Test Advanced...The final pressure in the two compartments are equal b. Volume of compartment I is 3...

PRACTICE SET 4 205

A to the Various IITs...S ingle Door Ent r y

Advanced

Paper 1

Duration : 3 Hours Max. Marks : 180

Test RIDER

Pra

cti

ce

Se

t4

Question Booklet Code 3

Please read the instructions carefully. You are allotted 5 minutes speciallyfor this purpose.

Question Paper Format

Marking Scheme

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This booklet is your question paper. Attempt all the questions.

Blank papers, clipboards, log tables, slide rules, calculators, cameras, cellular phones, pagers and

electronic gadgets are not allowed.

Write your name and roll number in the space provided on the bottom of this page.

The question paper consists of three parts (Physics, Chemistry and Mathematics). Each part consists

of three sections.

contains 10 multiple choice questions. Each question has four choices (a), (b), (c) and (d)

out of which only one is correct.

contains 5 multiple choice questions. Each question has four choices (a), (b), (c) and (d)

out of which one or more than one is/are correct.

contains 5 questions. The answer to each question is a single–digit integer, ranging

from 0 to 9 (both inclusive)

For each question in Section 1, you will be awarded for correct answer and zero mark for

unattempted questions. No negative marks will be awarded for incorrect answers in this section.

For each question in Section 2, you will be awarded for correct answer(s) and zero mark for

unattempted questions. In all other cases, minus one (–1) mark will be awarded.

For each question in Section 3, you will be awarded for the correct answer and zero mark for

unattempted questions. In all other cases, minus one (–1) mark will be awarded.

Section 1

Section 2

Section 3

2 marks

4 marks

4 marks

1. Imagine an atom made up of a proton and a hypothetical particle of double the mass of theelectron but having the same charge as the electron. The wavelength of the radiation that will beemitted when this particles jumps from first excited state to ground state (in terms of theRydberg constant R for the hydrogen atom) is equal to

a.2

3Rb.

1

3Rc.

1

2Rd. None of these

2. Let w be the angular velocity of earth’s rotation about its axis. Assume that the acceleration dueto gravity on earth’s surface has value at equator and the poles. An object weighed at the equatorgives same reading as taken at depth d below surface at pole ( )d R<< . The value of d is

a.w2 2R

gb.

w2 2

2

R

gc.

2 2 2w R

gd.

R

g

3. The potential difference between the points A and B is + 40 V. If the electric field is uniform then

the value of electric field at B

a. may be equal to 20 V/cm b. may be less than 20 V/cm

c. Both (a) and (b) are correct d. Both (a) and (b) are not correct.

4. Figure (a), shows a parallel-plate capacitor having square plates of edge a and plate-separation d .The gap between the plates is filled with a dielectric of dielectric constant K which varies parallelto an edge a, where K and a are constant and x is the distance from the left end.

K K xo= + aCalculate the capacitance?

a.e ao

o

a

dK

a2

2+é

ëêùûú

b.e ao

o

a

dK

a23

2

2

+éëê

ùûú

c.3

4 4

2e aoo

a

dK

a+éëê

ùûú

d.e ao

o

a

dK

a2

6 2+é

ëêùûú

5. A rigid body oscillates about point O. Distance between O and center of gravity is L and radiusof gyration is K, then what will be equivalent length?

a.K

Lb.

K

L

2

c. LL

K+

2

d. LK

L+

2

6. Two coherent light source radiate in phase. The distance between sources is d, where d/4 is thewavelength of light produced. The angles are measured from the line connecting the twosources. The distance from source to point of observation is significantly larger than wavelength,then

AdvancedTest RIDER

Part I

Section 1 Single Correct Option Type

10

A B2 cm

K

a

x

dx(b)(a)

a. at q = °60 , maxima at P b. at q = °30 , minima at P

c. at q = °60 , minima at P d. at q = °30 , maxima at P

7. The initial state of an ideal gas is represented by the point a on the p-V

diagram and its final state by the point e. The gas goes from the state a to the

state e by three quasi stationary processes represented by (i) abe (ii) ace (iii)

ade. The heat absorbed by the gas is

a. the same in all the processes

b. the same in processes (i) and (ii)

c. greater in process (i) than in (iii)

d. less in process (ii) than in (iii)

8. Time period of small oscillation of plate of mass m and given dimension

is

a. 2p m

kb. 2

3

2 2

2p m a b

ka

( )+

c. 2p ma

kbd. None of these

9. A particle moves with simple harmonic motion in straight line.

In 1st second, it covers d 1 and in 2nd second, it covers d 2 . What will be the amplitude of SHM

(particle start from rest)?

a. d d1 2 b.d

d1

2

c.d

d d12

1 22 -d.

2

312

1 2

d

d d-

10. A projectile is fired with velocity u at an angle q, so to strike a point on the inclined plane at an

angle a with the horizontal. The point of projection is at a distance d from the inclined plane on

the ground. The angle a is adjusted in such a way, so that the projectile can strike the inclined

plane in minimum time, find that minimum time.

a. tu u gd

gmin

tan

tan=

- -æ

èçç

ö

ø÷÷

2 2aa

b. tu u gd

gmin

sin

cos=

- -æ

èçç

ö

ø÷÷

2 2aa

c. tu u gd

gmin

sin

tan=

- -æ

èçç

ö

ø÷÷

2 2aa

d. tu u gd

gmin

cos

cos=

- -æ

èçç

ö

ø÷÷

2 2aa

PRACTICE SET 4 207

k

b

a

pre

ssu

re

volume (V)

e

a

cdb

()

p

d/2

d/2

S1

S2

P

θ

11. First overtone of open organ pipe (of length Lo ) beats with first overtone of a closed organ pipe

(of length Lc ) and beat frequency is 10 beats/s. Fundamental frequency of close organ pipe is

110 Hz and speed of sound is 330 m/s. Find out the correct choices for length Lo and Lc .

a. L c = 0.75 m b. L o = 33

34m or

33

35m

c. L o = 33

32m or 1 m d. L o = 33

32m or

33

34m

12. In a given diagram, detector D is stationary and source is oscillating with amplitude A and given

situation is mean position. Mass of block m and spring constant k. Assuming no damping, pick

correct options Tm

k=

æèç

öø÷2p

a. detector will detect maxima at intervals of T

b. detector will detect minima at intervals less than T

c. detector will detect frequency less than actual frequency for longer time interval

d. detector will detect the frequency greater than actual frequency for time intervalT A

v2

2-æèç

öø÷

,

where v speed of sound.

13. A block is connected to a spring in an elevator. Elongation in spring is x 1 4 2= mm and

x 2 3 2= mm, when it moves up and down with constant acceleration respectively. Then,

a. acceleration of elevator isg

7

2m s/

b. acceleration of elevator isg

3

2m s/

c. if same elevator moves horizontally with same acceleration then elongation will be 5 mm

d. if same elevator moves horizontally with same acceleration then elongation will be 10 mm

14. A partition divides a container having insulated walls into two

compartments I and II. The same gas fills the two compartments whose

initial parameters are given. The partition is a conduction wall which can

move freely without friction. Which of the following statements is/are

correct with reference to the final equilibrium position?

a. The final pressure in the two compartments are equal

b. Volume of compartment I is3

5

V

c. Volume of compartment II is12

5

V

d. Final pressure in compartment I is5

3

p

AdvancedTest RIDER

2p, 2V, TII

p, V, TI

m k

D

2A

Section 2 More Than One Correct Option

This section contains . Each question has four choices, (a), (b), (c)

and (d) out of which or is/are correct.

5 multiple choice questions

one more than one

15. Radiations of monochromatic waves of wavelength 400 nm are made incident on the surface of

metals Zn, Fe and Ni of work functions 3.4 eV, 4.8 eV and 5.9 eV respectively. Which of the

following is (are) correct?

a. Maximum KE associated with photoelectrons from the surface of any metal is 0.3 eV

b. No photoelectrons are emitted from the surface of Ni

c. If the wavelength of source radiation is doubled then KE of photoelectrons is also doubled

d. Photoelectrons will be emitted from the surface of all the three metals, if the wavelength of incident

radiations < 200 nm

16. A uniform vertical cylinder of cross-sectional area a floats, 90% submerged in an unknown liquid

inside a tank with cross-sectional area four times that of cylinder. When cylinder is pushed down

gently and released, it performs SHM. The maximum possible amplitude (in cm) for this SHM is

17. Massless springs, each with k = -1350 1Nm are attached

to the left and right walls as shown in the figure. A 1 kg

block is initially held against the left-hand spring,

compressing the spring by 0.1 m. The block is released

to move. The floor is frictionless except for the section AB. Coefficient of friction for AB is 0.3.

Find the distance (in cm) from point B, where block finally comes to rest.

18. The prism spectrum is spread out more at violet end than at the red end. Angular dispersion is

defined as f dl

= d

d, if deviation of small prism d is defined as d m= -( )1 0A and wavelength is

defined as ml

= +AB2

, where A and B are constant and A0 is prism angle then dispersion fl

µ 1N

.

Find value of N.

19. A body cools from 50°C to 40°C in 5 min. The surrounding temperature is 20°C. By how many °Cdoes the temperature decrease in the next 5 min? Round your answer to nearest integer.

20. A hydrogen like atom (atomic number Z) is in a higher excited state of quantum number n. The

excited atom can make a transition to the first excited state by successively emitting two photons

of energy 10.2 eV and 17.0 eV respectively. Alternatively, the atom from the same excited state

can make a transition to the second excited by successively emitting two photons of energies

4.25 eV and 5.95 eV respectively. Determine the value of n. (Ionisation energy of H-atom

= 13.6 eV)

PRACTICE SET 4 209

kk1 kg

µ = 0 µ = 0A Bµ = 0.3

2m

4 (Cross-sectional area)a

a

ll/10

l =80cm

Section 3 Integer Answer Type

This section contains . The answer to each question is a single-digit

integer, ranging from to (both inclusive).


0 9

Part II

21. Which of the following is correct pair of chiral isomers?

a. Cis [ ( ) ]CrCl O2 23X - and trans [ ( ) ]CrCl OX2 2

3-b. Cis [ ( ) ]CrCl O2 2

3X - and cis [ ( ) ]PtCl2 22en +

c. Trans [ ( ) ]CrCl O2 23X - and trans [ ( ) ]PtCl2 2

2en +d. Trans [ ( ) ]CrCl O2 2

3X - and cis [ ( ) ]PtCl2 22en +

22. CH CH

CH

CH CHCCl

Br

3

3

24

2

— —½

== ¾¾® P, Product P is

a. CH C

Br

CH

CH CH Br3

3

2 2— — — —½

½b. CH CH

CH

C

Br

Br

CH3

3

3— — —½ ½

½

c. CH C

Br

CH

CH

Br

CH3

3

3— — —½

½ ½d. CH CH

CH

CH

Br

CH

Br

3

3

2— — —½

½ ½

23. The pressure exerted by 1 mol of CO2 at 273 K is 34.98 atm. Assuming that volume occupied by

CO2 is negligible, the value of van der Waal’s constant for attraction of CO2 is

a. 3.59 dm atm mol6 2-b. 2.59 dm atm mol6 2-

c. 1.25 dm atm mol6 2-d. 1.59 dm atm mol6 2-

24. The pyrometallurgical process used for reduction of highly electropositive metal (such as

Ag, Au) is known as cyanide process. Why this process is chosen instead of carbon reduction

method?

P. Because this method is economically favourable

Q. Because carbon reduction method require high temperature

R. Because this process involves displacement reaction

S. Because this is done by using electrolysis

Choose the correct pair of statement(s) which explain the exact answer of the above question.

a. P and Q b. P and S c. P, Q and S d. R and Q

25. Acidic or basic impurity can be removed by using [ ]X . When a mixture of concentrated ore, [X ]

and carbon is heated, it produces a fusible slag. This process is known as [ ]Y

Here [X] and [Y] can be filled by

a. slag and roasting b. flux and smelting c. flux and calcination d. gangue and roasting

26. Which of the following is used for distinguishing carbinol and methyl carbinol?

a. NaOH b. Na c. NaOI d. CH COCl3

AdvancedTest RIDER


10

27. Helium crystallizes in FCC arrangement and the density of solid and liquid He is 1.59 g/cm 3 and

1.42 g/cm 3 respectively. Find the percentage of empty space in liquid He.

a. 33.92% b. 40.92% c. 38.92% d. None of these

28. Which of the following set of species have planer structures?

a. I , ICl Al Cl TeCl AlF3+

4–

2 6 4 3, , , b. I CH ClO SiF3–

3–

3–

22–, , ,

c. SCl ,N O , XeOF ,XeOF ,SF2 2 5 2 4 4 d. I Cl , XeF ,BrF , XeF2 6 2 4–

5–

29. The test used for detection of protein is

a. Silver mirror test b. Molisch test c. Biureate test d. Iodoform test

30. If p back bonding involves the vacant orbital of the central atom, then bond angle gets centered

due to

a. the increased bp-bp repulsions for the enhanced bond multiplicity

b. the decreased of lp-lp and lp-bp repulsion(s)

c. Both (a) and (b)

d. None of the above

31. The metal carbides of calcium and aluminium on hydrolysis produces

a. methane and ethane respectively b. ethyne and methane respectively

c. ethyne and ethyne respectively d. ethyne and propyne respectively

32. Which of them liberate a gas and turns lime water milky?

a. b. CH C H

COOH

COOH3 — —½

¾®D

c. d. HOOC CH COOHCa OH

—( ) —( )

2 5

2

¾¾¾¾®D

33. Which of the following represents wrong order of properties, that are written in bracket?

a. NH H AsH SbH BiH3 3 3 3 3< < < <P (acidic strength)

b. NH H AsH SbH BiH3 3 3 3 3< < < <P (reducing power)

c. NH H AsH SbH BiH3 3 3 3 3< < < <P (basicity order )

d. NH H AsH SbH BiH3 3 3 3 3< < < <P (bond angle)

34. Which of the following gives white turbidity with anhydrous ZnCl c HCl2 + onc. more faster

than

a. b. c. d.

PRACTICE SET 4 211

COOH

==O ∆

O

COOH∆





one more than one

—OH ?

OH

CH —OH2

OH

CH—CH3

OH

35. Which of the following statement is/are correct regarding [Fe(CN) ]63– complex?

a. It is an inner orbital complex b. It is an outer orbital complex

c. It is paramagnetic d. It is diamagentic

36. How many equivalent are there per mole of H S2 in its oxidation to SO2?

37. A person takes 6.1 g of an antacid tablet comprising bicarbonate ions at 20.8%. The volume of

CO2 evolved at 1 atm and 25°C in the stomach on neutralisation, multiplied by a factor of 10 will

be x L. Calculate approximate integer value of x .

38. Number of atoms bonded in cyclic fashion in the monomeric unit of nylon-6.

39. A solution of 1L has 0.6 g non-radioactive Fe 3+ with mass number 56. To this solution, 0.209 g of

radioactive Fe2+ is added with mass number 57, and the following exchange reaction occured.57 2 56 3 57 3 56 2Fe Fe Fe Fe+ + + ++ +r

At the end of 1h, it was found that 10 5- moles of non-radioactive 56 2Fe + was obtained and that the

rate of the reaction was 3.38 ´10 7- mol -1h -1 . The rate constant of the reaction is neglecting any

change in volume, calculate the activity of the sample at the end of 1 h t1 257 462/ ( ) .Fe = h and the

unit digit of answer will be

40. Calculate the solubility in 10 4- mol/L of TlN 3 in a solution prepared by shaking excess of TlN 3

and TlPO 4 . The solution contains/m mol of PO43– per 200 mL solution. K sp ( ) .TlN 3

456875 10= ´ - .

Suppose the answer comes as x then the value x /106 will be approximately

Part III

41. The number of values of y in [ , ]- 2 2p p satisfying the equation | sin | | cos | | sin |2 2x x y+ = are

a. 2 b. 3 c. 4 d. 5

42. A square ABCD is inscribed in a circle of radius 4 units. A point P moves inside the circle such

that d P AB d P BC( , ) min[ ( , ),£ d P CD d P DA( , ), ( , )] where d P AB( , ) denotes the distance of the

point P from the line AB. The area of the region covered by the point P is equal to

a. 4p b. 8p c. 8 16p - d. None of these

212 AdvancedTest RIDER






0 9

43. The sides of triangle ABC are in AP (order being a b c, , ) and satisfy2

1 9

2

3 7

1

5 5

8

2

!

! !

!

! ! ! ! ( ) !+ + =

a

b,

then the value of cos cosA B+ is

a.12

7b.

13

7c.

11

7d.

10

7

44. Let Sn be the sum of first n terms of an AP with non-zero common difference. IfS

S

n n

n

1 2

1

is

independent of n1 . The ratio of the first term and common difference is

a.1

3b.

1

2c.

1

4d.

1

5

45. If a vector r is equally inclined with the vectors a = cos $ sin $q qi j, b+ = - +sin $ cos $q qi j and c k= $ ,

then the angle between r and a is

a. cos- æèç

öø÷

1 1

2b. cos- æ

èçöø÷

1 1

3c. cos- æ

èçöø÷

1 1

3d. cos- æ

èçöø÷

1 1

2

46. The value of1

81

10

81

10

81

10

81

10

81

21

22

2

32

3

2

n n

n

n

n

n

nn

nC C C- + - + +... is

a. 2 b. 0 c.1

2d. 1

47. If a a a a1 2 3 100, , , ..., are the 100 roots of unity. The numerical value of1

5

£ £åi j

i j( )a a is

a. 0 b. ( ) /20 1 20c. 20 d. None of these

48. The number of real solution of the equation

( ) ( ) ( ) ( ) ( )x x x x x+ + + + + + + + - + =4 3 2 1 5 180 03 3 3 3 3 is

a. 0 b. 1 c. 2 d. 3

49. If D =

- +

- +

- + + + - +

1 1

1 1

2

2

2

2 2

z z

x y

zy z

x x xy y z

x z

x y z

xz

y x y

xz

( ) ( )

, where x, y and z RÎ , then

a. D = 1 b. D = 2 c. D = 0 d. D = 3

50. If f x f x( ) ( )- - =4020 0 and2009

2011

2011

2009

ò ò=x f x dx k f x dx( ) ( ) , then 10 -k will be

a. 2010 b. 2020 c. 2030 d. 2040

51. Let f x t dt x

x x

x

( ) ( | | ) ,

,= + - >

+ £

ìíï

îï

üýï

þïò 1 1 2

5 1 20 if y f x x

x x

x x= = = ì

íî

( ) | |–

if > 0

if < 0, then

a. f x( ) is not continuous at x = 2. b. f x( ) is continuous but not differentiable at x = 2

PRACTICE SET 4 213





one more than one

c. f x( ) is differentiable everywhere d. the right derivative of f x( ) at x = 2 does not exist

52. The linex y z- = + = -

-2

3

1

2

1

1intersects the curve xy c z= =2 0, if c equals to

a. 5 b. - 5 c. 1 d. - 1

53. If two tangents can be drawn to the different branches of hyperbolax y2 2

1 41- = from the point

( , )a a2 , then a can become equal to each and every point of which of the following intervals.

a. (– 2, 0) b. (0, 2)

c. (– ¥, – 2) d. (2, ¥)

54. Let a, b and c be three non-coplanar vectors and d be a non-zero vector, which is perpendicular to

( )a b c+ + . Now, d a b b c c a= ´ + ´ + ´( ) sin ( ) cos ( )x y 2 . Then,

a.a a c

a b c

× + = -( )

[ ]2 b.

d a c

a b c

× + =( )

[ ]2

c. minimum value of x y2 2+ isp2

4d. minimum value of x y2 2+ is

5

4

2p

55. sin sin3x x+ can be equal to (where x x= Îéëê

ùûú

Îsin–

, , [– , ]q, q p p2 2

1 1 )

a. - e

2b. - p

2c.

p2

d.e

2

56. Evaluate cos ln/

/x

x

xdx

2

21 3

1 3 -+

éëê

ùûú-ò

57. The minimum distance of the point (1, 1, 1) from the plane x y z+ + = 1 measured parallel to the

linex y z- = - = -2

1

3

2

4

3is k, where

3

14

kis

58. AB is a chord of the parabola y x2 8= with the vertex at A and BC is drawn perpendicular to AB

meeting the axis of parabola at C. The projection of BC on axis of parabola is equal to

59. If a b, ( )a b< are the two roots of the equation1 8

2110

2

10 102

--

=(log )

log (log )

x

x x, then find the value of

1

10

130000

2 3 2

4

( )a ba

a+ -é

ëê

ù

ûú

4 .

60. The number of real solutions of the equation 4 12 2log loglog (log )e x

e ex x= - + is






0 9

1. (a) Idea This problem is based on hydrogenspectrum. When an electron jumps from firstexcited state to ground state. Then extra energyE Em n− is emitted as a photon ofelectromagnetic radiation.

i.e.,1 1 12

2 2λ= −

RZn m

where, R is Rydberg constant and Z is atomicnumber.

∆E = Energy exchanged between states × me

12

1 12

12

22λ

= −

RZ

n n

= −

=2 111

14

32

2RR

( )

∴ λ = 23R

TEST Edge It is important according to JEEAdvanced. Every year, one question is asked fromthis topic. Students should concentrate onhydrogen spectrum, binding energy, ionisationpotential and wave function of an electron.

Remember Ionisation energy of the hydrogen is theamount of energy required to remove the electroncompletely. In ground state( )n = 1 , energy of atom is−13 6. eV and energy corresponding to n = ∞ is zero.

Hence, energy required to remove the electron fromthe ground state is 13 6. eV.

2. (a) Idea It is based on the concept of varition in thevalue of g due to rotation of the earth.

i.e., g g R′ = −λ ω λ2 2cos

Where g is the value of acceleration due togravity at a latitude.

At equator ′ = −g g Rω2

At depth, d on the pole ′′ = −

g g

d

R1

Both should be equal, then

′′ = ′g g

gR

gg

d

R1 1

2

−

= −

ω

So, the value of dR

g= ω2 2

TEST Edge This concept is important accordingto JEE Advanced. Every year, two questions areasked from this topic. Students should focus onvariation of g with altitude and depth, gravitationalpotential energy and escape velocity of particle.

Case 1 At the equator, λ = °0

then g g R′ = −λ ω2 [minimum value]

Case 2 At the pole ,λ = °90

then g g′ =λ [maximum value]

3. (d) Idea It is based on the relationE = −∆∆

∆V

rr, is the

shortest distance between the twoequipotential surfaces. Potential decreasessteeply in the direction of electric field.

So, every one must understand the relation between∆V and E.

E = −∆∆

V

r

Two cases are possible

Case 1

For this case, E = −20 Vcm

along A B→

Case 2

In this case ∆ ∆r x< , so E x < E.

∴ E > 20 Vcm

.

TEST Edge This concept is generally asked in JEEAdvanced/IIT. Student should concentrate on thisidea and relate with electric field and potentialenergy. It is given that two points A and B havingpotential difference between them is + 40 V.

PRACTICE SET 4 215

E

2 cm

BA

E

C

B

xA Ex

∆r ∆x

θθ

D

B

C

A

Ex

Ey

E

X

Y

∆y

∆r

E =−V

rAB

∆and E

V

yy

AC=−

∆

Here V VAB AC= but ∆ ∆r y<So, | | | |E Ey>

In the same way, we can take the component ofelectric field along the x-axis

ExADV

x=

−∆

| | | |E Ex <

4. (a) Idea This problem is based on a parallel platecapacitor with plate area A and separation dbetween the plates. Then, dielectric slab ofdielectric constant K is inserted in the spacebetween the plates.

i.e., CQ

V

K A

dKCo

o= = =ε

where CA

do= ε

is the capacitance without the

dielectric.

Consider a small strip of width dx at a separation x

from the left as figure (b).

This strip form a small capacitor of plate area x dx .Its capacitance is

dCK x adx

do o= +( )α ε

The given capacitor may be divided into such stripswith x varying from 0 to a.

All these strips are connected in parallel. Thecapacitance of the given capacitor is,

CK ax adx

do o

o

a

= +∫

( ) ε

= +

ε αoo

a

dK

a2

2

TEST Edge Students should concentrate on theenergy stored in a capacitor, energy density inelectric field and force between the plates of acapacitor.

Remember If instead of two plates, n uniform platesare placed at the same distance from each other andconnected successively, then capacitance of thisarrangement

i.e., Cn A

do=

−( )1 ε

5. (d) Idea This problem is asked about the concept ofmoment of inertia, torque and time period ofSHM. When the rigid body is oscillatingabout point O but it‘s weight is concentrated

on centre of gravity. Here, it is not a simplependulum, it is a physical pendulum, so firstwe have to find it‘s time period and then bycomparing it by time period formula, we canfind equivalent length.

Consider rigid body oscillates about O. Torque due togravity τ θ= −mgL sin

For small θτ θ= − ( )mg L

So, time period

TI

mg LO= 2π

where, IO is moment of inertia about O

I I mLO = +COM2

= +mK mL2 2

T

K

LL

g=

+2

2

π

So, LK

LLeq = +

2

TEST Edge The oscillation of pendulum is animportant topic in SHM. So one should practicethese types of questions as well. Let us consider anexample

A hollow bob initially filled with water is oscillating witha small hole at it’s bottom. Find the variation in it’s timeperiod, if any?

The length of simple pendulum is considered tosolve from point O to the centre of gravity of bob.Now try to solve this question.

6. (a) Idea This problem is based on Young‘s Double SlitExperiment in order to find out minima andmaxima of the interference pattern.

i.e., λ = d

4As λ is large in comparison with d so only 1st

assumption of YDSE is valid, so ∆x of any point

∆x d= =sin sinθ λ θ4

For θ = °60 , ∆x = × =43

22 3λ λ

Neither maxima nor minima.

For θ = °30 , ∆x = × =412

2λ λ maxima at P.


O

mg

θ

θ

O

TEST Edge Above concept is important speciallyfor JEE Advanced. Every year, one or twoquestions are asked. So, students should focus ondifferent cases of superposition of waves and YDSE.

Remember When a monochromatic light ofwavelength λ is incident on a circular aperture ofdiameter d, then angular width of dark fringe.

dsinθ λ= 1.22 or sinθλ

=1.22

d

Angular radius of central maximum

i.e., sin.

θλ

=122

ddiffraction pattern as shown in

figure,

Where Io is intensity of monochromatic light.

7. (c) Idea It is based on the concept of first law ofthermodynamics and thermodynamicsprocesses.

i.e., ∆ ∆ ∆Q U W= +where, ∆Q is change in heat energy, ∆U ischange in internal energy and ∆W is theamount of work done.

As initial and final condition is same that’s why ∆U willbe same for all three.

For abe, ∆ ∆ ∆Q U W= +Overall work done is positive as we can see that inupper part of graph, net volume increases.

So, ∆Q will be sum of ∆U ( )> 0 and ∆W ( ).> 0

In ae part,W = 0 so ∆ ∆Q U=In ade part,W < 0, ∆ ∆ ∆Q U W= −So, heat absorbed by gas is greater in process (i)than process (iii).

TEST Edge It is important according toJEE Advanced. Every year two or three questionsare asked in this concept. Student should relatethis idea with second law of thermodynamics ,

heat engines and Carnot cycle.

Remember

1. When work is done by the system, ∆W ispositive. If work is done on the system, ∆W isnegative.

2. When heat is given to the system, then ∆Q ispositive. If heat is given by the system, ∆Q isnegative.

3. A positive ∆W, decreases the internal energy

and positive ∆Q increase it.

8. (b) Idea This problem is based on time period ofsimple pendulum and parallel axes theorem onmoment of inertia.

i.e., TI

mgL= 2π

where, I is moment of inertia, and T is timeperiod of simple pendulum.

For small value of θ, spring force will be = k a( )θRestoring torque about hinge τ θ θ= =( ) ( )ka a ka2

So, tension on springTI

ka

H= 2 2π

IH = Moment of inertia about hinge

I I ma b

H = ++

COM

2 22

2

= + + +m a b m a b( ) ( )2 2 2 2

12 4= +m a b( )2 2

3

Tm a b

ka= +

23

2 2

2π ( )

TEST Edge This idea is generally asked in JEEAdvanced. It relates with conservation of energy inSHM, damped harmonic motion and resonance.

Remember If a simple pendulum is in carriagewhich is accelerating with an acceleration a, then

i e. . , g g a′ = −Case 1 If the acceleration a is upwards then

| |g g a′ = + and Tl

g a=

+2π

Case 2 If the acceleration a is downward, then( )g a>

i.e.,| |g g a′ = − and Tl

g a=

−2π

Case 3 In a freely falling left, g′ = 0 and T = ∞,

i.e., the pendulum will not oscillate.

9. (d) Idea This problem is based on SHM of a particle in astraight line. When a particle displaced from itsoriginal position, then to calculate value of itsamplitude.

By general equation of SHM with ω frequency

x A t= cos ( )ω

PRACTICE SET 4 217

–2 /λ d 2 /λ d– /λ d λ/d

I0

Principal maxima

Secondary maxima

minima

aθ θ

k

At t =1 s x A d= − 1

At t = 2 s x A d d= − −1 2

From equation,

A d A− =1 cos ( )ωA d d A− − =1 2 2cos ( )ω

= −A [ cos ]2 12 ω

A d d AA d

A− − = −

−

1 2

12

2 1

Ad

d d=

−2

312

1 2

TEST Edge Some important points related toSHM/oscillation.

1. The time period of SHM does not depend onamplitude.

2. As UCM and SHM are connected that is why ω(angular frequency) comes in the expression ofSHM as x A t= cos ω .

3. There is no basic difference between oscillationand vibration. In oscillation, frequency is lowwhile in vibration frequency is high.

10. (b) Idea This problem is based on projectile motion ofa particle i.e., equation of trajectory ofprojectile and to calculate the time of flight of aprojectile.

Suppose the coordinate of the point where projectilestrikes the inclined plane is ( , )x y .

Now, y u t gt= −( sin )θ 12

2 …(i)

and x t= (cos )θ …(ii)From the geometry of the figure,

d x y= + cotα …(iii)

From Eqs. (i), (ii) and (iii)

d u t u t gt= + −

( cos ) ( cos ) cotθ θ α1

22 …(iv)

For tdt

dmin,

θ= 0 which, gives tan cotθ α=

⇒ θ π α= −2

Now, substituting this value of θ in Eq. (iv)

tu u gd

gmin

sincos

=− −

2 2αα

TEST Edge Above idea is important according to JEEAdvanced. Every year, one question is asked fromthis concept. Students should focus on time offlight, range, maximum height and applications ofprojectile motion on an inclined plane.

Remember If air resist the projectile motion, then

1. Time taken by projectile during upward motion< time taken during downward motion.

2. The values of height attained and range of aprojectile decreases.

3. The projectile returns to the ground with lessspeed. At its trajectory, horizontal velocity alsodecreases.

11. (a,d) Idea The concept is based on the differencebetween the frequency of the standing

waves produced by one closed and one openorgan pipe will make a beat. Apply theformula, for the frequency of both closed andopen organ pipes and solve it.

First overtone of open organ pipe is

fN

Lv

Lv

v

Los

os

s

o1 2

22

= = × =

First overtone of close organ pipe

fv

Ls

c2

34

=

Q f f1 2 10− = ±Fundamental frequency of close organ pipe is110 Hz, so

1104

3304

= =v

L Lo

c c

Lc = 34

m = 0.75 m

Q So, its first overtone frequency must be 330 Hz andfrequency of open organ pipe should be 340 or320 Hz.

v

L Lo

o o

= =340330

⇒ Lo = 3433

m

v

L Lo

o o

= =320330

⇒ Lo = 3332

m

TEST Edge In wave, the topics of standing waves,beats and Doppler’s effect are important andquestions may come from these topics inJEE Advanced from this topic.

Remember The beat means intensity of the resultantwave will vary with time when two waves of nearlysame frequency will be superimposed. Thefrequency of the variation of beat amplitude is halfof the frequency of the variation of beat intensity.


θ0

θ θ> 1

θ1 θ

II

I

x

y

12. (a, c, d) Idea The given question is based on Doppler‘seffect. So, the observer is stationary but thesource is oscillating.

So, in the motion of oscillating source, thefrequency observed by the observer willalso vary. Just observing the motion ofsource and you can easily solve it.

Maxima will be detected when source will be at meanposition with maximum speed and after every timeinterval it will happen. Minima is detected, whensource is at maximum speed and moving away fromsource, this will also happen after every time interval.

Detector will having frequency less than actual, whensource is moving away as shown in figure.

Sound generated by source between 02

,T

will be

corresponding to less frequency.

Sound generated at t = 0, will reach D at tA

v1 =

and sound generated at tT=2

will reach D at

tT A

v2 2

3= +

So, time interval in which it hear less frequency is

t tT A

v

T2 1 2

22

− = − >

Similarly, greater frequency will be heared for time

intervalT A

v22−

.

TEST Edge Doppler’s effect is an important topicfrom sounds and waves. In doppler’s effect thereare same important points that one should notice.

1. If the source is at rest then the wavelength of thewave will not change but if the source is movingthen its wavelength and frequency (observed)both will change.

2. The motion of the medium will not affectvariation of frequency (observed), it is just therelative motion between source and observerdue to which Doppler’s effect occur.

13. (a, c) Idea The problem is based on the concept ofpseudo forces. When the body is on anaccelerating platform, then weight of thebody appears to be changed.

For upward moving elevator, kx mg ma1 − = …(i)

For downward moving elevator,

mg kx ma− =2 …(ii)

Dividing Eq. (i) by Eq. (ii), we getg a

g a

x

x

+−

= 1

2

= 4 23 2

∴ ag=7

For horizontal motion of elevator, spring will beinclined at some angle as given in figure.

kx mg3 cos θ = …(iii)

kx ma3 sin θ = …(iv)

Squaring and adding Eqs. (iii) and (iv), we get

( ) ( )kx m g a32 2 2 2= +

xm

kg a3

2 2= + …(v)

From Eqs. (i) and (v), we get

x

x

g a

g a3

1

2 2

=+

+

xg a

g a3

2 2

=+

+

, hence, x3 5= mm

TEST Edge Pseudo force is an important concept inlaws of motion and questions may come on pseudoforce in JEE Advanced.

PRACTICE SET 4 219

t = 0 t T= /2D

A A A

kx1

mg

ma

kx2

mg

ma

m

kx3 cos θ

kx3 sin θ

mg

kx3θ

θ

ma

Let us consider one example based on pseudo force.

A box is placed on the floor of the train. The trainstarts accelerating with an acceleration 5 m/s 2. Therelative acceleration of block with respect to traincan be find out.

Here, we can observed from the train’s frame, thenapply pseudo force and real friction force to solvethat problem.

14. (a,b,c,d) Idea The number of moles in each containerwill not change. The final pressures in bothcontainers will be same. Final temperaturewill also be same. Now this question couldeasily be solved.

At final equilibrium force must be equal from two side,so

p A p Af f1 2= ⇒ p pf f1 2

=

In final condition final temperature will also be samebecause conducting separator.

Conservation of moles in I and II componentspV

RT

p V

RTf= ( )1 …(i)

2 2 2p V

RT

pV

RTf( ) = …(ii)

VpV

pf1 =

VpV

pf2

4= …(iii)

V V V1 2 3+ =pV

p

pV

pV

f f

+ =43

pV

pV

f

× =5 3 ⇒ pp

f = 53

So, VV

135

= , VV

212

5=

TEST Edge Above question is based on kinetictheory of gases. In these types of questions, onemust understand if the partition is conducting thentemperature of both parts will be same, if thepartition is not fixed, then their final pressures willbe same. But concentrate on the number of moles ifthey are same or not for both the compartments.

15. (b, d) Idea The energy of photon must be greater thanthe work function of the material. If thewavelength is doubled then energy of photonwill be halved but the KEmax will not halvedbecause work function will remain constant.

Ephotonin Å)

eV= = =12400 124004000

31λ (

.

As W0 for Zn, Fe and Ni > 3.1 eV, there will be nophotoelectric emission from any surface.

To emit photoelectrons from all the three metals, λmaxshould corresponds to λmax for Ni (as it has highestW0 )

⇒ λ max (to start ejection from Ni)

= 12400

0W (eV)Å = =12400

5.92101.7 ÅÅ

If wavelength of the radiation is less than 2000 Å, thenphotoelectrons from all the metal surface will beemitted.

TEST Edge Photoelectric effect is an important partof modern physics. To solve the questions based onphotoelectric effect, it is necessary to understandthe graphs of intensity-current, current-stoppingpotential and frequency-stopping potential. Thenone should understand the photoelectricequation.

Remember In photoelectric effect, the kineticenergy of all the electrons are not same then KEmax

is obtained by surface electrons only.

16. (6) Idea As the cylinder goes inside the liquid, theliquid also comes up, so we have to see the netdisplacement of cylinder with respect to theliquid. Net displacement of the cylinder will beequal to the length of cylinder outside theliquid for maximum amplitude.

Cylinder can perform SHM only till it is partiallysubmersed. When cylinder goes down by x inside theliquid level comes up by ′x (say)

( )4a a x xa− ′ = ⇒ ′ =xx

3


p Af

p Af

V T1, V T2,

p Af1p Af2

x′

x

m

µs = 0.2

a = 5 m/s2

So, the centre of the cylinder goes down by (w.r.t. theliquid surface).

( )x x xl+ ′ = ≤4

3 10⇒ x

I≤ =340

6 cm

TEST Edge From mechanical properties of fluidssome good questions could be made. Let usconsider one such example

A block is attached with a spring (vertical) andsubmerged in a liquid. If it oscillates inside theliquid. One can ask to find the time period of theoscillation.

Here, spring force and Buoyant force both arevariable with l, so both will help in SHM.

17. (3) Idea It is based on conservation of mechanicalenergy for the two springs and thenon conservative force i.e., friction willdissipate same part of mechanical energy.

Velocity of block at A is according to energyconservation

mv kx2 2

2 2=

12

13502

2 2× = ×vA (0.1)

vA2 1350

100=

vA = 1350100

m s/

During motion of a block from A to B, we get

v v asB A2 2 2= +

[Here, a g= − = − × = −µ 0 3 10 3 2. /m s ]

= − × ×1350100

2 3 2

= −1350100

12

= −1350 1200100

= 150100

2m s/

Now with this KE, it will compress right spring.mv kxB f

2 2

2 2=

12

150100

13502

2× = × xf

xf2 150

13501

1001

900= × =

xf = = =130

10030

3.33 cm ~− 3 cm

TEST Edge Questions on this concept involvingconservative and non-conservative forces wereasked in IIT-JEE/JEE Advanced. Students shouldfocus on this concept and relate with laws of motionand its applications.

Let us consider an example

A box starts moving up a fixed inclined plane with aspeed 2 m/s. So, it has moved a distance 0.2 m alongthe inclined before coming to rest. The work doneby gravity can be find out.

Note, observe one thing that the mechanical energyis not conserved.

18. (3) Idea This problem is based on the relationbetween refractive index (µ) and the angle ofminimum deviation ( )δm .

i.e., δ µm A= −( )1

φ δλ

= d

d

This can be written as,

φ δµ

µλ

= ×d

d

d

d

δ µ= −( )1 0A

d

dA

δµ

= 0

µλ

= +AB

2 ⇒ d

d

Bµλ λ

= − 23

So, φλ

= − 2 03

BA ⇒ φλ

∝ 13

So, N = 3

TEST Edge It is important from the JEE Advancedpoint of view. Students should concentrate onprism, refraction at spherical surface, lens maker‘sformula and its magnification.

Remember It is found that the angle of deviation (δ)varies with the angle of incidence

i.e., δD r r r, 1 2= =

∴ rA=2

PRACTICE SET 4 221

K

ml

ρ

m θ = 30°

v=

2 m/s

An

gle

of

devi

atio

n

i

r r

= e

=1 2

i

Angle of incidence

So, δ δ= = + −m i i A( )

iA m= + δ

2

⇒ µ

δ

= =+

sin

sin

sin

sin

i

r

A

A

m

2

2

where, µ is a refractive index of a prism

Note For this prism, δ µm A= −( )1

19. (7) Idea It is based on the concept of Newton‘s law ofcooling in which temperature of body variesfrom T1 to T2.

i.e.,T T

tK

T TTo

1 2 1 2

2

− = + −

where t is the time in which temperature ofbody changes from T1 to T2

From Newton’s law of cooling, we getT T

tK

T TT1 2 1 2

02− = + −

50 405 60

50 402

20−×

= + −

K …(i)

405 60

402

20−

×= + −

TK

T…(ii)

Dividing Eq. (i) by Eq. (ii), we get10

4025

2−=

T T /

5 25 40 25T T= × −

T = × = °25 4030

33.33

So, ∆T ~ . .− °− = °40 33 33 6 67 ≈ °7

TEST Edge This idea is important according to JEEAdvanced. Every year, one or two questions areasked from this topic. Students should focus on thisidea and relate to perfectly black body, Stefan‘s lawand ideal gas equation.

Remember A surface or a medium which transmitsmost of the radiation is called diathermanous,substance.

e.g., dry air, rock salt etc.

A surface or a medium which does not transmitradiation at all is called opaque or a diathermanousmedium.

e.g., water, wood and solid.

20. (6) Idea This problem is based on the energy emissionof hydrogen atom and to calculate the excitedstate of an electron.

From the given conditions, we get

E En − = +2 10 2 17( . ) eV = 27.2 eV …(i)

E En − = +3 4 25 5 95( . . ) eV =10.2 eV …(ii)

So, Eq. (i) – Eq. (ii), we get,

E E3 2 17 0− = . eV

Z 2 13 614

19

17 0( . ) .−

=

Z 2 13 6536

17 0( . ) .

=

Z Z2 9= or = 3From Eq. (i) we get

Zn

2213 6

14

127 2( . ) .−

=

3 13 614

127 22

2( . ) .−

=n

14

10 2222− =

n.

10 02782n

= .

n2 36=n = 6

TEST Edge It is important according to JEEAdvanced. One question is asked in every year. So,students should concentrate on Bohr’s model,energy of hydrogen atom, hydrogen spectra andbinding energy.

Remember Total energy En, is the sum of the kinetic

and potential energies.

i.e., E K Ume

n hn n n= + = − 4

02 2 28ε

or En

n =−13 6

2

. eV

where, n is the shell number.

21. (b) Idea This problem includes conceptual mixing ofstructure and isomerism in coordinationcompound. While solving the problem,students are advised to draw the structures andmirror images of given isomers of coordinationcompound. Then analyse them either by thegiven pairs are superimposable or not, if theyare non-superimposable mirror image thenboth are optically active as well as chiral.

Keep in mind that almost all cis isomers areoptically active.

This problem involves the concept of structure andisomerism in coordination compound.

Structure of coordination compound


Cr

3 –OX

OX

Cl

Cl

Cr

OX

OX

Cl

Cl

OX

Cis X[CrCl (O ) ]2 23– Cis X[CrCl (O ) ]2 2

3–

Both are non-superimposable mirror image isomers.Hence, it will show optical activity and chirality.

Both isomers are non-superimposable mirror imageisomer, hence are optically active and will showchirality.

Overall, we can say all isomers of coordinationcompound will show optical activity due tonon-superimposable mirror relation with each other.

TEST Edge Similar problems based on the concept ofanalytical study of coordination compound andisomerism in coordination compound can also beasked in JEE Advanced, such as

What is the type of isomerism exhibited by the productwhen cis-diaminodichloronickel(I) bromide adds up withaqueous solution of silver bromide?

After following proper stepwise approach to solvethis question i.e., writing structural formula ofcomplex ion on reaction of coordination compoundwith silver bromide, you will get optical isomerismas answer.

22. (d) Idea This problem is based on the concept ofaddition reaction to alkene, while solving thisproblem students are advised to go throughproper mechanism i.e., electrophilic additionreaction to alkene via formation of cyclicbromonium ion.

Br Br Br BrNu

—( ) ( )

→ +++ −E

s

CH CH

CH

CH CH Br3

3

2— —

== + →⊕

CH CH

CH

CH CH

Br

Br3

3

2— — —

→

⊕

s

CH CH

CH

CH CH3

3

2— — —

Br Br

TEST Edge In JEE Advanced this type of problemsare included to judge the knowledge of students,regarding mechanism of organic reaction.

Similar problems based on nucleophilic additionreaction, Markovnikov addition reactions can alsobe asked.

e g. . , What will be the product when 2-methyl prop-1-eneis treated with deuteriumbromide in presence ofhydrogenperoxide?

This problem can be solved by using the concept offree radical addition reaction throughanti-Markovnikov’s addition and one can getanswer as tertiary butyl bromide.

23. (a) Idea This problem is based on conceptual mixingof van der Waal’s equation and calculation ofvan der Waal’s constant. The problem can besolved by undergoing following sequentialapproach.

• Write van der Waal’s equation and put thevalue of b as zero then calculate the value ofV using concept of quadratic equation.

• Calculate the value of a using considerationthat V is constant at given T and p followedby putting the value R, T and p.

pa

VV b RT+

− =2 ( )

∴ pa

VV RT+

=2

pV a

VV RT

2

2+

⋅ =

V p RTV a2 0− + =

VRT R T pa

p=

± −2 2 42

SinceV is constant at given p andT ,V can have onlyone value or discriminant = 0

∴ R T pa2 2 4= or aR T

p=

2 2

4

a = ××

( . ) ( ).

0 821 2734 34 98

2 2

= 3 59. dm 6 atm mol − 2

TEST Edge Similar problem based on concept of vander Waal’s equation, gas laws and unittransformation can also be asked, students are alsorecommended to go through in depth study of thesetopics. While solving these type of problems,students are advised to be careful regarding unitconversions.

24. (a) Idea This problem includes reduction of metal orewith its economic value. While solving theproblem students are advised to go through theconcept used in cyanide process and

characteristics of cyanide process.

The reduction of metal by carbon requires hightemperature which is economically not viable but thereduction using cyanide process requires lowtemperature and hence economically favourable.

Hence, statements P and Q correctly explain theexact reason.

PRACTICE SET 4 223

Pt

2+ 2+

en

Cl

Cl

Pt

Cl

Cl

Cis [Pt Cl (en) ]2 22+Cis [PtCl (en) ]2 2

2+

enen

en

TEST Edge Similar process including the concept ofcyanide process, along with its drawbacks,advantages using chemical transformations canalso be asked in JEE Advanced.

25. (b) Idea This problem includes purification of metalusing carbon reduction method and concept ofsmelting. While solving problems, students aresuggested to keep in mind the concept ofcarbon reduction method and smelting.

Impurities present in any ore may be acidic or basic.Impurity combines with flux to produce fusible slag,flux is chosen depending upon behaviour of impurity.If impurity is basic in nature then acidic flux is chosenand if impurity is acidic then basic flux is chosen.

Acidic flux SiO P O B O2 2 5 2 3, , etc.

Basic flux CaO, MgO etc.

Flux + Impurity → Fusible slag

This process of removing impurity by means ofconversion of impurity to fusible slag on treatmentwith flux is known as smelting.

TEST Edge Similar problems based on concept ofleaching, roasting, Baeyer’s process, Serpeck’sprocess, Hall and Heroults process can also beasked in JEE Advanced. In general the purpose toask these type of question is to judge the knowledgeregarding various viable metallurgical process andeconomical value of separate processes. Let us seeone such problem

Leaching of Ag S2 is carried out by heating Ag S2 with adilute solution of

(a) NaOH

(b) HCl

(c) NaOH in presence of O2

(d) NaCN only

NaOH in presence of O2

26. (c) Idea This problem is based on concept of namingof alcohol and haloform test. While solving theproblem students are suggested to keep inmind the key points (characteristics) ofhaloform reaction. Student must be carefulduring answering such question that substrateundergoing haloform test must have secondaryalcoholic group.

(NaOI) withCH OH3Carbinol

CH — CH — OH3 2methyl carbinol

↓

it does not givehaloform reaction.

↓

It gives haloformreaction with NaOl.

TEST Edge In JEE Advanced, these type of questionsare asked to judge the knowledge of studentsregarding the naming of organic compound andtest to distinguish them.

27. (a) Idea This problem is based on concept of FCCcrystal lattice structure, packing fraction andempty space determination. While solving theproblem, students are advised to calculate thepacking fraction of liquid He. By equating ratioof packing fraction of liquid He to density ofliquid He and packing fraction of solid He todensity of solid He first. Then calculate emptyspace by using formula

empty space =100−packing fraction of liquid He

Packing fraction ofLiquid HeDensity of Liquid He

= Packing fraction of solid Hedensity of solid He

x

142074159.

..

= ⇒ x = 66 8. %

Empty space = (100 – 66.08) = 33.92%

TEST Edge In JEE Advanced, these type of problemare asked to judge the basic subject knowledge aswell as analytical thinking of students. So whilesolving such problems in JEE Advanced, studentsare advised to keep their analytical thinking as wellas basic need to solve the problem in front of him.Similar problem based on packing efficiency, size ofvoid, can also be asked in JEE Advanced e.g.,

Which of the following expression represent the packingfraction of NaCl correctly if ions along an edge diagonalare absent?

By using the ratio of volume of effective number ofcations and anions to volume of unit cell. We can getanswer as

P.F =

4

3r r

r

+3

–3

–3

π 5

24

16 2

+

28. (b) Idea This problem is based on concept ofhybridisation and structure of inorganicspecies. While solving this problem, studentsare advised to keep in mind the concept ofVSEPR theory and bent rule.

Only set (d) has all planar structures

Look at the atoms only, all the atoms in abovestructures are in same plane, Hence set (d) denoteplanar species.


Xe XeBr

I I

F F F

F F

F F

F F

F F

Linear Pentagonal pyramidalsquare pyramidal

Cl ClCl

Cl ClCl

sp d3 2 planar

TEST Edge Similar problems based on concept ofisoelectronic, isostructural and isolobal species canalso be asked in JEE Advanced, such as

Which is the correct pair of isoelectronic and isostructuralspecies.

(a) BF andCF3 3

s

(b) NO CO3–

32–,

(c) NH and PH3 3 (d) BF NH4–

4+,

After using concept of isoelectronic andisostructural species one can get answer as (b).

29. (c) Idea This problem is based on the principleinvolved in detection of protein and can besolved by using test for the detection ofpeptide bond.

Biureate test is used for test of protein.

All compounds containing — —C

O

H

N peptide bond

gives this test.

Protein →∆

CuSO NaOH4 ,Violet colour

TEST Edge Similar problems based on distinguishingtest for primary, secondary and tertiary amine canalso be asked. So, students are advised to gothrough study of these topics and remember the keypoint used to distinguish these three.

30. (c) Idea This problem is based on conceptualmixing of back bonding, VSEPR theory andbent rule. While solving this problem studentsare advised to look at the problem withthinking that how many reasons are possiblefor bond angle to be centered.

Since during back bonding transfer of electron takesplace from filled orbital to vacant orbital which createa double bond character between linking atoms.The linking between atoms create double bondcharacter mainly due to two reasons(a) increased bp-bp repulsion and(b) decreases in lp-lp repulsion and lp-bp repulsion

TEST Edge Similar problems based on concept of πback bonding to explain reactivity, adduct formingability structural deformation etc can also be asked.So students are advised to go through in depthstudy of these topics in comparative manner.

31. (b) Idea This problem is based on hydrolysis ofdifferent metal carbides.

Aluminium carbide i e. ., Al4 3C on hydrolysis producesmethane.

Al C H O Al OH CH4 3 2 3 44 3+ → +( )Calcium carbide i e. ., CaC2 on hydrolysis producesethyne.

CaC H O Ca OH CH CH2 2 2+ → + ≡≡( )

TEST Edge Similar problems based on nature of byproduct on hydrolysis of carbides and theirchemical reaction can also be asked, so students areadvised to go through the study and analysis ofchemical reaction involved in hydrolysis of metalcarbides and metal halides, such as

A tetrachloride of molecular mass 170 when undergoeshydrolysis produces

(a) silicon (b) silicone

(c) silicate (d) silicic acid

The answer will be silicic acid.

32. (a,b,d) Idea This problem is based on concept ofdecarboxylation of carboxylic acid. Whilesolving this problem students must have twokey points that which gas turns lime watermilky and which molecule will produce thisgas on heating.

Beta keto acid give CO2 on heating.

α α, -dicarboxylic acid gives CO2 on heating.

CH C

COOH

H — COOH CH CH — COOH32

3 2– CO→

CHCH CH

O

OH

CH CH C

O

OH2

2 2

2 2

— — —

— — —

C Ca(OH)2

→

CHCH CH C

O

O

CH CH C

O

OC2

2 2

2 2

— — —

— — —

→aCaCO3∆-

According to Bridth rule does not evolveCO2.

TEST Edge Similar type of problems includingconceptual mixing of nature of product and thechemical properties of product can also be asked,such as

When an organic dicarboxylic acid is treated withcalcium hydroxide, its salt is produced which whenheated produces a cyclic compound [ ]X having molecularformula C H O6 10 . X on heating with dilute alkaliproduces Y. What are X and Y?

PRACTICE SET 4 225

O

COOH∆

—CO2

O

O

COOH

==O

After making of all steps required one can getanswer as

33. (c,d) Idea Problem is based on conceptual mixing ofacidic strength, reducing power, basicity orderand bond angle of hydrides of nitrogen family.While solving these type of problems studentsmust have the knowledge of acidic strength,reducing power, basicity order and bent rule.

This problem includes concept of variouscharacteristics of hydrides of nitrogen.

NH3 • removal of H capacity increases

PH3 • acidic strength increases

AsH3 • basic strength decreases

SbH3

BiH3

• bond angle decreases

(H—A—H bond angle)

TEST Edge Similar problems based on characteristicsof hydrides of chalcogens, icosagens can also beasked.

34. (a,b,d) Idea This problem is based on concept of Lucas,test, while solving this problem students areadvised to understand the key conceptthrough which alcohols undergo formationof turbidity during Lucas test.

Rate of reaction towards Lucas reagent ∝ stability ofcarbocation

TEST Edge Similar problems based on the concept ofvictor Mayer test, oxidation test, dehydration testetc. to distinguish 1°, 2° and 3° alcohol can also beasked.

35. (a,b,d) Idea This problem involves conceptual mixingof hybridisation of coordination compoundand their magnetic properties.

Hints Hybridisation in [ ] –Fe(CN)63 complex ion

GSVSEC of Fe in [ ] –Fe(CN)63

Hybridisation = d sp2 3

Here, there is one unpaired electron, hence it is aparamagnetic complex.

TEST Edge Similar problems based on conceptualmixing of hybridisation, isomerism, magneticproperties and characteristics of coordinationcompounds can also be asked in JEE Advanced.

36. (6) Idea Problem is based on concept of oxidation andchange in oxidation number. While solving thisproblem, students are advised to go throughcalculation of change in oxidation number thento calculate number of moles of H S2 used.

H S SO22

24–

→+

Change in oxidation number = +4 – (– 2) = + 6

∴ 1 mole H S2 6= equivalents of H S2

TEST Edge Similar problems based on redox reactionand its application can also be asked. So studentsare advised to go through study of redox reactionand its application in calculation of equivalentweight.

37. (7) Idea This problem is based on stoichiometry andgeneral concept used in quantitativecalculations. While solving this problem,students are advised to follow the givenstepwise approach

Calculate number of moles of CO2 first

Calculate total value of CO2 produced

HCO CO3 2– →

mole of HCO36161

20 8100

0 0286– . ..= × = mole of CO2

1 mole of CO2 at 25° and 1 atm = 24.4 L

∴ 0.0286 mol CO2 24 4 0 0286= ×. .

= =0 6947 07. . L

VCO L2

07 10 7= × =.

TEST Edge Similar problems based, on conceptualmixing of gaseous law, stoichiometry, limitingreagent can also be asked.

38. (7) Idea This problem is based on the concept ofmonomeric unit of polymer Nylon 6. To solvethese types of problems, students must becareful because the question is asked aboutnumber of atoms present on cycle of monomerunit i.e., excluding H. Hence count themembers of the cycle including N andexcluding H and O.

TEST Edge Similar problems, with slightmodification can also be asked in JEE Advanced, sostudents are advised to go through the study ofmonomeric unit present in various polymers.

39. (2) Idea This problem is based on concept of kineticstudy of radioactive reaction. While solvingthis problem, student must have knowledge ofrate and activity of radioactive decay.


O

X = Y =

O

s s

× × × × × × × × × × × ×

CN

3d 4s 4p

CN CN CN CN CN

H

NO

Caprolactame

Initial moles of non-radioactive56Fe3+ = 0 6

560 0107

..=

Initial moles of radioactive 57 2 0 20957

0 0036Fe + = =..

Moles of non-radioactive 56 2Fe + = −10 5

R k= + +[ ] [ ]57 2 56 3Fe Fe

3.38 ×10− 7= −10 2 (0.0036 − − −−10 0 0107 105 5x ) ( . )–

where x is the amount of 57 2Fe + decayed in one hourdue to radioactivity

x = × −4 4 10 4. moles(0.0036 – x y− ) = 0.0036 e c− λ 1

(where y is the number of moles of nucleidisintegrated from 57 3Fe +)

λ = =0 693 4 62 015. / . . h−1

⇒ y = × −714 10 5. molesActivity = λ N = − × −015 0 0036 4 4 10 4. ( . .

− × −714 10 5. ) × ×6 023 1023.A = ×274 1020. dis/h

Hence, the unit number is 2.

TEST Edge Similar problems including conceptualmixing of radioactive disintegration, rate law ofradioactive decay, half life period, average lifeperiod etc. can also be asked, such as

X 227 has a half life of 22 years, the decay of X 227 followstwo parallel paths

What will be the value of decay constant for Th and Frrespectively?

Ans. λ1 = 0.00063 yr−1

λ 2 =0.03087 yr−1

Which comes via taking the ratio of yields.

40. (2) Idea This problem is based on solubility andsolubility principle. While solving thisproblem, students must have the knowledge ofsolubility product principle.

TlN l3 3= ++ −T N

(s+ 1.5 ×10 − 2) s

Tl ls

3 41 5 10

43

5 10

32 3

PO T POr+

× +

−

×− −

+( . ) ( )

For TlN3 , s s( . ) .+ × = ×− −15 10 5 6875 102 4

⇒ s s2 40 015 5 6875 10 0+ − × =−. .

⇒ s =− ± + × −0 015 0 015 2275 10

2

2 4. ( . ) .

= 175 × 10− 4 mol/LAs per the given condition

1.75 1.75× = ≈− −10 10 24 6/

TEST Edge In JEE Advanced, generally similartypes of problems with slight conceptual mixing ofelectrochemistry can also be asked.

The value of the molar solubility of Ag CO2 3 in 0.1 M.Na CO2 3 will be ............... . If

Ksp = × −16 10 13

On solving you will get = 4 10–6×

41. (c) Idea Here apply concept of modulus function,

periodic function such as y x= =| |x x

x xif > 0

if− < 0

sin [ , ]x ∈ − 1 1 and cos [ , ]x ∈ − 1 1 and draw thegraph of y x=|sin |and y x x= +|sin | |cos |2 2

Given function is

f x x x( ) | sin | | cos |= +2 2

Now,

f x x xπ π π4

24

24

+

= +

+ +

sin cos

= +| cos | | sin |2 2x x

= f x( )

Thus, the function is periodic with periodπ4

.

Let y x x= +| sin | | cos |2 2

y x x x x2 2 22 2 2 2 2= + +sin cos | sin | | cos |

= +1 4| sin |x

Now, | sin | [ , ]4 0 1x ∈1 4 1 2+ ∈| sin | [ , ]x

y 2 1 2∈ [ , ]

y ∈ [ , ]1 2

Thus, | sin | | cos | [ , ]2 2 1 2x x+ ∈and | sin |y ≤1

So, solution is possible if both the sides are equal to 1.

From the graphs, solutions are

x = ± ±π π2

32

,

∴ Number of solutions are 4.

PRACTICE SET 4 227

λ1

λ2λ

X

227Th % yield = 2227

Fr % yield = 98223

π/2 3 /2ππ 2πy x= |sin |

1

π/4 3 /4π πf x x x( ) = |sin 2 | + [cos 2 ]

y

xπ/2

1

√2

0

TEST Edge Domain and range of algebraic functions,greatest integer function and solution oftrigonometric function, related questions are asked.To solve such type of questions, students areadvised to understand the concept of the functionssuch as, the function f R R: → defined as

f x

x

xx

x( )

| |

=≠

=

for

for

0

0 0

is called the signum function

42. (a) Idea To solve this problem, use the concept ofarea of given region in 2D, such as the area of

sector ( )OABO isθ π

3602r

If the pointP lies in the sector created by ∠ AOB, then

d P AB d P BC( , ) ( , )≤d P AB d P CD( , ) ( , )≤d P AB d P DA( , ) ( , )≤

⇒ d P AB d P BC( , ) min [ ( , ),≤ d P CD d P DA( , ), ( , )]

Hence, the required area is equal to sector created

by ∠ =

AOB

π2

.

So, the required area = Shaded area

= = =π π πr 2 4 2

424

4( )

TEST Edge Generally in JEE Advanced, area of thegiven region such as area of sector, area of polygonrelated questions are asked to solve such type ofquestions. Students are advised to understand theconcept of two dimensional geometry, such as, thearea of a polygon of n sides with vertices.A x y A x y A x yn n n1 1 1 2 2 2( , ), ( , )... ( , ) is

= + + + − + + +1

21 2 2 3 1 1 2 2 3 1[{( ... ) ( ... )}]x y x y x y y x y x y xn n

43. (a) Idea Here apply the concept of binomialcoefficient, cosine formula and AP to solve this

problem. Such as nr

nn rC C

n

r n r= =

−−!

!( )!, In

∆ ABC, cos , cosAb c a

bcB

a c b

ac= + − = + −2 2 2 2 2 2

2 2

and if a b c, , are in AP then 2b a c= +

21 9

23 7

15 5

!! !

!! !

!! !

+ +

= +

+

210

101 9

103 7

110

105 5!

!! !

!! ! !

!! !

= + +

2

101

1010

110

3

10

5! !C C C

= + +110

2 2101

103

105!

[ ( ) ( ) ( )]C C C

= + + + +110

101

109

103

107

105!

[ ]C C C C C

[Q nr

nn rC C= − ]

= = =−110

22

108

210 1

9

!( )

! ( )!

a

b

⇒ 210

22

9 3

! ( )!=

a

b

⇒ 9 3= a, 2 10b = ⇒ a = 3, b = 5

Since a b c, , are in AP

2b a c= +10 3= +c

c = 7

cos Ab c a

bc= + − = + −

× ×=

2 2 2

225 49 9

2 5 71314

cos Bc a b

ca= + − =

2 2 2

21114

cos cosA B+ = + = =13 1114

2414

127

TEST Edge Binomial expression, properties oftriangle and AP, GP related question are asked. Tosolve such type of questions, students are advisedto understand the concept of binomial expressionproperties of triangle also acquainted yourself withthe properties of triangle such as, the area of triangle

is ∆ = =1

2

1

2bc A ca Bsin sin

= = − − − = =1

2 4ab c s s a s b s c

abc

Rrssin ( )( )( )

where R and r are the radii of the circumcircle andin circle of the ∆ ABC respectively.

44. (b) IdeaQ The first term of an AP be a and common

difference be d, then Sn

a n dn = + −2

2 1[ ( ) ] and

apply the condition forSn n

Sn1 2

1

is independent of

n1.

Given that Sn is sum of first n terms of an AP withnon-zero common difference.

So, Sn n

A n n dn n1 21 2

1 222 1= + −[ ( ) ]

and Sn

A n dn11

122 1= + −[ ( ) ]


BA

O

CD

AB

O

θ

⇒S

S

n nA n n d

nA n d

n n

n

1 2

1

1 21 2

11

22 1

22 1

=+ −

+ −

[ ( ) ]

[ ( ) ]

⇒S

Sn

A d n n d

A d n d

n n

n

1 2

1

21 2

1

22

= − +− +

( )( )

S

S

n n

n

1 2

1

will be independent from n1, if 2 0A d− =

⇒ 2A d= ⇒ A

d= 1

2⇒ 1 : 2

TEST Edge In JEE Advanced, properties of AP, nth

terms of an AP, arithmetic mean related questionsare asked. To solve such type of questions, studentsare advised to understand the concept of AP suchas, if common difference d, number of terms n and

the last term of AP are given then Sn

l n dn = − −2

2 1[ ( ) ]

45. (b) Idea If three vectors a, b cand are mutually

perpendicular to each other then

a. b b.c c.a= = = 0

Q a. b a b=| || |cosθGiven, a i j= +cos $ sin $θ θ

b i j c k= − + =sin $ cos $, $θ θ

It can be easily observed that,| | | | | |a b c= = =1 and a b b c c a⋅ = ⋅ = ⋅ = 0

If r is equally inclined with the mutually perpendicularvectors a i j= +cos $ sin $θ θ , b j j= − +sin $ cos $θ θ andc k= $, then r must be of type r a b c= + +λ ( ).

Now, substituting the values of a,b,c in above, we getr a b c= + +λ ( )

= − + + +λ θ θ θ θ[(cos sin )$ (sin cos ) $ $ ]i j k

To find angle between a and r, we find a ⋅ r.

∴ cos ( , )r ar a

r a= ⋅ =

| | | |13

∴ ( , ) cosr a =

−1 13

TEST Edge Properties of dot products of two vectorssuch as the vectors are perpendicular to each otherand angle between two vectors related questionsare asked. To solve such type of questions, studentsare advised to understand the concept of dotproduct of two vectors. Such as, the component of a

vector b along to vector a isa. b

|a|a2

46. (d) Idea

Q ( ) ...1 0 1 22

33− = − + − + +x C C x C x C x C xn n n n n n

nn

and nrC

n

r n r=

−!

!( )!. Now use the relation

between binomial coefficient to solve thisproblem.

Given that

181

1081

1081

1081

1081

21

22

2

32

3

2

n n

n

n

n

n

nn

nC C C− + − ....

So, we can write it as follow

⇒ 181

10 10 1020

21

22

2 23

3n

n n n nC C C C[ − + −

+ +... ]22

210nn

nC

⇒ 181

1 10 2n

n[ ]−

⇒ ( )−981

2n

n⇒ 81

811

n

n=

TEST Edge In JEE Advanced, general term,greatest term and the properties of binomialcoefficient related questions are asked. To solvesuch type of questions, students are suggested thatto understand the concept of binomial coefficientsuch as, the coefficient of x y zn n n1 2 3 in the expansion

of ( )x y z n+ + isx

x x x

!

! ! !1 2 3

where n n n n= + +1 2 3

47. (a) Idea Z i n= + =cos sin ( ) /θ θ 1 1

then, Z = (cos 0 + i sin 0)1/ n

( ) cos sin ,/12 0 2 01 n r

ni

r

n= + + +

π π

r = 0 1, , 2...( )n− 1

Z e r ni r

n= = −2

0 1 1π

, , ......( )

Z = 1, ei n2π/ , ei n4π/ e i n n( ( )/ )2 1−

then 1, α ,α α2 3, ...α n−1 where α π= ei n2 /

are the nth roots of unity

then, 1 02 1+ + + =−α α α n

α α α α αi j5 5

15

25

1005 21

2∑∑ = + + +[( ... )

− + + + +( ... )]α α α α110

210

310

410

= − =12

0 0 0[ ]

Because ( ... ),

,α α α1 2 100

100 1000 100

P P P P K

P K+ + + =

=≠

ifif

TEST Edge Cube roots of unity, properties of nth

roots of unity related questions are asked. To solvesuch type of questions understand the concept ofcomplex number such as, the product of nth roots of

unity i e. . , 1.α α α. ... ( )2 1 11n n− −= − .

48. (b) Idea To solve this problem, apply the concept ofnumber of real solution in theory of equationand also use the concept of AM such as ifa a a1 2 5... are positive real numbers thena a a a a1 2 3 4 5

5

+ + + +will be AM of these

numbers.

PRACTICE SET 4 229

( ) ( ) ( ) ( )x x x x+ + + + + + +4 3 2 13 3 3 3

+ − + =( )x 5 180 03

First AM of { , , , , }x x x x x+ + + + −4 3 2 1 5

⇒ x x x x x+ + + + + + + + −4 3 2 1 55

= +5 55

x

⇒ 5 15

1( )x

x+ = +

Let y x= +1

Now, ( ) ( ) ( )y y y y+ + + + + +3 2 13 3 3 3

+ − + =( )y 6 180 03

( )A B A B A B AB+ = + + +3 3 3 2 23 3

y y y y y y3 2 3 227 27 9 12 6 8+ + + + + + +

+ + + + + + −y y y y y3 2 3 31 3 3 216

+ − +108 18 1802y y

5 150 03y y+ =

5 30 02y y( )+ =

y 2 30 0+ ≠

So, 5 0y = y = 0

y x= +1, so x + =1 0 or x = −1

TEST Edge Common roots, symmetric functions ofthe roots, nature of roots with respect to two realnumbers related questions are asked. To solve suchtype of questions, students are advised tounderstand the concept of quadratic equations andexpressions such as, if a < 0, then the quadraticexpression y ax bx c= + +2 has no least value but it

has greatest value4

4

2ac b

a

−at x

b

a= −

2.

49. (c) Idea To solve this problem use the concept ofproperties of determinants. If each element of acolumn or row of determinant is zero, then itsvalue is zero.

Given that

∆ =

− +

− +

− + + + − +

1 1

1 1

2

2

2

2 2

z z

x y

z

y z

x x x

y y z

x z

x y z

xz

y x y

xz

( ) ( )

Multiply C1 by x C, 2 by y and C3 by z, we get

∆ =

− +

− +

− + + + −

1

2xyz

x

z

y

z

x y

zy z

x

y

x

z

xy y z

xz

y x y z

xz

( )

( )

( ) ( ) (x y y

xz

+ ⋅)

Now, C C C1 2 3+ + → C1

∆ =

+ − +

− + + +

− − + + + − −

1

22 2 2xyz

x y x y

zy z y z

xy yz yx y yz xy y

x

( )

( )

z

y

zy

xy x y z

xz

x y

zz

xx y y

xz

( )

( )

( )+ +

− +

− + ⋅2

∆ =

− +

+ + − +

0

0

02

y zx y

z

y xz

xy x y z

xz

x y y

xz

/( )

/

( ) ( )

If there is any one column is 0, then determinant willbe zero.

So, ∆ = 0

TEST Edge Product of determinants, solution oflinear equation by determinants related questionsare asked. To solve such type of questions, studentsare advised to understand the concept ofdeterminants, such as

Let ∆( )( ) ( )

( ) ( )x

a x b x

a x b x= 1 1

2 2

then ∆′ = ′ ′+

′ ′( )

( ) ( )

( ) ( )

( ) ( )

( ) (x

a x b x

a x b x

a x b x

a x b x1 1

2 2

1 1

2 2 )where

dash ( )′ denotes derivative with respect to x.

50. (b) Idea f x dx f a b x dx kf x dxa

b

a

b

a

b( ) ( ) , ( )= + −∫ ∫ ∫

= ∫k f x dxa

b

( )

f x dx F b F a ca

b

( ) ( ) ( )∫ = − + also apply the concept

of integration to solve this problem.

Let I x f x dx x= = + −∫ ∫2009

2011

2009

20112009 2011( ) ( )

f x dx( )2009 2011+ −

⇒2009

20114020 4020∫ − −( ) ( )x f x dx

From propertya

b

a

bf x dx f a b x dx∫ ∫= + −( ) ( )

[Q f x f x( ) ( )= −4020 ]

⇒2009

20114020∫ −( ) ( )x f x dx

⇒ 40202009

2011

2009

2011

∫ ∫−f x dx x f x dx( ) ( )

= −∫40202009

2011f x dx I( )


2 40202009

2011I f x dx= ∫ ( )

⇒ = − ∫I f x dx20102011

2009( )

⇒ I k f x dx= ∫2011

2009( )

∴ k = − 2010

⇒ 10 2020− =k

TEST Edge Evaluation of definite integral bysubstitution, properties of definite integral anddifferentiation under integral sign related questionsare asked. To solve such type of questions, studentsare advised to understand the concept of definiteintegral such as, for any two functions f x( )and g x( ),integrable on the intervals [a b, ], the Schwarz -Bunyakovsky in equality holds.

f x g x dx f x dx g x dxa

b

a

b

a

b( ) ( ) ( ) ( )∫ ∫ ∫≤ −2 2

51. (a, d) Idea

f x dx f x dx f x dxa

c

c

b

a

b( ) ( ) ( )= +∫ ∫∫ , if y f x x= =( ) | |

is a modulus function then,| |xx x

x x=

>− <

if

if

0

0. A

function f x( ) is said to be continuous at x a= ifLHL = RHL = f a( ) and a function is said todifferentiable if LHD = RHD.

For x > 2

f x t dt t dtx

( ) ( | | ) ( | | )= + − + + −∫ ∫0

1

11 1 1 1

= + − + + −∫ ∫0

1

11 1 1 1{ ( )} ( )t dt t dt

x

= − +∫ ∫0

1

12( )t dt t dt

x

= −

+

2

2 2

2

0

1 2

1

tt t

x

= −

− −

+ −

2

12

0 02

12

2

( )x

= + − = +32 2

12 2

12 2x x

∴ f xx

x

x x

( ) ,

,.= + >

+ ≤

2

21 2

5 1 2

f ( )242

1 3+ = + =

f ( ) ( )2 5 2 1 11− = + =∴ f x( ) is not continuous at x = 2.

Now, f ( )2 11=f ( )2 3+ =

i e. ., RHL ≠ f ( )2

∴ RHD at x = 2 does not exist.

TEST Edge Continuity of a function at a point,continuity in an open interval, continuity in a closedinterval and continuity of composite functions,related questions are asked.

To solve such type of questions, students areadvised to understand the concept of continuity of afunction. Such as, a function f x( ) is said to becontinuous in an open interval ( , )a b if it iscontinuous at each point of ( , )a b .

52. (a,c) Idea To solve this problem, first of all find thepoint of intersection then put these values inthe curve xy c= 2

For the points, where the line intersects the curve, wehave z = 0.

∴ x y− = + = −−

23

12

0 11

⇒ x y− = + =23

12

1

⇒ x − =23

1,y + =1

21

⇒ x = 5, y =1

Putting these values in xy c= 2, we have

5 2= c

⇒ c = ± 5

TEST Edge Coplanar lines, the lines areperpendicular, parallel, equation of a line in variousforms related questions are asked. To solve suchtype of questions, students are advised tounderstand the concept of line in 3-D such as, theangle between the two lines r a b= +1 1 and

r a b= +2 2 is given by cos.

| || |θ = b b

b b1 2

1 2

.

53. (c,d) Idea Draw the graph of hyperbolax y2 2

1 41− = and

draw the tangent to it. The linesx

a

y

b± = 0 are

the asymptotes to the hyperbolax

a

y

b

2

2

2

2 1− = .

P( , )α α 2 lies on the parabola y x= 2

PRACTICE SET 4 231

Hyperbola

x2

1y2

4– = 1

Hyperbola

x2

1y2

4– = 1

O

Asym

pto

tes

=2

yx

y

x

Parabola=y

x2Parabola =y x2

P a, a( )2P a, a( )2

If two tangents can be drawn to the hyperbolax y2 2

1 41− = from the pointP( , )α α 2 , thenP( , )α α 2 must

lie outside the hyperbolax y2 2

1 41− =

If two tangents can be drawn to the different

branches of hyperbolax y2 2

1 41− = from the point

( , )α α 2 , then P( , )α α 2 must lie in the shaded region inthe adjoining figure.

Since, the asymptotes are

y x= ± 2 or ( )y x2 24 0− =∴ 2 2α α< and − <2 2α α⇒ α α( )− >2 0 and α α( )+ >2 0

⇒ α < 0 or α > 2 and α < − 2 or α > 0

Taking intersection of the above two results, we getthe final range of α as

α ∈ − ∞ −( , )2 or α ∈ ∞( , )2

TEST Edge Tangent and normal to the hyperbola,properties of hyperbola, conjugate hyperbolarelated questions are asked.

To solve such type of questions, students areadvised to understand the concept of hyperbolasuch as the straight line y mx c= + is a tangent to thecurve, if e a m b2 2 2 2= −

54. (a,d) Idea If a, b,c are given as a i a j k,= + +a ai$ $ $

2 3

b i j k= + +b b b1 2 3$ $ $ and c i j k= + +c c c1 2 3

$ $ $

then (a b) c = abc× ⋅ =[ ]

a a a

b b b

c c c

1 2 3

1 2 3

1 2 3

Q a a b b c c× = × = × =0

Also use the concept of maximum andminimum.

Now,

d a a b b c c a a⋅ = × + × + × ⋅{( ) sin ( ) cos ( )}x y 2

[Q( )a b a× ⋅ = 0, ( )c a a× ⋅ = 0]

= [ ] cosa b c y …(i)

also d a b c⋅ + + =( ) 0

d a = d b+c⋅ − ⋅( ) …(ii)

From Eqs. (i) and (ii), we get

cos( )

[ ]y = − ⋅ +d b c

a b c

Similarly, sin( )

[ ]x = − ⋅ +d a b

a b cand − = ⋅ +

2d a c

a b c

( )[ ]

∴ sin cos( )

[ ]( )

[ ]x y+ + = − ⋅ + − ⋅ +

⋅2

d a b

a b c

d b c

a b c

− ⋅ + = − − ⋅ ⋅ + ⋅ + ⋅d a c

a b c

a d b d c d

a b c

( )[ ]

[ ][ ]

2

also d a b c⋅ + + =( ) 0 [Qd a b c⊥ + +( )]

∴ sin cosx y+ + =2 0

⇒ sin cosx y+ = − 2

⇒ sin x = −1, cos y = −1

⇒ x = − π2

, y = π

[Q we want the minimum value of x y2 2+ ]

∴ x y2 22

22

45

4+ = + =π π π

.

TEST Edge Scalar triple product, properties of scalartriple product and geometrical application of scalartriple product related questions are asked. To solvesuch type of questions, students are advised tounderstand the concept of scalar triple product,such as for any three vectors a, b cand[ ] [ ]a b b c c a abc× × × = 2 = × ⋅| |( )a b c 2

55. (a,d) Idea To solve this problem, use the concept ofvalue of trigonometric expression. Also applythe concept of differentiation, maxima, minimaand range of the function.

Let us assume that y x x= +sin sin3

⇒ y x x x= − +( sin sin ) sin3 4 3

⇒ y x x= −4 3(sin sin )

which is a cubic equation.

On differentiating both the sides w.r.t.x , we getdy

dxx x x= −4 3 2(cos sin cos )

⇒ dy

dxx x= −4 1 3 2cos ( sin )

∴ dy

dx= 0 ⇒ cos x = 0

and sin x = ± 13

⇒ sin x = ± 13

or m 1

On substituting sin x = ± 1

y x x= −4 1 2sin ( sin ).

We get, y = ± − =4 1 1 0( )

On substituting x = ± 13

in

y x x= −4 1 2sin ( sin ), we get

y = ± −

= ±4

13

113

83 3

ymax = 83 3

and ymin = − 83 3


x

y

⇒ Range of y is−

83 3

83 3

,

⇒ (sin sin ) ,38

3 38

3 3x x+ ∈ −

Now,8

3 3154= . ;

e

2136

2157≈ =. ; .

π

TEST Edge Solution of trigonometric equation,principal value, range and domain, and solution oftrigonometric inequalities, related questions areasked. To solve such type of questions, students aresuggested to understand the concept oftrigonometric equations and also acquaintedyourself with the concept of maxima or minima.Such as, f x x( ) ,= cosec domain is ( , )−∞ ∞ − ∈{ : }n n Iπand range is ( , ] [ , )−∞ − ∪ ∞1 1

56. (0) IdeaQ f x dx

f x f x

f x dx f x f xa

a a( )

( ) ( )

( ) ( ( ) ( )−∫ ∫=− = −

− =

0

20

if

if

Also apply the concept of limit to solve thisproblem.

Let f x xx

x( ) cos ln= −

+22

f x xx

x( ) cos ( ) ln− = − +

−22

f x xx

x( ) cos ln− = − ⋅ −

+22

⇒ f x f x( ) ( )− = −In function

−∫−+

1 3

1 3 22/

/cos lnx

x

xdx = 0

Property If

So, in odd0

a

0

a−∫

∫ ∫= + −a

af x dx

f x dx f x dx

( )

( ) ( )

function,it will be zero.

TEST Edge Evaluation of limits of algebraicfunctions, modulus function and greatest integerfunctions related questions are asked. To solve suchtype of question, students are advised tounderstand the concept of limit such as,

lim ,x a

n nnx a

x ana

→

−−−

= 1 where n ∈ 9

57. (1) Idea To solve this problem use the concept of thedistance of point from plane. The distancebetween two points A x y z( , , )1 1 1 and φ( , , )x y z2 2 2

is ( ) ( ) ( )x x y y z z1 22

1 22

1 22− + − + −

Let the given point is A ( , )1 1 and the plane isP x y z: + + =1 and the given line is

Lx y z

:− = − = −21

32

43

We have to find the minimum distance of the point A

from the planeP measured parallel to lineL. Supposea line parallel to L passing through A intersects theplane P at B, then we have to find distance AB.

Let’s find equation of AB. Now, AB passes throughA ( , , )1 1 1 and is parallel to L. So, direction ratios of AB

should be same asL i e. .,< >1 2 3, , Hence, equation of

AB isx y z− = − = −1

11

21

3.

Now, we should find the coordinates of B to find AB.Coordinates of any point on line

ABx y z

k:− = − = − =11

12

13

are of type

( , , )k k k+ + +1 2 1 3 1 .also point B k k k( , , )+ + +1 2 1 3 1 lies on the planeP x y z: + + =1.

∴ k k k+ + + + + =1 2 1 3 1 1

k = − = −26

13

⇒ B23

13

0, ,

Hence, the length

AB =

+

+ =2

313

1143

2 22

∴ 314

314

143

1k = =

TEST Edge Distance between two plane, the planeare parallel or perpendicular the angle between twoplane related questions are asked. To solve suchtype of questions, students are advised tounderstand the concept of plane, such as, the lengthof the perpendicular from a point having positionvector a to the plane r. n = d is given by

pd= −| |a. n

|n|

58. (8) Idea First of all draw the graph of the given curvethen apply the concept of projection of line onanother line, to solve this problem.

Let us assume that AB makes angle θ with positivex-axis.

Also, let the coordinates of B are ( , )x y

∴ tan θ = y

xand y x2 8=

∴ Projection of BC on the x-axis

= =° −

=LCy

ytan ( )

tan90 θ

θ = =y

x

2

8

So, the required projection is 8 units.

PRACTICE SET 4 233

A

( , )x yB

L Cx

90°–θyθ

y x2 = 8

TEST Edge The equation of the bisector of the anglewhich contains a given point, the equation ofreflected ray and equation of a line in various forms,related questions are asked. To solve such type ofquestions, students are advised to understand theconcept of straight line in 2-D . Such as, thehomogeneous second degree equationax xhy by2 22 0+ + = represent a pair of straight linethrough the origin if h ab2 ≥ .

59. (10) IdeaQ log e x = 2 ⇒ x e= 2

Apply the concept of quadratic equation tosolve this problem, also use the concept offactorisation.

It is given that α β, are the roots of equation

1 82

1102

10 102

−−

=(log )log [log ]

x

x x

Let log10 x y=

So,1 8

21

2

2−−

=y

y y( )

1 8 22 2− = −y y y

⇒ − + − + =8 2 1 02 2y y y

⇒ − − + =6 1 02y y

⇒ 6 1 02y y+ − =6 3 2 1 02y y y+ − − =

3 2 1 1 2 1 0y y y( ) ( )+ − + =3 1 0y − = 2 1 0y + =

y = 13

y = −12

log1012

x = − and log1013

x =

x =−

1012 or x =10

13

x = 1101 2/ or x =101 3/

So, α = 1101 2/ β =101 3/

Now, α β2 31 2

21 3 31

110

10 1+ =

⋅ +/

/( )

⇒ 110

10 1× +

⇒ 1 1 2+ =∴ ( )α β2 3 21+ = =2 42

again α 4 210= −

⇒ α 4 1100

= or1

1004α=

Now,( )α β

αα

2 3 2

41

30000+ −

4

⇒ 110

2 100 300001

1002( ) − ×

⇒ 110

400 300[ ]−

⇒ 10010

10=

TEST Edge Interval in which the roots lie andquadratic in equations related question are asked.To solve such type of questions, students areadvised to understand the concept of quadraticequation such as, let f x ax bx c( ) = + +2 , wherea b c R, , ∈ and a ≠ 0. Suppose N N N R, ,1 2 ∈ andN N1 2< , then interval N N1 2, will be containedbetween the roots of f x( ) = 0 if a f N( ) ,1 0<a f N( )2 0< , a f N( )2 0<

60. (1) Idea log b alog

logc

c

a

b

y f x xa= =( ) log is defined when x > 0 an a > 0.Also apply the concept of factorisation, to solvethis problem.

Now, 4 42

4

4 2log loglog (log )

loge

e

x

x

= (using logloglogb

c

c

aa

b= )

= 44

222

log (log )

loge x

Q log logab

amm

b=

1

= 4 42log (log )e x

= =4 42 2log (log ) (log )e x

e x

∴ The equation is

(log ) log (log )e e ex x x2 2 1= − +2 1 02t t− − = [where, t xe= log ]

⇒ t = −112

,

loge x =1

⇒ x e=

But t = − 12

⇒ loge x = − 12

Then, log (log )2 e x is not defined.

∴ There is only one real solution.

TEST Edge Solution of algebraic function,exponential functions, trigonometric functionrelated questions are asked. To solve such type ofquestions, students are advised to understand theconcept of solution of the function and alsoacquainted yourself with the properties of function.

e g. . ,Find the number of integral solution of the inequality( ) ( ) ( )

( )( )

x x x

x x

− − −− −

≤1 4 7

2 60

2 3

.

To solve this problem, use the concept of curvemethod and will get the answer i e. . , f x( ) ≤ 0 forx = 1 2 3 4, , ,


Test Advanced...The final pressure in the two compartments are equal b. Volume of compartment I is 3...

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