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78 Electromagnetism Success Magnet-Solutions (Part-II)
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Section - E : Matrix-Match Type
1. Answer A(p, q), B(p, r, s, t), C(p, r, s, t), D(r, s, t)
(A) At (a, 0, 2a ) and (a, 2a , 0) ipE 0. Ep
(B) At (a, 0, 0) and (0, 0, a) jpE and hence Ep is maximum
(C) E at (a, 0, 0) is in +x direction
E at (0, 0, a) is in x direction
(D) At (a, 0, 0) and (0, 0, a) E is parallel to ip , hence Ep is zero.
(E) At (0, a, 0)
E
is in xdirection
2. Answer A(r), B(r), C(q), D(q)
Case I : Saturation charge = C 2E= 2 CE
Maximum energy stored in capacitor = 2 CE2
Heat generated in resistor = 2 CE2
Energy supplied by battery = 4 CE2
Case II : Saturation charge = 2 CE
Maximum energy in capacitor = 2 CE2
Energy supplied by cells = CE2+ CE. 2E = 3 CE2
Heat generated in resistor = 3 CE2 2 CE2= CE2
3. Answer A(p, t), B(q, s), C(p, s), D(r, s)
R
OA
Work done in rotating the dipole equals change in potential energy of the system. Potential of the centre should
be zero, as metal is earthed. Electrons will flow from or to earth to satisfy this.
4. Answer A(p), B(p), C(r, s), D(p)
C2= kC
2(increases)
V2< V
2, V
1> V
1
E1> E
1. p
Force between plates =0
2
2 A
Q
C> CQ> Q
Force increases
Potential difference between pointsXand Yis equal to e.m.f. of cell
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Ratio of energy stored =2
1
C
C
C2> C
2 Ratio increases
5. Answer A(p, r, s), B(t), C(p, r), D(p, q, r)(A) Plane is equatorial
30
1 14 2
b
a
kp qd r dr
a br
(B) E ds
(C) y= 0 is equatorial
(D) x= 0 is axial and some points are between charges.
5a. Answer (1) [JEE (Advanced)-2014]
P. Not along +y Q. Not along +x
Q1
Q2
Q3
Q4
Q1
Q2
Q3
Q4
R. Not along y S. Not along x
Q1
Q2
Q3
Q4 Q
1 Q
2 Q
3 Q
4
6. Answer A(p, q, s), B(r, t), C(s), D(p, r, t)
RC
t
RC
t
eR
EeII
0
RC
tlI 0lnln
RC
1cot
(A) > RC> RC
(B) < RC< RC r, t
I0> I
0 R< Rs
I0< I
0 R> Rp, r, t
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80 Electromagnetism Success Magnet-Solutions (Part-II)
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7. Answer A(q, r, t), B(q, s, t), C(q, s, t), D(q, s, t)
(A) Electric field inside parallel plate capacitor is uniform. So, energy density 20
2
1EUE is non-zero and
uniform.
(B) Electric field between the shells is due to charge on inner sphere which behaves
like a point charge. So, its electric field is non-uniform.
(C) Electric field between the shells is due to charge which has flowed from earth to
inner sphere. It also behaves like a point charge whose electric field is non-uniform.
q q 1+q
1
q1
(D) In the region outside the outer shell, there is electric field due to charge (q q1)
which behaves like a point charge.
8. Answer A(p), B(p, q, r, t), C(p, t), D(p, q, r, t)
(A) VA
= VB
BA
R1 i
1
V1
V3
V2
C1
R2 i
2
R3
R4
(B) Depending on values of parameters shown VA V
B
and VB VCmay be +ve, ve, zero.(C) As potential difference across the plates is same, V
A V
B= 0 is V
B V
C> 0
(D) Depending on values of k1, k
2, k
3V
A V
Bmay be +ve, ve, zero V
B V
C> 0
8a. (1, 4) [JEE (Advanced)-2014]
1 2C C C
0 0
1 2
/ 3 2 / 3,
K A AC C
d d
0( 2)
3
K AC
d
1
2C K
C K
Also,1 2
,
VE E
d where Vis potential difference between the plates.
9. Answer A(p, q, r, t), B(p, r, t), C(p, q, r, s), D(q, t)
5
R
20
2.5 V
Q
i
P
ai 1.025
5.2
V251.05.2 QP VV
p, q, r, t
2 V
R
i
Q
P
i= 0
VP V
Q= 2V
p, r, t
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10
0 V0 V2 V 2 V
Q+ Q2 3 1R1
PQ
i
3 V 1 V
Ai 1.010
12
VP V
Q= 2V
Current through R= 0
Q= CV = 2 1 = 2 C
p, q, r, s
+ Q Q
Q
i
P
i= 0.1 A
VP V
Q= 4 4
0 0.1 = 0 V
Q = CV = 0
= 2 2 = 4C
10. Answer A(r, s), B(q, r, s), C(q, r, s), D(p, q, r, s, t)
(A) is property of material, for conductors it decreases with rise in temperature.
(B) Thermal energy generated in unit volume =
2E
(C) J= E
(D)
AVAEI
11. Answer A(q, r), B(q, r), C(q), D(q)
(A) In parallel = 2x
In series =x
(B) In parallel =x
In series = 2x
(C) Key closed = 0
Key open =x
(D) Key closed =x
Key open =x
12. Answer A(p, s, t), B(s), C(p, q, r, t), D(q)
(A) For current in E2to be zero 1
2
1
E RER r
(B) For maximum power in R, 1 2
1 2
r rR
r r
(C) V1= V
2always
(D) For current in E1to be zero 21
2
E RE
R r
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12a. Answer (1, 2, 4) [JEE (Advanced)-2014]
Using KVL, inABCDEFA, we get
1 2 3 1
0iR V iR V
1 2
1 3
V Vi
R R
Using KVL inABCDA,
1
1 1
1
0 0 V
V iR i R
V1
V2
R1
R2
R3
B
A D
F E
iC
1 1 2
1 1 3
V V V
R R R
1 1 1 3 1 1 2 1
V R V R V R V R
1 2
1 3
V V
R R
Now possible answers are 1, 2, 4.
13. Answer A(r, s, t), B(r, t), C(q), D(p, q, r, s)Thermal power in maximum when R = r.
Maximum power =r
E
4
2
r R= R
P
For any other value Power mg sin respectively.
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(C) jilBF 0
A component of this force is down along the incline. Case is similar to part (A).
(D) jilBF
0
A component of this force is up along the incline. Case is similar to part (B).
23. Answer A(p), B(s, t), C(q), D(r)
(A)
2
0 .4
tan.4
sin.4 L
L
(B)L
L2
2
0
(C) 3
2
2
2
10
2
RRL
(D) L= does not depend on
(p)2
2
0
2
LL
(q)3
2
2
2
10
4
RRL
(r) L= does not depends on
(s)
L
L
2
2
0
(t) Area of cross section of smaller solenoid l2
24. Answer A(p, t), B(p, t), C(q, s), D(q, r)
(p) Flux through the loop is not changing
(q) Flux through the loop is not changing
(r) Flux through the loop is increasing
(s) Flux through the loop is decreasing
(t) Area of cross section of smaller solenoid 2
25. Answer A(p, q, r, s, t), B(q, r, t), C(p, q, r, s, t), D(q)
26. Answer A(p), B(p, q, r, t), C(p), D(q, t)
(A) VA V
C= Zero
(B) Field is nonconservative
Field Lines
a b
cd
(C) Electrostatic field = 0
(D) Electric field at C, positive
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86 Electromagnetism Success Magnet-Solutions (Part-II)
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26a. Answer (4) (IIT-JEE 2009)
According to Lenz law I1 is from ato band I
2from cto d.
27. Answer A(p, s), B(r), C(q), D(r, s)Since no current flows, rod is in uniform motion
28. Answer A(p, r, s, t), B(p, r, s, t), C(p, r, s, t), D(p, r, s, t)
In all the physical situations given in columnI, all the possibilities arise.
(A) = 60, VA
VB= 0
> 60 VA
VB= +ve
< 60 VA
VB= ve
(B) Depending on field is uniform or non-uniform induced current imay be positive, negative, zero.
(C) The rod may be moving in any direction or may be moving with constant speed.dt
dqmay accordingly
be +ve, ve or zero.
(D) Induced emf is such a coil varies sinusoidally, may be zero at an instant and may have any polaritydepending on sense of rotation of the loop.
29. Answer A(p), B(q, t), C(r), D(s)
(A) 104
40
R
VI
R
AI 2100
(B) V0= I
0. Z= 250 V
(C) V502
0 V
V
VL= 40
VC
VR= 40
V2 = (VL V
C)2+ V
R
2
VC= 10 V
(D) 110
10
I
VX
c
c
Vc= 10 volts because V= 50 volts
12 2,
2L C
X X
2 2
100L
,
1
'C
XC
1 100
2C
2
100C
1 1
2 2 2 22
100 100
CL
125Hz
22
100
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30. Answer A(p, t), B(q), C(p, t), D(q)
e.m.f. leads current
XL> X
C
CL
1
On decreasing ,XLX
Cdecreases
Zdecreases
(A)Z
VI
0
0 increases
(B)
Z
R1cos decreases
(C) P =Z
V2
increases
(D) Zdecreases
Incase of resonance, current amplitude and therefore power developed is maximum
31. Answer A(p, r, s, t), B(q, r, s), C(p, r, s, t), D(p, r, s, t)
(A) XL=X
C resonance
A110
10
Z
VIRZ , I
1= 1A
VL= I X
L= 10 V
VC= IX
C= 10 V
(B) ZXX CL
VR= 0 But V
L=VC= 10 V
I= 0, I= 0
(C) All potential is dropped across Land CVL= V
C= 10 V
22
11
1
2
1
RXX CL
XL= X
C
Z = R
A110
10
Z
VI
VL= V
C= V
R= 10 V, 1
101
10I A
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(D) XL= X
C
Equivalent to case in part - A
I1= 1A
31(a) Answer A(r, s, t), B(q, r, s, t), C(p, q), D(q, r, s, t) IIT-JEE 2010
(p) V1= 0, V
2= V(XL= 0)
(q) XL= 0, V
1= 0, V
2= V
(r) XL= L= 2fL= 1.884
V1= IX
L
V2= IR
V1< V
2
(s) XL= 1.884
XC=
31 1010
3L
XC>> X
L
V2= IX
C, V
1= IX
L
(t) R= 1 k
31010
3CX
XC> RV
2> V
1
Also, V2= IXL, V1= IR.