Download section-a - Dr. Sangeeta Khanna

Subject-Physics
Part : 1
SECTION-A
(Single Answer is Correct)
Q.1

A frustum of cone having end radii a and b and slant length  is
wound with N turns of wire carrying current I, as shown in the
figure. Assume continuous distribution of wires over the
surface. Find the magnetic field of induction B at the vertex O
of the cone.
(a.)
(b.)
(c.)
(d.)
B
 0 NI(b  a)
2
ln
b
a
O
N
b
a
0 NI
(b  a) ln b / a
2L2
2 0 NI
b
(b  a) ln  
B=
2
L
a
None
B=
Ans. A
Q.2
The magnetic flux through each of five faces of a neutral playing dice is given by = ± N Wb, where
N (= 1 to 5) is the number of spots on the face. The flux is positive (out-ward) for N even and
negative (inward) for N odd. What is the flux through the sixth face of the die ?
(a.)
[Sol.
Q.3
(b.)
3 Wb
4 Wb
(c.)
(d.)
– 3 Wb
– 4 Wb
By Gaus's law of magnetism.
T = 0 = – 1 + 2 – 3 + 4 – 5 + 
 = 3 Wb
]
The diagram shows a solenoid carrying time varying current I = I0t. (I0 is constant) On the axis of
this solenoid a conducting ring is being placed as shown in figure. The mutual inductance of the ring
and the solenoid is M and self inductance of the ring is L. If the resistance of the ring is R then the
maximum current which can flow through the ring is
(a.)
(2M + L)
I0
R
(b.)
MI 0
R
(c.)
(2M – L)
I0
R
(d.)
(M + L)
I0
R
Ans. B
Q.4
Sol.
A coil is connected to an alternating emf of voltage 24 V and of frequency 50 Hz. The reading on
the ammeter connected to the coil in series is 10 mA. If a 1µF capacitor is connected to the coil in
series the ammeter shows 10 mA again. Find the inductance of the coil.
(a.) 2 H
(b.) 5H
(c.) 10 H
(d.) none
B
24
 z  (L) 2  R 2
3
10 10
1 

R   L 

C 

R  (L) =
2
(L) = – L +
2
2
2
1
C
1
1
=
= 5H
2
2 C
2 100100106
Q.5 The graph shows the current versus the voltage in a driven
RLC circuit at a fixed frequency. The arrow indicates the
direction that this curve is drawn as time progresses. In this
plot, the drawn as time progresses. In this plot, the
L=
(a.)
(b.)
(c.)
(d.)
Ans. B
Q.6
Current lags the voltage by about 90 degrees
Current leads the voltage by about 90 degrees
Current and voltage are in phase
Current and voltage are 180 degrees out of phase
In figure, A sphere suspended from the ceiling starts from rest at position (1) and ends its upward
swing at position (2). Ignoring friction and air resistance, the correct relation among the following
is
l

L
nail
Position (1)
(a.)
(c.)
[Sol.
Position (2)
cos  =
L sin   l cos 
Ll
(b.)
cos  =
L cos   l cos 
Ll
(d.)
sin  =
L cos   lsin 
Ll
sin  =
L sin   lsin 
Ll
By conservation of Energy = h1 = (h2 + h3)
L


l
h1 h2
(L–)
h3
L cosq = cos() + (L – ) cos (  )
L cos   l cos 
cos  =
Ll
SECTION-B
(Multiple Answer is Correct)
Q.1
A very long solenoid perpendicular to the page generates a magnetic field into the page whose
magnitude increases with time. This induces an emf in a conducting wire loop around the solenoid
which lights two identical bulbs connected in series along the wire. Now two points diametrically
opposed on the wire loop are shorted with another wire lying to the right of bulb B in the plane of
the page. After the shorting wire is inserted.
(a.)
bulb A goes out
(b.)
(c.)
bulb B goes out
(d.)
bulb B gets brighter
bulb A gets brighter
Ans. CD
Q.2
Figure shows crossection view of a infinite cylindrical wire with a cavity, current density is uniform
j   j0kˆ as shown in figure.
y
×× ×
× ×
× ×

×
× × × a
× ×
× × ×
(a.)
(b.)
(c.)
(d.)
x
magnetic field inside cavity is uniform.
magnetic field inside cavity is along a .
magnetic field inside cavity is perpendicular to a
If an electron is projected with velocity v0 ˆj inside the cavity it will move undeviated.
Ans. ACD
0 ( j  a)
2
Q.3 A series RLC circuit is driven by a generator at frequency 1000 Hz. The inductance is 90.0 mH;
Sol.
Magnetic field is given by B 
capacitance is 0.500 µF; and the phase constant has a magnitude of 60.0° (Take 2 = 10)
(a.)
(b.)
Here current leads the voltage in phase
Here voltage leads the current in phase
(c.)
Resistance of circuit is
(d.)
At resonance  =
Ans. BCD
Sol.
80

3
2
10 4 rad/sec.
3
L = 90 × 10–3 × 2 × 1000 = 180 
1
1
1000
 21000 =
=
6
C
0.5 10

 circuit is inductive VL > VR
 voltage leads the current
1
L 
C = 80 
tan  =
R
R
80 
R=
3
1
1
at resonance,  =
=
LC
90 103  0.5 106
=
Q.4
1
45 199
=
105
2
10 4
=
3
15  2
The plates of a parallel plate capacitor are completely filled by a solid dielectric. This capacitor is
connected to a battery. Now the plates of the capacitor are pulled slightly apart. At the steady state,
(a.)
(b.)
(c.)
(d.)
the p.d. across the plates has increased.
the energy stored by the capacitor has increased
the capacitance of the capacitor has decreased.
the charge on capacitor has decreased
Ans. CD
Q.5
A jumper jumps upwards . Choose the correct statement.
(a.)
(b.)
(c.)
(d.)
the force exerted by the ground on him while he is attempting to jump is greater than his
weight
work is done by normal force on him while he attempts to jump
Since the feet touching ground remain at rest while he is attempting to jump, force exerted by
ground on him is equal to his weight
Since the feet touching ground remain at rest while he is attempting to jump, work done by
the force exerted by ground on him is zero
Ans. AD
SECTION-C – (Comprehension Type Questions)
Comprehension – 1 ( Next 3 Questions)
A long thin, vertical wire has a net positive charge  per unit length. In addition, there is a current I in the
wire. A charged particle moves with speed u in a straight-line trajectory, parallel to the wire and at a
distance r from the wire. Assume that the only forces on the particle are those that result from the charge
on and the current in the wire and that u is much less than c, the speed of light.
Q.
1
Choose the correct option
(a.)
(b.)
(c.)
If the charge of the particle is positive, the current in wire should be in the same direction as
that of the velocity of charge
If the charge of the particle is negative, the current in wire should be in the same direction as
that of the velocity of charge
The direction of current in wire is opposite to the velocity of charge irrespective of the sign
of charge.
(d.)
The direction of current in wire is same as the velocity of charge irrespective of the sign of
charge.
Ans. D
Sol.
q (E  v  B)  0

qU × B
qE
 V is inside
 i and v are parallel irr. of charge.
Q.
Suppose that the current in the wire is reduced to I/2. Which of the following changes, made
2
simultaneously with the change in the current, can be done if the same particle is to remain moving
parallel to the wire.
(a.)
(b.)
(c.)
(d.)
Ans. D
Sol.
Doubling the charge per unit length on the wire only
Doubling the charge on the particle only
Doubling the distance from the wire.
Doubling the speed of the particle
µi
2k
–v× 0 =0
2 r
r
 speed to be doubled.
PASSAGE – 2 ( Next 3 questions)
A loudspeaker system uses alternating current to amplify sound of certain frequencies. It consists of 2
speakers.
R
Tweeter
R
C
Woofer
L
~
V0 sin (t)
Tweeter - which has smaller diameter produces high frequency sounds. Woofer- which has larger diameter
produces low frequency sound. For purpose of circuit analysis, we can take both speakers to be of equal
resistance R. The equivalent circuit is shown in the figure. The 2 speakers are connected to the amplifier
via capacitance and inductance respectively. The capacitor in tweeter branch blocks the low frequency
sound but passes the high frequency. The inductor in woofer branch does the opposite.
Q.3
What is the frequency which is sounded equally loudly by both speakers
(a.)
1 R2
1

2
2 L LC
(b.)
1 4R 2
1

2
2 L
LC
(c.)
1
1
R2
 2
2 LC 4L
(d.)
1
2 LC
Ans. D
Q.4
For a combination of L,R & C the current in woofer & tweeter are always found to have a phase
difference of /2. What is the relation between L,R & C.
(a.)
L = 2R2C
(b.)
L=
2 R2 C
(c.)
L = R2 C
(d.)
L=
R 2C
2
Ans. C
PASSAGE – 3 ( Next 3 questions)
In the RLC series circuit shown, the readings of voltmeters are V1 = 150 V, V2 = 50 V and the source has
emf 130 V
Q.5
Find the power factor of the circuit
(a.)
(b.)
3/5
4/5
(c.)
12/13
(d.)
5/13
(c.)
70 V and 20 V
(d.)
10 V and 40 V
Sol.rr (VL – VC) = 50
VR2 + (VL – VC)2 = 1302

VR2 = 1302 – 502
V
120 12

power factor = R =
=
V
130 13
Q.6 Values of VL and VC are respectively
(a.)
2
Sol.rr VR +


90 V and 40 V
VL2
=
(b.)
100 V and 50 V
1502
VL = 1502  1202 = 90 V
VL – VC = 50
VC = 90 – 50 = 40 V
SECTION-D
Matrix Match Type Questions
Q.1
Match column – I with column II. (Single match)
Column-I
(i)
(ii)
(iii)
(iv)
A circular conducting loop rotating about
its axis in a uniform constant magnetic
field perpendicular to its plane.
A circular conducting loop moving along
its plane in pure translation in a uniform
constant magnetic field perpendicular to
its plane.
A circular conducting loop placed in a
magnetic field perpendicular to its plane
which is decreasing with time.
A square conducting loop rotating about
its side in a uniform constant magnetic
field perpendicular to its axis. Consider a
time when the plane of the loop is
parallel to the magnetic field.
Column- II
(a.)
Net induced emf in the loop is nonzero
(b.)
Net induced emf is zero but a small
part of the loop may have some emf
induced across it.
(c.)
Induced emf in any small part of the
loop is zero.
(d.)
Induced electric field outside the
conductor is zero.
(e.)
Current flows in the loop.
Ans.
[Ans. (A) R,S (B) Q,S (C) P,T (D) P, S,T ]
Q.2
Four different circuit components are given in each situation of column-I and all the
components are connected across an ac source of same angular frequency  = 200rad/sec. The
information of phase difference between the current and source voltage in each situation of
column-I is given in column-II. Match the circuit components in column-I with corresponding
results in column-II.
Column-I
Column- II
(i)
(a.)
(ii)
(b.)
(iii)
(c.)
(iv)
(d.)
the magnitude of required phase

difference is
2
the magnitude of required phase

difference is
4
the current leads in phase to
source voltage
the current lags in phase to source
voltage
Ans. (A) q,r (B) p,s (C) p,r (D) q, s
1/ C

Sol. (a) tan  =
 =
, current leads source voltage because reactance is capacitive
R
4
(b) Pure inductive circuit  = / 2 , current lags behind source voltage because reactance
is inductive
(c) as R = 0, tan 
/ 2 , current leads source voltage because reactance is capacitive
L

(d) tan  =
= 1  =
, current lags behind source voltage because reactance is
4
R
inductive
SECTION-E
(Integer Type Questions)
Q.1
Side rail of length 2L are fixed on a horizontal plane at a distance  from each other. These ends are
connected by two identical ideal batteries with emf E by resistanceless wires (see figure). On the
rails is a rod of mass m, which may slide along them. The entire system is placed in a uniform
vertical magnetic field B. Assuming that the resistance of the rod is R and the resistance per unit
length of each of the rails equal to , find the period of small oscillations (in sec.) arising from
shifting the rod from the equilibrium along the rails. Neglect friction, internal resistance of batteries
and induced emf in the rod.
[Take : B = T ,  =  volt,  = 0.5m, L = 1m,  = 1/m, R = 0.25 , m = 100 gm. ]
B

m, R

2L
Ans. 0001
Sol. If rod is in middle, i = 0

F=0



 2x 

 2
2(L  x) 3(L  x)
 x 2 
Eq. emf =
= 
1
1
2

2
2(L  x) 2(L  x)
L  x2
2L
1
1
1
1
=
× 2


L  x2
2
R eq. 2(L  x) 2(L  x)

Req. =
(L2  x 2 )
L
(L + x)

x
L
(L – x)
B


(L + x)
(L – x)
x
x
L
i=
=
2
2
2
(L  x )
(L  x 2 )  RL
R
L
x B
 B
ma = F = –iB =
 2
x
2
2
(L  x )  RL L  RL
a=
Q.2
 B
x
m(L2  RL)
 T  2
m(L2  RL)
 B

Consider the circuit shown in the figure. Find the resonance
1
frequency () for the circuit. (Given R = L and
= 5 units).
LC
T = 1 sec.
R
L
C
~
Ans. 2
Q.3
A conducting ring of mass m and radius r has a weightless conducting rod PQ of length 2r and
resistance 2R attached to it along its diameter. It is pivoted at its centre C with its plane vertical, and
two blocks of mass m and 2m are suspended by means of a light in-extensible string passing over it
as shown in the figure. The ring is free to rotate about C and the system is placed in a magnetic field
B (into the plane of the ring). A circuit is now completed by connecting the ring at A and C to a
battery of emf V. Find the value of V so that the system remains static.
Ans. V =
Q.4
mg
R
Br
For the circuit shown, what is the current (in mA) in the ideal inductor when the current in the
battery is 0.50 A in the direction shown.
10mH
i2
i1
10
i
20
12V
Ans. 300 mA
Sol
P.D across battery = 12 v– ir = 12 – 20 × 0.5 = 2
= 10 i1  i1 = 0.2 A
 i2 = i – i1 = 0.3 A
 300 mA
Q.5
An electric dipole of length 2 cm is placed with its axis making an angle of 60° to a uniform electric
field of 105 î N/C. If it experiences a torque of 8 3 k̂ Nm, calculate the magnitude of potential
energy of the dipole.
Ans. 8 J
Subject-Chemistry
Part : 1
SECTION-A
SECTION – A (Single Correct Choice Type)
CH3
1.
H3C – CH2
H
Ag2O/H2O
Major product:

Ph
NMe3
CH3
CH3
Ph
a.
b.
CH3
H
c.
H
d.
Ph
Ph
Ph
CH2
C
Sol. Reaction is Hofmann elimination reaction.
2. The structure of (E)-1-chloro-3-methyl-3-hexene is
a.
Cl
Cl
b.
c.
d. None of these
Cl
3.
C
In the dehydrohalogenation of 2-bromobutane, which conformation leads to formation of cis-2-butene?
CH3
CH3
H
Br
a.
H
A
CH3
H
H
CH3
CH3
b.
Br
H
H
CH3
H
H
Br
H
c.
CH3
H
H
Br
d.
CH3
H
4.
1. Li
Br
Br
a. X is
X is
c.
X is
d.
X is
2,7-Dimethyl octane
2. CuI
x & y are
b.
y
x
CuLi ; y is
CuLi ; y is Br
2 CuLi ;
y is
Br
CuLi ; y is Br
C
Br
1. Li
Br
Sol.
CuLi
2
2. CuI
2,7-dimethyloctane
5.
What is the decreasing stability order (most stable  least stable) of the following carbocations?
+
+
+
+
+
b. 1  4 > 2  5 > 3
a. 3 > 2 > 1 > 4 > 5
c. 3 > 2  5 > 1  4
d. 3 > 1  4 > 2  5
C
6.
What are the products obtained from the following reaction?
Br
HC  CNa
Produc
t
Et2O
H
H
C
C
C
C
+
a.
+
b.
20%
80%
80%
20%
H
C
C
c.
d.
C
100%
B
C
H
SECTION – B (More than One Answer)
NH2
Cl
1.
1. Ph3CK+

2. NH3, NH2
100%
Which of the following statement is true for this reaction?
a. It is elimination cum addition reaction
b. An Aromatic intermediate is formed
c. One of the ‘’ bond of intermediate is formed by sp2 – sp2 overlapping
d. Carbocation intermediate is formed in 2nd reaction
A, B, C
Br
O
1. H3C – C – CH3
Mg/ether
2.
A

2. H 3 O
3. , H+
B
NBS
Ag2O(moist)
C
O3
Zn + H2O
D
E+F
[give H2 gas
with Na]
OH
a. Product C is
b. Product D is
O
OH
H
c. Product E is
d. Product F is
C=O
H
B,C,D
CH2  Br
CH3
Sol. A = PhMgBr
B = Ph  C
C = Ph  C
C = Ph  C
CH2
CH2
E = Ph – C – CH2 – OH
CH2
F=H–C–H
O
3.
CH2  OH
O
Which of the following compounds will give E1 reaction?
Br
Br
a.
C
CH2CH3
b.
Cl
Ph – CH – CH3
H2C = CH – CH – CH3
c.
d.
Cl
A,B,C,D
4.
Among following anions which are more stable than
CH2
CH2
O
Θ
a. CH2  CHCH2 C H2
||
Θ
c. CH3  C C H2
b.
C,D
O
||
Sol.
NO2
Θ
CH3  C C H2O,2N
CH2
-M
5.
d.
In the following reaction, the product(s) formed is(are)
OH
CHCl3
OH–
?
CH3
OH
OH
O
OHC
OH
CHO
CHO
H3C
CH3
P
CHCl2
Q
a. P(major)
B,D
H3C
CH3
CHCl2
R
b. Q (minor)
S
c. R (minor)
d. S (major)
SECTION – C (Comprehension Type)
Comprehension – 1
Type of elimination reaction in which least substituted alkene is major product known as Hofmann’s
elimination. Such reaction occur in following condition:
(X) when base is bulky

(Y) when leaving group is very poor such as fluoride, ammonium group (  NR 3 ) etc.
(Z) when alkyl halide contain one or more double bonds.
1.
What is the major product of the following reaction?
CH3 CH3
|
|
H3 C  C H  C H  C  CH3
|
Me CO ΘK 
 3  

|
Br
CH3
CH3 CH3
|
|
a. H 2 C  CH  C H  C  CH3
|
CH3 CH3
|
|
CH3
CH2 CH3
||
|
b. H3 C  CH  C  C  CH3
CH3
|
c. H3 C  H 2 C  C  C  CH3
d. None of these
|
2.
CH3
A
Which of the following will give product of elimination with LiAlH4.
CH3
|
a. CH3  C H  Cl
b. CH3  C  Br
|
|
CH2  CH3
CH3
CH3
c. CH3  CH2  CH  CH2
|
Br
B
|
CH3
|
d. CH3  C H  CH2  CH2
|
Cl
CH3
C2H5O
CH2 – CH – CH – CH3
3.

Br
CH3
CH3
CH = CH – CH
a.
CH2 – CH = C
b.
CH3
CH3
CH3
CH3
CH2 – CH – CH
c.
OC2H5
A
CH2 – CH2 – C – CH3
d.
CH3
OC2H5
Comprehension - 2
Reimer-Tiemann reaction introduces an aldehyde group on to the aromatic ring of phenol, ortho to the
hydroxyl group. This reaction involves electrophilic aromatic substitution. This is a general method for
the synthesis of substituted salicylaldehydes as depicted below.

OH
ONa
OH
CHO
CHO
aq. HCl
[I]
(intermediate)
CH3
CH3
(I)
CH3
(II)
(III)
4.
Which one of the following reagents is used in the above reaction?
a. NaOH + CH3Cl
b. NaOH + CH2Cl2
c. NaOH + CHCl3
C
Sol. NaOH + CHCl3
d. NaOH + CCl4

OH
 Cl3 C  
 Cl2 C 
Cl3C – H 

Cl
H2O
5.
The electrophile in this reaction is

a.  CHCl
C
c.  CCl 2
b. C Cl3
In Reimer Tiemann reaction : OH  CHCl3 

 Θ  CCl3 


H O
Sol.
Cl
2
base
6.
d.  CCl 3

CCl2
chloro carbene
(electrophile)
The structure of the intermediate I is
 
 
 
 
O Na
O Na
O Na
O Na
CH2Cl
a.
CHCl2
CCl3
b.
B
c.
CH3
CH3
O–
O–
d.
CH3
O–
H
CCl2
Sol.
CH2OH
CH3
O–
CHCl2
CCl2
OH
CHO
+ 2OH-
CHO
aq. HCl

CH3
CH3
CH3
I (intermediate)
CH3
CH3
SECTION – D (Matrix Type)
1.
Match List – I with List – II (Single Match)
Column – I (Statement)
(A)
(B)
(C)
(D)
Θ
(p)
H2S is weaker nucleophile than S H
Θ
Column – II (Reason)
Bulky group present on nucleophilic centre
decreases nucleophilicity
Nucleophilicity decreases on going from
left to right in the period of periodic table
A species with negative charge is a
stronger nucleophile than a similar species
without a negative charge
Nucleophilicity decreases on going down
in the group of the periodic table
Θ
(q)
C H3 is stronger nucleophile than NH2


(r)
R 3 P is stronger nucleophile than R 3 N
Θ
CH3  C H  O is weaker nucleophile than
(s)
|
CH3
Θ
CH3 O
Sol. A – r; B – q; C – s; D – p
2. Match Column – I with Column – II.
Column – I
CH3OH
/ CH3 
Ph

CH

(A)
\ Br

Column – II
(p) E1
CH3
(B)
|
(q) E2
NaNH
2
H3 C  C  Br  
|
CH3
Ph
H2S
(C)
Br
(r)
1st order kinetics
(s)
IInd order kinetics
C2H5O
(D)
F
A  p, r; B  q, s; C  r; D  s
Sol.
SECTION – E (Integer Type)
1.
How many chiral centers are present in tetracycline?
CH3
H
OH
N
H
CH3
OH
C – NH2
O
OH
O
Sol. 5
2. How many chain and position isomers can be obtained from the alkane C6H14?
Sol. 5
3. How many of the following reactants give primary alcohol as a major product when reacts with RMgX
followed by acidification?
O
(i)
CH2 – CH2
(ii) H – C – H
O
(iii) CH2 – CH – CH3
(iv) CH3 – C – H
(v)
(vi)
O
O
O
O
O
||
(viii) CH3 COO  CH/
(vii) CH3  C  CH3
CH3
\ CH
(ix) HCOOC2H5
(x) CH3COCl
3
Sol. 4 ; (i), (ii), (v), (ix)
4. The number of isomeric alkenes (structural as well as stereo) that can be formed by
dehydrohalogenation of the following alkyl halide under E1 condition is
CH3
|
E1
CH3  C  CH  CH2  CH3 
 Pr oducts
|
|
CH3 Br
Sol.
5
CH3
CH3
H
|
|
|
CH2  C  C H  CH2  CH3
CH3  C  C  C CH3
|
|
;
|
|
CH3 H
5.
CH3


|
CH3
CH3
d& 
cis & trans
How many of the following are activating groups for Benzene ring.
CH3
O

CH3  C  CH  CH  CH3

|
||
O
||

O
||
 CH3 ;  N H2 ;  N  O;  Cl ;  C  CH3 ;  NO 2 ;  O  C  R;  C  N H2 ;  C  CH3

|
CH3

Sol.
O
||
4; ( CH3 ;  N H2 ;  C(CH3 )3 ;  O  C  R)