WT-Sankalp-NM MCP Solution Final Code-1 AIEEE[1]
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Transcript of 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
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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)
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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
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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
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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.
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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.
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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.
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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
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(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
´=
\ (b) is correct answer.
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.
\ (b) is correct answer.
(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.
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(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)
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(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)
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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
explanation 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.
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)
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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
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(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
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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
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(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
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(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.
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(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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