WT-Sankalp-NM MCP Solution Final Code-1 AIEEE[1]

8/3/2019 WT-Sankalp-NM MCP Solution Final Code-1 AIEEE[1]

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Value of p

-ò 0

1 sinx dx is

(a) ( )-3 3 3 3 (b) ( )-2 2 2 2

(c) 1 (d) 0

1 – sin x =

2 x

cos sin2 2

pæ ö-ç ÷è ø

Þ x x

1 sin x cos sin2 2

- = -

If

x0, Then 1 sin x

2 4

pæ öÎ ® -ç ÷è ø is positive.

If x

, Then 1 sinx2 4 2

p pæ öÎ ® -ç ÷è ø

is negative.

Þ0

1 sinx

p-ò dx

2

0

2

x x x xcos sin dx sin cos dx

2 2 2 2

pp

p

æ ö æ ö= - + -ç ÷ ç ÷è ø è øò ò

2

02

x x x x2 sin cos 2 cos sin

2 2 2 2

pp

pæ ö= + + - -ç ÷è ø

2 (2 2 2)= -

Evaluate-

+ -ò 2

2

dx

1 |x 1|.

(a) log 2 (b) 2 log 2

(c) 3 log 2 (d) 4 log 2

2 1 2

2 2 1

1 1 1dx dx dx

1 |x 1| 1 (1 x) 1 x 1- -

= ++ - + - + -ò ò ò

In the int erval ( 2, 1), |x 1| 1 x

In the int erval (1, 2), |x 1| x 1

- - = -ìí

- = -î

1 2

2 1

1 1dx dx

2 x x-

= +-ò ò

=1 2

2 1log (2 x) log x-- - +

= log 4 + log 2 = 3 log 2

p

+ò

2n

n0

dx

1 cot nx is equal to

(a) 0 (b)p

4n

(c)p

2n(d)

p2

nx = t Þ n dx = dt

Þ2

n0

1 dtI

n 1 cot t

p

=

+ò

2 n

n n0

1 sin t dt 1

n n 4sin t cos t

p

p= = ´

+ò

since,2 n

n n0

sin d

4sin cos

p

q q p=

q + qò

p

++ò

4

0

sin x cos x

3 sin 2x dx is equal to

(a) -1

log 34

(b)1

log 34

(c) -1

log 43

(d) None of these

sin x – cos x = t, Þ (cos x + sin x) dx = dt

On squaring 1 – sin 2x = t2

Þ

0 0

2 21 1

1 1

I dt dt3 1 t 4 t-= =+ - -ò ò

0

1

1 2 tlog

4 2 1 -

+æ ö= ç ÷-è ø

1log 3

4=

®¥ = +å2n

2 2nr 1

1 rlim

n n requals

(a) +1 5 (b) - +1 5

(c) - +1 2 (d) +1 2








2n 2n

2 2 2n nr 1 r 1

2

r1 r 1 nlim limn nn r r

1n

®¥ ®¥= ==

++

å å

(On taking n2 common from radical)

2

20

ndx etc.

1 x

=+

ò

®=

ò 2 x

2

0

x 0

cos t dt

lim x sin x

(a) 1 (b) 2

(c) 4 (d) None of these

On applying LH rule,

4

x 0

cos x .2xLimit lim

1.sin x x cos x®=

=

4

x 0

2 cos x 2lim 1

sinx 2cos x

x®

= = =+

p

ò 2

3

0

|sin x|dx is equal to

(a) 0 (b) p

(c)8

3(d)

3

8

3 3

0 0

I 2 |sin x|dx 2 sin x dx

p p= =ò ò

The value of -ò 1000

x [x]

0

e dx is where [.]

represents Greatest Integer Function.

(a)-1000e 1

1000(b)

--

1000e 1

e 1

(c) 1000 (e – 1) (d)-e 1

1000

1000 x [x]

0

e dx--ò

1 2 1000 x [x] x [x] x [x]

0 1 999

e dx e dx ... e dx- - -= + + +ò ò ò

1 2 3 x x 1 x 2

0 1 2

e dx e dx e dx- -= + +ò ò ò

1000 x 999

999

..... e dx-+ + ò

= (e1 – e0) + (e1 – e0) + .... + (e1 – e0)

= 1000 (e – 1)

The value of the integral

-ò 2e

e

1e

log xdx,

xis

(a)

3

2 (b)

5

2

(c) 3 (d) 5

elog x0

x=

Þ loge x = 0 Þ x = 1

Þ1log x

0 if x (e ,1) x

-< Î

and2log x

0 if x (1, e ) x

³ Î

Þ

21 e

1 1e

log x log x 5I dx dx

x x 2-= + =ò ò

Area of the region bounded by y = x2and

y2 = x is

(a) 1 (b)1

2

(c)1

3(d)

1

4

Points of intersection are

2 x x x 0,1= Þ =

Þ Points are (0, 0), (1, 1)








Required area is

11 32 3/2

0 0

2 x 1 x x dx x sq.units

3 3 3

é ùé ù- = - =ê úê úë û ê úë û

ò

Let

ì <ï= í ³ïî

2 x : x 0

f(x) x : x 0

Area bounded by the curve y = f(x),

x2=9a2 and x-axis is9a

2. Then the value

of a is

(a)1

4(b)

1

3

(c)1

2(d) 1

Following graph show the required area in

shaded form

Required area

0 3a

2

3a 0

x dx x dx-

= +ò ò

23 9a 9a

9a2 2

= + =

Þ 2a2 + a – 1 = 0

Þ1

a 1,2

= -

Now, a > 0 i.e.,1

a2

=

Area bounded by the curves

3y = –x2 + 8x – 7 and + =+4 y 1

x 3is

(a) 7 – 6 log 2 (b) 8 – 7 log 2

(c) 9 – 8 log 2 (d) None of these

Shifting the origin to (3, –1) as area is invariant

under any co-ordinate system. i.e., X = x –3, Y

= y + 1

These two equations become

3 (Y – 1) = – (X + 3)2 + 8 (X + 3) – 7

4Y

X=

Smplification will get us

3y = – x2 + 2x + 11

and4

y x

=

These two curves intersect at points when

212 x 2x 11

x= - + +

Þ x3 – 2x2 – 11x + 12 = 0

Solutions of this equation are x = 1, – 3, 4.

But x = – 3 corresponds to intersection in 3rd

quadrant.

Þ x = 1, 4 are the points in Ist quadrant

Required area4 42

1 1

x 2x 11 4dx dx

3 x

- + += -ò ò

= 9 – 8 log 2 sq. units

Using the following passage to

solve Questions no. 13 to 15

Consider the parabola y = x2 + 1 and the

line x + y = 3.

The line cuts the parabola at A and B. Let

the abscissa of A and B be a and b.

The values of a + b and ab must be

respectively

(a) 1 and 2 (b) 2 and 1

(c) –1 and –2 (d) None of these

On eliminating y between line and curve we

get

x2 + x – 2 = 0Þ a + b = – 1, ab = – 2








Þ (c) is correct.

The area bounded between the parabola

and the line in terms of a and b must be

(a)a + b - ab a + b

a - b + -2( )

| | 23 2

(b)a +b a + b

a - b + +2 2

| | 23 2

(c) |a – b| |2–(a + b) – (a2 + b2)|

(d) None of these

Area bounded between the parabola and line

2 1(y y ) dx

b

a

= -ò

2(3 x) (x 1) dxb

a

é ù= - - +ê úë ûò

2(2 x x ) dx

b

a

= - -ò

3 3 2 2

2 ( )3 2

a - b a - b= + - a - b

2 2

| | 23 2

a + b + ab a + b= a - b + -

2( )| | 2 sq. units

3 2

a + b - ab a + b= a - b + -

Þ (a) is correct.

The area bounded between the parabola

and the line must be

(a) 2 sq. units

(b)35

6sq. units

(c) 4 sq. units

(d) None of these

Put a + b = – 1, ab = – 2

in the correct answer of Q. we get area = 4 sq.

units

Þ (c) is correct.

The area of the plane region bounded

by the curves x + 2y2 = 0 and x + 3y2=1 is

equal to

(a)1

3(b)

2

3

(c) 43

(d) 53

Here x + 3y2 = 1 and x + 2y2 = 0 intersect at

x = – 2 and y = ± 1

Hence required area1

2 2

1

(1 3y 2y )dy

-

= - +ò

1 12 2

1 0

(1 y )dy 2 (1 y ) dy

-

= - = -ò ò

13

0

y 12 y 2 1

3 3

é ù æ ö= - = -ê ú ç ÷è øê úë û

4sq.units.

3=

(c) is correct.

The area bounded by

|y| = |{|sin x|}–1 | " 0 < x < p and line

x = p, where {x} represents fractional

part of x

(a) p – 2 (b) 2 ( p – 2)

(c) 3 (p – 2) (d) 4 ( p – 2)

Required graph is

Area = 2

0

(1 sin x) dx 2 ( 2) sq. units

p

- = p -ò

Area of the curve =+ 2

x y

1 xin the first

quadrant between x = 0 and at the point

where it attains maxima is

(a)1

log 52

(b)1

log 22

(c) log 5 (d)1

log 3

3








Graph of the curve 2

x y

1 x=

+is

and

2

2 2

dy 1 x

dx (1 x )

-=

+

Þ x = 1 is a point of local as well as absolute

maxima

Þ We have to find

1 1

2

0 0

xArea y dx dx

1 x= =

+ò ò

12

0

1log (1 x )

2é ù= +ê úë û

1(log 2 log1)

2= -

1

2= log 2 sq. units

The expressionò

ò

n

0

n

0

[x]dx

,

{x} dx

where [x] and

{x} are integral and fractional parts of x

and n Î N is equal to

(a)-1

n 1(b)

1

n

(c) n (d) n – 1

1 2 n

0 1 n 11 2 3 n

0 1 2 n 1

[x] dx [x]dx .... [x] dx

xdx (x 1) dx (x 2) dx .... (x (n 1))dx

-

-

+ + +=

+ - + - + + - -

ò ò ò

ò ò ò ò

n(n 1)

2 n 11

n2

-

= = -æ öç ÷è ø

If n > 1, then

( )

¥

+ +ò n

21

1dx

x 1 x

must

be equal to

(a)-2

1

n 1(b)

-

-2

1

n 1

(c)-2

2

n 1(d)

+2

1

n 1

Put 2 x x 1 t+ + =

then 2 x 1 t x+ = -

Þ x2 + 1 = t2 + x2 – 2tx

Þ2t 1

x2t

-=

2

2

t 1 x

2t

-= dt

When x ® ¥, t ® ¥ when x ® 1, t ® 1

Þ2

2 n1

t 1 1I . dt

2t t

¥ += ò

Þ n n 2 21

1 1 1 1dt

2 t t n 1

¥

+æ ö

+ =ç ÷-è ø

ò

If A = ò 2

2 x

1

e dx then ò 4e

e

log x dx must be

(a) 2e4 – e – A (b) e 4 – e – A

(c) 2e4 + e + A (d) None of these

Let log x = t2 or2t x e=

Þ1

dx 2t dt x

=

Þ

4e 2 2t

e 1

log x dx 2t.e t dt=ò ò

22 22 2 2t t t

11 1

t e .2t dt t . e e dtæ ö

= = -ç ÷è ø

ò ò

{Integration by parts}

= 2 . e4 – e – A

If f (x) is monotonically increasing and

is differentiable on [a, b ], then

bb-

a a

+ò ò f( )

1

f( )

f(x) dx f (x) dx is equal to

(a) f (a) – f (b) (b) bf(a)– af(b)

(c) bf(b)– af(a) (d) f(a) + f(b)

Since f (x) is monotonically increasing in

[a, b]

Þ f –1 exist.








Let g (x) = f –1 (x)

In second integral put f –1 (x) = t

Þ x = f (t) dt, dx = f 1 (t) dt

when x = f (a), t = f –1 [f (a)]=aand when x = f (b), t = f –1 [f(b)] =b

Þ Given expression = 1

f(x)dx t . f (t) dt

b b

a a

+

ò ò

f(x) dx t . f(t) f(t) dt

b bba

a a

= + -ò ò

= bf (b) – a f (a)

If d

f(x)dx

= g(x) for a £ x £ b then,

b

a

f(x)ò g(x) dx equals

(a) f (b) – f (a)

(b) g(b) – g(a)

(c)2 2[f(b)] [f (a )]

2

-

(d)2 2[g(b)] [g(a )]

2

-

bb b b

a a aaf(x) g(x)dx f(x). g(x) dx f '(x) . g(x) dx= -ò ò ò ò

Now g(x) dx f(x)=ò

Þ

bb b

2

a aa

f(x)g(x)dx (f(x) g(x) f(x) dx= - ´ò ò

Þ

b2 2

a

2 f(x) g(x) dx [f(b)] [f(a)]= -ò

Þ

b 2 2

a

[f(b)] [f (a )]f(x) g(x) dx 2

-=ò

Area of the curve y = x 4 –2x3+x2+3

between two local minimas in the first

quadrant is

(a)71

3(b)

91

3

(c)71

30(d)

91

30

y’

= 4x3

– 6x2

+ 2x To get local minima of the function

y’ = 0

Þ 4x3 – 6x2 + 2x = 0

Þ 2x (2x – 1) (x – 1) = 0

Þ1

x 0, ,12

= are the points of local extrema

Now, y’’ = 12x2 – 12x + 2

y’’ > 0 for x = 0, 1

y’’ < 0 for1

x2

=

Þ Local minima at x = 0, 1

Þ Required area1

4 3 2

0

(x 2x x 3) dx= - + +ò

15 4 3

0

x 2x x3x

5 4 3

é ù= - + +ê ú

ê úë û

91

30= sq. units

Let P be a variable point on the ellipse

2 2

2 2

x y1

a b+ = with foci F

1and F

2. If A is

the area of the triangle PF1F

2, then the

maximum value of A is

(a)ea

b(b) aeb

(c)ab

e(d)

e

ab

a cos bsin 11

Area ae 0 12

ae 0 1

f f= -

1( 1) b sin ( 2ae)

2= - f -

= abe sin fFor maximum value of A = abe as sin f can go

upto I.

Questions number 26 to 30 are

Assertion–Reason type questions.

If both assertion and reason are

correct and reason is the correctexplanation of assertion.








If both assertion and reason are true

but reason is not the correct

explanation of assertion.

If assertion is true but reason is false.

If assertion is false but reason is true.

32

1

2

sin

p

-

pò (sin x) dx ¹ 0

a

a

f(x)

-ò dx = 0 if f(x) is odd

function.(a) A (b) B (c) C (d) D

Statement 2 is true but Statement 1 is wrong.

Since sin –1 (sin x) = p – x

If 3

x2 2p p£ £

3 32 2 21

22

xsin sin(x) dx x

2

p p

-

pp

é ùÞ = p -ê ú

ê úë ûò

2 2 23 9.

2 8 2 8

p p p p= p - - +

= 0

Area of the curves

2 2

2 2

x y1

a b+ = and

2 2

2 2

(x ) (y )1

a b

- a - b+ =

must be sameArea remains same if the origin is shifted.

(a) A (b) B (c) C (d) D

Statement 1 is true and follows from Statement

2. On putting X = x – a, Y = y – b the second

curve becomes2 2

2 2

X Y1

a b+ = whose area is same

as that of

2 2

2 2

x y1

a b+ =

The curve y = – 2 x

2+x+1

is symmetric about theline x = 1.A parabola is symmetricabout its axis.

(a) A (b) B (c) C (d) D

Statement 1 is true and follows from Statement

2 since equation of line of symmetry is given

bydy

0dx

=

Þ – x + 1 = 0

Þ x = 1

Area of the curve x

2+2y

2+4x+4y=0 is

3 2 p .

Area of any curvepassing through origin isirrational.

(a) A (b) B (c) C (d) D

The equation can be written as

(x + 2)2 + 2 (y + 1)2 = 6

Þ2 2(x 2) (y 2)

16 3

+ ++ =

This curve is an ellipse whose area is

pab= 3 2 p

Þ Statement 1 is correct.

Statement 2 is false since the curve

221 1

x yæ ö

- + =ç ÷ ppè øpasses through origin

1A circle of radius

æ öç ÷pè ø

.

But it area2

11

æ ö= p =ç ÷pè ø

(Rational)

Let f(x) = (log x)2, g(x) = (log x)

3be two

curves. Let A1

and A2

respectivelyrepresent area of these curves from

x = 1 to x = e.A

1> A

2

(log x)2 £ (log x)

3over [1, e]

(a) A (b) B (c) C (d) D

Statement 2 is false since over [1, e], log xÎ[0,1]

Þ (log x)2 > (log x)3

Þ

e e2 3

1 1

(log x) dx (log x) dx³ò ò

Þ A1> A

2

Thus statement 1 is true.








In which of the following molecules

substituent show +R effect and –I effect ?

(a) Phenol (b) Nitrobenzene

(c) Phenyl cyanide (d) None of these

(a)

In phenol OH-gg

gg

, oxygen is more electronegative

then carbon, therefore, it is electron

withdrawing, i.e., –I effect but + R (resonance

effect) due to lone pair of electrons.

(b) has –I and –R effect.

(c) has –I and –R effect.

(d) is not possible.

Carbocations are involved in all

reactions except

(a) Dehydration of alcohols

(b) Dehydrohalogenation of R–X

(c) Aldol condensation

(d) SN1 reactions

(c)In aldol condensation carbanions are involved.

Which of the following cation is most

stable ?

(a)

(b)

(c)

(d) Tropylium cation.

(d) Tropylium cation is most stable because

aromaticity is being followed i.e., planar

structure and resonance stabilization.

(a) ( )6 5 3C H C

Åsecond most stable due to 9

resonating structures, i.e., +ve charge is

delocalised over 3 benzene rings.(b) is less stable due to 6 resonating structures.

(c) is less stable due to 3 resonating structures.

.

These two structures are :

(a) Optical isomers

(b) Tautomers

(c) Geometrical isomers

(d) Resonating structures (mesomers)

(d) They are resonating structures because of

identical arrangement of atoms, same number

of pair electrons.

(a) is not possible due to absence of chiral ‘C’

atom.

(b) is not possible Q shifting of H+ is not taking

place.

(c) is not possible because conditions of

geometrical isomerism are not satisfied (a=b).

The correct order of stability of

conformations of NH2 —CH

2 —CH

2 —OH is

(a) Anti > Gauch > Eclipsed

(b) Gauch > Anti >Eclipsed

(c) Eclipsed > Gauch > Anti

(d) Gauch > Eclipsed > Anti

(b)

So gauch form stabilized by intramolecular

hydrogen bonding hence gauch is more stable

tham anti.

\ (b) is correct answer.

The heat of hydrogenation of hex–1–ene

is 126 kJ mol –1. When a second double

bond is introduced int he molecule, the

heat of hydrogenation of the resulting

compound is found to be 230 kJ mol –1.

The resulting compound will be

(a) 1, 5–hexadiene

(b) 1, 4–hexadiene(c) 1, 3–hexadiene








(d) 1, 2–hexadiene

(c)(c) is correct because it is conjugated diene and

resonance stabilized.

(a) and (b) are not correct answers because they

are isolated diene, therefore their heat of

hydrogenation should be double i.e. 252 kJ

mol –1.

(d) is not correct because it is cumulative diene.

The configuration of chiral centers in

the compound is

(a) 2S, 3R (b) 2R, 3S(c) 2S, 3S (d) 2R, 3R

(c)

\ (c) is correct answer.

A 0.1 M solution of an enantiomerically

pure chiral compound (X) has anobserved rotation of 0.20 in a 1 dm

sample tube. The molecular mass of the

compound is 150. The specific rotation

of (A) is

(a) 13.3 (b) 0.015

(c) 1.33 (d) 0.10

(b)Concentration of solution = 0.1 M

The specific rotation =0.1 150

0.0151000

´=


Which is not correct about isomers ?

(a) They have same molecular formule

(b) They have same vapour density

(c) They have same empirical formulae

(d) All are correct

(d)

Which of the following is more

reasonable resonance form of

(a) (b)

(c) (d)

(b)(a), (c) and (d) are incorrect because in these

structures both rings are antiaromatic while

in (b) both rings are aromatic due to 6 and 2pelectrons in delocalisation.


(a)

(b)

(c)

(d)

(c)

(c) is possible becuse group

is o and p–directing for benzene on right hand

due to hyperconjugation by circled group.

(a) is not possible because it is an electrophilic

substitution reaction and not free radical

substitution.








(b) is not possible because

group is meta–directing.

(d) is not possible because is

electron releasing for benzene on right side

which will direct the incoming electrophile

(Br+) at ortho or para position.

The reactivity order of

CH3 –F, CH

3– Cl, CH

2= CH – CH

2 –Cl is

(a) CH3 –F>CH

3 –Cl > CH

2=CH–CH

2 – Cl

(b) CH2=CH–CH

2 –Cl>CH

3 –Cl>CH

3 –F

(c) CH3 –Cl>CH

2=CH–CH

2 –Cl>CH

3 –F

(d) CH2=CH–CH

2 –Cl>CH

3 –F>CH

3 –Cl

(b) The reactivity of compounds depends upon

bond energy, which in turn depends upon,

(i) size of atom attached to carbon.

(ii) type of intermediate formed after cleavage.

Out of CH3 –X and CH

2=CH–CH

2 –X after

heterolytic cleavage.

CH3 –X gives

3CHÅ and CH2=CH–CH

2 –X give

CH2=CH– 2CHÅ

when the intermediate formed is stable, then

compound forming the intermediate is highly

reactive.

Thus CH2=CH– 2CHÅ is resonance stabilized

hence more stable than 3CHÅ.

So CH2=CH–CH

2 –X is more reactive than CH

3 –X.

So most reactive is CH2=CH–CH

2 –Cl.

Now out of CH3 –F and CH

3 –Cl the C–F bond is

stronger than C–Cl bond, hence CH3 –Cl is more

reactive than CH3 –F.

Thus order of reactivity CH2=CH–CH

2 –Cl>CH

3 –

Cl>CH3 –F

Hence choic (b) is correct while (a), (c), (d) are

incorrect.

(a) ‘A’ is conc. HNO3/conc. H

2SO

4, ‘B’ is

, ‘C’ is Cl2/FeCl

3.

(b) ‘A’ is Cl2/FeCl

3, ‘B’ is ,

‘C’ is conc. HNO3/conc. H

2SO

4.

(c) ‘A’ is Br2/FeBr

3, ‘B’ is ,

‘C’ is conc. HCl/conc. HNO3.

(d) None of these.

(a)

, when reacts with

alc. KOH gives mainly

(a) 2–phenylpropene

(b) 1–phenylpropene

(c) 3–phenylpropene

(d) 1–phenylpropan–2–ol

(b)

In SN2 the reactivity follows the order

(a)








(b)

(c)

(d) C2H

5Cl>CH

3Cl>CH

3CH

2CH

2Cl

(a) The reactivity of alkyl halide in S

N2 reaction

is in the order of CH3 —X > 1° > 2° > 3°.

The correct order of reactivity of

following compounds in SN1 reaction is

(a) IV > III > I > II

(b) I > II > IV > III

(c) I > III > IV > II

(d) III > II > I > IV

(c)In (I) carbocation is more stable than (II), (III)

and (IV) while in (II) and (III) carbocation

stabilized by resonace but number of resonating

structures in (II) is more than (III) while in (IV)

lone pair of electrons of bromine takes part in

resonace so it is very difficult to remove Br

from (IV) hence it is less reactive.

\ (c) is correct answer.

Which of the following is correct ?

(a) SN1 reaction involves TS and

completed in polar protic solvents

(b) SN2 reaction is one step process.

(c) Walden inversion is SN1 reaction

(d) All are correct

(b)

(a) is wrong because SN1 reaction involvescarbocation as the intermediate.

(b) is correct answer because all stereoselective

reactions must be the stereospecific reaction.

(c) is wrong because Walden inversion is SN2

reaction.

(d) is also wrong.

The false statement about SN1 reaction is

(a) polar solvents favour the formation

of carbonium ion

(b) stronger nucleophile increases the

carbocation formation

(c) more stable carbocation favour SN

1

reaction

(d) racemic mixture is formed with

chiral substrate

(b)

Which of the following statements is/

are correct about SN2 reaction ?

(A) The reaction proceeds with

simultaneous bond formation and

bond fission

(B) The rate of reaction is independent

of concentration of nucleophile

(C) The nucleophile attacks the carbon

atom on the side of the molecule

opposite to that of leaving group

(a) A, B (b) A, C

(c) B, C (d) A, B, C

(b)

Aryl halides are less reactive towards

nucleophilic substitution reaction as

compared to alkyl halides is due to(a) the inductive effect

(b) resonance stabilisation

(c) larger C – Cl bond length

(d) sp3 hybridized carbon attached to the

chlorine atom

(b)

Select incorrect statement :

(a) Reactivity of alkyl halides towards E1

or E2 reaction is 3° > 2° > 1° alkyl

halide

(b) E1 reaction has carbonium ion asintermediate

(c) Dehydrohalogenation is trans–

elimination

(d) E1 and SN1 reactions are competitive

(b)

In the following reaction final product

is :

(a)

(b)

(c) both (a) and (b)

(d) None of these

(b)








Identify the product with proper

configuration.

(a) retention

(b) inversion

(c) recemisation

(d) no reaction

(b)

In E1–CB reaction, intermediate is :

(a)

(b)

(c)

(d)

(b)

Leaving tendency of the following in

increasing order is :

(a) VI < V < III < II < IV < I

(b) I < II < III < VI < V < IV

(c) I < III < V < II < IV < VI

(d) cannot be decided

(a)

Questions number 56 to 60 are

Assertion–Reason type questions.

If both assertion and reason are

correct and reason is the correct


If both assertion and reason are true

but reason is not the correct


If assertion is true but reason is false.

If assertion is false but reason is true.

Cyclopropane has the

highest heat of

combustion per

methylene groupIts potential energy is

raised by angle strain

(a) A (b) B

(c) C (d) D

(a)

In proteic solvent

order of

nucleophilicity of

halide ions is I (–) > Br(–)

> Cl(–) > F(–)

In polar proteic

medium ion–dipole

attraction takes

places. F(–) is strong

base hence due to

more ion dipole

attraction it is

arrested by solvent

molecules.

(a) A (b) B

(c) C (d) D

(a)

In polar proteic medium ion–dipole attractiontakes places. F(–) is strong base hence due to

more ion dipole attraction it is arrested by

solvent molecules.

Solvents like DMF,

DMSO, DMA etc.

solvate cation very

well.

It is due to their

negative ends orient

around the cation and

donate unshared

electron pairs tovacant orbital of

cations.

(a) A (b) B

(c) C (d) D

(a)








and

can

produce and

respectively

The mechanism for

the given reaction is

E1

i.e. through

carbocation formation

in pressence of base.

(a) A (b) B

(c) C (d) D

(c)

Branched alkanes

have lower boiling

point than theirunbranched isomers

Branched alkanes has

relatively small

surface area, so there

are less forces of

attraction act among

molecules

(a) A (b) B

(c) C (d) D

(a)

Two coaxial solenoids are made by

winding thin insulated wire over a pipe

of cross-sectional area A = 10 cm2 and

length = 20 cm. If one of the solenoids

has 300 turns and the other 400 turns,

their mutual inductance is

(m0

= 4p × 10 –7 TmA –1)

(a) 2.4 p × 10 –5 H (b) 4.8 p × 10 –4 H

(c) 4.8 p × 10 –5 H (d) 2.4 p × 10 –4 H

A horizontal overhead powerline is at a

height of 4m from the ground and carries

a currnet of 100 A from east to west. The

magnetic field directly below it on the

ground is (m0 = 4p × 10 –7

TmA –1

)

(a) 2.5 × 10 –7

T, southward

(b) 5 × 10 –6

T, northward

(c) 5 × 10 –6

T, southward

(d) 2.5 × 10 –7

T, northward

For protecting a sensitive equipment

from the external magnetic field, itshould be

(a) placed inside an aluminium can

(b) placed inside an iron can

(c) wrapped with insulation around it

when passing current through it

(d) surrounded with fine copper sheet

An ideal coil of 10 H is connected in

series with a resistance of 5W and a

battery of 5 V. 2 s after the connection

is made, the current flowing (in ampere)in the circuit is

(a) (1 – e) (b) e

(c) e (d) (1 – e –1)

Two identical conducting wires AOB and

COD are placed at right angles to each

other. The wire AOB carries an electric

current I1 and COD carries a current

I2. The magnetic field on a point lying

at a distance d from O, in a direction

perpendicular to the plane of the wires

AOB and COD, will be given by








(a)+m æ ö

ç ÷p è ø

1/21 20 I I

2 d

(b) ( )m

+p

1/22 201 2I I

2 d

(c) ( )m

+p0

1 2I I2 d

(d) ( )m

+p

2 201 2I I

2 d

If a long hollow copper pipe carries a

current, then magnetic field is produced

(a) inside the pipe only

(b) outside the pipe only

(c) both inside and outside the pipe(d) no where

An inductor (L = 100 mH) a resistor (R =

100W) and a battery (E = 100 V) are

initially connected in series as shown in

the figure. After a long time the battery

is disconnected after short circuiting the

points A andB . The current in the circuit

1 ms after the short circuit is

A B

E

R

L

(a) 1/e A (b) eA

(c) 0.1 A (d) 1 A

A long straight wire of radius a carries

a steady current I. The current is

uniformly distributed across its cross-

section. The ratio of the magnetic field

at a/2 and 2a is

(a) 1/4 (b) 4

(c) 1 (d) 1/2

A straight wire of diameter 0.5 mm

carrying a current of 1A is replaced by

another wire of diameter 1 mm carrying

the same current. The strength of

magnetic field far away is

(a) twice the earlier value(b) one-half of the earlier value

(c) one quarter of the earlier value

(d) same as earlier value

The flux linked with a coil at any instant

t is given by f = 10 t2 – 50t + 250

The induced emf at t = 3 s is

(a) –190 V (b) -10 V

(c) 10 V (d) 190 V

A charged particle with charge q enters

a region of constant, uniform and

mutually orthogonal fields Er

and Br

with a velocity vr

perpendicular to both

Er

and Br

, and comes out without any

change in magnitude or direction of vr

.

Then

(a) = ´

rrr

2

Bv E

B(b) = ´

rrr

2

Ev B

B

(c) = ´

rrr

2

Bv E

E(d) = ´

rrr

2

Ev B

E

Magnetic field due to 0.1A current

flowing through a circular coil of radius

0.1m and 1000 turns at the centre of

the coil is

(a) 0.2 T (b) 2 × 10 –4

T (c) 6.28 × 10

–4T (d) 9.8 × 10

–4T

A long solenoid has 200 turns/cm and

carries a current I. The magnetic field

at its centre is 6.28 × 10 –2 Wb/m2.

Another long solenoid has 100 turns/

cm and it carries a current I/3. The

value of the magnetic field at its centre

is

(a) 1.05 × 10 –2 Wb/m2

(b) 1.05 × 10 –5 Wb/m2

(c) 1.05 × 10 –3 Wb/m2

(d) 1.05 × 10 –4 Wb/m2

A current I flows along the length of an

infinitely long, straight, thin walled pipe.

Then

(a) the magnetic field is zero only on the

axis of the pipe

(b) the magnetic field is different at

different points inside the pipe

(c) the magnetic field at any point inside

the pipe is zero(d) the magnetic field at all points inside








the pipe is the same, but not zero

An electron moves with a velocity

1 × 103

m/s in a magnetic field of

induction 0.3 T at an angle 30°. If e/m

of electron is 1.76 × 1011

C/kg, the

radius of the path is nearly

(a) 10 –8

m (b) 2 × 10 –8

m

(c) 10 –6

m (d) 10 –10

m

One conducting U-tube can slide inside

another as shown in figure,

maintaining electrical contacts

between the tubes. The magnetic field

B is perpendicular to the plane of the

figure. If each tube moves towards the

other at a constant speed v , then the

emf induced in the circuit in terms of B , l and v , where l is the width of each

tube, will be

v

v

A

D C

B× × × ×××

×

×

×

××××××

×

×

× ×

×

× ×

×

(a)Blv (b) –Blv

(c) zero (d) 2Blv

A charged particle moves through a

magnetic field perpendicular to its

direction. Then

(a) the momentum changes but the

kinetic energy is constant

(b) both momentum and kinetic energy

of the particle are not constant

(c) both momentum and kinetic energy

of the particle are constant

(d) kinetic energy changes but themomentum is constant

Current is flowing in a coil of area A

and number of turns N, then magnetic

moment of the coil M is equal to

(a) NiA (b) Ni/A

(c) Ni/ A (d) N2Ai

A coil having n turns and resistance RWis connected with a galvanometer of

resistance 4RW. This combination is

moved in time t seconds from a magnetic

field W1weber to W

2weber. The induced

current in the circuit is

(a)-2 1W W

5 Rnt(b)

( )-- 2 1n W W

5 Rt

(c) ( )-- 2 1W W

Rnt(d) ( )-- 2 1n W W

Rt

In a region, steady and uniform electric

and magnetic fields are present. These

two fields are parallel to each other. A

charged particle is released from rest

in this region. The path of the particle

will be a

(a) helix (b) straight line

(c) ellipse (d) circle

A charged particle of charge q and mass

m enters perpendicularly in a magnetic

field Br

. Kinetic energy of the particle is

E; then frequency of rotation is

(a)p

qB

m(b)

pqB

2 m

(c)p

qBE

2 m(d)

pqB

2 E

In a uniform magnetic field of induction

B , a wire in the form of semicircle of

radius r rotates about the diameter of

the circle with angular frequency w. If

the total resistance of the circuit is R ,

the mean power generated per period

of rotation is

(a)p w2B r

2 R(b)

( )p w2

2B r

8 R

(c)( )p w 2B r

8 R(d)

( )p w22B r

8 R

Two thin, long, parallel wires, separated

by a distance d carry a current of I

ampere in the same direction. They will

(a) attract each other with a force of ( )m

p

20I

2 d

(b) repel each other with a force of ( )m

p20I

2 d








(c) attract each other with a force of ( )m

p

20

2

I

2 d

(d) repel each other with a force of ( )

m

p

20

2

I

2 d

The magnetic field of a given length of

wire carrying a current for a single turn

circular coil at centre is B, then its value

for two turns for the same wire when

same current passing through it is

(a) B/4 (b) B/2

(c) 2B (d) 4B

Two coils are placed close to each other.

The mutual inductance of the pair of

coils depends upon

(a) the rates at which currents are

changing in the two coils

(b) relative position and orientation of

the two coils

(c) the materials of the wires of the coils

(d) the currents in the two coils

Two concentric coils each of radius

equal to 2p cm are placed at right angles

to each other. 3 A and 4A are thecurrents flowing in each coil

respectively. The magnetic induction in

Wb/m2

at the centre of the coils will be

(m0 = 4p × 10 –7

Wb/Am)

(a) 12 × 10 –5

(b) 10 –5

(c) 5 × 10 –5

(d) 7 × 10 –5

A charge q moves in a region where

electric field Er

and magnetic field Br

both exist, then the force on it is

(a) ´ rrq v B (b) + ´r rrqE q v B

(c) + ´r r r

qB qB v (d) ( )+ ´r r r

q B q E v

When the current changes from + 2 A

to –2 A in 0.05 s, an emf of 8 V is induced

in a coil. The coefficient of self-

induction of the coil is

(a) 0.2 H (b) 0.4 H

(c) 0.8 H (d) 0.1 H

A charged particle of mass m and charge

q travels on a circular path of radius r

that is perpendicular to a magnetic field

B. The time taken by the particle to

complete one revolution is

(a)p2 mq

B(b)

p 22 q B

m

(c)p2 qB

m(d)

p2 m

qB

The magnetic flux through a circuit of

resistance R changes by an amount Dfin a time Dt. Then the total quantity of

electric charge Q that passes any point

in the circuit during the time Dt is

represented by

(a)

Df

= D

1

Q .R t (b)

Df

=Q R

(c)Df

=D

Qt

(d)Df

=D

Q R.t

The inductance between A and D is

3H 3H 3HDA

(a) 3.66 H (b) 9H

(c) 0.66 H (d) 1 H

A current i ampere flows along an

infinitely long straight thin walled tube,

then the magnetic induction at any

point inside the tube is

(a) infinite (b) zero

(c) mp0 2i. tesla

4 r(d) 2i tesla

r

A coil in the shape of an equilateral

triangle of side l is suspended between

the pole pieces of a permanent magnet

such that Br

is in plane of the coil. If

due to a current ‘i’ in the triangle a

torque t acts on it, the side l of the

triangle is

(a) tæ öç ÷è ø

1/22Bi3

(b) tæ öç ÷è ø

2Bi3








(c)æ ötç ÷è ø

1/2

23 Bi

(d)t1

Bi3

If a current is passed through a spring

then the spring will(a) expand

(b) compress

(c) remain same

(d) None of the above

A long wire carries a steady current. It

is bent into a circle of one turn and the

magnetic field at the centre of the coil

is B. It is then bent into a circular loop

of n turns. The magnetic field at the

centre of the coil will be(a) nB (b) n

2B

(c) 2nB (d) 2n2B

A very long straight wire carries a current

I. At the instant when a charge +Q at

point P has velocity vr , as shown, the

force on the charge is

(a) opposite to ox (b) along ox

(c) opposite to oy (d) along oy

A conducting square loop of side L and

resistance R moves in its plane with a

uniform velocity v perpendicular to oneof its sides. A magnetic induction B

constant in time and space, pointing

perpendicular and into the plane at the

loop exists everywhere with half the

loop outside the field, as shown in

figure. The induced emf is

×

××

×

×

×

×

×

×

××

×

×

×

×

×

×

××

×

×

×

×

×

×

××

×

×

×

×

×

×

××

×

×

×

×

×

×

××

×

×

×

×

×

×

××

×

×

×

×

×

L v

(a) zero (b)RvB

(c)vBL

R(d)vBL

The magneitc field due to a current

carrying circular loop of radius 3 cm at

a point on the axis at a distance of 4 cm

from the centre is 54 m T. What will beits value at the centre of the loop ?

(a) 250 m T (b) 150 m T

(c) 125 m T (d) 75 m T

If the angle between the vectors Ar

and

Br

is q, the value of the product ( )B A . A´r rr

is equal to

(a) BA2

cos q (b) BA2

sin q(c) BA

2sin q cos q (d) zero

Two long conductors, separated by a

distance d carry currents I1 and I2 in

the same direction. They exert a force

F on each other. Now the current in one

of them is increased to two times and

its direction is reversed. The distance

is also increased to 3d. The new value

of the force between them is

(a) – 2F (b) F/3

(c) – 2F/3 (d) – F/3

An electron moves in a circular orbit

with a uniform speed v. It produces a

magnetic field B at the centre of the

circle. The radius of the circle is

proportional to

(a) B/v (b) v/B

(c) v/B (d) B/v

If the radius of a coil is changing at the

rate 10 –2

units in a normal magnetic

field 10 –3

units, the induced emf is1 mV. Find the final radius of the coil.








(a) 1.6 cm (b) 1.4 cm

(c) 1.2 cm (d) 1.5 cm

When a charged particle moving with

velocity rv is subjected to a magnetic

field of inductionr

B , the force on it isnon-zero. This implies that

(a) angle betweenrr

v and B is

necessarily 90°

(b) angle betweenrr

v and B can have any

value other than 90°

(c) angle betweenrr

v and B can have any

value other than zero and 180°

(d) angle betweenrr

v and B is either zero

or 180°

A particle of mass M and charge Q

moving with velocity vr describes a

circular path of radius R when subjected

to a uniform transverse magnetic field

of induction B. The work done by the

field when the particle completes one

full circle is

(a) (Mv2/R) 2pR (b) zero

(c) BQ2pR (d) BQv2pR

Two infinite conductors are

perpendicular to each other carrying

current I as shown in figure. Find the

magnetic field at that point P at a

distance r along one of the conductors.

(a)mp0I

4 r(b)

mp0I

2 r

(c)mp0

2 r(d) None of these

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WT-Sankalp-NM MCP Solution Final Code-1 AIEEE[1]