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NARAYANA IIT ACADEMY

INDIA

JEE MAIN – 2013

QUESTION PAPER, KEY & SOLUTIONS

NARAYANA IIT ACADEMY JEE MAIN – 2013

Important Instructions :

Time : 3 hr. CODE : P Maximum Marks – 360. 1. Immediately fill in the particulars on this page of the Test Booklet with Blue/Black Ball

point Pen. Use of pencil is strictly prohibited. 2. The Answer Sheet is kept inside this Test Booklet. When you are directed to open the Test

Booklet, take out the Answer Sheet and fill in the particulars carefully. 3. The test is of 3 hours duration. 4. The test Booklet consists of 90 questions. The maximum marks are 360. 5. There are three parts in the questions paper A, B, C consisting of Physics, Chemistry and

mathematics having 30 questions in each part of equation weightage. Each question is allotted 4 (four) marks for correct response.

6. Candidates will be awarded marks as stated above in instruction No. 5 correct response of each question 1/4 (one fourth) marks will be deducted for indicating incorrect response of each questions. No deduction from the total score will be made if no response is indicated for an item in the answer sheet.

7. There is only one correct response for each question. Filling up more than one response in any question will be treated as wrong response and marks for wrong response will be deducted accordingly as per instruction 6 above.

8. Use Blue/ Black Ball Point Pen only for writing particulars/marking responses on Side -1 and Side – 2 of the Answer Sheet. Use of Pencil is strictly prohibited.

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12. The CODE for this Booklet is P. Make sure that the CODE printed on Side – 2 of the Answer Sheet is the same as that on this booklet. In case of discrepancy, the candidate should immediately report the matter to the invigilator for replacement of both the Test Booklet and the Answer Sheet.

13. Do not fold or make any stray marks on the Answer Sheet.


PHYSICS 1. A uniform cylinder of length L and mass M having cross – sectional area A is suspended,

with its length vertical, from a fixed point by a massless spring, such that it is half submerged in a liquid of density at equilibrium position. The extension x0 of the spring when it is in equilibrium is :

(1) Mgk

(2) Mg LA1k M

(3) Mg LA1k 2M

(4) Mg LA1k M

(Here k is spring constant) Key. 3

Sol. 0

Lkx Ag Mg2

0

Lkx Mg Ag2

LAMg 12M

Mg

F = AgB L2

kx0 F B

2. A metallic rod of length ‘1’ is tied to a string of length 2l and made to rotate with angular

speed on a horizontal table with one end of the string fixed. If there is a vertical magnetic field ‘B’ in the region, the e.m.f. induced across the ends of the rod is :

l2l

(1) 22B l

2 (2)

23B l2 (3)

24B l2 (4)

25B l2

Key. 4


Sol. 2AC

1 B (3 )2

2AB

1 B (2 )2

2

BC AC AB

5B2

2lA B C

l 3. This question has Statement I and Statement II. Of the four choices given after the

Statements, choose the one that best describes the two Statements. Statement – I : A point particle of mass m moving with speed v collides with stationary point

particle of mass M. If the maximum energy loss possible is given as 21 mf m then f

2 M m

.

Statement – II : Maximum energy loss occurs when the particles get stuck together as a result of the collision. (1) Statement – I is true, Statement – II is true, Statement – II is a correct explanation of Statement – I. (2) Statement – I is true, Statement – II is true, Statement – II is Not a correct explanation of Statement – I. (3) Statement – I is true, Statement-II is false. (4) Statement – I is false, Statement – II is true.

Key. 4 Sol. mV + M(0) = mV1 + MV2

2 1

2 1

V Ve O

2 1

mVV Vm M

2 2 21 2

1 1 1Loss mV mV MV2 2 2

2 2

22

1 (m M) m VmV2 2 (m M)

2

2

MmV2(m M)

; So statement I is wrong.

4. Let 0 denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = time and A = electric current, then :

(1) 1 3 20[ ] [M L T A] (2) 1 3 4 2

0[ ] [M L T A ] (3) 1 2 1 2

0[ ] [M L T A ] (4) 1 2 10[ ] [M L T A]


Key. 2 Sol. a b c d

0[ ] [M] [L] [T] [A] 1 3 2 4 a b c dM L A T [M] [L] [T] [A]

5. A projectile is given an initial velocity of ˆ ˆi 2 j m/s, where i is along the ground and j is along the vertical. If g = 10 m/s2, the equation of its trajectory is :

(1) y = x – 5x2 (2) y = 2x – 5x2 (3) 4y = 2x – 5x2 (4) 4y = 2x – 25x2 Key. 2 Sol. x = (1) t

21y 2t gt2

2gxy 2x

2

6. The amplitude of a damped oscillator decreases to 0.9 times its original magnitude in 5s. In another 10s it will decrease to times its original magnitude, where equals :

(1) 0.7 (2) 0.81 (3) 0.729 (4) 0.6 Key. 3 Sol. kt

0A A e

0

0

AA 1ln kt t lnA k A

1 105 lnk 9

…(1)

1 115 lnk

….(2)

31 10

9

3(0.9) 0.729 7. Two capacitors C1 and C2 are charged to 120 V and 200 V respectively. It is found that by

connecting them together the potential on each one can be made zero. Then (1) 1 25C 3C (2) 1 23C 5C (3) 1 23C 5C 0 (4) 1 29C 4C Ans. (2)

Sol.

1 2C 120 200C 1 23C 5C


8. A sonometer wire of length 1.5 m is made of steel. The tension in it produces an elastic strain of 1%. What is the fundamental frequency of steel if density and elasticity of steel are 7.7 × 103 kg/m3 and 2.2 × 1011 N/m2 respectively?

(1) 188.5 Hz (2) 178.2 Hz (3) 200.5 Hz (4) 770 Hz Ans. (2) Sol. Stress = Y × strain

11 12.2 10100

9F 2.2 10A

0v 1 Ff2 2 A

l l

9

0 31 2.2 10f

2 1.5 7.7 10

51 20 103 7

30

1 2f 103 7

0f 178.17 Hz 9. A circular loop of radius 0.3 cm lies parallel to a much bigger circular loop of radius 20 cm.

The centre of the small loop is on the axis of the bigger loop. The distance between their centres is 15 cm. If a current of 2.0 A flows through the smaller loop, then the flux linked with bigger loop is

(1) 119.1 10 weber (2) 116 10 weber (3) 113.3 10 weber g (4) 96.6 10 weber Ans. (1)

Sol.

2

20 123/22 2

1 1

i rr Mi

2 r r

2 20 1 2

3/22 21

r rM

2 r x

Mi

2 20 1 2

3/22 21

r ri

2 r x


2 22 27

3/22 22 2

3.14 20 10 0.3 104 10 2

2 20 10 15 10

119.1 10

10. Diameter of a plano-convex lens is 6 cm and thickness at the centre is 3 mm. If speed of light in material of lens is 2 × 108 m/s, the focal length of the lens is

(1) 15 cm (2) 20 cm (3) 30 cm (4) 10 cm Ans. (3) Sol. 2 22R R 0.3 3

R 15.15cm

2 1 2 1

v u R

3 / 2 1 3 / 2 1v 15

3 12v 2 15

V = 45 cm

v 45f3 / 2

f = 30 cm 11. What is the minimum energy required to launch a satellite of mass m from the surface of a

planet of mass M and radius R in a circular orbit at an altitude of 2R?

(1) 5GmM6R

(2) 2GmM3R

(3) GmM2R

(4) GmM3R

Ans. (1)

Sol. GMm GMm 5GMmE6R R R

12. A diode detector is used to detect an amplitude modulated wave of 60% modulation by using a condenser of capacity 250 pico farad in parallel with a load resistance 100 kilo ohm. Find the maximum modulated frequency which could be detected by it.

(1) 10.62 MHz (2) 10.62 kHz (3) 5.31 MHz (4) 5.31 kHz Ans. NO OPTION SOL. QUESTION AMBIGUITY


13. A beam of unpolarised light of intensity I0 is passed through a Polaroid A and then through another Polaroid B which is oriented so that its principal plane makes an angle of 45° relative to that of A. The intensity of the emergent light is:

(1) I0 (2) I0/2 (3) I0/2 (4) I0/8 Ans. (3)

Sol. 0(1st polarisation)

II2

20 0(2nd polarisation)

I II cos 452 4

14. The supply voltage to a room is 120 V. The resistance of the lead wires is 6 . A 60 W bulb is already switched on. What is the decrease of voltage across the bulb, when a 240 W heater is switched on in parallel to the bulb?

(1) Zero Volt (2) 2.9 Volt (3) 13.3 Volt (4) 10.04 Volt Ans. (4) Sol.

R160W

R = 60

S

240 W

120 V

= 12060

2

=240

R2

=

120240

2

=60

R = 60

120 V

240

60

When S is open

bulb120 240V 117.073

246

When S is closed

2

eq eq120P 300W R300

48 (Heater and Bulb)

bulb120 48V 106.66

54

V 10.4 V no option is correct The closest option is (4) only 15.

2p0

p p0

v0v

2v0

15.


The above p-v diagram represents the thermodynamic cycle of an engine, operating with an ideal monatomic gas. The amount of heat, extracted from the source in a single cycle is:

(1) 0 0p v (2) 0 013 p v2

(3) 0 011 p v2

(4) 0 04p v

Ans. (2)

Sol. Only positive q is to be considered hence o o13q p v2

16. A hoop of radius r and mass m rotating with an angular velocity 0 is placed on a rough horizontal surface. The initial velocity of the centre of the hoop is zero. What will be the velocity of the centre of the hoop when it ceases to slip?

(1) 0r4 (2) 0r

3 (3) 0r

2 (4) 0r

Ans. (3) Sol. Applying conservation of angular momentum about point of contact 0 cI I MV R

v0=0

0

vc

initial final 2 2 c

0 cvMR MR Mv RR

20 cMR 2MRv

0c

Rv2

17. An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass M. The piston and the cylinder have equal cross sectional area A. When the piston is in equilibrium, the volume of the gas is V0 and its pressure is P0. The piston is slightly displaced from the equilibrium position and released. Assuming that the system is completely isolated from its currounding, the piston executes a simple harmonic motion with frequency:

(1) 0

0

A P12 V M

(2) 0 02

V MP12 A

(3) 2

0

0

A P12 MV

(4) 0

0

MV12 A P

Ans. (3)

Sol.

A M

P0 ,V0


If displaced by x then restoring0

dVF A.dP ABV

2

2restoring

0 0

Ax BA xF A.B M xV V

2

0

0

P A xV

2 2

0 0

0 0

A P A P12MV 2 MV

18. If a piece of metal is heated to temperature and then allowed to cool in a room which is at temperature 0 , the graph between the temperature T of the metal and time t will be closed to :

1.T

0 t 2.

T

0 t0

3.

T

0 t0

4.

T

0 t0

Ans. (3)

Sol. As per Stefan’s Law 4 40

ddt which decrease with t

19. This question has Statement – I and Statement – II. Of the four choices given after the Statements, choose the one that best describes the two Statements.

Statement – I : Higher the range, greater is the resistance of ammeter. Statement – II : To increase the range of ammeter, additional shunt needs to be used across

it. (1) Statement – I is true, Statement – II is true, Statement – II is the correct explanation of

Statement – I (2) Statement – I is true, Statement – II is true, Statement – II is not the correct explanation

of Statement – I (3) Statement – I is true, Statement – II is false (4) Statement – I is false, Statement – II is true Ans. (4) Range of the ammeter is given by

gg

RI I 1

R

and resistance of the ammeter is given by

g

g

R RR

R R

Hence to have higher range, is the resistance of ammeter should be made smaller.


20. In an LCR circuit as shown below both switches are open initially. Now switch S1 is closed, S2 kept open. (q is charge on the capacitor and RC is Capacitive time constant). Which of the following statement is correct?

V

C

L

R S1

S2 (1) Work done by the battery is half of the energy dissipated in the resistor

(2) At CVt , q2

(3) At 2t 2 , q CV(1 e ) (4) At 1t , q CV(1 e )2

Ans. (3) As during the charging t /q CV(1 e ) Hence at t 2 2q CV(1 e ) 21. Two coherent point sources S1 and S2 are separated by a small distance ‘d’ as shown. The

fringes obtained on the screen will be

S1 S2

d

dScreeenO

(1) points (2) straight lines (3) semi-circles (4) concentric circles Ans. (4) The waves reaching from S1 and S2 at any point on the screen which is equidistant from the

central points O will have same path difference. So locus of such points will be concentric circles on the Screen.

22. The magnetic field in a traveling electromagnetic wave has a peak value of 20 nT. The peak value of electric field strength is

(1) 3 V/m (2) 6 V/m (3) 9 V/m (4) 12 V/m Ans. (2) Sol : As, E BC Hence, 9 8E 20 10 3 10 6 V / m


23. The anode voltage of a photocell is kept fixed. The wavelength of the light falling on the cathode is gradually changed. The plate current I of the photocell varies as follows :

(1) O

I

(2)

O

I

(3) O

I

(4)

O

I

Ans. (4)

When is increased the cutoff value of current will stop. 24. The I – V characteristic of an LED is

(1)

O

I(R) (Y) (G) (B)

RedYell

owGre

enBlu

e

V

(2)

O

BGYR

I

V

(3)

O

I

V

(4)

RYGB

V O

I Ans. (1) I – V characteristic of an LED is similar to forward I – V characteristics of P – N junction

diode and threshold voltage increases on decreasing the wavelengths. 25. Assume that a drop of liquid evaporates by decrease in its surface energy, so that its

temperature remains unchanged. What should be the minimum radius of the drop for this to be possible? The surface tension is T, density of liquid is and L is its latent heat of vaporization.

(1) L/T (2) T / L (3) T/L (4) 2T/L Key: 4 Sol: (4) Let the radius of drop is x and ‘dx’ thickness of drop evaporates. Change is Area = A = 4x2

– 4(x – dx)2


= 4x2 2dx1 1 8 xdx

x

So energy released = T. 8xdx Volume of liquid evaporates = 4x2dx

So, T . 8xdx = . 4x2dx . L 2TxL

26. In a hydrogen like atom electron makes transition from an energy level with quantum number n to another with quantum number (n – 1). If n > > 1, the frequency of radiation emitted is proportional to

(1) 1n

(2) 2

1n

(3) 3/ 2

1n

(4) 3

1n

Key : 4 Sol: (4)

22 2

1 1RCZ(n 1) n

= 22 2

22 2

RCZ 1 RCZ 11 1 1n n n11

n

= 2 2

2 3 3RCZ 2 2RCZ 11 1

n n n n

27. The graph between angle of deviation () and angle of incidence (i) for a triangular prism is represented by

(1) iO

d

(2) iO

d

(3) iO

d

(4) iO

d

Key : 3 Sol: (3) = i + e – A, when i = e, is minimum. 28. Two charges, each equal to q, are kept at x = – a and x = a on the x-axis. A particle of mass

m and charge 0 2

qq is placed at the origin. If charge q0 is given a small displacement (y < <

a) along the y – axis, the net force acting on the particle is proportional to


F sin F sin

2F cos Y

F

q a a q X

Y

F

q0

(1) y (2) – y (3) 1

y (4) 1

y

Key : 1 Sol: (1) Force due to each charge is

02 2

kq qF(a y )

From figure resultant force is 02 2 2 2

kqq y2Fcos 2.(a y ) a y

Y < < a, 0r 3

2k q q yF F ya

(towards +ve y-axis)

29. Two short bar magnets of length 1 cm each have magnetic moments 1.20 Am2 and 1.00 Am2 respectively. They are placed on a horizontal table parallel to each other with their N poles pointing towards the South. They have a common magnetic equator and are separated by a distance of 20.0 cm. The value of the resultant horizontal magnetic induction at the mid-point O of the line joining their centres is close to (Horizontal component of earth’s magnetic induction is 3.6 10–5 Wb/m2).

(1) 3.6 10–5 Wb/m2 (2) 2.56 10–4 Wb/m2 (3) 3.50 10–4 Wb/m2 (4) 5.80 10–4 Wb/m2

Key : 2 Sol: (2) Field at centre (O) due to both magnet

= 701 23 1 3

1 1(M M ) 10 (1.20 1.00)4 r (10 )

= 2.2 10–4 (towards North) Earth’s horizontal component of magnetic field at that point is 3.6 10–5 towards north. So Net field at that point = 2.2 10–4 + 3.6 10–5 = 2.56 10–4 Wb/m2 30. A charge Q is uniformly distributed over a long rod AB of length L as shown in the figure.

The electric potential at the point O lying at a distance L from the end A is L L

O A B

(1) 0

Q8 L

(2) 0

3Q4 L

(3) 0

Q4 L ln 2

(4) 0

Qln 24 L


Key : 4 Sol: (4)

Let’s consider as elementary length ‘dx’ at a distance x from O. Qdq dx.L

Potential due to

dq at O is 0 0

1 dq 1 Q dxdv4 x 4 L x

So, net potential at O

2L

0 0L

1 Q dx QV dV ln (2)4 L x 4 L

CHEMISTRY 31. Which of the following complex species is not expected to exhibit optical isomerism ?

(1) [Co(en)3]3+ (2) [Co(en)2Cl2]+

(3) [Co(NH3)3Cl3] (4) [Co(en)(NH3)2Cl2]+ Key. 3 Sol: Meridional and facial distribution do not show optical isomerism 32. Which one of the following molecules is expected to exhibit diamagnetic behaviour ?

(1) C2 (2) N2 (3) O2 (4) S2 Key. (1,2) Sol: C2 1s2 < *1s2 < 2s2 < *2s2 < 2 22 2x yp p : No unpaired electron hence diamagnetic.

N2 1s2 < *1s2 < 2s2 < *2s2 < 2 22 2x yp p < 22 zp : No unpaired electron hence diamagnetic O2 1s2 < *1s2 < 2s2 < *2s2 < 22 zp < 2 22 2x yp p < 1 1*2 *2x yp p : This is paramagnetic S2 is also paramagnetic since like O2 it has unpaired electrons in ABMO

33. A solution of (-)-1- chloro-1- phenylethane in toluene racemises slowly in the presence of a small amount of SbCl5 due to the formation of : (1) carbanion (2) carbene (3) carbocation (4) free radical

Key. (3) Sol:

H C3H C3

Toluene, SbCl5H

H+

Cl

CC

In presence of SbCl5, carbocation will be formed.

34. Give 3 2

4

0 0/ /

0.74 ; 1.51Cr Cr MnO Mn

E V E V


2 32 7

0 0/ /

1.33; 1.36Cr O Cr Cl Cl

E E V Based on the data given above, strongest oxidizing agent will be: (1) Cl (2) Cr3+ (3) Mn2+ (4) 4MnO

Key. (4) Sol: Cr3+ + 3e Cr(s) : 0 0.74redE V

( 6)2

2 7Cr O

+ 6e 2Cr3+ : 0 1.33redE V ( 7)

4MnO

+ 5e Mn2+ : 0 1.51redE V Cl + 1 e Cl : 0 1.36redE V Thus tendency of reduction will be most in MnO4

into Mn2+. Hence this will be the strongest oxidizing agent in given species.

35. A piston filled with 0.04 mole of an ideal gas expands reversibly from 50.0 mL to 375 mL at a constant temperature of 37.00C. As it does so, it absorbs 208J of heat. The values of q and w for the process will be : (R = 8.314 J/mol K) (ln 7.5 = 2.01) (1) q = +208 J, w = -208 J (2) q = -208J, w = -208 J

(3) q = 208 J, w = +208J (4) q = +208J, w = +208 J Key. (1) Sol: In isothermal expansion process U = 0

q = + 208 J Thus w = 208 J (expansion)

36. The molarity of a solution obtained by mixing 750 mL of 0.5 (M) HCl with 250 mL of 2 (M) HCl will be (1) 0.875 M (2) 1.00 M (3) 1.75 M (4) 0.975 M

Key. (1) Sol: Molarity of mixture of identical species is given by

Mm (V1 + V2) = M1V1 + M2V2 Mm (750 + 250) = 0.5 750 + 2 250

Mm = 875 0.8751000

M

37. Arrange the following compounds in order of decreasing acidity

(1) II > IV > I > III (2) I > II > III > IV (3) III > I > II > IV (4) IV > III > I > II Ans. (3) Sol. –R, –I effects increase but +I, +R effects decrease acidic strengths.


38. For gaseous state, if most probable speed is denoted by C*, average speed by C and mean square speed by C, then for a large number of molecules the ratios of these speeds are

(1) C* : C : C 1.225 : 1.128 : 1 (2) C* : C : C 1.128 : 1.225 : 1 (3) C* : C : C 1 : 1.128 : 1.225 : 1 (4) C* : C : C 1 : 1.225 : 1.128 Ans. (3)

Sol. Ratio is 82 : : 3 .

39. The rate of a reaction doubles when its temperature changes from 300 K to 310 K. Activation energy of such a reaction will be

(R = 8.314 JK–1 mol–1 and log 2 = 0.301) (1) 53.6 kJ mol–1 (2) 48.6 kJ mol–1 (3) 58.5 kJ mol–1 (4) 60.5 kJ mol–1 Ans. (1)

Sol. a1

2 2 1

Ek 1 1logk 2.303R T T

1log 2 2.303 8.314 300 310Ea kJmol

10 1000

.

40. A compound with molecular mass 180 is acylated with CH3COCl to get a compound with molecular mass 390. The number of amino groups present per molecule of the former compound is

(1) 2 (2) 5 (3) 4 (4) 6 Ans. (2) Sol. –OH groups of carbohydrates get acctylated 3COOCH

No of –OH groups = 390 180 542

41. Which of the following arrangements does not represent the correct order of the property started against it?

(1) 2 2 2 2V Cr Mn Fe : paramagnetic behavior (2) 2 2 2 2Ni Co Fe Mn : ionic size (3) 3 3 3 3Co Fe Cr Sc : stability in aqueous solution (4) Sc < Ti < Cr < Mn : number of oxidation states Ans. (1) Sol. Greater the number of unpaired electrons, higher is the paramagnetic behaviour . 42. The order of stability of the following carbocations

is

(1) III > II > I (2) II > III > I (3) I > II > III (4) III > I > II Ans. (4)


Sol. Benzyl and allyl carbocations are more stable due to resonance. 43. Consider the following reaction:

24 2 4xMnO yC O zH 2

2 2zxMn 2yCO H O2

The values of x, y and z in the reaction are, respectively: (1) 5, 2 and 16 (2) 2, 5 and 8 (3) 2, 5 and 16 (4) 5, 2 and 8 Ans. (3) Sol. 2

4 2 42MnO 5C O 16H 2

2 22Mn 10CO 8H O x = 2 y = 5 z = 16. 44. Which of the following is the wrong statement? (1) ONCl and ONO– are not isoelectronic (2) O3 molecule is bent (3) Ozone is violet-black in solid state (4) Ozone is diamagnetic gas Ans. No Answer Sol. All the four given statements are correct. 45. A gaseous hydrocarbon gives upon combustion 0.72 g. of water and 3.08 g. of CO2. The

empirical formula of the hydrocarbon is: (1) C2H4 (2) C3H4 (3) C6H5 (4) C7H8 Ans. (4)

Sol. No. of mole of H2O = 0.72 0.04 mole18

No. of mole of H = 2 × 0.04 = 0.08 mole

No. of mole of CO2 = 3.08 0.07 mole44

No. of mole of ‘C’ = 0.07 mole

No. of mole of C 0.07 7No. of mole of H 0.08 8

Empirical Formula = C7H8. 46. In which of the following pairs of molecules / ions, both the species are not likely to exist? (1) 2

2 2H , He (2) 22 2H , He (3) 2

2 2H , He (4) 22 2H , He

Ans. (3) Sol.

Species Bond order

2H 1 [1 0] 0.52

22He 1 [4 2] 1

2


22H 1 [2 1] 0.5

2

22H 1 [0 0] 0

2

2He 1 [2 2] 02

Both 22H and He2 have zero bond order so they are not likely to exist.

47. Which of the following exists as covalent crystals in the solid state? (1) Iodine (2) Silicon (3) Sulphur (4) Phosphorous Ans. (2) Sol. Silicon crystalises as diamond type covalent solid. 48. Synthesis of each molecule of glucose in photosynthesis involves: (1) 18 molecules of ATP (2) 10 molecules of ATP (3) 8 molecules of ATP (4) 6 molecules of ATP Ans. (1) Sol. 18 molecules of A.T.P are required. 49. The coagulating power of electrolytes having ions Na , 3Al and 2Ba for arsenic sulphide

sol increases in the order (1) 3 2Al Ba Na (2) 2 3Na Ba Al (3) 2 3Ba Na Al (4) 3 2Al Na Ba Ans. (2) Coagulating power is directly proportional to charge density. 50. Which of the following represents the correct order of increasing first ionization enthalpy for

Ca, Ba, S, Se and Ar? (1) Ca < S < Ba < Se < Ar (2) S < Se < Ca < Ba < Ar (3) Ba < Ca < Se < S < Ar (4) Ca < Ba < S < Se < Ar Ans. (3) (i) Ar being a noble gas has highest IE (ii) Down the group ionization enthalpy decreases. (iii) ‘S’ block elements will have lower ionization enthalpy than ‘P’ block.

51. Energy of an electron is given by E = 2

182

Z2.178 10 Jn

. Wavelength of light required to

excite an electron in an hydrogen atom from level n = 1 to n = 2 will be (h = 346.62 10 Js and c = 8 13.0 10 ms ) (1) 71.214 10 m (2) 72.816 10 m (3) 76.500 10 m (4) 78.500 10 m Ans. (1) 2 1E E E

18 18hc 2.178 10 2.178 10

4 1


34 8

186.62 10 3 10 32.178 104

71.214 10 m 52. Compound (A), 8 9C H Br , gives a white precipitate when warmed with alcoholic 3AgNO .

Oxidation of (A) gives an acid (B), 8 6 4C H O . (B) easily forms anhydride on heating. Identify the compound (A).

(1)

CH Br2

CH3

(2)

C H2 5

Br (3)

CH Br2

CH3

(4)

CH Br2

CH3

Ans. (4)

Aryl bromide cannot give ppt. as aryl cations are unstable. 53. Four successive members of the first row transition elements are listed below with atomic

numbers. Which one of them is expected to have the highest 3 20M /M

E (1) Cr(Z = 24) (2) Mn(Z = 25) (3) Fe(Z = 26) (4) Co(Z = 27) Ans. (4) As it is expected on the basis of electronic configuration.

3 2

4 0 5 0

Mn e Mn more stable

3d 4s 3d 4s

(as it is completely half filled subshell) However experimental data suggests that 3 2 oCo e Co ,E is greater than other 3 2M / M couples. 54. How many litres of water must be added to 1 litre of an aqueous solution of HCl with a pH of

1 to create an aqueous solution with pH of 2? (1) 0.1 L (2) 0.9 L (3) 2.0 L (4) 9.0 L


Ans. (4) 1 1 2 2M V M V 1 2

210 1lit 10 V 2V 10lit volume of water to be added is 9 litre. 55. The first ionization potential of Na is 5.1 eV. The value of electron gain enthalpy of Na+ will

be (1) – 2.55 eV (2) –5.1 eV (3) –10.2 eV (4) +2.55 eV Key : 2 Sol: (2) ionization potential = – electron gain enthalpy of metal cation 56. An organic compound A upon reacting with NH3 gives B. On heating B gives C. C in

presence of KOH reacts with Br2 to give 2 2 2 ,CH CH NH A is

(1) 3CH COOH (2) 3 2 2CH CH CH COOH (3) 3

3

|CH CH COOH

CH

(4) 3 2CH CH COOH

Key : 4 Sol: (4)

3 2CH CH COOH

A

3 43 2

NH CH CH C OON H

B

2 2 2 2

H O CH C H CONH

C

2 /3 2 2Br KOH CH CH NH

57. Stability of the species 2 2,Li Li and 2Li increases in the order of (1) 2 2 2Li Li Li (2) 2 2 2Li Li Li (3) 2 2 2Li Li Li (4) 2 2 2Li Li Li Key : 2 Sol: (2) Normally Stability Bond order but in case of species with same bond order more the

number of electrons in anti bonding molecular orbitals lesser the stability 58. An unknown alcohol is treated with the “Lucas reagent” to determine whether the alcohol is

primary, secondary or tertiary. Which alcohol reacts fastest and by what mechanism? (1)secondary alcohol by 1NS (2) tertiary alcohol by 1NS (3) secondary alcohol by 2NS (4) tertiary alcohol by 2NS Key : 2 Sol: (2) Tertiary alcohol reacts fastest with Lucas reagent by 1NS mechanism 59. The gas leaked from a strong tank of the Union Carbide plant in Bhopal gas tragedy was


(1) Methylisocyanate (2) Methylamine (3) Ammonia (4) Phosgene Key : 1 Sol: (1) MIC was responsible for Bhopal gas tragedy. 60. Experimentally it was found that a metal oxide has formula 0.98 .M O Metal M, is present as

2M and 3M in its oxide. Fraction of the metal which exists as 3M would be (1) 7.01 % (2) 4.08 % (3) 6.05% (4) 5.08 % Key : 2 Sol: (2) Let the number of 3M and 2M ions be x and y respectively 0.98x y ….(i) 3 2 2x y ….(ii) No. of 3M ion = 0.04 And 2M ion = 0.94

So percentage of 3M ion 0.04 100 4.08%0.98

MATHEMATICS

61. Distance between two parallel planes 2x + y + 2z = 8 and 4x + 2y + 4z + 5 = 0 is

(1) 32

(2) 52

(3) 72

(4) 92

Key. (3)

Sol: Required distance = 2 2 2

5 87222 1 2

62. At present, a firm is manufacturing 2000 items. It is estimated that the rate of change of

production P w.r.t. additional number of workers x is given by 100 12 .dP xdx

If firm

employs 25 more workers, then the new level of production of items is (1) 2500 (2) 3000 (3) 3500 (4) 4500 Key. (3) Sol: (100 12 )x dx dp

3/2241003

x x P C

when x = 0 P = 2000

P = 2000 + 100 x - 3/2243

x

As x is 25 P = 3500


63. Let A and B be two sets containing 2 elements and 4 elements respectively. The number of subsets of A B having 3 or more elements is:

(1) 256 (2) 220 (3) 219 (4) 211 Key. (3) Sol: As A B has 8 elements

Number of subsets of A B having 3 or more elements is 8C3 + 8C4 + . . . . . + 8C8 = 28 – 1 – 8 – 28 = 219

64. If the line 2 3 41 1

x y zk

and 1 4 5

2 1x y z

k

are coplanar, then k can have

(1) any value (2) exactly one value (3) exactly two values. (4) exactly three values Key. (3)

Sol: 1 1 11 1 0

2 1

k

k

2k 3k 0 k 0, 3 65. If the vectors ˆˆAB 3 4i k

and ˆˆ ˆAC 5 2 4i j k

are the sides of a triangle ABC, then the

length of the median through A is (1) 18 (2) 72 (3) 33 (4) 45

Key. (3) Sol: Let A (0,0, 0) then B (3, 0, 4) and C (5, –2, 4) so length of median through A will be

16 1 16 33 66. The real number k for which the equation, 2x3 + 3x + k = 0 has two distinct real roots in [0,

1] (1) lies between 1 and 2 (2) lies between 2 and 3

(3) lies between -1 and 0 (4) does not exist. Key. (4) Sol: Let f(x) = x3 + 3x + k

f ' x = 6x2 + 3 As f ' x > 0 f(x) is S.I. function f(x) = 0 will have at most one real root in [0, 1] so no real number k exist.

67. The sum of first 20 terms of the sequence 0.7, 0.77, 0.777, ….. is

(1) 207 179 1081

(2) 207 99 109

(3) 207 179 1081

(4) 207 99 109

Ans. (3)


Sol. 1 2 207 1 10 1 10 ....... 1 109

1 2 207 20 10 10 ...... 109

1 2010 1 107 20 19 110

s

207 179 1081

68. A ray of light along x 3 y 3 gets reflected upon reaching x-axis, the equation of the reflected ray is

(1) y x 3 (2) 3 y x 3 (3) y 3x 3 (4) 3 y x 1 Ans. (2)

Sol. Slop of reflect ray is 13

and it passing through 3, 0 is

y 0 1x 3 3

3 y x 3

69. The number of values of k, for which the system of equation k 1 x 8y 4k

kx k 3 y 3k 1 Has no solution, is (1) infinite (2) 1 (3) 2 (4) 3 Ans. (2) Sol. k 1 k 3 8k

2k 4k 3 0 k 1, 3 but for k = 1 we have infinite solution

70. If the equation 2x 2x 3 0 and 22x bx c 0, z, b, c R , have a common root, then a : b : c is

(1) 1 : 2 : 3 (2) 3 : 2 : 1 (3) 1 : 3 : 2 (4) 3 : 1 : 2 Ans. (1) Sol. Here D < 0


Both roots is common

a b c1 2 3

a : b : c = 1 : 2 : 3 71. The circle passing through (1, –2) and touching the axis of x at (3, 0) also passes through the

point (1) (–5, 2) (2) (2, –5) (3) (5, –2) (4) (–2, 5) Ans. (3) Sol. 2 2 2x y k3 k

It passes (1, –2), then 2 2 21 3 2 k k

2 24 4 k 4k k k 2 Centre (3, –2) Equation of circle 2 2x 3 y 2 4 Clearly circle passes (5, –2) 72. If x, y, z are in A.P. and 1 1 1tan x, tan y and tan z are also in A.P., then (1) x = y = z (2) 2x = 3y = 6z (3) 6x = 3y = 2z (4) 6x = 4y = 3z Ans. (1) Sol. x, y, z are in AP 2y = x + z ……..(i)

1 1 12 tan y tan x tan z

22y x z

1 xz1 y

21 y 1 xz

2y xz .……(ii) From (i) and (ii) x = y = z Remark : Here we have assumed that x, y and z are of the same sign other wise x = – , y = 0,

z = for any positive will be the solution and then no option will be correct 73. Consider : Statement – I : (p q) ( p q) is a fallacy. Statement – II : (p q) (~ q p) is a tautology.


(1) Statement-I is true; Statement – II is true; Statement – II is a correct explanation for Statement – I.

(2) Statement-I is true; Statement – II is true; Statement – II is not a correct explanation for Statement – I.

(3) Statement – I is true; Statement – II is false. (4) Statement – I is false; Statement – II is true. Ans. (2) Sol.

p q ~ ~ (p ~ q)

(~ p q)

(p ~ q)(~ p q)

T T F F F F F T F F T T F F F T T F F T F F F T T F F F

(p q) (~ q ~ p) (p q)(~ q ~ p)

T T T F F T T T T T T T

74. If 5 3f (x)dx (x), then x f (x )dx is equal to :

(1) 3 3 2 31 x x 3 x x dx C3 (2) 3 3 3 31 x x 3 x x dx C

3

(3) 3 3 2 31 x x x x dx C3

(4) 3 3 3 31 x x x x dx C3

Ans. (3) Sol. Let I = 5 3x f (x )dx

I = 2 3 31 3x .x f (x )dx3

put x3 = t

1I t.f (t)dt3

1I t f (t)dt f (t)dt dt3

3 31I x . (x ) (t)dt3

3 3 3 21I x . (x ) (x ).3x dx3


3 3 2 31I x (x ) x . (x )dx3

.

75. x 0

(1 cos 2x)(3 cos x)limx tan 4x

is equal to

(1) 14

(2) 12

(3) 1 (4) 2

Ans. (4)

Sol. 2

2x 0

2 sin x 1lim . . .(3 cos x)tan 4x4 x4x

= 2 4 24 .

76. Statement – I : The value of the integral /3

/6

dx1 tan x

is equal to 6 .

Statement – II : b b

a a

f (x)dx f (a b x)dx .

(1) Statement-I is true; Statement – II is true; Statement – II is a correct explanation for Statement – I.

(2) Statement-I is true; Statement-II is true; Statement – II is not a correct explanation for Statement – I.

(3) Statement – I is true; Statement-II is false. (4) Statement – I is false; Statement – II is true. Ans. (4)

Sol. Let /3

/6

dxI1 tan x

/3

/6

b b

a a

dxI1 cot x

using f (x)dx f (a b x)dx

/3

/6

2I dx

2I = I6 12 .

77. The equation of the circle passing through the foci of the ellipse 2 2x y 1

16 9 , and having

centre at (0, 3) is : (1) x2 + y2 – 6y – 7 = 0 (2) x2 + y2 – 6y + 7 = 0


(3) x2 + y2 – 6y – 5 = 0 (4) x-2 + y2 – 6y + 5 = 0 Ans. (1) Sol. Foci of the ellipse are 7,0

radius of the circle = 4 equation of circle is x2 + y2 – 6y – 7 = 0. 78. A multiple choice examination has 5 questions. Each question has three alternative answers

of which exactly one is correct. The probability that a student will get 4 or more correct answers just by guessing is :

(1) 5

173

(2) 5

133

(3) 5

113

(4) 5

103

Ans. (3) Sol. Required probability = P (4 correct answer by guessing) + P (5 correct answer by guessing)

= 4 5

4 5

C C1 2 15 . . 53 3 3

= 5

113

.

79. The x-coordinate of the incentre of the triangle that has the coordinates of mid points of its sides as (0, 1), (1, 1) and (1, 0) is

(1) 2 2 (2) 2 2 (3) 1 2 (4) 1 2 Ans. (2) x coordinate of incentre

= 0 4 0 4 2 22 2 2 2 4 2 2

2(1, 0)A(0, 0) B(2, 0)

(1, 1)(0, 1)2

C(0, 2)

2 2Ö

80. The term independent of x in the expansion of 10

2/3 1/3 1/2x 1 x 1

x x 1 x x

is

(1) 4 (2) 120 (3) 210 (4) 310 Ans. (3)


Given expression =

10

1/3x 1 x 1

x 1x x 1

= 101/3 1/2x 1 1 x

= 101/3 1/2x x

General term, 10 r r10 1/23

r 1 rT C x x

For term independent of x

10 r r 03 2

20 – 2r – 3r = 0 5r = 20 r = 4 Hence the term independent of x = 10

4C = 210

81. The area (in square units) bounded by the curves y x , 2y x 3 0 , x – axis, and lying in the first quadrant is

(1) 9 (2) 36 (3) 18 (4) 274

Ans. (1) y x (1) 2y – x + 3 = 0 (2) On solving, (1) and (2) we get A = (9, 3) Hence, area of the shaded region

9

0

x dx Area of ABC

93/2

0

2 1x 6 33 2

= 9

C(3, 0)O

(9, 3)A

B(9, 0)


82. Let nT be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If n 1 nT T 10 , then the value of n is

(1) 7 (2) 5 (3) 10 (4) 8 Ans. (2) n 1 n

3 3C C 10

n n 1n 1 n 2 10

6

n(n – 1) = 20 n = 5

83. If z is a complex number of unit modulus and argument , then 1 zarg1 z

equals

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

Ans. (3) Put, z = cos isin

z 1 cos i sin1 z 1 cos i sin

= i

2

i2

2cos cos isine2 2 2

2cos cos i sin e2 2 2

= ie

so, 1 zarg1 z

84. ABCD is a trapezium such that AB and CD are parallel and BC CD . If ADB , BC = p and CD = q, then AB is equal to

(1) 2 2p q sinp cos qsin

(2)

2 2p q cosp cos q sin

(3) 2 2

2 2p q

p cos q sin

(4)

2 2

2

p q sin

pcos qsin

Ans. (1)

AB BDsin sin

2 2p q sin

ABsin .cos cos .sin

2 2

2 2 2 2

p q .sinAB q psin cos

p q p q

2 2 2 2

p qsin , cosp q p q


AB = 2 2p q sinp cos qsin

C

B A

D

pÖp + q

2

2

q

85. If 1 31 3 32 4 4

P

is the adjoint of a 3 × 3 matrix A and |A| = 4, then is equal to

(1) 4 (2) 11 (3) 5 (4) 0 Key : 2 Sol: (2) P = Adj A |P| = | Adj A| |P| = |A|2

2 6 16 11

86. The intercepts on x-axis made by tangents to the curve, 0

| | , ,x

y t dt x R which are parallel

to the line 2 ,y x are equal to (1) 1 (2) 2 (3) 3 (4) 4 Key : 1 Sol: (1)

| |dy xdx

| | 2x 2x If x = 2

222

00

22ty t dt

Point is (2, 2) Equation of tangents 2 2 2y x 2 2y x x intercept is 1


If x = –2, y = –2 Equation of tangents is 2 2 2y x 2 2y x x intercept is –1 87. Given : A circle 2 22 2 5x y and a parabola, 4 5y x Statement – I : An equation of a common tangent to these curves is 5.y x

Statement - II : If the line 5 0y mx mm

is their common tangent, then m satisfies 4 23 2 0m m

(1) Statement – I is true; Statement – II is true; statement – II is a correct explanation for statement - I

(2) Statement – I is true; Statement – II is true; statement – II is not a correct explanation for statement - I

(3) Statement – I is true; Statement – II is false (4) Statement – I is false; Statement – II is true Key : 2 Sol: (2) Eq. of tangent to 2 4 5y x is

5 y mx

m ….(1)

(1) is also the tangent of 2x2 + 2y2 = 5

then,

2

2

50 0521

m

m

m4 + m2 – 2 = 0 m = ± 1, which satisfy m4 – 3m2 + 2 = 0, and so an equation of common tangent is

y = x + 5 .

88. If 1sec tan ,y x then dydx

at x=1 is equal to

(1) 1/ 2 (2) 2 (3) 1 (4) 1 / 2 Key : 1 Sol: (1) y = sec (tan–1 x)

= 1 2sec sec 1 x

21 y x


2

1 22 1

dy xdx x

( 1)

12

at x

dydx

89. The expression tan cot1 cot 1 tan

A AA A

can be written as

(1) sin A cos A + 1 (2) secA cosec A + 1 (3) tan A + cot A (4) sec A + cosec A Key : 2 Sol: (2)

cossin sincos sin1cos cos1

sin

AA AA A

A AA

= sin sin cos coscos sin cos sin cos sin

A A A AA A A A A A

= 3 3sin cos

sin cos (sin cos )

A A

A A A A

= 1 sin cossin cos A A

A A

= sec A cosec A + 1 90. All the students of a class performed poorly in Mathematics. The teacher decided to give

grace marks of 10 to each of the students. Which of the following statistical measures will not change even after the grace marks were given ?

(1) mean (2) median (3) mode (4) variance Key : 4 Sol: (4) Variance

22 ix x

N

After given 10 grade marks to each 10i ix x

And 10x x No. change in variance.

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