ONLINE PRACTICE TEST-1 (STREAM SA) QUESTIONS & SOLUTIONS OF KISHORE VAIGYANIK PROTSAHAN YOJANA (KVPY) 2012 Date : 18-09-2011
Max. Marks : 100
One-Mark Questions One Mark SCQ (7)
Five Marks (2)
Physical/Inorganic
One Mark SCQ (7)
1.
So l .
S1 : Potential : Potential energy of the two opposite charge system increases with the decrease in distance. S2 : When : When an electron make transition from higher orbit to lower orbit it's kinetic energy increases. S3 : When an electron make transtition from lower energy to higher energy state its potential energy increases. increases. st + S4 : 11eV : 11eV photon can free an electron from the 1 excited state of He -ion. (A) T T T T (B*) F T T F (C) T F F T (D) F F F F S1 : Potential energy of the two opposite charge system decreases with decrease in distance, S4 : The energy of I st excited state of He + ion = – 3.4 Z2 = – 3.4 × 22 – 13.6 eV. S2 and S3 are correct statement.
2.
Find the minimum possible atomic number of the element in which all subshell's corresponding to n + ! = 4 are completely filled : (A*) 20 (B) 18 (C) 36 (D) 30
3.
Find incorrect statement regarding the position of the element having atomic number = 62. (A*) It belongs to 'd'-block (C) It belongs to 3 rd gr group
4.
(B) It belongs to 'f'-block (D) It belongs to 6 th period
If ratio of mole fractions of solute and solvent is same for 2 different solutions, 1 prepared in H 2O and otehr prepared in solvent 'S' having molar mass = 162 g/mol, is equal. Find ratio
(A) 1 5.
(B*) 9
(C) 8
(molality ) H2O (Molality)solvent 'S'
=?
(D) 4
In which of the following reactions is there a change in the oxidation number of nitrogen atom: !" N O (A) 2 NO2 ! 2 4
(C) N2O5 + H2O
! !"
(B) NH3 + H2O 2HNO3
! !"
NH4+ + OH –
(D*) None of these
Sol. In 1st reaction, oxisdation number of N is + 4, in 2 nd reaction it is –3 & in 3rd reaction it is +5, none of which changes during reaction. KVPY-1_Page # 1
6.
Assuming that O 2 molecule is spherical in shape with radius 2 Å, the percentage of the volume of O 2 molecules to the total volume of gas at S.T.P. is : (A*) 0.09 %
(B) 0.9 %
Sol.
7.
For the equilibrium at constant temperature : A(aq.)
(D) 0.045 %
(2 $ 10 8 )3 $ 6 $ 10 23 × 100. 22400
4 × Required % = 3
%$
(C) 0.009 %
&
B(aq.)
following graph has been obtained in the molar conc. of "A" vs molar conc. of "B". Find correct statement .
(A) Value of equilibrium constant at points X and Y is different. (B) Value of reaction quotient at points X and Y is different. (C*) Points X and Y represents two different equilibrium states. (D) Points X and Y represents same equilibrium states. Organic (3) 8. For which of the following reaction K f is greater than K b ?
(A*) HCOOH
(B) CH3COOH
–
(C) CH3 –ONa+
+
+
+
H2O
(D) Ans. (A) Sol. In acid base reaction, strong acid displaces weak acid from their salt and HCOOH is stronger acid than benzoic acid. 9.
Which of the following is correct set of aromatic compounds ? (A) Benzene, Cyclopropenyl anion and Furan.
(B*) Flurrene (C 60), Benzene and Pyrene
KVPY-1_Page # 2
Sol. 10.
(C) Cyclopropenyl cation, Cyclopentadienyl cation and Flurrene (D) Cycloheptatrienyl anion, Cyclopropenyl cation and Benzene Flurrene (C60), Benzene (C 6H6) and pyrene all are aromatic. If 2-Chloropentane is further monochlorinated with Cl 2 in presence of diffused sunlight, how many isomers of dichloro-pentane are formed ? (A*) 12 (B) 5 (C) 4 (D) 8
Sol.
Five-Marks Questions Physical/Inorganic 11.
Five Marks SCQ (2)
Equal moles of BaCl 2 and Mg(ClO4)2 are present in two different solutions. As increasingly different masses of K2SO4 are added to these solutions in two independent experiments, following graph is obtained. Which of the following can be concluded from this graph ? (Ba = 137, K = 39, Mg = 24.3, S = 32, O = 16, N = 14, Cl = 35.5)
(A) BaSO4 and MgSO4 are precipitated in (1) and (2) (B*) BaSO4 and KClO4are precipitated in (1) and (2) (C) KClO4 and KCl are precipitated in (1) and (2) (D) KCl and MgSO 4 are precipitated in (1) and (2)
KVPY-1_Page # 3
12.
Match the molecules listed in column-I with the characteristics given in column-II using the codes given. (1) Column - I
Column - I
(a) XeF4
(p) Two lone pairs trans to each other with central atom sp3d2 hybridised.
(b) XeF2
(q) One lone pair at the apex of trigonal pyramid with central atom sp 3 hybridised.
(c) IF 5
(r) One lone pair with central atom sp 3d2 hybridised
(d) XeO3
(s) Three lone pairs occupying equatorial positions with central atom sp 3d hybridised.
(a) (b) (A*) (p) (s) (C) (q) (r) Sol.
(c) (d) (r) (q) (s) (p)
(a) (b)
(c) (d)
(B) (s) (p)
(r) (q)
(D) (p) (s)
(q) (r)
(A)
(a)
(b)
(c)
(d)
Organic (1) 13.
An organic compound "A" C 15H24 on ozonolysis gives C 6H6O3 (B) and 3 moles of C3H6O (C). "C" gives yellow precipitate with iodine and NaOH. One of the tautomeric form of "B" is aromatic and gives violet colour with FeCl3 solution. "B" gives only two form of oximes when treated with NH 2OH and HCl aqueous. Structural formula of A is :
(A)
(B*)
(C)
(D)
Sol.
KVPY-1_Page # 4
ONLINE PRACTICE TEST-1 (STREAM SA) QUESTIONS & SOLUTIONS OF KISHORE VAIGYANIK PROTSAHAN YOJANA (KVPY) 2012 Date : 09-09-2012
Max. Marks : 100
SCQ.(15) 14.
A box is put on a weighing scale which is adjusted to read zero when the box is empty .Tiny balls each of mass m are released from height h above its base at a rate n balls/sec. If the balls collide with the base such that they immidiately come to rest after collision then reading after time t . (B*) mn [ 2gh ! gt ]
(A) mnt Sol.
15.
Sol.
16.
R= "
dm + (mnt) g dt
= mn
2gh ! mn gt
(C) mngt
(D) zero
10g of ice at –20ºC is added to 10 g of water at 50ºC. Then the resulting temperature is (specific heat of ice = 0.5 cal g –1 ºC –1 and latent heat of fusion of ice = 80 calg –1 ) (A) –20ºC (B) 15ºC (C*) 0º C (D) 50ºC Heat required by ice to come to 0 ºC of water = ms#t + mL f = 10 × 0.5 × 20 + 10 × 80 = 900 cal Heat given by water = 10 × 1 × 50 = 500 cal So there will be ice and water in the mixture Hence temprature is 0 º C The K.E. of a satellite is 2 MJ ,the total energy satellite is (A*) – 2MJ
(B) 1 MJ
(C) –
1 MJ 2
(D) – 4 MJ
Sol.
TE = – K.E. = – 2 MJ
17.
Two similar thin palno convex lenses are combined together in two different ways as shown . Then the ratio of focal length of a to b will be
(A) 3 : 2
(B*) 1 : 1
(C) 1 : 2
(D) 2 : 1
2(n – 1) = fb R
Sol.
fa =
18.
Magnetic flux $ (in weber ) linked with a close circuit of resistance 10 % varies with time t (in seconds) as $ = 5t 2 – 4t + 1 then the induced EMF in the circuit at t = 0.2 second (A) .04 V (B) – 0.4 V (C) – 2.0 V (D*) 2.0 V KVPY-1_Page # 1
Sol. 19.
d$ = – (10 t – 4) = – (10 × 0.2 – 4) = 2.0 V dt The ratio of minimum to maximum wavelength in Balmer series in hydrogen atom spectrum is (A*) 5 : 9 (B) 5 : 36 (C) 1 : 4 (D) 3 : 4
E = –
1 Sol.
1
–
& min imum 2 32 & max imum = 1 – 1 2 2 2
2
20.
'
In a double slit experiment , the separation between the slits is d and distance of the screen from slit is D . If the wavelength of light used in & and ( is the intensity of central bright fringe , Then intensity of light at distance x from the central maxima on the screen is
. /2 xd + , ) (A) 4(cos , &D ) - * ( ( (0 = 4 + 4 + 2
. /xd + , ) (C) 4(sin ,- 2&D *)
. /xd + ) (B*) (cos , - &D *
2
Sol.
= 5: 9
2
2
. /xd + , ) (D) (sin ,- &D *) 2
( ( . 2/ xd + 1 cos, ) 4 4 - & D *
. /xd + ) = (cos2 , - &D * 21.
The time period of simple pendulum of infinite length is (R e = radius of earth) (A*) T = 2 /
Re g
(B)T = 2/
2 Re g
(C)T = 2 /
Re 2g
(D) T = '
1 Sol.
T = 2/
T = 2 /
22.
7 1 14 ! 2 g5 6 Re ! 3
8'
Re g
Two large parallel conduting plates are a distance d apart and have a potential difference V between them. A unifrom magnetic induction B is also present in the region between the plates along a direction parallel to the plates. A particle of charge +q and mass m is released from rest in the crossed electric and magnetic fields from the surface of one plate as shown in figure.. The trajectory of the particle (neglect gravitational force) will be :
(A) straight line Sol.
!
(B) circle
(C*) cycloid
(D) helical path
Speed v of +q is not constant, path in cycloid. It is a well known result.
KVPY-1_Page # 2
23.
A cat C is running towards O with constant velocity 6 m/s and a rat R is running towrards O with constant velocity 8 m/s as shown in figure. Initially RO = 40 m and CO = 50 m. Minimum distance between rat and cat will be:
(A) 30 m Sol.
(B) 15 m
(C) 20 m
tan9 =
6 , 8
9 = 37º, It is direction of VR,C with horizontal line RO.
tan9 =
6 , 8
9 = 37º,
(D*) 16 m
RO VR,C
MO , MO = 30, CM = 20 10 Minimum distnace = 20 cos 37 º = 16m
tan 37º =
24.
If all the batteries are ideal then potential at junction C is :
(A) +8V Sol. 25.
Vc ; 40 5
(B) +2V
!
Vc ; 2 2
!
Vc ; 4 3
:0
(C) +6V
(D*) +10V
< VC = 10 volt
Three indentical bulbs each of resistance 2 % are connected as shown. The maximum power that can be consumed by individual bulb is 32W, then the maximum power consumed by the combination is :
KVPY-1_Page # 3
Sol.
26.
(A*) 48 W (B) 96 W 2 P = i R 32 = i2 (2) i = 4A Pmax = (2)2 (2) + (2 2)2 + (4) 2 (2) = 8 + 8 + 32 = 48 W
(D) 160 W
Two coaxial rings are separated by a distance x. R is radius of larger ring having charge Q uniformly distributed on it.The other ring is having very small radius r (<
Sol.
(C) 128 W
E=
E=
(B) KQx 2
2 3 / 2
(x ! R ) KQx 2
2 3 / 2
(x ! R )
R 3
, E is maximum for x =
,E
x=
(C) R 2
R 3
(D*)
R 2
.
R 2
KVPY-1_Page # 4
Part - A 1.
(1-10)
A cancer which is not a tumor is : (A) Lymphoma
2.
KVPY Test Paper - 1 Biology
(B*) Leukemia
(C) Prostate cancer
(D) Oral cancer
Sometimes urea is fed to ruminates to improve their health. It works by (A*) Helping growth of gut microbes that break down cellulose (B) Killing harmful microorganisms in their gut (C) Increasing salt content in the gut (D) Directly stimulating blood cell proliferation
3.
The hypersensitive reaction in response to a type I allergen which produces characteristic symptoms is described as allergic response. Which of the following immunoglobulin is mostly involved in this type of cellular response ? (A) IgA (B) IgM (C) IgD (D*) IgE
4.
The process of cell death involving DNA cleavage in cells is known as (A) Necrosis (B*) Apoptosis (C) Cytokinesis (D) Endocytosis
5.
Patients who have undergone organ transplants are given anti-rejection medications to (A) Minimize infection (B) Stimulate B- macrophage cell interaction (C*) Prevent T- lymphocyte proliferation (D) Adopt the HLA of donor
6.
In a diploid organism, there are three different alleles for a particular gene. Of these three alleles one is recessive and the other two alleles exhibit co-dominance. How many phenotypes are possible with this set of alleles? (A) 3
7.
(B) 6
(C*) 4
(D) 2
Bacteriochlorophylls are photosynthetic pigments found in phototrophic bacteria. Their function is distinct from the plant chlorophylls in that they (A*) Do not produce oxygen
(B) Do not conduct photosynthesis
(C) Absorb only blue light
(D) Function without a light source
8.
The distance between two consecutive DNA base pairs is 0.34 nm. If the length of a chromosome is 1 mm the number of base pairs in the chromosome is approximately (A*) 3 million (B) 1.5 million (C) 30 million (D) 6 million
9.
ELISA, the standard screening test for HIV, detects which of the following ? (A) HIV DNA (B) HIV RNA (C) HIV proteins (D*) Antibodies to HIV proteins
10.
Conversion of the Bt toxin produced by Bacillus thuringienesis to its active form in the gut of the insects is mediated by (A) Acidic pH of the gut (B*) Alkaline pH of the gut (C) Lipid modification of the protein (D) Cleavage by chymotrypsin
ONLINE PRACTICE TEST-1 (STREAM SA) QUESTIONS & SOLUTIONS OF KISHORE VAIGYANIK PROTSAHAN YOJANA (KVPY) 2012 Date : 09-09-2012
Max. Marks : 100
SCQ.(15) 14.
A box is put on a weighing scale which is adjusted to read zero when the box is empty .Tiny balls each of mass m are released from height h above its base at a rate n balls/sec. If the balls collide with the base such that they immidiately come to rest after collision then reading after time t . (B*) mn [ 2gh ! gt ]
(A) mnt Sol.
15.
Sol.
16.
R= "
dm + (mnt) g dt
= mn
2gh ! mn gt
(C) mngt
(D) zero
10g of ice at –20ºC is added to 10 g of water at 50ºC. Then the resulting temperature is (specific heat of ice = 0.5 cal g –1 ºC –1 and latent heat of fusion of ice = 80 calg –1 ) (A) –20ºC (B) 15ºC (C*) 0º C (D) 50ºC Heat required by ice to come to 0 ºC of water = ms#t + mL f = 10 × 0.5 × 20 + 10 × 80 = 900 cal Heat given by water = 10 × 1 × 50 = 500 cal So there will be ice and water in the mixture Hence temprature is 0 º C The K.E. of a satellite is 2 MJ ,the total energy satellite is (A*) – 2MJ
(B) 1 MJ
(C) –
1 MJ 2
(D) – 4 MJ
Sol.
TE = – K.E. = – 2 MJ
17.
Two similar thin palno convex lenses are combined together in two different ways as shown . Then the ratio of focal length of a to b will be
(A) 3 : 2
(B*) 1 : 1
(C) 1 : 2
(D) 2 : 1
2(n – 1) = fb R
Sol.
fa =
18.
Magnetic flux $ (in weber ) linked with a close circuit of resistance 10 % varies with time t (in seconds) as $ = 5t 2 – 4t + 1 then the induced EMF in the circuit at t = 0.2 second (A) .04 V (B) – 0.4 V (C) – 2.0 V (D*) 2.0 V KVPY-1_Page # 1
Sol. 19.
d$ = – (10 t – 4) = – (10 × 0.2 – 4) = 2.0 V dt The ratio of minimum to maximum wavelength in Balmer series in hydrogen atom spectrum is (A*) 5 : 9 (B) 5 : 36 (C) 1 : 4 (D) 3 : 4
E = –
1 Sol.
1
–
& min imum 2 32 & max imum = 1 – 1 2 2 2
2
20.
'
In a double slit experiment , the separation between the slits is d and distance of the screen from slit is D . If the wavelength of light used in & and ( is the intensity of central bright fringe , Then intensity of light at distance x from the central maxima on the screen is
. /2 xd + , ) (A) 4(cos , &D ) - * ( ( (0 = 4 + 4 + 2
. /xd + , ) (C) 4(sin ,- 2&D *)
. /xd + ) (B*) (cos , - &D *
2
Sol.
= 5: 9
2
2
. /xd + , ) (D) (sin ,- &D *) 2
( ( . 2/ xd + 1 cos, ) 4 4 - & D *
. /xd + ) = (cos2 , - &D * 21.
The time period of simple pendulum of infinite length is (R e = radius of earth) (A*) T = 2 /
Re g
(B)T = 2/
2 Re g
(C)T = 2 /
Re 2g
(D) T = '
1 Sol.
T = 2/
T = 2 /
22.
7 1 14 ! 2 g5 6 Re ! 3
8'
Re g
Two large parallel conduting plates are a distance d apart and have a potential difference V between them. A unifrom magnetic induction B is also present in the region between the plates along a direction parallel to the plates. A particle of charge +q and mass m is released from rest in the crossed electric and magnetic fields from the surface of one plate as shown in figure.. The trajectory of the particle (neglect gravitational force) will be :
(A) straight line Sol.
!
(B) circle
(C*) cycloid
(D) helical path
Speed v of +q is not constant, path in cycloid. It is a well known result.
KVPY-1_Page # 2
23.
A cat C is running towards O with constant velocity 6 m/s and a rat R is running towrards O with constant velocity 8 m/s as shown in figure. Initially RO = 40 m and CO = 50 m. Minimum distance between rat and cat will be:
(A) 30 m Sol.
(B) 15 m
(C) 20 m
tan9 =
6 , 8
9 = 37º, It is direction of VR,C with horizontal line RO.
tan9 =
6 , 8
9 = 37º,
(D*) 16 m
RO VR,C
MO , MO = 30, CM = 20 10 Minimum distnace = 20 cos 37 º = 16m
tan 37º =
24.
If all the batteries are ideal then potential at junction C is :
(A) +8V Sol. 25.
Vc ; 40 5
(B) +2V
!
Vc ; 2 2
!
Vc ; 4 3
:0
(C) +6V
(D*) +10V
< VC = 10 volt
Three indentical bulbs each of resistance 2 % are connected as shown. The maximum power that can be consumed by individual bulb is 32W, then the maximum power consumed by the combination is :
KVPY-1_Page # 3
Sol.
26.
(A*) 48 W (B) 96 W 2 P = i R 32 = i2 (2) i = 4A Pmax = (2)2 (2) + (2 2)2 + (4) 2 (2) = 8 + 8 + 32 = 48 W
(D) 160 W
Two coaxial rings are separated by a distance x. R is radius of larger ring having charge Q uniformly distributed on it.The other ring is having very small radius r (<
Sol.
(C) 128 W
E=
E=
(B) KQx 2
2 3 / 2
(x ! R ) KQx 2
2 3 / 2
(x ! R )
R 3
, E is maximum for x =
,E
x=
(C) R 2
R 3
(D*)
R 2
.
R 2
KVPY-1_Page # 4
KVPY TEST-1 09-09-2012 1. Sol.
3
#
(3x – 4) (3x – 1) = 0
Since x " R+, only x = !n 2. Sol.
3.
3
The number of solution of the equation 2 x log2 + 2 2 ! x log2 – 5 = 0 where x " R+ is (A) 2 (B*) 1 (C) 0 (D) more than 2 The equation is equivalent to 3x + 4.3 –x – 5 = 0
#
x = !n
4 and 0 3
4 is acceptable. 3
The 288th term of the series a, b, b, c, c, c, d, d, d, d, e, e, e, e, e, f, f, f, f, f, f, .............. is (A) u (B) v (C) w (D*) x a, bb, c,c,c, d,d,d,d, .... consider the arrangement of n letters. n(n $ 1) % 288 Then 1 + 2 + 3 + ... + n = 2 i.e., n2 + n – 576 % 0 & n % 23 & If we write up to 23 letters i.e. upto w, the total number of terms 23 ' 24 = 1 + 2 + 3 + ... + 23 = = 276 2 & terms from 276 to 300 are the letter x each & 288th letter is x. 1 2
Three pieces of cakes of weights 4 Ibs, 6
1 3 Ibs and 7 Ibs respectively are to be divided into parts of 5 4
equal weights. Further, each must be as heavy as possible. If one such part is served to each guest, then what is the maximum number of guests that could be entertained ? (A) 54 (B) 72 (C) 20 (D*) 41 Sol.
- 9 27 36 *
HCF + 2 , 4 , 5 ( = , )
HCF (9, 27, 36) 9 = Ibs 20 LCM(2, 4, 5)
= weight of each piece. 9 27 36 $ $ = 18.45 2 4 5 18 .45 ' 20 Maximum number of guests = = 41 9
Total weight =
4.
Given n(A) = 11, n(B) = 13, n(C) = 16, n( A . B ) = 3, n( B . C ) = 6, n( A . C ) = 5 and n( A . B . C ) = 2, then the value of n[ A c . ( B/C) ] = (A) 4
(B) 7
(C*) 13
(D) 23
Sol. A
a
b c e d f g
B
C
Given : n(A) = 11 # a + b + d + e = 11 n(B) = 13 # b + c + e + f = 13 n(C) = 16 # d + e + f + g = 16 n(A . B) = 3 # b + e = 3 n(B . C) = 6 # e + f = 6 n(A . C) = 5 # d + e = 5 n(A . B. C) = 2 # e = 2 Solving all equations Page # 1
e = 2, d = 3, f = 4, b = 1, a = 5, c = 6, g = 7 To find : n[Ac. (B /C)] B/C = b + c + d + g Ac = c + f + g Ac.(B/C) = c + g Ac.(B/C) = c + g = 6 + 7 = 13
5.
Number of solutions of the equation, [ y + [y] ] = 2 cos x is ( where y = (1/3) 0sin x $ 0sin x $ 0sin x1 1 1 and [ ] = greatest integer function )
Sol.
(A*) 0 (B) 1 (C) 2 (D) 2 [y] = cos x and y = [sin x] = [sin x] = cos x not possible for any value of x.
6.
If x =
n4 2
, satisfies the equation sin
(A) n = !1, 0, 3, 5 Sol.
2
–
3 4 x 4 34 % – % 4 2 2 4
#
–
4 54 %x% 2 2
#
But given x =
x 2
= 1 – sin x and the inequality (C*) n = 0, 2, 4
x 2
!
4 2
%
34 4
, then
(D) n = !1, 1, 3, 5
n4 2 n = – 1, 0, 1, 2, 3, 4, 5
#
x x – cos = 1 – sin x. 2 2
#
x * - x x x sin – cos = + sin – cos ( 2 ) 2 2 , 2
#
x * - x + sin – cos ( 2 ) , 2
#
either tan
if tan
if sin
#
! cos
(B) n = 1, 2, 4, 5
"
sin
x
x = 0 2
2
x : x 7 8sin 2 – cos 2 – 15 = 0 9 6
x = 0 2
or
#
x = 0, 2 4
#
n = 0, 4
sin
x x – cos = 1. 2 2
x x – cos = 1 2 2
- x 4 * sin + – ( = , 2 4 )
1 2
= sin
4 4
4 4 4 – = K 4 + ( –1) k 2 4 4 for odd values of k, x = 2k 4 # & n = 0, 4 x = 0, 2 4 , for even values of k, x = 2k 4 + 4 # &n=2 x = 4, Hence overall n = 0, 2, 4 #
Page # 2
7.
In the figure below, AL is perpendicular to BC and CM is perpendicular to AB. If CL = AL = 2BL, find A MC/AM. (A) 2 (B*) 3
M
(C) 4 C
(D) Cannot be determined Sol.
B
L
In the figure, if BL = 1, LA = LC = 2
&
AB = 5 and AC = 2 2
Let MA = x CM =
8 ! x2
We have (AB)(CM) = (CB)(AL) = 5 8 ! x 2 = 2(3) = 6
&
8 ! x 2 =
# x=
0.8
MC =
&
8.
8 ! x 2 =
MC ; AM
7. 2 0. 8
7.2
= 3
A rhombus and rectangle have a common diagonal of length a units. If area of rhombus and rectangle are 8 and 4 respectively, then a equals to (A*)
Sol.
36 # 8 – x2 = 7.2 5
4 4
(B)
3
Area of rhombus = d1 = a, d 2 = 2 ×
3 2
(C)
1 4 2
(D)
3
1 × d1 × d2 2
a tan < = a tan < 2
1 a 2 tan < × a × a tan < = 2 2 2 A2 = a cos < × a sin < = a sin < cos < A1 =
a 4 sin < a 4 sin 2 < sin cos ' ' < < A1A2 = = 2 cos < 2 sin2< =
2A 1 A 2 a4
A 2 a 2 sin < cos < ; A1 a 2 sin < = 2 cos2< 2 cos < cos2< =
A2 2A 1
Page # 3
sin < + cos2< = 1
#
2
2A 1 A 2 A 2 $ a4 2A 1 = 1
2'8' 4 4 $ = 1 a4 2' 8 4
256 a = 3
#
4
9.
a=
;
1 4 3
4 4
3
In the given figure. Find radius of smaller circle.
5
5
3
3 (A)
(B*)
5 – 3
=
3 3 – 5 4
>
(C)
3$ 5 8
(D)
3$2 5 12
Sol.
AC = R cot < = 3 # R = 3 tan < AC = AB + BC AC = r cot < + (R + r) cos < (R – r) cot < = (R + r) cos < R – r sin < = R $ r
#
cos 2< =
3 5
sin < =
1 – cos 2< 1 ; 2 5
r =
5 – 1 5 $1
#
r ; R
5 – 1 5 $1
× 3 tan <
- 5 – 1 * - 1 * + ( + ( 3 = + (× , 5 $ 1 ) , 2 ) 3 = 2
1 – sin < r ; 1$ sin < R
- 1 1 * +"sin < , & tan < ; (( + 2 ) 5 ,
=
- 5 – 1 * 3 - 6 – 2 5 * 3 3 – 5 + (; + (; + 5 $1( 2 + ( 4 4 , ) , )
>
Page # 4
10.
If y = ax 2 + bx + c represents the curve given in the figure and b 2 = 2(b + 2ac), where a ? 0 and AP = 3 units, then OP = A
O
3 3 (B) 2 4 2 Given relation b = 2(b + 2ac) # b2 = 2b + 4ac # – b2 + 4ac = – 2b (A)
Sol.
D b = – = OP 4a 2a OP = 3 (" AP = 3)
#
AP = –
& 11.
P
(C*) 3
(D) 6
& AP = OP
The number of 5 digit numbers of the form x y z y x in which x < y is: (A) 350 (B*) 360 (C) 380
(D) 390
Sol.
The first digit ' x ' can be any one of from 1 to 8 where as ' z ' can be any one from 0 to 9 when x = 1, y = 2, 3, 4,......, 9 x = 2, y = 3, 4,......, 9 and so on Thus the total = (8 + 7 + 6 +...... + 2 + 1) 10 = 360
12.
The number of ways in which 5 X's can be placed in the squares of the figure so that no row remains empty is :
Sol.
13.
Sol.
(A) 56 (B*) 44 (C) 98 8C ! [ only top remains exmpty + bottom empty] 5 8C – [ 6C + 6C ] = 56 – 12 = 44 5 5 5
(D) 40
A straight line 2x + 3y = 1 is rotated 45 º in anticlock wise direction about the origin, the equation of line in its new position is (A*) – x + 5y = 2x + 3y = 1 2x 13
+
3
(B) x – 5y =
2
y=
(C) –x + 5y = 1
2
(D) x – 5y = 1
1
13 13 let the normal drawn from the origin to the line make an angle @ with the positive direction of x axis. Then cos @ =
2 13
, sin @ =
3 13
1 1 4 * - cos + @ $ ( = cos @ sin @ = – – 4 ) 2 2 , 1 1 4 * - sin + @ $ ( = sin @ + cos @ = 4 ) 2 2 ,
1 26 5 26
&
equation of the line in new position is –
i.e.
– x + 5y =
1 26
x +
5 26
y =
1 13
2
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