1505 lts paper 1 aiot dlp

7/25/2019 1505 Lts Paper 1 Aiot Dlp

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READ THE INSTRUCTIONS CAREFULLY / œi; b u un ' d ; u l i<

GENERAL / l e ;:

1. This sealed booklet is your Question Paper. Do not break the seal till you are instructed to do so.

; g e k s g j c U / k i q f L r d k i z ' u i = g S A b ld h e q g j r c r d u r k s M + s tc r d b ld k f u n s Z ' k u f n ; k tk ; s A2. Use the Optical Response sheet (ORS) provided separately for answering the questions.

i z ' u k s a d k m Ÿ k j n s u s d s f y , v y x ls n h x ; h v k W I V h d y f j L i k a l ' k h V (v k s - v k j- , l-)(ORS) d k m i ; k s x d j s a A

3. Blank spaces are provided within this booklet for rough work.

d P p s d k ; Z d s f y , b l i q f L r d k e s a [ k k y h L F k k u f n ; s x ; s g S a A

4. Write your name and form number in the space provided on the back cover of this booklet.

, d i q f L r d k d s f i N y s i ` " B i j f n , x , L F k k u e s a v i u k u k e rF k k Q k W e Z u E c j f y f [ k ,A

5. After breaking the seal of the booklet, verify that the booklet contains 40 pages and all the 20 questions in each

subject and along with the options are legible.

b l i q f L r d k d h e q g j r k s M + u s d s c k n œ i ; k tk ° p y s f d b le s a40 i ` " B g S a v k S j v k S j i z R ; s d f o " k ; d s lH k h 20 i z ' u v k S j m u d s m Ÿ k j f o d Y i

B h d ls i < + s tk ld r s g S A

QUESTION PAPER FORMAT AND MARKING SCHEME / i 'ui= d i :i v v du ; tu:

6. The question paper has three parts : Physics, Chemistry and Mathematics. Each part has three sections.

b l i z ' u i = e s a r h u H k k x g S a % H k k S f r d f o K k u] j lk ; u f o K k u v k S j x f . k rA g j H k k x e s a r h u [ k . M g S a A

7. Carefully read the instructions given at the beginning of each section.

i z R ; s d [ k . M d s i z k j E H k e s a f n ; s g q , f u n s Z ' k k s a d k s / ; k u ls i < + s A

8. Section-I contains 10 multiple choice questions with one or more than one correct option.

Marking scheme : +4 for correct answer, 0 if not attempted and –2 in all other cases.

[ .M-I e s a10 c g q f o d Y i h ; i z ' u g S A f tu d s,d ; ,d l v d

f o d Y i lg h g S a A

v d ; tu: +4 lg h m Ÿ k j d s f y ,]0 i z ; k l u g h a d j u s i j rF k k –2 v U ; lH k h v o L F k k v k s a e s a A

9. Section-II contains 2 ‘match the following’ type questions and you will have to match entries in Column-I with the

entries in Column-II.

Marking scheme : for each entry in column-I. +2 for correct answer, 0 if not attempted and –1 in all other cases.

[ k . M -II e s a2 ‘ lq e s f y r ’ i z k :i d s i z ' u g S a f tle s a v k id ye-I d h i z f o " V ; k s a d k s

d ye-II d h i z f o f " V ; k s a ls lq e s f y r d j s a x s A

v d ; tu: d k W y e -I d h i z R ; s d i z f o f " V d s f y , +2 lg h m Ÿ k j d s f y ,]0 i z ; k l u g h a d j u s i j rF k k –1 v U ; lH k h v o L F k k v k s a e s a A

10. There is no questions in SECTION-III / [ k . M –III e s a , d H k h i z ' u u g h a g S

11. Section-IV contains 8 questions. The answer to each question is a single digit integer ranging from

0 to 9 (both inclusive)

Marking scheme : +4 for correct answer and 0 in all other cases.

[ k . M -IV e s a8 i z ' u g S a A i z R ; s d i z ' u d k0 ls 9 r d ( n k s u k s a ' k k f e y ) d s c h p d k , d y v a d h ; i w . k k ± d g S A

v d ; tu: +4 lg h m Ÿ k j d s f y , rF k k0 v U ; lH k h v o L F k k v k s a e s a A

PAPER – 1

Test Type : ALL INDIA OPEN TEST (MAJOR) Test Pattern : JEE-Advanced

TEST # 12 TEST DATE : 15 - 05 - 2016

TARGET : JEE (ADVANCED) 2016

LEADER TEST SERIES / JOINT PACKAGE COURSE

Paper Code : 0000CT103115008

DISTANCE LEARNING PROGRAMME(Academic Session : 2015 - 2016)

Time : 3 Hours Maximum Marks : 264

Please see the last page of this booklet for rest of the instructions / œ i; ' u n ' d y, b l i rd d v re i B d i

D O N O T

B R E A K T H E S E A L S W I T H O U T B E I N G

I N S T R U C T E D T O D O S O B Y T H E I N V I G I L A T O R / u

d

d

v

u

n

'

d

c

u

e

g

u

r

M



ALL INDIA OPEN TEST/JEE (Advanced)/15-05-2016/PAPER-1

SOME USEFUL CONSTANTS A tomic No. H = 1, B = 5, C = 6, N = 7, O = 8, F = 9, Al = 13, P = 15, S = 16, Cl = 17, Br = 35,

Xe = 54, Ce = 58,

A tomic masses : H = 1, Li = 7, B = 11, C = 12, N = 14, O = 16, F = 19, Na = 23, Mg = 24,

Al = 27, P = 31, S = 32, Cl = 35.5, Ca=40, Fe = 56, Br = 80, I = 127,

Xe = 131, Ba=137, Ce = 140,

LTS-2/40 0000CT103115008

Note : In case of any correction in the test paper, please mail to [email protected] within 2 days along with Paper Code

& Your Form No.( u V

; f n b l i z ' u i = e s a d k s b ZCorrection g k s r k s œ i ; kPaper Code , o a v k i d s Form No. , o a i w . k ZTest Details d s lk F k 2 f n u d s v U n j [email protected] i j mail d j s a A)

Space for Rough Work / dP p d ; d y, L u

Boltzmann constant k = 1.38 × 10 –23 JK –1

Coulomb's law constant

1

= ×1

4

Universal gravitational constant G = 6.67259 × 10 –11 N–m2 kg –2

Speed of light in vacuum c = 3 × 108 ms –1

Stefan–Boltzmann constant

= 5.67 × 10 –8 Wm –2 –K –4

Wien's displacement law constant b = 2.89 × 10 –3 m–K

Permeability of vacuum µ0 = 4 × 10 –7 NA –2

Permittivity of vacuum

0 =

2

0

1

c

Planck constant h = 6.63 × 10 –34 J–s




P H Y S I C S

LTS-3/400000CT103115008

PART-1 : PHYSICS

H x

-1 :H rd oK u

SECTION–I : (Maximum Marks : 40)

[ .M–I : ( v dre v d: 40)

This section contains TEN questions.

Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these

four option(s) is (are) correct.

For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS

Marking scheme :

+4 If only the bubble(s) corresponding to all the correct option(s) is (are) darkened

0 If none of the bubbles is darkened

–2 In all other cases

b l [ k . M e s anl i z ' u g S a i z R ; s d i z ' u e s a p k j f o d Y i(A), (B), (C) rF k k (D) g S a A b u p k j f o d Y i k s a e s a ls,d ; ,d l v d f o d Y i lg h g S a A i z R ; s d i z ' u e s a ] lH k h lg h f o d Y i ( f o d Y i k s a ) d s v u q :i c q y c q y s ( c q y c q y k s a ) d k s v k s -v k j- , l- e s v a d u ; k s tu k :

+4 ; f n f lQ Z lH k h lg h f o d Y i ( f o d Y i k s a ) d s v u q :i c q y c q y s ( c q y c q y k s a ) d k s d k y k f d ; k tk ; 0 ; f n d k s b Z H k h c q y c q y k d k y k u f d ; k g k s

–2 v U ; lH k h v o L F k k v k s a e s a1. Two identical blocks are resting on supports as shown. Two identical bullets travelling with the same

speed hit the blocks & get embedded in them in very short t ime. The blocks leave the contact immediately

after bullet strikes them. In case (i) bullet's velocity is toward centre of mass and in case (ii) bullet's

velocity is off centre. The kinetic energies after collision are K 1 & K

2 and maximum height reached by

the centre of mass are h1 & h

2respectively. Then :

n k s , d tS ls C y k W d f p = k u q lk j v k / k k j k s a i j f o j k e k o L F k k e s a g S A n k s , d tS lh x k s f y ; k ° le k u p k

V d j k r h g S rF k k c g q r v Y i le ; e s a b u e s a / k a l tk r h g S A x k s f y ; k s a d s V d j k u s d s B h d i ' p k r ~ C y k W

(i) e s a x k s y h d k o s x Ê O ; e k u d s s U Ê d h v k s j g S rF k k i z d j . k(ii) e s a x k s y h d k o s x d s U Ê ls i j s g S A V D d j d s c k n x

ÿ e ' k % K 1 rF k k K

2 rF k k Ê O ; e k u d s U Ê k j k i z k I r v f / k d r e ≈ ° p k b Z ÿ e ' k %h

1 o h

2 g S r k s % &

Case (i) Case (ii)

(A) K 1 < K

2(B) K

1 = K

2(C) h

1 > h

2(D) h

1= h

2

BEWARE OF NEGATIVE MARKING

HAVE CONTROL HAVE PATIENCE HAVE CONFIDENCE 100% SUCCESS

Space for Rough Work /dP p d ; d y, L u



P H Y S I C S


LTS-4/40 0000CT103115008


2. A barometer is made of glass. It contains mercury (density 13.6 × 103 kg/m3, surface tension

0.085 N/m, angle of contact 143°). The barometer scale has a least count of 1mm. If the radius of

barometer tube is 0.5 mm, and height of mercury column recorded is 762 mm, what can be true

atmospheric pressure (in mm of mercury column) :-

d k ° p ls c u s c S j k s e h V j e s a ? k u R o13.6 × 103 kg/m3, i ` " B r u k o0.085 N/m o lE i d Z d k s . k143° o k y k i k j k H k j k g q v k g S b ld s i S e k u s d k v Y i r e k a d1mm g S A ; f n c S j k s e h V j u y h d h f = T ; k0.5 mm o i k j n ~ L r E H k d h n tZ ≈ ° p k b Z762 mm g k s r k s

o k L r f o d o k ; q e . M y h ; n k c ( i k j n ~ L r E H k i jmm e s a ) g k s ld r k g S % &

(A) 760 mm (B) 761 mm (C) 763 mm (D) 765 mm

3. For a certain transverse standing wave on a long string, an antinode is formed at x = 0 and next to it, a

node is formed at x = 0.10 m. The displacement y(t) of the string particle at x = 0 is shown in figure.

(A) Transverse displacement of the particle at x = 0.05m and t = 0.05 s is – 2 2 cm.

(B) Transverse displacement of the particle at x = 0.04 m and t = 0.025 s is – 2 2 cm.

(C) Speed of the travelling waves that interfere to produce this standing wave is 2 m/s.

(D) The transverse velocity of the string particle at x =1

15 m and t = 0.1 s is 20 cm/s

y(cm)0.05 0.1 0.15 0.2

t(s)

4

–4

f d lh y E c h j L lh i j , d v u q i z L F k v i z x k e h r j a x d s f y ; s , d i z L i U nx = 0 i j rF k k b ld s c k n , d f u L i U nx = 0.10 m

i j c u r k g S Ax = 0 i j j L lh d s d . k d k f o L F k k i uy(t) f p = e s a n ' k k Z ; k x ; k g S % &(A) x = 0.05m rF k k t = 0.05 s i j d . k d k v u q i z L F k f o L F k k i u – 2 2 cm g S A

(B) x = 0.04 m rF k k t = 0.025 s i j d . k d k v u q i z L F k f o L F k k i u – 2 2 cm g S A

(C) b lv i z x k e h r j a x d k s m R i U u d j u s d s f y ; s O ; f r d f j r g k s u s o k y h i z x k e h r j a x k s a d h p k y2 m/s g S A

(D) x =1

15 m rF k k t = 0.1 s i j j L lh d s d . k d k v u q i z L F k o s x20 cm/s g S A




P H Y S I C S

LTS-5/400000CT103115008


4. In the figure shown is the surface charge density on the upper metallic plate.

(A) The ratio of energy density in st dielectric to second dielectric is5

3

(B) The ratio of energy density in st dielectric to second dielectric is3

5

(C) Total induced surface charge density on the interface of the two dielectric is –3

15

(D) Total induced surface charge density on the interface of the two dielectric is –2

15

d/3

2d/3

Area of plate = A

A

B

I Dielectricst

IInd

Dielectric

K = 31

K = 52

i z n f ' k Z r f p = e s a ≈ i j h / k k f R o d I y s V i j i ` " B h ; v k o s ' k ? k u R o d k e k u g S % &

(A) i z F k e rF k k f r h ; i j k o S |q r e s a ≈ tk Z ? k u R o d k v u q i k r5

3 g S A

(B) i z F k e rF k k f r h ; i j k o S |q r e s a ≈ tk Z ? k u R o d k v u q i k r3

5 g S A

(C) n k s u k s a i j k o S |q r k s a d s v a r j k i ` " B i j d q y i z s f j r i ` " B h ; v k o s ' k ? k u R o –

3

15

g S A

(D) n k s u k s a i j k o S |q r k s a d s v U r j k i ` " B i j d q y i z s f j r i ` " B h ; v k o s ' k ? k u R o – 2

15

g S A



P H Y S I C S


LTS-6/40 0000CT103115008


5. A parallel beam of light is incident on a lens of focal length 10 cm. A parallel slab of refractive index

1.5 and thickness 3 cm is placed on the other side of the lens. The Nature of the final image from the

lens slab combination is: (Assume rays to be paraxial) :-

(A) At Distance of 11 cm and Image is virtual if x = 12 cm.

(B) At Distance of 11 cm and Image is real if x = 14 cm.(C) At the Distance of 11 cm and Image is Virtual if x = 16 cm

(D) At Distance of 11 cm and Image is real if x = 5 cm.

x

Q k s d l n w j h10 cm o k y s y s a l i j , d le k U r j i z d k ' k i a q tv k i f r r g k s r k g S A y s a l d s n w lj h v k s j1.5 v i o r Z u k a d rF k k3 cm

e k s V k b Z o k y h , d le k U r j i f Í d k j [ k n h tk r h g S A y s a l& i f Í d k f u d k ; ls v a f r e i z f r f c E c ( f d j . k k s a d k s

(A) v k H k k lh g k s x k rF k k11 cm d h n w j h i j g k s x k ; f nx = 12 cm g k s A(B) o k L r f o d g k s x k rF k k11 cm d h n w j h i j g k s x k ; f nx = 14 cm g k s A

(C) v k H k k lh g k s x k rF k k11 cm d h n w j h i j g k s x k ; f nx = 16 cm g k s A

(D) o k L r f o d g k s x k rF k k11 cm d h n w j h i j g k s x k ; f nx = 5 cm g k s A

6. A bead is connected at one end of an inextensible massless string whose other end is fixed to a fixed

vertical cylinder. Cylinder does not rotate. The bead can move on a horizontal smooth surface. The

bead is given a velocity v0 perpendicular to the string. The bead moves on a curve and consequently

collides on the cylinder after sometime. Then, before it collides:

(A)Work done by string on the bead is positive(B) The speed of the bead remains v

0.

(C) The tension in the string remains constant.

(D) Kinetic energy of bead increases gradually.

R

V0

f d lh d . k d k s v f o r k U ; Ê O ; e k u g h u j L lh d s , d f lj s i j tk s M + k tk r k g S f tld k n w lj k f lj k , d f L F k j ≈ / o k Z / k

g q v k g S A c s y u ? k w . k Z u u g h d j r k g S A ; g d . k k S f r t f p d u h lr g i j x f r d j ld r k g S A d . k d k s j L lh d sv0 o s x f n ; k

tk r k g S A d . k o ÿ i j x f r d j r k g q v k d q N le ; i ' p k r ~ c s y u i j V d j k r k g S A r c V d j k u s ls i g y s % &

(A) j L lh k j k d . k i j f d ; k x ; k d k ; Z / k u k R e d g S A(B) d . k d h p k yv0 c u h j g r h g S A

(C) j L lh e s a r u k o f u ; r c u k j g r k g S A (D) d . k d h x f r t ≈ tk Z / k h j s & / k h j s c < + r h g S A




P H Y S I C S

LTS-7/400000CT103115008


7. In the figure shown the block of mass 6 kg is slowly released till the rod is in equillibrium of the

following figures, which figure can be equillibrium position of the rod :-

i z n f ' k Z r f p = e s a6 kg Ê O ; e k u d s C y k W d d k s N M + d s lk E ; k o L F k k e s a j g u s r d / k h j s & / k h j s N k s M +

f o d Y i N M + d h lk E ; k o L F k k f L F k f r d k s n ' k k Z ld r k g S \

10kg

4m

smooth

6kg 8m

(A) (B) (C) (D)



P H Y S I C S


LTS-8/40 0000CT103115008


8. A small bead of mass m is threaded on a smooth circular ring of radius a in xy plane as shown. The

bead experiences a force whose potential energy is given by2 2

2 2

x yV C

a b

b > a . The bead is at

x = a at rest as shown. Neglect gravity. Choose the CORRECT statement(s):-

(A) At initial position, normal force =2c

a(B) At initial position, there is no acceleration.

(C) If the bead was at y = a at rest , it's normal force is 2

2ca

b(D) If the bead was at y = a at rest, there is no acceleration.

x

y

Ê O ; e k um d k , d N k s V k d . k f p = k u q lk jxy r y e s aa f = T ; k d h f p d u h o ` Ÿ k k d k j o y ; i j c a / k k g S A d . k i j , d c y y

g S f tld h f L F k f r t ≈ tk Z2 2

2 2

x yV C

a b

b > a k j k n h tk r h g S A d . k f p = k u q lk jx = a i j f o j k e k o L F k k e s a g S A x q

d k s u x . ; e k u s a Alg dF k u@ dF k u k s a d k s p q f u ; s % &

(A) Á k j f E H k d f L F k f r i j v f H k y E c c y2c

a g S A

(B) i z k j f E H k d f L F k f r i j d k s b Z R o j . k u g h a y x r k g S A

(C) ; f n d . k y = a i j f o j k e k o L F k k e s a g S r k s b ld k v f H k y E c c y2

2ca

b g S A

(D) ; f n d . k y = a i j f o j k e k o L F k k e s a g S r k s d k s b Z R o j . k u g h y x r k g S A




P H Y S I C S

LTS-9/400000CT103115008


9. A particle is subjected to two simultaneous SHM

x = A sin t

y = A sin (t + )

Choose the correct path(s) followed by particle.

, d d . k n k s lj y v k o r Z x f r ; k ° , d lk F k d j r k g S %

x = A sin t

y = A sin (t + )

d . k k j k r ; lg h iF k p q f u ; s % &

(A) 2

x

y

(B) = x

y

(C)3

x

y

(D)3

x

y



P H Y S I C S


LTS-10/40 0000CT103115008

10. In a toy gun, we have a horizontal hollow smooth pipe. Two blocks of different masses are fitted in it.

The spring can be compressed by a certain distance and then released. It was seen that (case-1) when

m1 is fixed & m

2was released, it comes out with a speed of 4 m/s. (case-2) when m

2 was fixed & m

1

was released, it comes out with a speed of 2m/s. In all cases, initial compression of spring was same &

finally it remains at natural length (spring is not connected to either blocks).

(A) If both blocks are released simultaneously, v1 =

2

5 m/s and 2

8v

5 m/s.

(B) If both blocks are released simultaneously, 1

20v

4 m/s and

2v 20 m/s.

(C) m2 took lesser time to come out in case-1 than m

1 in case-2.

(D) m2 took more time to come out in case-1 than m

1 in case-2.

m1 m2

, d f [ k y k S u s e s a , d k S f r t [ k k s [ k y k f p d u k i k b i f y ; k tk r k g S A b le s a v y x& v y x Ê O ; e k u o k y s n k s

tk r s g S A f L i z a x d k s d q N n w j h r d lE i h f M + r d j f o j k e k o L F k k ls N k s M + f n ; k tk r k g S A ; g n s [ k k-1 e s a tcm1

f L F k j g S rF k km2 d k s f o j k e k o L F k k ls N k s M + k tk r k g S r k s ; g4 m/s d h p k y ls c k g j v k r k g S A i z d j . k -2 e s a tcm

2 f L F k j g S rF k k

m1 d k s f o j k e k o L F k k ls N k s M + k tk r k g S r k s ; g2m/s p k y ls c k g j v k r k g S A lH k h i z d j . k k s a e s a f L i a z a x d k i z k j f E H

F k k rF k k v a r e s a ; g v i u h e w y y E c k b Z e s a j g r h g S A ( f L i z a x f d lh H k h C y k W d ls tq M + h g q b Z

(A) ; f n n k s u k s a C y k W d k s d k s , d lk F k f o j k e k o L F k k ls N k s M + k tk ; s r k sv1 =

2

5 m/s o 2

8v

5 m/s g k s x k A

(B) ; f n n k s u k s a C y k W d k s a d k s , d lk F k f o j k e k o L F k k ls N k s M + k tk ; s r k s1

20v

4 m/s o

2v 20 m/s g k s x k A

(C) i z d j . k -1 e s am2 ] i z d j . k -2 e s am

1 d h r q y u k e s a c k g j f u d y u s e s a d e le ; y s x k A

(D) i z d j . k -1 e s am2 ] i z d j . k -2 e s am

1 d h r q y u k e s a c k g j f u d y u s e s a v f / k d le ; y s x k A





P H Y S I C S

LTS-11/400000CT103115008

SECTION–II : (Maximum Marks : 16)

[ .M

–II : ( v dre v d

: 16)

This section contains TWO questions.

Each question contains two columns, Column-I and Column-II.

Column-I has four entries (A), (B), (C) and (D) Column-II has five entries (P), (Q), (R), (S) and (T)

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.

The ORS contains a 4 × 5 matrix whose layout will be similar to the one shown below :

(A) (P) (Q) (R) (S) (T)

(B) (P) (Q) (R) (S) (T)

(C) (P) (Q) (R) (S) (T)

(D) (P) (Q) (R) (S) (T)

For each entry in column-I, darken the bubbles of all the matching entries. For example, if entry

(A) in Column-I matches with entries (Q), (R) and (T), then darken these three bubbles in the ORS.

Similarly, for entries (B), (C) and (D).

Marking scheme :

For each entry in Column-I

+2 If only the bubble(s) corresponding to all the correct match(es) is (are) darkened



b l [ k . M e s an i z ' u g S a i z R ; s d i z ' u e s a n k s d k W y e g S a ]d ye-I rF k kd ye-II

d ye-I e s ap i z f o " V ; k °(A), (B), (C) rF k k (D) g S a d ye-II e s ai p i z f o f " V ; k °(P), (Q), (R), (S) rF k k (T) g S a d ye-I d h i z f o f " V ; k s a d kd ye-II d h i z f o f " V ; k s a ls lq e s f y r d j s a d ye-I d h , d ; k , d ls v f / k d i z f o f " V ; k ° ]d ye-II d h , d ; k , d ls v f / k d i z f o f " V ; k s a ls lq e s f y r g k s ld r h v k s -v k j- , l- e s a s u h p s n ' k k Z ; h x ; h tS lh4 × 5 v k O ; w g n h x ; h g S:

(A) (P) (Q) (R) (S) (T)

(B) (P) (Q) (R) (S) (T)

(C) (P) (Q) (R) (S) (T)

(D) (P) (Q) (R) (S) (T)

d ye

-I d h i z R ; s d i z f o f " V d s f y , lH k h l q e s f y r i z f o f " V ; k s a d s c q y c q y k s a d k s d k y k d j s a Ad ye-I d h i z f o f " V(A) i z f o f " V ; k s a(Q), (R) rF k k (T) ls lq e s f y r g k s ] r k s b u r h u k s a c q y c q y k s a d k s v k s -v k

d j s a A b lh i z d k j i z f o f " V ; k s a(B), (C) rF k k (D) d s f y ; s H k h d j s a v a d u ; k s tu k :

d ye-I d h i z R ; s d i z f o f " V d s f y , +2 ; f n f lQ Z lH k h f o d Y i ( f o d Y i k s a ) d s v u q :i c q y c q y s ( c q y c q y k s a ) d k s d k y k f d ; k tk ;

0 ; f n d k s b Z H k h c q y c q y k d k y k u f d ; k g k s –1 v U ; lH k h v o L F k k v k s a e s a



P H Y S I C S


LTS-12/40 0000CT103115008


1. A double slit interference pattern is produced on a screen as shown in the figure, using monochromatic

light of wavelength 500nm. Point P is the location of the central bright fringe, that is produced when

light waves arrive in phase without any path difference. A choice of three strips A, B and C of transparent

materials with different thicknesses and refractive indices is available, as shown in the table, these are

placed over one or both of the slits, singularly or in conjunction, causing the interference pattern to be

shifted across the screen from the original pattern. In the column-I, how the strips have been placed, ismentioned whereas in the column-II, order of the fringe at point P on the screen that will be produced

due to the placement of the strip (s), is shown. Correctly match both the column.

Column-I Column-II

(A) Only strip B is placed over slit-I (P) First Bright

(B) Strip A is placed over slit-I and strip C is (Q) Fourth Dark

placed over slit-II

(C) Strip A is placed over the slit-I and strip (R) Fifth Dark

B and strip C are placed over the slit-II

in conjunction.

(D) Strip A and strip C are placed over slit-I (in (S) Central Brightconjuction) and strip B is placed over slit-II. (T) First dark

Slit I

Slit II

P

Screen

Film A B C

Thickness5 1.5 0.25

in m

Re fractive index 1.5 2.5 2

, d f f L y V O ; f r d j . k i z f r :i d k s f p = k u q lk j i n s Z i j500nm r j a x n S / ; Z d s , d o . k h Z ; i z d k ' k k j k m R i U u f d ; k k f c U n qP d s U Ê h ; p e d h y h f ›U t d h f L F k f r g S tk s d y k e s a i z d k ' k r j a x k s a d s f c u k iF k k U r j i g q ° p u s i j

e k s V k b Z rF k k v i o r Z u k a d , o a i k j n ' k h Z i n k F k k s Z o k y h r h u i f Í d k v k s aA, B o C d k s u h p s i z n f ' k Z r r k f y d k e s a n ' k k Z ; k x ; kv d s y s ; k , d& n w lj s d s ≈ i j f d lh , d vF k o k n k s u k s a f L y V k s a i j j [ k u s i j i n s Z i j O ; f r d j . k i z f r :i] e w y i z f r

g k s ld r k g S A d k W y e -I e s a n ' k k Z ; k x ; k g S f d b u i f Í d k v k s a d k s f d l i z d k j j [ k k tk r k g S tc f d d k W y e -II e s a b u i f Í d k v k s a d k s j [ k u s d s d k j . k i n s Z i j f c U n qP i j m R i U u f ›U tk s a d k ÿ e n ' k k Z ; k x ; k g S A d k W y e lq e s y u d h f t; s s A

d ye-I d ye-II

(A) d s o y i f Í d k B d k s f L y V -I i j j [ k k tk r k g S (P) i z F k e p e d h y h (B) i f Í d k A d k s f L y V -I i j rF k k i f Í d kC d k s (Q) p r q F k Z v n h I r

f L y V -II i j j [ k k tk r k g S A(C) i f Í d k A d k s f L y V -I i j rF k k i f Í d kB o C (R) i k ° p o h v n h I r

d k s , d d s ≈ i j , d f L y V -II i j j [ k u s i j (D) i f Í d k A o C d k s , d d s ≈ i j , d f L y V -I i j (S) d s U Ê h ; p e d h y h

rF k k i f Í d kB d k s f L y V -II i j j [ k u s i j (T) i z F k e v n h I r




P H Y S I C S

LTS-13/400000CT103115008


2. A square loop of conducting wire is placed near a long straight current carrying wire as shown. Match

the statements in column-I with the corresponding results in column-II.

Column-I Column-II

(A) If the magnitude of current I is increased (P) Induced current in the loop will be

clockwise(B) If the magnitude of current I is decreased (Q) Induced current in the loop will

be anticlockwise

(C) If the loop is moved away from the wire (R) wire will attract the loop

(D) If the loop is moved towards the wire (S) wire will repel the loop

(T) Torque about centre of mass of loop

is zero due to magnetie force

p k y d r k j ls c u s , d o x k Z d k j y w i d k s f p = k u q lk j , d y E c s lh / k s / k k j k o k g h r k j d s u tn h d j [ k k tk r k

d h f t; s A

d ye-I d ye-II

(A) ; f n / k k j kI d s i f j e k . k d k s c < + k ; k tk ; s r k s (P) y w i e s a i z s f j r / k k j k n f k . k k o r h Z g k s x h

(B) ; f n / k k j kI d s i f j e k . k d k s ? k V k ; k tk ; s r k s (Q) y w i e s a i z s f j r / k k j k o k e k o r h Z g k s x h A

(C) ; f n y w i d k s r k j ls n w j y s tk ; k tk ; s (R) r k j y w i d k s v k d f " k Z r d j s x k A(D) ; f n y w i d k s r k j d h v k s j y s tk ; k tk ; s (S) r k j y w i d k s i z f r d f " k Z r d j s x k A

(T) p q E c d h ; c y d s d k j . k y w i d s Ê O ; e k u

d s U Ê d s lk i s k c y k ? k w . k Z ' k w U ; g k s

SECTION–III : Integer Value Correct Type

[ .M–III : i . d e u lg i d

No question will be asked in section III / [ .M III e d b i 'u ug g A



P H Y S I C S


LTS-14/40 0000CT103115008


SECTION–IV : (Maximum Marks : 32)

[ .M–IV : ( v dre v d: 32)

This section contains EIGHT questions.

The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive

For each question, darken the bubble corresponding to the correct integer in the ORS

Marking scheme :

+4 If the bubble corresponding to the answer is darkened

0 In all other cases

b l [ k . M e s av B i z ' u g S a

i z R ; s d i z ' u d k m Ÿ k j0 ls 9 r d] n k s u k s a ' k k f e y] d s c h p d k , d,dy v d ; i . d g S

i z R ; s d i z ' u e s a ] v k s -v k j- , l- i j lg h i w . k k ± d d s v u q :i c q y c q y s d k s d k y k d j s a

v a d u ; k s tu k :

+4 ; f n m Ÿ k j d s v u q :i c q y c q y s d k s d k y k f d ; k tk ; 0 v U ; lH k h v o L F k k v k s a e s a

1. An electron moving with velocity1v 1i

m/s at a point in a magnetic field experiences a force

1F ejN

, where e is the charge of electron. If the electron is moving with a velocity2

ˆ ˆv i j

m/s

at the same point, it experiences a force 2ˆ ˆF e i j N

. Find the magnitude of force (in N) the

electron would experience if it were moving with a velocity 3 1 2ˆv v v

at the same point.

p q E c d h ; k s = e s a f d lh f c U n q i j o s x1v 1i

m/s ls x f r ' k h y b y s D V ™ k W u i j c y1F ejN

y x r k g S ] tg k °e b y s D V ™ k W u

v k o s ' k g S A ; f n b y s D V ™ k W u b lh f c U n q i j o s x2 ˆ ˆv i j

m/s ls x f r ' k h y g k s r k s b l i j c y 2 ˆ ˆF e i j N

y x r k g S A ; f n b y s D V ™ k W u b lh f c U n q i j o s x3 1 2

ˆv v v

ls x f r ' k h y g k s r k s b y s D V ™ k W u i j y x u s o k y s c y d k i f j e k . k (N e s a) K k r

d h f t; s A

2. Find current in the branch CD of the circuit (in ampere).

i z n f ' k Z r i f j iF k e s a ' k k [ k kCD e s a / k k j k d k e k u ( , f E i ; j e s a ) K k r d h f t; s A

2 2

2

B

A

C

D

10V




P H Y S I C S

LTS-15/400000CT103115008


3. Figure shows a conducting disc rotating about its axis in a perpendicular uniform and constant magnetic

field B. A resistor of resistance R is connected between the centre and the rim. The radius of the disc

is 5.0 cm, angular speed = 40 rad/s, B = 0.1 T and R = 1 . The current through the resistor in

milliampere is.

f p = e s a , d p k y d p d r h f d lh y E c o r ~ le :i f u ; r p q E c d h ; k s =B e s a b ld h v k d s lk i s k ? k w . k Z u d j j g h g S A f j e d s e/ ; , d i z f r j k s / kR tk s M + f n ; k tk r k g S A p d r h d h f = T ; k5.0 cm ] d k s . k h ; p k y = 40 rad/s, B = 0.1 T

o R = 1 g S A i z f r j k s / k e s a i z o k f g r / k k j k f e y h , f E i ; j e s a K k r d h f t; s A

× ×

R

×

×

×

×××××

×

×

×

×

4. A continuous flow geyser has water flowing in it at a continuous rate. It comes in at a rate of

0.08 litres/sec and leaves at a rate of 0.081 litres/sec. The temperature coefficient of thermal expansion

of water is 2 × 10 –3/°C. The specific heat capacity of water is 4000 J/kg°C. If the inlet temperature is

7°C at which the density of water is 103 kg/m3, what is the power consumed by geyser (in kW) ?

, d lr r ~ i z o k g x h tj e s a ty , d f u ; r n j ls i z o k f g r g k s j g k g S A ty b le s a0.08 y h V j@ ls d . M n j ls i z o s s ' k d j r k g S r

0.081 y h V j@ ls d . M n j ls c k g j f u d y r k g S A ty d k r k i h ; i z lk j . k r k i x q . k k a d2 × 10 –3/°C g k s r k g S A ty d h f o f ' k " V

≈ " e k / k k f j r k4000 J/kg°C g k s r h g S A ; f n i z o s ' k h r k i e k u7°C g k s f tl i j ty d k ? k u R o103

kg/m3

g k s r k s x h tj k j k O ; f ; r ' k f D r (kW e s a ) D ; k g k s x h \



P H Y S I C S


LTS-16/40 0000CT103115008


5. Two positive and two negative point charges (|Q| = 0.01 µC) are placed at the corners of a square

insulator frame of side 15 m and the side of the frame are hinged together. The frame is deformed into

a rhombus as shown. The new potential at the centre is given by volt. Fill in OMR sheet.

, d 15m H k q tk o k y s o x k Z d k j d q p k y d ›s e d s ' k h " k k s ± i j n k s / k u k R e d rF k k n k s Ω . k k R e d f c U n q v k(|Q| = 0.01 µC) j [ k s

tk r s g S a rF k k ›s e d h H k q tk d k s f p = k u q lk j , d lk F k d h y d h r d j n s r s s g S A b l ›s e d k s f p = k u q lk j , d

e s a c n y f n ; k tk r k g S A v c d s U Ê i j u ; s f oH k o d k e k u o k s Y V g k s r k s d k e k u K k r d j s a A

+Q –Q

–Q +Q

Q –Q

–Q +Q

74º

6. While expanding a mixture of gas did work P1V

1 – P

2V

2on the piston. Here P

1 and V

1 are initial

pressures and volumes and P2 and V

2 are final pressures and volumes. Find molar heat capacity of the

gas in this process. Take for mixture as 1.5. Express your answer in terms of xR where R is universal

gas constant. Fill x in OMR sheet.

, d x S lh ; f eJ . k i z lj . k d s n k S j k u f i L V u i jP1V

1 – P

2V

2 d k ; Z d j r k g S ] ; g k °P

1 o V

1 ÿ e ' k % i z k j f E H k d n k c o v k ; r u rF k k

P2 o V

2 ÿ e ' k % v a f r e n k c o v k ; r u g S A f eJ . k d s f y ; s = 1.5 y h f t; s A b l i z ÿ e e s a x S l d h e k s y j ≈ " e k / k k f j r k ; f

xR g k s r k sx d k e k u K k r d h f t; s AR lk o Z f = d x S l f u ; r k a d g S A




P H Y S I C S

LTS-17/400000CT103115008


7. The body of mass m = 0.5 kg is held on a vertical wall by a force F of 15 N as shown in the figure is

applied on the body at an angle = 53º to the vertical (Figure). The coefficient of friction between the

body and the wall is µ = 0.5. Determine acceleration of the body in m/s2.

Ê O ; e k um = 0.5 kg o k y s f i . M d k s f p = k u q lk j , d ≈ / o k Z / k j n h o k j d s lg k j s15 N d k c yF y x k d j j k s d d j j [ k k x ; k

g S A c y d k s ≈ / o k Z / k j ls = 53º d k s . k i j y x k ; k tk r k g S A ; f n f i . M o n h o k j d s e/ ; ? k " k Z . k x q . k k a dµ = 0.5 g k s r k s f i . M

d k R o j . k(m/s2 e s a) K k r d h f t; s A

F

8. An electron of mass m when accelerated through a potential difference V, has de Broglie wavelength

. The de-Broglie wavelength associated with a particle of mass m/4 and charge e accelerated through

the same potential difference is n. Then find the value of n.

m n z O ; e k u d k , d b y s D V ™ k W u tcV f oH k o k U r j ls R o f j r g k s r k g S r k s b ld h M h c z z k W X y h r j a x n S / ; Z g S A le k u f oH k o k U r j ls R o f j

n z O ; e k um/4 rF k k v k o s ' ke o k y s , d d . k ls lE c a º M h & c z k W X y h r j a x n S / ; Zn g S r k sn d k e k u K k r d h f t,A



C H E M I S T R Y


LTS-18/40 0000CT103115008

PART-2 : CHEMISTRY

H x-2 : l ;u oK u


[ .M

–I : ( v dre v d

: 40)





Marking scheme :




b l [ k . M e s anl i z ' u g S a

i z R ; s d i z ' u e s a p k j f o d Y i(A), (B), (C) rF k k (D) g S a A b u p k j f o d Y i k s a e s a ls,d ; ,d l v d f o d Y i lg h g S a A

i z R ; s d i z ' u e s a ] lH k h lg h f o d Y i ( f o d Y i k s a ) d s v u q :i c q y c q y s ( c q y c q y k s a ) d k s v k s -v k j- , l-


+4 ; f n f lQ Z lH k h lg h f o d Y i ( f o d Y i k s a ) d s v u q :i c q y c q y s ( c q y c q y k s a ) d k s d k y k f d ; k tk

0 ; f n d k s b Z H k h c q y c q y k d k y k u f d ; k g k s

–2 v U ; lH k h v o L F k k v k s a e s a

1. Select the correct statement(s) for an ideal gas-(A) All the molecules move with same speed

(B) Kinetic energy of a molecule remains unchanged due to collision with other molecule

(C) Kinetic energy of molecules depends only on temperature

(D) Speed of molecules depend on both molecular mass & temperature

, d v k n ' k Z x S l d s f y , l g h dF k u d k p ; u d h f t, -

(A) lH k h v . k q le k u o s x ls x f r d j r s g S

(B) , d v . k q d h x f r t ≈ tk Z ] v U ; v . k q d s lk F k V D d j d s d k j . k v i f j o f r Z r j g r h g S

(C) v . k q v k s a d h x f r t ≈ tk Z d s o y r k i i j f uH k Z j d j r h g S

(D) v . k q d k o s x] v . k q d s Ê O ; e k u rF k k r k i n k s u k s a i j f uH k Z j d j r k g S





C H E M I S T R Y

LTS-19/400000CT103115008


2. If heat of formation of PCl3(g) and PH

3(g) are 140 and 8 kJ /mole respectively and heat of atomisation

of phosphorous[P4(s)], chlorine and hydrogen are 320, 120 and 216 kJ/mole respectively then.

(A) Bond energy of P–Cl is 63 kJ/mole

(B) Bond energy of P–H is 210 kJ/mole

(C) Heat of formation of isolated gaseous hydrogen atom is 108 kJ/mole

(D) Heat of formation of isolated gaseous chlorine atom is 60 kJ/mole

; f n PCl3(g) rF k kPH

3(g) d s f u e k Z . k d h ≈ " e k ÿ e ' k %140 rF k k8 kJ / e k s y g S rF k k Q k W L Q k s j l [P

4(s)] ] D y k s j h u rF k k

g k b M ™ k s tu d h i j e k f . o ; d j . k d h ≈ " e k ÿ e ' k %320, 120 rF k k216 kJ/ e k s y g S r k s

(A) P–Cl d h c a / k ≈ tk Z63 kJ/ e k s y g S

(B) P–H d h c a / k ≈ tk Z210 kJ/ e k s y g S

(C) f o y f x r x S lh ; g k b M ™ k s tu i j e k . k q d h f u e k Z . k d h ≈ " e k108 kJ/ e k s y g S

(D) f o y f x r x S lh ; D y k s j h u i j e k . k q d h f u e k Z . k d h ≈ " e k60 kJ/ e k s y g S3. 100 L aqueous solution of KBr is electrolysed by using current of 9.65 ampere for 10 sec. using

Pt electrodes. Which of the following species will be produced due to electrolysis ?

KBr d s 100 L ty h ; f o y ; u d k s Pt b y s D V ™ k s M d k m i ; k s x d j d s10 sec. d s f y ; s 9.65 , s f E i ; j d h / k k j k d s i z ; k s x

k j k o S |q r v i ? k f V r f d ; k tk r k g S A o S |q r v i ? k V u d s d k j . k f u E u e s a ls d k S u lh L i h ' k h t m R i k f n

(A) H2

(B) O2 (C) Br 2 (D) KOH

4. Which of the following molecule(s) is/are non planar -

f u E u e s a ls d k S u ls v . k q v le r y h ; g S a-

(A) BF3

(B) SF6

(C) IF5

(D) PCl5

5. Which of the following option is/are CORRECT for solution of liquid NH3 with alkali metal -

(A) It is paramagnetic (B) It is blue in colour

(C) It is good conductor of electricity (D) It is nonconductor of electricity

f u E u e s a ls d k S u ls f o d Y i] k k j / k k r q d s n z o NH3 d s lk F k f o y ; u d s f y , lg h g S @ g a S \

(A) ; g v u q p q E c d h ; g S (B) ; g u h y s j a x d k g S

(C) ; g f o |q r d k lq p k y d g S (D) ; g f o |q r d k d q p k y d g S



C H E M I S T R Y


LTS-20/40 0000CT103115008


6. Which of the following anion(s) give 'x' gas with dil. HCl, which turns lime water milky :-

f u E u e s a ls d k S u ls Ω . k k ; u r u qHCl d s lk F k x S l 'x' n s r s g S a r k s p w u s d s i k u h d k s n q f / k ; k d j n s r h g S A

(A) SO3

2– (B) CO3

2– (C) S2– (D) CH3COO –

7. Which of the following complex/ion is/are paramagnetic.

f u E u e s a ls d k S u ls la d q y@ v k ; u v u q p q E c d h ; g S @ g S a \

(A) Ni(CO)4

(B) [MnCl4]2– (C) K

4[Fe(CN)

6] (D) [CoCl

4]2–

8. Which statement(s) is/are correct for the compounds X and Y :

H–C C–CH –C–CH –CH2 2 3

O(X)

CH –C C–CH –CH –C–H3 2 2

(Y)

;

O

(A) X and Y are chain isomers

(B) X and Y are functional isomers

(C) X and Y can be differentiated by Tollen's reagent

(D) X and Y both have same degree of unsaturation

; k S f x dX rF k kY d s f y ; s f u E u e s a ls d k S u ls dF k u lg h g S a @ g S &

H–C C–CH –C–CH –CH2 2 3

O(X)

CH –C C–CH –CH –C–H3 2 2

(Y)

;

O

(A) X rF k kY J ` a [ k y k le k o ; o h g S

(B) X rF k kY f ÿ ; k R e d le k o ; o h g S

(C) X rF k kY d k s V k W y s U lv f H k d e Z d k j k f oH k s f n r f d ; k tk ld r k g S

(D) X rF k kY n k s u k s a e s a v la r ` I r r k d h d k s f V le k u g k s r h g S




C H E M I S T R Y

LTS-21/400000CT103115008


9. C = CH C3

H H

CH3 OsO / NaHSO4 3 (A)

(B)(i) CF C3 –

(ii) Aq. KOH

O OH

O

Select correct statement(s) for above reaction sequence :

(A) A is pair of enantiomer and A and B are distereomers

(B) B is pair of enantiomer and A and B are distereomers

(C) A represents meso compound

(D) B represents meso compound

C = CH C3

H H

CH3 OsO / NaHSO4 3 (A)

(B)

(i) CF C3 –

(ii) Aq. KOH

OO

H

O

m i j k s D r v f H k f ÿ ; k ÿ e d s f y ; s lg h dF k u k s a d k p ; u d h f t; s &

(A) A i z f r f c E c :i le k o ; o h d k ; q X e g S rF k kA , o a B f o o f j e le k o ; o h g S

(B) B i z f r f c E c :i le k o ; o h d k ; q X e g S rF k kA , o a B f o o f j e le k o ; o h g S(C) A e h lk s ; k S f x d d k s i z n f ' k Z r d j r k g S

(D) B e h lk s ; k S f x d d k s i z n f ' k Z r d j r k g S

10. Which of the following compound(s) will form stable (isolable) hydrate on addition of H2O ?

f u E u e s a ls d k S u ls ; k S f x dH2O d k ; k s x d j k u s i j L F k k ; h ( i ` F k D d j . k h ;) g k b M ™ s V c u k ; s a x s a &

(A) CCl –C–H3

O

(B)

O

(C)

O

O O

(D)

O



C H E M I S T R Y


LTS-22/40 0000CT103115008


[ .M–II : ( v dre v d: 16)







(A) (P) (Q) (R) (S) (T)

(B) (P) (Q) (R) (S) (T)

(C) (P) (Q) (R) (S) (T)

(D) (P) (Q) (R) (S) (T)

For each entry in column-I, darken the bubbles of all the matching entries. For example, if entry(A) in Column-I matches with entries (Q), (R) and (T), then darken these three bubbles in the ORS.


Marking scheme :





b l [ k . M e s an i z ' u g S a

i z R ; s d i z ' u e s a n k s d k W y e g S a ]d ye

-I rF k kd ye

-II d ye-I e s ap i z f o " V ; k °(A), (B), (C) rF k k (D) g S a

d ye-II e s ai p i z f o f " V ; k °(P), (Q), (R), (S) rF k k (T) g S a

d ye

-I d h i z f o f " V ; k s a d kd ye

-II d h i z f o f " V ; k s a ls lq e s f y r d j s a

d ye-I d h , d ; k , d ls v f / k d i z f o f " V ; k ° ]d ye-II d h , d ; k , d ls v f / k d i z f o f " V ; k s a ls lq e s f y r g k s ld r h

v k s -v k j- , l- e s a s u h p s n ' k k Z ; h x ; h tS lh4 × 5 v k O ; w g n h x ; h g S:

(A) (P) (Q) (R) (S) (T)

(B) (P) (Q) (R) (S) (T)

(C) (P) (Q) (R) (S) (T)

(D) (P) (Q) (R) (S) (T)

d ye-I d h i z R ; s d i z f o f " V d s f y , lH k h l q e s f y r i z f o f " V ; k s a d s c q y c q y k s a d k s d k y k d j s a A

d ye-I d h i z f o f " V(A) i z f o f " V ; k s a(Q), (R) rF k k (T) ls lq e s f y r g k s ] r k s b u r h u k s a c q y c q y k s a d k s v k s -v k

d j s a A b lh i z d k j i z f o f " V ; k s a(B), (C) rF k k (D) d s f y ; s H k h d j s a

v a d u ; k s tu k : d ye-I d h i z R ; s d i z f o f " V d s f y , +2 ; f n f lQ Z lH k h f o d Y i ( f o d Y i k s a ) d s v u q :i c q y c q y s ( c q y c q y k s a ) d k s d k y k f d ; k tk ;





C H E M I S T R Y

LTS-23/400000CT103115008


1. Column-I Column-II

(A) pH of mixture of 0.1N HCN (K a = 10 –10) and 0.1M NaCN (P) 10

(B) pH of mixture of 0.1N BOH (pK b = 6) and 0.05M B

2SO

4(Q) 7

(C) pH of mixture of 0.1M salt of weak acid (pK a = 5) and weak (R) 6

base (pK b = 7)

(D) pH of mixture of 500 litre of 0.02 M HNO3 and 500 litre of (S) 8

0.01M Sr(OH)2

(T) Solution is acidic

d ye

-I d ye

-II

(A) 0.1N HCN (K a = 10

–10) rF k k0.1M NaCN d s f eJ . k d h pH (P) 10

(B) 0.1N BOH (pK b = 6) rF k k0.05M B

2SO

4 d s f eJ . k d h pH (Q) 7

(C) n q c Z y v E y d s0.1M y o . k (pK a = 5) rF k k n q c Z y k k j (R) 6

(pK b = 7) d s f eJ . k d h pH

(D) 500 y h V j0.02 M HNO3 rF k k500 y h V j0.01M Sr(OH)

2(S) 8

d s f eJ . k d h pH (T) f o y ; u v E y h ; g S



C H E M I S T R Y


LTS-24/40 0000CT103115008


2. Column-I (Species) Column-II (Characteristic properties)

(A) XeO2F

2(P) Non-planar

(B) XeOF4 (Q) Utilisation of d-orbital(s) in bonding

(C) PH4

+ (R) Planar

(D) ICl4

– (S) Polar molecule

(T) Tetrahedral geometry

d ye-I ( L i ' t) d ye-II ( v H y . d x .)

(A) XeO2F

2(P) v le r y h ;

(B) XeOF4

(Q) c a / k u e s ad- d k d k s a d k i z ; k s x f d ; k tk r k g S

(C) PH4

+ (R) le r y h ;

(D) ICl4

– (S) / k q z o h ; v . k q

(T) p r q " Q y d h ; T ; k f e f r

SECTION–III : Integer Value Correct Type

[ .M

–III :i . d e u lg i d





C H E M I S T R Y

LTS-25/400000CT103115008


[ .M–IV : ( v dre v d: 32)




Marking scheme :








1. If equivalent weight of Ba(SCN)2 when it reacts with oxidising agent and forms product,

SO42– , CO3

2– & NO3 – is

Molecularweight

X.Then calculate value of

X

4.

tc Ba(SCN)2 ] v k W D lh d k j d d s lk F k f ÿ ; k d j d sSO42– , CO3

2– rF k k NO3 – m R i k n c u k r k g S ] ; f n b ld s r q Y ; k a

H k k j d k e k uX

v . k q H g S ] r k s

X

4 d s e k u d h x . k u k d h f t; s \

2. If for the reaction at 27ºC , 2A(g) + B(g) 2C(g), Ur = 25.6 kcal and entropy of A, B, C

are 200, 100 , 500 cal/mole resepectively , then calculate magnitude of G (in Kcal) of reaction.

[R = 2cal/mole-K]

Fill your answer as sum of digits (excluding decimal places) till you get the single digit answer.

27ºC i j v f H k f ÿ ; k 2A(g) + B(g) 2C(g), d s f y ; s Ur = 25.6 kcal rF k kA, B, C d h , s V ™ k W i h ÿ e ' k %

200, 100 , 500 cal/mole g S r k s v f H k f ÿ ; k d sG (Kcal e s a) d s i f j e k . k d h x . k u k d h f t; s A

[R = 2cal/mole-K]

viu m d v d d

( n ' eyo L u d N M d

)rc rd ; x d t, tc rd v id b d b v d i r

u g t ,A




C H E M I S T R Y


LTS-26/40 0000CT103115008


3. Among the following find out total number of reagent(s), which give white precipitate with aqueous solu-

tion of Zn+2.

NH4OH, NaOH, (NH

4)

2S, K

4[Fe(CN)

6]

f u E u e s a ls , s ls v f H k d e Z d k s a d h d q y la [ ; k c r k b , s a tk sZn+2 d s ty h ; f o y ; u d s lk F k ' o s r v o k s i n s r s g a S A

NH4OH, NaOH, (NH

4)

2S, K

4[Fe(CN)

6]

4. Find the total number of species that do not exist.

H2, HeH – , He

2

+, Li2, O

2

2– , FeI3

f u E u e s a ls , s lh L i h ' k h t d h d q y la [ ; k c r k b ; s f tu d k v f L r R o u g h a g S A

H2, HeH – , He

2

+, Li2, O

2

2– , FeI3

5.

HOOC COOH

(A)

For the above compound how many statement(s) are true ?

(i) Total number of stereoisomers possible for (A) are 3

(ii) Total number of meso compounds possible for (A) is 1

(iii) Number of plane of symmetry in cis form of A are 2

(iv) cis form of above compound (A) is meso

(v) Number of enantiomeric pair possible for (A) is 1

HOOC COOH

(A)

m i j k s D r ; k S f x d d s f y ; s f d r u s dF k u lg h g S A

(i) (A) d s f y ; s f = f o e ~ le k o ; f o ; k s a d h lE H k k f o r d q y la [ ; k3 g S

(ii) (A) d s f y ; s e h lk s ; k S f x d k s a d h lE H k k f o r d q y la [ ; k1 g S

(iii) A d s f ll :i e s a le f e f r d s r y d h la [ ; k2 g S

(iv) m i j k s D r ; k S f x d(A) d k f l l :i e h l k s g S

(v) (A) d s f y ; s i z f r f c E c :i le k o ; o h ; q X e k s a d h lE H k k f o r la [ ; k1 g S




C H E M I S T R Y

LTS-27/400000CT103115008


6. Total number of compound(s) having higher bond order of (C–O) than CH –C3

O

O –

:

, s ls ; k S f x d k s a d h d q y la [ ; k c r k b ; s f tu e s aCH –C3

O

O –

d h r q y u k e s a(C–O) d k c a / k ÿ e m P p g k s r k g S A

(i)

O

(ii)

O

(iii)

OH

(iv) CH3 –OH (v) CH

3 –CH

2 –OH

(vi) CH –CH –C3 2

O

H(vii)

O

7. Total number of compounds which react faster than

NO

F

on reaction with NaOH :

f u E u e s a ls , s ls ; k S f x d k s a d h d q y la [ ; k c r k b ; s tk s NaOH d s lk F k v f H k f ÿ ; k d j k u s i j

NO2

F

d h r q y u k e s a r s th

ls f ÿ ; k d j r s g S a A

(i)

NO2

Cl

(ii)

NO2

F

NO2

(iii)

F

NO2

(iv)

NO2

F

NO2O N2

(v) Br NO2(vi) I NO2



C H E M I S T R Y


LTS-28/40 0000CT103115008


8. Total number of compounds which can show both aldol condensation and iodoform test :

, s ls ; k S f x d k s a d h d q y la [ ; k c r k b ; s tk s , s Y M k s y la ? k u u rF k k v k ; k s M k s Q k e Z i j h k . k n k s u k s a

(i) CH –C–CH3 3

O

(ii)CH –C–H3

O

(iii)

O

(iv)

O OH

(v)

O

O

(vi) CH3 –CH

2 –OH




M A T H E M A T I C S

LTS-29/400000CT103115008

PART-3 : MATHEMATICS

H x

-3 :x . r


[ .M

–I : ( v dre v d

: 40)





Marking scheme :




b l [ k . M e s anl i z ' u g S a i z R ; s d i z ' u e s a p k j f o d Y i(A), (B), (C) rF k k (D) g S a A b u p k j f o d Y i k s a e s a ls,d ; ,d l v d f o d Y i lg h g S a A i z R ; s d i z ' u e s a ] lH k h lg h f o d Y i ( f o d Y i k s a ) d s v u q :i c q y c q y s ( c q y c q y k s a ) d k s v k s -v k j- , l- e s v a d u ; k s tu k :

+4 ; f n f lQ Z lH k h lg h f o d Y i ( f o d Y i k s a ) d s v u q :i c q y c q y s ( c q y c q y k s a ) d k s d k y k f d ; k tk ; 0 ; f n d k s b Z H k h c q y c q y k d k y k u f d ; k g k s

–2 v U ; lH k h v o L F k k v k s a e s a

1. Consider the plane P 2x – 2y + z = 9 and a point A(13,–11,15) not on the plane. Now which of the

following holds good ?

(A) foot of perpendicular from A on plane P is (1,1,9)

(B) image of point A in plane P is (–11,13,3)

(C) perpendicular distance of A from plane P is 18.

(D) The volume of tetrahedron formed by xy,yz,zx planes with plane P is 35/23.

e k u k le r yP 2x – 2y + z = 9 rF k k , d f c U n qA(13,–11,15) g S ] tk s le r y i j u g h a g S A v c f u E u e s a ls d k

lR ; g k s a x s \

(A) le r y P i j f c U n qA ls y E c d k i k n(1,1,9) g k s x k A

(B) le r y P e s a f c U n qA d k i z f r f c E c(–11,13,3) g k s x k A(C) le r y P ls f c U n q A d h y E c o r n w j h18 g k s x h A

(D) le r y P d s lk F kxy,yz,zx le r y k s a k j k f u f e Z r p r q " Q y d k v k ; r u35/23 g k s x k A






LTS-30/40 0000CT103115008


2. If z4 + 1 = z2, then-

(A) Least positive argument of z is6

(B) Greatest positive argument of z in [0,2] is11

6

(C) The four possible values of z forms square on argand plane

(D) The four possible values of z forms rectangle on argand plane.

; f n z4 + 1 = z2 g S ] r c -

(A) z d k U ; w u r e / k u k R e d d k s . k k a d6

g k s x k A

(B) v U r j k y[0,2] e s az d k e g Ÿ k e / k u k R e d d k s . k k a d11

6

g k s x k A

(C) v k x Z . M le r y e s az d s p k j lE H k o e k u o x Z d k s f u f e Z r d j s a x s A

(D) v k x Z . M le r y e s az d s p k j lE H k o e k u v k ; r d k s f u f e Z r d j s a x s A

3. If ƒ : R R be a continuous function satisfying the equation 2xƒ(x) + (1 + x2)ƒ'(x) = 1 and ƒ(0) = 0,

then-

(A) ƒ(x) is an odd function (B) Number of integers in the range of ƒ(x) is 1.

(C) Number of solutions of 1

ƒ x4

are 2 (D) Number of solutions of 1

ƒ ƒ x2

are 2.

; f n ƒ : R R , d la r r ~ Q y u g S ] tk s le h d j . k2xƒ(x) + (1 + x2)ƒ'(x) = 1 rF k k ƒ(0) = 0 d k s lU r q " V d j r k g S ] r c -

(A) ƒ(x) f o " k e Q y u g k s x k A (B) ƒ(x) d s i f j lj e s a i w . k k ± d k s a d h la [ ; k1 g k s x h A

(C) 1

ƒ x4

d s g y k s a d h la [ ; k2 g k s x h A (D) 1

ƒ ƒ x2

d s g y k s a d h la [ ; k2 g k s x h A

4. Let A be invertible square matrix whose inverse is given as 1

1 2 3

A 1 3 6

2 5 7

. Now which of the

following holds good (|A| represents determinant of A)-

(A) A is idempotent matrix (B) Trace (adjA) 112

(C) Trace (A adjA)3

2 (D) |A –1| = –2





LTS-31/400000CT103115008


e k u kA O ; q R ÿ e . k h ; o x Z v k O ; w g g S ] f tld k i z f r y k s e1

1 2 3

A 1 3 6

2 5 7

k j k f n ; k x ; k g S A v c f u E u e s a ls d k S u ls lR

(|A|, A d s lk j f . k d d k s n ' k k Z r k g S)-

(A) A o x Z le v k O ; w g g k s x k A (B) v u q j s [ k(adjA)11

2 g k s x k A

(C) v u q j s [ k (A adjA)3

2 g k s x k A (D) |A –1| = –2

5. Figure shows a parabola with vertex at origin and axis of symmetry along x-axis. S(1,0) is the focus of

parabola & A(t1) and B(t

2) are the extremeties of a focal chord making an angle of 30º with

positive x-axis. Which of the following holds good ?

(A) Latus rectum of parabola is 4 (B)2

1 1t 2 3t 1 (C) Length AB is 1 (D) 2

2 2t 2 3t 1

y

0

B(t )2

A(t )1

30º

S (1,0) x

f p = k u q lk j , d i j o y ; f tld k ' k h " k Z e w y f c U n q i j g S rF k kx- v k d h le f e r d s v u q f n ' k i j o y ; d h u k f H kS(1,0) g S rF k k

u k H k h ; th o k d s v f U r e f lj sA(t1) rF k kB(t

2) / k u k R e dx- v k d s lk F k30º d k d k s . k c u k r s g S A f u E u e s a ls d k S u ls lR

(A) i j o y ; d k u k f H k y E c 4 g k s x k A (B) 2

1 1t 2 3t 1

(C) AB d h y E c k b Z1 g k s x h A (D) 2

2 2t 2 3t 1





LTS-32/40 0000CT103115008


6. Let a,b R and ab 1. If 6a2 + 20a + 15 = 0 and 15b2 + 20b + 6 = 0 and

3

32

b p

qab 9 ab 1

(where p,q are coprime naturals then)-

(A) The digit at units place of 'p' is 6 (B) Digit at unit's place of q is 6

(C) The digit at units place of p is 5 (D) Digit at unit's place of q is 5.

e k u ka,b R rF k kab 1 g S A ; f n6a2 + 20a + 15 = 0 , o a 15b2 + 20b + 6 = 0 rF k k

3

32

b p

qab 9 ab 1

( tg k ° p,q lgvH k k T ; i z k d ` r la [ ; k ; s a g S ) r c -

(A) p d s b d k b Z L F k k u k s a i j v a d 6 g k s x k A (B) q d s b d k b Z L F k k u k s a i j v a d 6 g k s x k A

(C) p d s b d k b Z L F k k u k s a i j v a d5 g k s x k A (D) q d s b d k b Z L F k k u k s a i j v a d5 g k s x k A

7. Let ƒ : R +

R be a function satisfying the functional rule ƒ(a.b) = ƒ(a) + ƒ(b) a,bR +

. Then which

of the following always holds good for1

x

0

I ƒ(3 )dx is -

e k u kƒ : R + R , d Q y u g S ] tk s Q y u f u ; eƒ(a.b) = ƒ(a) + ƒ(b) a,bR + d k s la r q " V d j r k g S A r c f u E u e s a

d k S u lk ln S o1

x

0

I ƒ(3 )dx d s f y ; s lR ; g k s x k &

(A)(3)

I

2

(B)log3

I

2

(C) ( 3) (D)3

ƒ( 2)

2

8. If m

mk 0

1 k ƒ x lim x

m m

; then-

; f n m

mk 0

1 k ƒ x lim x

m m

g S ] r c -

(A) 1

ƒ 02

(B) 1

ƒ 12

(C) 3

ƒ 12

(D) 3

ƒ 22





LTS-33/400000CT103115008

9. For which of the following values of k the equation4 2

6 3

x xk

x 2x 1

has no solution in [1,] ?

k d s f d u e k u k s a d s f y , v U r j k y[1,] e s a le h d j . k4 2

6 3

x xk

x 2x 1

d k d k s b Z g y u g h a g S ] g k s a x s

(A)1

6(B)

1

5(C)

1

4(D)

1

3

10. Let ƒ be a continuous non-algebraic twice derivable increasing function in (0,1) & ƒ"(x) > 0.

If ƒ(0) = 0, ƒ(1) = 1 and g(x) is inverse of ƒ(x), then which of the following holds good ?

e k u k(0, 1) e s aƒ n k s c k j v o d y u h ; , d la r r ~ v c h tx f . k r h ; o/ k Z e k u Q y u g S rF k kƒ"(x) > 0 g S A ; f nƒ(0) = 0,

ƒ(1) = 1 rF k kg(x), ƒ(x) d k i z f r y k s e g S ] r k s f u E u e s a ls d k S u ls lR ; g k s a x s \

(A)

ƒ x

' x x 0,1x

(B)

ƒ x

' x x 0,1x

(C)

ƒ x xx 0,1

x g x (D)

ƒ x x

x 0,1x g x






LTS-34/40 0000CT103115008


[ .M–II : ( v dre v d: 16)







(A) (P) (Q) (R) (S) (T)

(B) (P) (Q) (R) (S) (T)

(C) (P) (Q) (R) (S) (T)

(D) (P) (Q) (R) (S) (T)

For each entry in column-I, darken the bubbles of all the matching entries. For example, if entry

(A) in Column-I matches with entries (Q), (R) and (T), then darken these three bubbles in the ORS.


Marking scheme :





b l [ k . M e s an i z ' u g S a i z R ; s d i z ' u e s a n k s d k W y e g S a ]d ye-I rF k kd ye-II

d ye

-I e s ap i z f o " V ; k °(A), (B), (C) rF k k (D) g S a d ye-II e s ai p i z f o f " V ; k °(P), (Q), (R), (S) rF k k (T) g S a d ye-I d h i z f o f " V ; k s a d kd ye-II d h i z f o f " V ; k s a ls lq e s f y r d j s a d ye-I d h , d ; k , d ls v f / k d i z f o f " V ; k ° ]d ye-II d h , d ; k , d ls v f / k d i z f o f " V ; k s a ls lq e s f y r g k s ld r h v k s -v k j- , l- e s a s u h p s n ' k k Z ; h x ; h tS lh4 × 5 v k O ; w g n h x ; h g S:

(A) (P) (Q) (R) (S) (T)

(B) (P) (Q) (R) (S) (T)

(C) (P) (Q) (R) (S) (T)

(D) (P) (Q) (R) (S) (T)

d ye-I d h i z R ; s d i z f o f " V d s f y , lH k h l q e s f y r i z f o f " V ; k s a d s c q y c q y k s a d k s d k y k d j s a Ad ye-I d h i z f o f " V(A) i z f o f " V ; k s a(Q), (R) rF k k (T) ls lq e s f y r g k s ] r k s b u r h u k s a c q y c q y k s a d k s v k s -v k

d j s a A b lh i z d k j i z f o f " V ; k s a(B), (C) rF k k (D) d s f y ; s H k h d j s a v a d u ; k s tu k :

d ye-I d h i z R ; s d i z f o f " V d s f y , +2 ; f n f lQ Z lH k h f o d Y i ( f o d Y i k s a ) d s v u q :i c q y c q y s ( c q y c q y k s a ) d k s d k y k f d ; k tk ;






LTS-35/400000CT103115008


1. Consider an experiment of tossing a fair of coin six times. Let A and B be two events defined as

follows

A : Number of heads greater than number of tails.

B : The third toss results in head. Now match the columns.

Column-I Column-II

(A) If a

P A b

where a and b are relatively prime then 'a' is (P) an even number

(B) If p

P Bq

, where p and q are relatively prime then 'q' is (Q) an odd number

(C) IfA u

PB v

where u and v are relatively prime then 'u' is (R) a prime number

(D) IfB m

P

A n

where m and n are relatively prime then (S) a composite number

3(m + n) is (T) neither prime nor

composite number

e k u k , d i j h k . k e s a , d f u " i k i k r h f lD d s d k s N % c k j m N k y k tk r k g S A e k u k n k s ? k V u k ; s aA rF k kB b l i z d k j i f jH k k f " k r g S

A : f p r k s a d h l a [ ; k ] i V k s a d h la [ ; k l s v f / k d g S A

B : r h lj h m N k y i j i f j . k k e f p r v k r k g S A v c L r E H k k s a d k f e y k u d h f t,

d ye-I d ye-II

(A) ; f n a

P A

b

g S ] tg k °a rF k k b i j L i j vH k k T ; l[ ; k ; s a g S ] r k sa g k s x k (P) , d le la [ ; k

(B) ; f n p

P Bq

g S ] tg k ° p rF k kq i j L i j vH k k T ; l[ ; k ; s a g S ] r k sq g k s x k (Q) , d f o " k e la [ ; k

(C) ; f nA u

PB v

g S ] tg k °u rF k k v i j L i j vH k k T ; la [ ; k ; s a g S ] r k su g k s x k (R) , d vH k k T ; la [ ; k

(D) ; f nB m

PA n

g S ] tg k ° m rF k kn i j L i j vH k k T ; la [ ; k ; s a g S ] (S) , d la ; q D r la [ ; k

r k s3(m + n) g k s x k (T) u k r k s vH k k T ; u k g h

la ; q D r la [ ; k





LTS-36/40 0000CT103115008


2. Column-I Column-II

(A) The number

4991000

2k 1

k 0

2k 1 C

100

is divisible by (P) 2

(B) 4

3 1

is divisible by (Q) 4

(where [.] denotes greatest integer function)

(C) A non-constant geometric series has sum of its terms (R) 8

added to infinity are 12 & if every term is squared this sum

becomes144

7. If common ratio is

p

q where p & q are

relatively prime then 2q + p is divisible by (S) 9

(D) Let xi 1,2,3,4 where i = 1,2,3,4 and x

i x

j if

i j. If |xi – i| > 0, then number of 4 digit numbers of the (T) 11

form 'x1x

2x

3x

4' is

d ye

-I d ye

-II

(A) la [ ; k

4991000

2k 1

k 0

2k 1 C

100

, ........ ls f oH k k f tr g k s x h (P) 2

(B) 4

3 1

, ........ ls f oH k k f tr g k s x k (Q) 4

( tg k ° [.] e g Ÿ k e i w . k k ± d Q y u d k s n ' k k Z r k g S)

(C) , d p j x q . k k s Ÿ k j J s . k h d s v u U r i n k s a r d tk s M + u s i j (R) 8

; k s x Q y12 g S rF k k ; f n i z R ; s d i n d k o x Z f d ; k tk ; s r k s ; k s x Q y

; k s x Q y144

7 g k s tk r k g S A ; f n lk o Z v u q i k r

p

q g S ] tg k ° p rF k k q

i j L i j vH k k T ; la [ ; k ; s a g S ] r k s(2q + p), ....., ls f oH k k f tr g k s x k (S) 9

(D) e k u kxi 1,2,3,4 ( tg k ° i = 1,2,3,4 rF k kx

i x

j,

; f n i j g S ) ; f n |xi – i| > 0 g S ] r k sx

1x

2x

3x

4 :i o k y h p k j (T) 11

v a d k s a d h la [ ; k v k s a d h la [ ; k g k s x h

SECTION–III : Integer Value Correct Type [ .M–III : i . d e u lg i d






LTS-37/400000CT103115008



[ .M–IV : ( v dre v d: 32)




Marking scheme :








1. Let ƒ(x) = (2x – )3 + 2x – cosx. If 1dx

dx

at x = is

1,a N

a , then the value of 'a' is

e k u kƒ(x) = (2x – )3 + 2x – cosx g S A ; f n 1dx

dx

, x = i j

1,a N

a g S ] r k sa d k e k u g k s x k

2. The least positive integral value of k for which the equation2

2

x x 1k

x x 1

has no real solution is

k d k U ; w u r e / k u k R e d i w . k k ± d e k u f tld s f y ; s le h d j . k2

2

x x 1k

x x 1

d k d k s b Z o k L r f o d g y u g h a g S ] g k s x k

3. A point P lies on the graph

2 2y 1 x 2

164 49

. If distance of P from one focus is 8 and its distance

from other focus is 'd', then sum of digits of number 'd' is

2 2

y 1 x 21

64 49

d s v k j s [ k i j , d f c U n qP f L F k r g S A ; f n , d u k f H k lsP d h n w j h8 g S rF k k n w lj h u k f H k ls b l

n w j hd g S ] r k s la [ ; kd d s v a d k s a d k ; k s x Q y g k s x k





LTS-38/40 0000CT103115008


4. If log2y + (21+x + 21–x) (log y) + 22x+1 + 21–2x = 0; where x,y R, then the value of |log(x + y)| is

; f n log2y + (21+x + 21–x) (log y) + 22x+1 + 21–2x = 0, ( tg k ° x,y R g S )] r k s|log(x + y)| d k e k u g k s x k

5. Let a,b,c

be non-zero non-collinear vectors such that c a c b

. The value of

21 a c b a a b a

is

e k u ka,b,c

v ' k w U ; v la j s [ k h ; lf n ' k b l i z d k j g S f d c a c b

g S ] r k s 21 a c b a a b a

d k

e k u g k s x k

6. Ifx

1 x 1 alim tan

x 1 2x 2 b

(a,b N); then the value of a + b is

; f n x1 x 1 alim tan

x 1 2x 2 b

(a,b N) g S ] r k sa + b d k e k u g k s x k

7. Derivative of n2(nx) with respect to n(nx) at x = ee (which e is Napier's constant)

x = ee i j n(nx) d k n2(nx) d s lk i s k v o d y t g k s x k( tg k ° e u s f i ; j v p j d k s n ' k k Z r k g S)

8. The maximum integral value of 'k' for which 23 x , x 1

ƒ xx k , x 1

has local minima at x = 1 is

k d k v f / k d r e i w . k k ± d e k u ] f tld s f y ,

23 x , x 1

ƒ x x k , x 1

d k x = 1 i j L F k k u h ; f u f E u " B e k u g S ] g k




LTS-39/400000CT103115008




OPTICAL RESPONSE SHEET / v V dy i l ' V:

12. The ORS is machine-gradable and will be collected by the invigilator at the end of the examination.

v k s - v k j- , l- e ' k h u& tk ° P ; g S rF k k ; g i j h k k d s le k i u i j f u j h k d d s k j k , d = d j f y ; k tk ; s x k A13. Do not tamper with or mutilate the ORS. / v k s - v k j- , l- d k s g s j& Q s j@ f o œ f r u d j s a A14. Write your name, form number and sign with pen in the space provided for this purpose on the original. Do not write

any of these details anywhere else. Darken the appropriate bubble under each digit of your form number.

v i u k u k e] Q k W e Z u E c j v k S j v k s - v k j- , l- e s a f n , x , [ k k u k s a e s a d y e ls H k j s a v k S j v i u s g L r k k j d j s a A b uv k S j u f y [ k s a A Q k W e Z u E c j d s g j v a d d s u h p s v u q :i c q y c q y s d k s d k y k d j s a A

DARKENING THE BUBBLES ON THE ORS / v v ,l i c yc y d d y d u d o:

15. Use a BLACK BALL POINT PEN to darken the bubbles in the upper sheet.

≈ i j h e w y i ` " B d s c q y c q y k s a d k sd y c y o b V dye ls d k y k d j s a A16. Darken the bubble COMPLETELY / c q y c q y s d k si . :i l d k y k d j s a A17. Darken the bubbles ONLY if you are sure of the answer / c q y c q y k s a d k sr d k y k d j s a tc v k i d k m Ÿ k j f u f ' p r g k s A

18. The correct way of darkening a bubble is as shown here :

c q y c q y s d k sd y d j u s d k m i ; q D r r j h d k ; g k ° n ' k k Z ; k x ; k g S:

19. There is NO way to erase or "un-darken" a darkened bubble

d k y s f d ; s g q ; s c q y c q y s d k s f e V k u s d k d k s b Z r j h d kug g S A20. The marking scheme given at the beginning of each section gives details of how darkened and not darkened

bubbles are evaluated.

g j [ k . M d s i z k j E H k e s a n h x ; h v a d u ; k s tu k e s ad y d; x; rF k kd y u d; x, c q y c q y k s a d k s e w Y ; k a f d r d j u s d k r j h d k f n x ; k g S A

21. Take g = 10 m/s2 unless otherwise stated.

g = 10 m/s2 i z ; q D r d j s a ] tc r d f d v U ; d k s b Z e k u u g h a f n ; k x ; k g k s A

I HAVE READ ALL THE INSTRUCTIONS

AND SHALL ABIDE BY THEM

e u l un ' d i< y; g v e m ud

vo'; i yu d: x d: x A

____________________________

Signature of the Candidate / i d g r

I have verified the identity, name and roll

number of the candidate, and that question

paper and ORS codes are the same.

e u i d i p; u e v Q e u c d i

y; d i 'u i= r v v ,l d M n u le u g A

____________________________

Signature of the invigilator / u d d g r

NAME OF THE CANDIDATE / i d u e.................................................................................

FORM NO / Q e u c..............................................

1505 lts paper 1 aiot dlp

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