0°C. The absolute temperature of the black body is
doubled and its surface area halved and the experiment repeated for the same time. Which of the followingstatements would be most nearly correct?
d f".kdk fofdj.k ls mRlftZr dqy Å tkZ dks 5 feuV rd ,df=kr djds ikuh xeZ fd;k tkrk gSA ikuh dk rki 10.0°C ls
nksgjk;k trk gSA buesa ls dkSulk dFku lcls lgh yxrk gS \(A) The temperature of the water would increase from 10.0°C to a final temperature of 12°C
ikuh dk rki 10.0°C ls 12°C gks tk;sxkA(B) The temperature of the water would increase from 10.0°C to a final temperature of 18°C
ikuh dk rki 10.0°C ls 18°C gks tk,xkA(C) The temperature of the water would increase from 10.0°C to a final temperature of 14°C
ikuh dk rki 10.0°C ls 14°C gks tk,xkA(D) The temperature of the water would increase from 10.0°C to a final temperature of 11°C
ikuh dk rki 10.0°C ls 11°C gks tk,xkASol. H
1 = AT4 × 5 = Ms
1
H2 =
2A
(2T)4 × 5 = msq2
2 = 8
1
2 = 8ºC
The temperature of water would increase from 10ºC to a final temperature of 18ºC
92. A small asteroid is orbiting around the sun in a circular orbit of radius r0 with speed V0. A rocket is launchedfrom the asteroid with speed V = V0 where V is the speed relative to the sun. The highest value of forwhich the rocket will remain bound to the solar system is (ignoring gravity due to the asteroid and effect ofother planets)
,d xzfgdk V0 osx ls lw;Z ds pkjksa vksj r0 f=kT;k ds oÙkh; iFk esa pDdj yxkrh gSA bl xzfgdk ls V = V0 osx ls ,d jkdsV
Rocket will remain band to the solar system its B moving energies negative.
0mv21
rGMm 2
0
2
0
mv21
rGMm
RESONANCE PAGE - 33
KVPY QUESTION PAPER - STREAM (SB / SX)
20
2
0
vm21
rGMm
mv02 =
21
m2v02
= 2 .
93. A radioactive nucleus A has a single decay mode with half life A. Another radioactive nucleus B has twodecay modes 1 and 2. If decay mode 2 were absent, the half life of B would have been A/2. If decay mode 1
were absent, the half life of B would have been 3 A. If the actual half life of B is B , then the ratio A
B
is
,d jsfM;ksa lfØ; ukfHkd A ,dy&{k; iz.kkyh single decay mode) ls {kf;r gksrk gS vkSj bl izfØ;k dh v/kZ vk;q A gSA
A/2 gksrh gSA ;fn {k; iz.kkyh 1 miyC/k ugha gksrh rks B dh v/kZ vk;q 3 A gksrh gSA ;fn B dh okLrfod v/kZ vk;q B gS rc
vuqikr A
B
dk D;k eku gksxk \
(A) 73
(B) 27
(C) 37
(D) 1
Sol. B = 1 + 2
AAAB 32n7
32n2n22n
= 73
A
B
.
94. A stream of photons having energy 3 eV each impinges on a potassium surface. The work function ofpotassium is 2.3 eV. The emerging photo-electrons are slowed down by a copper plate placed 5 mm away.If the potential difference between the two metal plates is 1V, the maximum distance the electrons can moveaway from the potassium surface before being turned back is
3 eV Å tkZ ds QksVku d.k ,d iksVsf'k;e ds IysV dh lrg ij iM+rs gSaA iksVsf'k;e dk dk;Z Qyu (work function) 2.3
eV gSA mRlftZr QksVks bysDVªkWuksa dks 5 mm nwj j[ks rkacs ds IysV ls /khek fd;k tkrk gSA ;fn nksuksa IysVksa ds chp dk foHkokarj
1V gks rks] ihNs ykSVus ds igys bysDVªkWu iksVksf'k;e IysV ls vf/kdre~ fdruh nwjh rd tk ldrs gSa \(A) 3.5 mm (B) 1.5 mm (C) 2.5 mm (D) 5.0 mm
Sol. 3eV = 2.3 eVK
max = 3 � 2.3 = 0.7 eV
5 mm 1V1V 5mm0.7 V 5 × 0.7 mm
= 3.5 mm.
95. Consider three concentric metallic spheres A, B and C of radii a, b, c respectively where a<b<c. A and B areconnected whereas C is grounded. The potential of the middle sphere B is raised to V then the charge on thesphere C is
/kkrq ds rhu ladsUnzh xksys A, B ,oa C dh f=kT;k,a a, b, c gSa tgkW a<b<cA A ,oa B tqM+s gSa] tcfd C dks HkwlaifdZr fd;k x;k
gSA ;fn chp okys xksys B dk foHko c<+kdj V fd;k tkrk gS rc C xksys ij vkos'k D;k gksxk \
(A) � 40V bcbc
(B) + 40V bcbc
(C) � 40V acac
(D) zero 'kwU;
RESONANCE PAGE - 34
KVPY QUESTION PAPER - STREAM (SB / SX)
Sol. V = c
KQb
Kq
0c
)Qq(k
q + Q = 0q = �Q
Vc
KQ)Q(
bK
Vb
1
c
1KQ
Q = 04)cb(
bcV
.
96. On a bright sunny day a diver of height h stands at the bottom of a lake of depth H. Looking upward, he cansee objects outside the lake in a circular region of radius R. Beyond this circle he sees the images of objectslying on the floor of the lake. If refractive index of waler is 4/3, then the value of R is :
,d mTtoy fnu esa h yackbZ dk ,d xksrk[ksj H xgjkbZ okys ,d rkykc ds ry ij [kM+k gSA Å ij dh vksj ns[kus ij og
rkykc ds ckgj dk n'; R f=kT;k ds oÙk ds nk;js esa ns[k ldrk gSA bl oÙk ds ckgj og rkykc ds ry ij fLFkr oLrqvks
ds izfrfcEc gh ns[k ikrk gSA ;fn ty dk viorZukad 4/3 gS rks R dk eku D;k gksxk \
(A) 7
)hH(3 (B) 7h3 (C)
37
)hH( (D)
35
)hH(
Sol.
34
(sinC) = 1 sinC = 43
tanC = 17
3 =
hHR
R = )hH(7
3 .
RESONANCE PAGE - 35
KVPY QUESTION PAPER - STREAM (SB / SX)
97. As shown in the figure below, a cube is formed with ten indentical resistance R (thick lines) and two shortingwires (dotted lines) along the arms AC and BD.
uhps n'kkZ, fp=k esa ,d ?ku 10 leku ifjek.k okys izfrjks/k R ¼eksVh js[kk,a½ rFkk Hkqtkvksa BD ,oa AC ds vuqfn'k nks y|qifFkr
rkjksa ls cuk gqvk gSA
Resistance between point A and B is
A ,oa B ds chp dk izfrjks/k D;k gksxk \
(A) 2R
(B) 6R5
(C) 4R3
(D) R
Sol.
W.S. BridgeRAB = R/2.
98. A standing wave in a pipe with a length L = 1.2 m is described y
,d L = 1.2 ehVj yackbZ ds ikbi eas ,d vizxkkeh rjax dks bl Qyu ls fu:fir fd;k tkrk gSA
y (x, t) = y0 sin
x
L2
sin
4x
L2
Based on above information, which one of the following statement is incorrect.(Speed of sound in air is 300 ms�1)
Å ijh nh xbZ tkudkjh ds vk/kkj ij fuEufyf[kr esa ls dkSu lk dFku xyr gS \ ¼/;ku ns fd ok;q esa /ofu dh xfr 300
ms�1 gSA)(A) A pipe is closed at both ends
ikbi nksuksa fljksa ij can gSA(B) The wavelength of the wave could be 1.2 m
rjax dk rjaxnS/;Z 1.2 m gks ldrk gSA(C) There could be a nods at x = 0 and antinode at x = L/2
x = 0 ij fuLian (node) ,oa x = L/2 ij izLian (antinode) gks ldrk gSA
RESONANCE PAGE - 36
KVPY QUESTION PAPER - STREAM (SB / SX)
(D) The frequency of the fundamental mode of vibrations is 137.5 Hz
daiu dh ewy fo/kk (fundamental mode of vibrations) dh vkofÙk 137.5 Hz gSASol. From equation
2 =
L4
= 2L
= 0.6 m.
Ans. (B)
99. Two block (1 and 2) of equal mass m are connected by an ideal string (see figure below) over a frictionlesspulley. The blocks are attached to the ground by springs having spring constants k1 and k2 such thatk1 > k2 .
fp=k esa ,d vkn'kZ jLlh ls ca/ks leku nzO;eku m ds nks xqVds (1 ,oa 2) ,d ?k"kZ.k jfgr f?kjuh ls yVds gSaA nksuksa xqVdksa
Initially, both springs are unstretched. The block 1 is slowly pulled down a distance x and released. Just afterthe release the possible value of the magnitudes of the acceleration of the blocks a1 and a2 can be
'kq: esa nksuksa dekfu;ka esa dksbZ f[kapko ¼ruko½ ugha gSA xqVds 1 dks /khjs ls uhps x nwjh rd [khapdj NksM+ fn;k tkrk gSA NksM+us
100. A simple pendulum is released from rest at the horizontally stretched position. When the string makes anangle with the vertical, the angle which the acceleration vector of the bob makes with the string is givenby
,d ljy yksyd dks {kSfrt ls [khaph gqbZ fLFkfr ls fLFkj voLFkk esa NksM+k tkrk gSA tc jLlh Å /oZ ls dks.k cukrh gS] ml
101. The final major product obtained in the following sequence of reactions is :fuEufyf[kr vuqØfed vfHkfØ;kvksa dk eq[; vafre mRikn D;k gS \
(A) (B) (C) (D)
Ans. (D)
Sol.
102. In the DNA of E. Coli the mole ratio of adenine to cytosine is 0.7. If the number of moles of adenine in the DNAis 350000, the number of moles of guanine is equal to :E Coli ds DNA esa ,sMsuhu vkSj lkbVkslhu dk eksyj vuqikr 0.7 gSA ;fn DNA esa ,sMsuhu ds eksyksa dh la[;k 350000 gSrc Xokuhu ds eksyksa dh la[;k D;k gksxh \(A) 350000 (B) 500000 (C) 225000 (D) 700000
Ans. (B)
Sol.G
3500007.0
CA
GT
G = 7.0
350000 = 5,00,000 Ans.
103. (R)-2-bromobutane upon treatment with aq. NaOH gives(R)-2-czkseksC;qVsu vkSj tyh; NaOH vfHkfØ;k dk mRikn D;k gksxk \
(A) (B)
(C) (D)
Ans. (C)
RESONANCE PAGE - 39
KVPY QUESTION PAPER - STREAM (SB / SX)
Sol.
104. Phenol on treatment with dil. HNO3 gives two products P and Q. P is steam volatile but Q is not. P and Q are
respectivelyQhukWy dk ruq HNO
3 (dil. HNO
3) ds lkFk foospu djus ij nks mRikn P vkSj Q curs gSaA P Hkki }kjk ok"i'khy (steam
o�nitrophenol has intra molecular H�bonding hence steam volatile ie P, where as �Q� is p-nitrophenol, Q has
inter molecular H-bonding.
105. A metal is irradiated with light of wavelength 660 nm. Given that the work function of the metal is 1.0 eV, thede-Broglie wavelength of the ejected electron is close to :,d /kkrq dks 660 nm rjax}S/;Z ds izdk'k ls fdjf.kr (irradiated) fd;k tkrk gSA Kkr gS fd /kkrq dk dk;Z&Qyu (Work
function) 1.0 eV gSA fu"dkflr bysDVªku dh Mh&czksXyh (de-Broglie) rjax}S/;Z buesa ls fdl ds fudV gS :(A) 6.6 × 10�7 m (B) 8.9 × 10�11 m (C) 1.3 × 10�9 m (D) 6.6 × 10�13 m
Ans. (C)Sol. Kinetic energy xfrt Å tkZ = h � work function dk;Z Qyu
KE = 660
1240 � 1
KE = 0.878 eV
= V
150Å
= 878.0
150
= 13.07 Å = 1.3 × 10�9 m
106. The inter-planar spacing between the (2 2 1) planes of a cubic lattice of length 450 pm is :,d ?ku tkyd (Cubic lattice) dh yEckbZ 450 pm gSA blds (2 2 1) ryksa dh varj&ry nwjh D;k gS \(A) 50 pm (B) 150 pm (C) 300 pm (D) 450 pm
Ans. (B)
Sol. 222 Kh
a
= 222 )1()2()2(
450
=
3450
= 150 pm
RESONANCE PAGE - 40
KVPY QUESTION PAPER - STREAM (SB / SX)107. The H for vaporization of a liquid is 20 kJ/mol. Assuming ideal behaviour, the change in internal energy for
the vaporization of 1 mole of the liquid at 60ºC and 1 bar is close to :
108. Among the following, the species that is both tetrahedral and diamagnetic is :buesa ls dkSu prq"Qydh; vkSj izfrpqEcdh; nksuksa gh gS \(A) [NiCl
4]2� (B) [Ni(CN)
4]2� (C) Ni(CO)
4(D) [Ni(H
2O)
6]2+
Ans. (C)Sol. As in Ni(CO)
4 hybridisation of Ni is sp3 and CO is strong field effect ligand therefore, it is diamangetic.
Ni(CO)4 esa Ni dk ladj.k sp3 gS rFkk CO izcy {ks=k fyxs.M gS blfy, ;g izfrpqEcdh; gSA
109. Three moles of an ideal gas expands reversibly under isothermal condition from 2 L to 20 L at 300 K. Theamount of heat-change (in kJ/mol) in the process is :,d vkn'kZ xSl ds rhu eksy 300 K ij lerkih voLFkk esa 2 L ls 20 L rd mRØe.kh; rjhds ls izlkfjr gksrs gSaA bl izfØ;kesa Å "ek ifjorZu dk ifjek.k (kJ/mol bdkbZ esa) D;k gS :(A) 0 (B) 7.2 (C) 10.2 (D) 17.2
Ans. (D)
Sol. W = � nRT ln1
2VV
W = � 3R × 300 ln 10
= 1000
314.8900� × 2.3 = � 17.2 kJ/mol.
q = � W = 17.2 kJ/mol. (For isothermal process lerkih; izØe ds fy, E = 0).
110. The following data are obtained for a reaction, X + Y Products.Expt. [X
0]/mol [Y
0]/mol rate/mol L�1 s�1
1 0.25 0.25 1.0 × 10�6
2 0.50 0.25 4.0 × 10�6
3 0.25 0.50 8.0 × 10�6
The overall order of the reaction is :(A) 2 (B) 4 (C) 3 (D) 5vfHkfØ;k X + Y mRIkkn ds fy, fuEufyf[kr vkadM+s izkIr gq,Aiz;ksx vuqØe [X
0]/eksy [Y
0]/eksy nj/eksy L�1 s�1
1 0.25 0.25 1.0 × 10�6
2 0.50 0.25 4.0 × 10�6
3 0.25 0.50 8.0 × 10�6
bl vfHkfØ;k dh lexz dksfV (overall order of the reaction) D;k gS \(A) 2 (B) 4 (C) 3 (D) 5
Ans. (D)Sol. Rate nj = K . [X]p [Y]q
From expt. 1 and 2, p = 2 and from expt. 1 and 3, q = 3. Therefore, over all order = 5.iz;ksx 1 o 2 ls p = 2 rFkk iz;ksx 1 o 3 ls q = 3Ablfy, lEiw.kZ dksfV = 5 gSA
RESONANCE PAGE - 41
KVPY QUESTION PAPER - STREAM (SB / SX)
BIOLOGY111. When hydrogen peroxide is applied on the wound as a disinfectant, there is frothing at the site of injury,
which is due to the presence of an enzyme in th skin that uses hydrogen peroxide as a substrate to produce(A) hydrogen (B) carbon dioxide (C) water (D) oxygen
112. Persons suffering from hypertension (high blood pressure) are advised a low-salt diet because(A) more salt is absorbed in the body of a patient with hypertension(B) high salt leads to water retention in the blood that further increases the blood pressure(C) high salt increases nerve conduction and increases blood pressure(D) high salt causes adrenaline release that increases blood pressure
Ans (B)mPp jDrpki ls ihfMr O;fDr dks ued de ysus dh lykg nh tkrh gSA D;ksafd](A) mPp jDrpki ds dkj.k ued dk vo'kks"k.k T;knk gksrk gS
113. Insectivorous plants that mostly grow on swampy soil use insects as a source of(A) carbon (B) nitrogen (C) phosphorous (D) magnesiumdhV&Hk{kh ikS/ks nyny esa ik, tkrs gSA bu ikS/kksa dks dhMksas ls D;k feyrk gS\(A) dkCkZu (B) ukbVªkstu (C) QkLQksjl (D) eSXuhf'k;e
Ans (B)
114. In cattle, the coat colour red and white are two dominant traits, which express equally in F1 to produce roan(red and white clour in equal proportion). If F1 progeny are self-bred, the resulting progeny in F2 will havephenotypic ratio (red:roan:white) is -(A) 1 : 1 : 1 (B) 3 : 9 : 3 (C) 1 : 2 : 1 (D) 3 : 9 : 4,d izdkj ds i'kqvksa dh [kky ds nks izcy fo'ks"kd jax] Hkwjk vkSj lQsn gSa tks F1 larfr esa cjkcj vuqikr esa izdV gkrs gS]tks fd gYds Hkwjs jax ds gksaxsA vxj F1 larfr dk Loiztuu fd;k tk, tc F2 larfr ds Hkwjk % gYdk Hkwjk% lQsn jaxksa dsley{k.kh; (phenotypic) vuqikr D;k gksxk\(A) 1:1:1 a (B) 3:9:3 (C) 1:2:1 (D) 3:9:4
Ans (C)
115. The restriction endonuclease EcoR-I recognises and cleaves DNA sequence as shown below -5� � G A A T T C � 3�
3� � C T T A A G � 5�
What is the probable number of cleavage sites that can occur in a 10 kb long random DNA sequence?(A) 10 (B) 2 (C) 100 (D) 50
Ans (B)
EcoR-1 ,d fjfLVªD'ku ,UMksU;wDyh,l fd.od gS tks fd fuEufyf[kr DNA vuqØe dks igpku dj mldk foHkktu djrkgS5� � G A A T T C � 3�
KVPY QUESTION PAPER - STREAM (SB / SX)116. Which one of the following is true about enzyme catalysis ?
(A) the enzyme changes at the end of the reaction(B) the activation barrier of the process is lower in the presence of an enzyme(C) the rate of the reaction is retarded in the presence of an enzyme(D) the rate of the reaction is independent of substrate concentration
117. Vibrio cholerae causes cholera in humans. Ganga water was once used successfully to combat the infec-tion. The possible reason could be -(A) high salt content of Ganga water(B) low salt content of Ganga water(C) presence of bacteriophages in Ganga water(D) presence of antibiotics in Ganga water
Ans (C)euq";ksa esa dkWysjk jksx fofcz;ksa dkWysjs dhVk.kq ls gksrk gSA xaxkty ds lsou ls bl laØe.k dks jksdk tk ldrk FkkA bldk D;kdkj.k gk ldrk gS\(A) xaxkty esa ued dh ek=kk vf/kd gSA (B) xaxkty esa ued dh ek=kk de gS
(C) xaxkty esa thok.kqHkksth (bacteriophages) gSA (D) xaxkty esa izfrtSfodh (antibiotics ) gSaAAns (C)
118. When a person begins to fast, after some time glycogen stored in the liver is mobilized as a source ofglucose. Which of the following graphs best represents the change of glucose level (y-axis) in his blood,starting from the time (x-axis) when he begins to fast ?
(A) (B)
(C) (D)
Ans (C)
RESONANCE PAGE - 43
KVPY QUESTION PAPER - STREAM (SB / SX)
miokl 'kq: djus ds dqN le; (time x v{k½ ds ckn ;d r ds Xykbdkstu ds HkaMkj dk Xywdkst (glucose y v{k½
esa ifjorZu 'kq: gksrk gSA fuEufyf[kr fdl xzkQ esa miokl djus okys ds jDr esa Xywdkst dh ek=kk dks lgh n'kkZ;k
x;k gS\
(A) (B)
(C) (D)
Ans (C)
119. The following sequence contains the open reading frame of a polypeptide. How many amino acids will thepolypeptide consist of ?5� � AGCATATGATCGTTTCTCTGCTTTGAACT�3
120. Insects constitute the largest animal group on earth. About 25-30% of the insect species are known to beherbivores. In spite of such huge herbiore perssure, globally, green plants have persisted. One possiblereason for this persistence is :(A) food preference of insects has tended to change with time(B) herbivore insects have become inefficient feeders of green plants(C) herbivore population has been kept in control by predators(D) decline in reproduction of herbivores with time