1 electrostatics chapter 01 : electrostatics · 1 0 chapter 01 : electrostatics 1. q = ne n = 18 19...
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1
Chapter 01 : Electrostatics
1. q = ne
n = 18
19
5.8 10
1.6 10
−
−
××
= 36.25
As n is not an integer, such a charge cannot
exist.
2. Fnet = 1 2
2
0
1 q q
4 rπε(110/100) (90/100) times
i.e., 99
100 times
∴ Net force = 99
100× 100 = 99 N
3. Force on – q1 due to q2 is
1 212 2
Cq qF
b= along X-axis
Force on – q1 due to − q3 is
1 313 2
Cq qF
a= at ∠θ with negative direction of
Y-axis.
∴ x component of force on – q1 is
Fx = F12 + F13 sinθ = Cq12 3
2 2
q qsin
b a
+ θ
i.e., Fx ∝ 2 3
2 2
q qsin
b a
+ θ
4. Asq
Aσ = , to increase the charge density, area
‘A’ should be made very small. 5.
Since the centre of the square lies at origin,
hence each quadrant will have the charge that
cancels the charge of diagonally opposite
quadrant. This results in the zero net charge on
square. 6. At neutral point,
2 2 2 2
0 0
1 20 1 Q
4 (20 10 ) 4 (40 10 )− −× = ×
πε × πε ×
∴ Q = 80 C 7. τ = pE sin θ
∴ p = 26
4
9 10
Esin 10 sin30
−τ ×=
θ °
p = 1.8 × 10–29
Cm
8. According to Gauss’ law,
φ = 0
q
ε
If the radius of the gaussian surface is
increased the outward flux remains constant as
it depends only on the charge enclosed by the
surface.
9. V = 0
1
4πε
q
R
V1 = 0
1
4πε 1
1
q
R and V2 =
0
1
4πε 2
2
q
R
Both the sphere having same potential
V1 = V2
1
1
q
R = 2
2
q
R
∴ 1
2
q
q = 1
2
R
R
10. V = 0
1 q
4 rπε i.e., V ∝ q …Q r is constant
∴ He
H
V
V= He
H
q
q=
2
1
∴ VHe = 2(VH) = 54.4 V 11. φE = E × 4πr
2
Also φE = 0
q
ε
∴ 0
q
ε= E × 4πr
2
q = 4πε0Er2 = 4πε0(Ar)r
2 = 4πε0Ar
3
Substituting values,
q = 3
9
100 (0.20)
9 10
××
C = 8.89 × 10–11
C
12. Work done = change in potential energy
= 0
1
4πε 1 2
1
q q
r –
0
1
4πε1 2
2
q q
r
= 0
1
4πε q1q2
1 2
1 1
r r
−
= 9 × 109 × 12 × 10
–6 × 8 × 10
–6
× 2 2
1 1
4 10 10 10− −
− × ×
= 9 × 96 × 10–3
× 100 100
4 10
−
= 39 96 10 600
40
−× × ×
= 13 J 13. C = 4πε0R
C = 4 × 3.142 × 8.85 × 10−12
× 0.9
= 100.10 × 10−12
∴ C = 100 pF
P
σ0(xy)
(0, 0) O
Electrostatics01
1
Chapter 02 : Current Electricity
1. I = q
t =
n e
t
n = 1 million = 106
∴ I = 6 19
3
10 1.6 10
10
−
−
× ×
= 1.6 × 10–10
A. 2. Resistivity depends only on the material of the
conductor.
3. 1 2R R 9+ = and 1 2
1 2
R R2
R R=
+ ⇒ 1 2R R 18=
2
1 2 1 2 1 2R R (R R ) 4R R− = + −
∴ R1 – R2 = 81 72 3− =
on solving,
R1 = 6 Ω, R2 = 3 Ω 6. Let I1 and I2 be the current through AB and
AD respectively.
To find current through galvanometer Ig,
applying Kirchhoff’s 2nd
law to loop ABDA,
− 5I1 − 10 Ig + 15 I2 = 0
∴ − I1 − 2Ig + 3I2 = 0 ….(i)
Applying Kirchhoff’s 2nd
law to loop BCDB,
−10 (I1 − Ig) + 20 (I2 + Ig) + 10 Ig = 0
∴ −10I1 + 10Ig + 20I2 + 20 Ig + 10Ig = 0
∴ −10I1 + 20I2 + 40Ig = 0
∴ −I1 + 2I2 + 4Ig = 0 ….(ii)
Subtracting equation (ii) from equation (i),
−I1 + 3I2 − 2Ig = 0
−I1 + 2I2 + 4Ig = 0
+ − −
I2 − 6Ig = 0 ∴ I2 = 6Ig ….(iii)
From equations (ii) and (iii),
−I1 + 2(6Ig) + 4Ig = 0
∴ −I1 + 12 Ig + 4Ig = 0
∴ I1 = 16Ig ….(iv)
Adding equations (iii) and (iv),
I1 + I2 = 6Ig + 16Ig
∴ I1 + I2 = 22Ig
∴ 2 = 22 Ig (Q I1 + I2 = 2)
∴ Ig = 2
22=
1
11 A
7. I = 4E E 3E 3 12
5r R 5r R 5(0.2) 20
− ×= =
+ + += 1.7 V
8. In balance condition, no current will flow
through the branch containing S.
9. R1000 = V2/750 and R200 = V
2/P;
Now, R1000 = R200 (1 + α × 800)
So, 2
V
750 =
2V
P(1 + 4 × 10
–4 × 800)
∴ P = 750 (1 + 0.32) = 990 W.
10. Drift velocity vd = I
nAe =
2
I 4
n D e
×π
i.e., vd ∝ 2
1
D
∴ d1
d2
v
v =
2
2
2
1
D
D =
21
2
= 1
4 12. When connected in series,
total resistance of 5 resistors = 5R
∴ Power dissipated = 2V
5R = 5
∴ 2V
R = 25 ….(i)
When connected in parallel,
total power = 5 2V
R
= 5 × 25 = 125 W
13. Drift velocity, vd = µE
µ = dv
E = d
v
V / l
VE
= Q
l
µ = 0.2 2
100
×
= 4.0 × 10
−3 m
2V−1
s−1
14. Let current through 6 volt battery be I1 and
5 volt battery be I2. The circuit can be drawn
as:
Applying Kirchhoff’s second law to loop
ABCDEFA,
8 (I1 + I2) + 2 I1 = 6
∴ 10 I1 + 8 I2 = 6
∴ 5 I1 + 4 I2 = 3 ….(i)
Applying Kirchhoff’s second law to loop
BCDEB,
8 (I1 + I2 ) + 2 I2 = 5
8 I1 + 10 I2 = 5 ….(ii)
I2
I1
8 ΩDC
B E
A Fr = 2 Ω
E = 6 volt
r = 2 Ω
E = 5 volt
Current Electricity02
2
2
Physics Vol-II (Med. and Engg.)
Multiplying equation (i) by 5 and equation (ii)
by 2 then subtracting,
1 2
1 2
1
25I 20I 1516I 20I 10
9I 5
+ =+ =
− − −=
∴ I1 = 5
A9
From equation (i),
I2 =
53 5
9
4
− × =
2
36 =
1
18A
∴ I1 = 5
9A and I2 =
1
18A
Current through external resistance,
I = I1 + I2 = 5
9+
1
18=
11
18A
18. P = I
2R. Current is same so P ∝ R.
In the first case, it is 3R, in second case it is
2R
3, in third case it is
R
3and in fourth case
the net resistance is 3R
2.
i.e., III II IV I
R R R R< < <
∴ III II IV I
P P P P< < <
19. I
A = nevd
∴ vd = I
Ane =
6 26 19
5
4 10 5 10 1.6 10− −× × × × ×
∴ vd = 1
40 1.6× =
1
64m/s
20. P
Q =
1
3 ⇒ Q = 3P
and P 40
Q 40
+
+ =
3
5
∴ P 40
3P 40
++
= 3
5
∴ P = 20 Ω and Q = 3P = 60 Ω 21. S = X || 10 = 8 Ω
∴ 10X
10 X+ = 8
∴ 10X = 80 + 8X
∴ 2X = 80
∴ X = 40 Ω 23. 0.9 (2 + r) = 0.3 (7 + r)
∴ 6 + 3r = 7 + r
∴ r = 0.5 Ω
24. R = 2
8
2 2
50 1050 10
A (50 10 )
−−
−
ρ ×= × ×
×
l= 10
–6 Ω
25. 3
1 1
3
2 2
R (1 t ) 50 (1 3.92 10 20)
R (1 t ) 76.8 (1 3.92 10 t)
−
−
+ α + × ×= ⇒ =
+α + ×
∴ t ≈ 167 °C
26. V
E =l
; E is constant (voltage gradient)
∴ 1 2
1 2
V V=
l l ⇒
1.1 V
140 180=
∴ 180 1.1
V 1.41V140
×= =
1
Chapter 03 : Magnetic Effect of Electric Current
2. The magnetic field B = µ0nI
= 4π × 10−7
× 4000
1/ 2 × 2
Nn
= Q
l
= 64 π × 10−4
T
= 2.01 × 10−2
T ≈ 2 × 10−2
T 3. The field is non-zero only inside the core
surrounded by the windings of the toroid. The
field outside the toroid is zero. 5. F = IlB sin θ = 1.2 × 0.5 × 2 × sin (90°) = 1.2 N 6. The force per unit length is,
2
0 2 IF
4 R
µ= ×
π
If R is increased to 2R and I is reduced to I / 2,
the force per unit length becomes,
2
0 2(I / 2)F
4 2 R
µ′ = ×
π=
2
0 2I 1 F.
4 R 8 8
µ× =
π
7. τ = N I A B sin θ
τ = 1 × 10 × 0.01 × 0.1 × sin 90° = 0.01 Nm 8. Area of coil, A = 8 cm × 6 cm = 48 cm
2
= 48 × 10−4
m2
Maximum torque τ = NBIA
= 2000 × 0.2 × 200 × 10−3
× 48 × 10−4
Nm
= 0.384 Nm ≈ 0.4 Nm.
9. S = G
n 1−, n =
50
10 = 5
12 = G
5 1− =
G
4
∴ G = 48 Ω 10. τ = NIABcos (60°) ⇒ τ = NIAB sin (90°− 60°)
= 500 × 0.2 × 4 × 10–4
× 10–3
×1
2
τ = 2 × 10–5
N-m
11. S = NAB
k
To increase the sensitivity of MCG, N, A and
B should be large but k should be small.
12. Current sensitivity = I
θ=
NAB
k
= 4
8
80 5 10 5
10
−
−
× × ×
= 20 × 106 rad/A
= 20 rad/µA
13. gI 10 S
I 100 S G= =
+
∴ 10 1
10 G 10=
+
∴ G = 100 − 10
∴ G = 90Ω
14. R = g
V
I – G
∴ 0 = 3
V
3 10−× – 100
∴ V = 100 × 3 × 10–3
= 0.3 V 15. The magnetic induction at the centre of a
circular coil of N turns, radius R and carrying
a current I is given by,
B = 0 NI
2R
µ
Here B1 = 0 NI
2R
µand B2 =
( )0N 3I
2R
µ
Now B = 2 2
1 2(B ) (B )+
= 0 NI
2R
µ(1 3)+
∴ B = 02 NI
2R
µ=
0 NI
R
µ
16. r = mv
qB =
v
(q / m)B
= 5
7
2 10
2.5 10 0.05
×
× ×
= 0.16 m = 16 cm
18. S = I
θ =
1
K =
NAB
k
S ∝ N
S
S
′ =
N
N
′
N′ = 125
100 × 48 =
5
4 × 48 = 60
19. RS = g
V
I − G =
15
0.5 − 2
= 150
5− 2
= 30 − 2
= 28 Ω
Magnetic Effect of Electric Current 03
2
2
Physics Vol-II (Med. and Engg.)
20. B1 = 0 I
.4 a
µ θπ
, B2 = 0 I
.4 b
µ θπ
,
as b < a
B2 > B1
∴ Field due to ABCD = B2 − B1 = 0I 1 1
4 b a
µ θ − π
21. The field induction at O due to straight part of
conductor is, B1 = 0
4
µπ
2I
R
The field induction at O due to circular coil is,
B2 = 0
4
µπ
2 I
R
π
Both the fields will act in opposite direction,
hence the total field of induction at O will be,
B = B2 – B1 = 0
4
µπ
2I
R (π – 1)
∴ B = 0I
2 R
µπ
(π – 1)
1
Chapter 04 : Magnetism
1. M ∝ iA 2. M = iA = 12 × 7.5 × 10−4 = 9 × 10−3 Am2 As current in loop is clockwise, direction of
magnetic dipole moment vector is downward. 3. The magnetic dipole moment of the earth, M = iA ∴ M = i(πR2)
∴ i = 2
M
Rπ =
21
12
6.4 10
3.14 6.4 6.4 10
×× × ×
∴ i = 4.97 × 107A ∴ ≈ 5 × 107 A 4. Magnetic dipole moment, M = NiA = Ni.(πr2)
= 5 × 10 × 22
7 ×
7
100 ×
7
100
∴ M = 0.77 Am2 The direction of M is perpendicular to the
plane of the coil. Hence it is along the Z-axis.
5. Mo = e v r
2, Lo = m v r
∴ e
2 m= gyromagnetic ratio = o
o
M
L
6. ν = 6.8 × 109 MHz = 6.8 × 1015 Hz r = 1.06/2 = 0.53 Å = 0.53 × 10–10 m M = i A = e ν π r2
= (1.6 × 10–19) × (6.8 × 1015) ×22
7
× (0.53 × 10–10)2 ∴ M = 9.7 × 10–24 A-m2
7. Magnetic field intensity,
Ba = 0
4
µπ
3
2M
rand Be = 0
4
µπ
3
M
r
∴ B ∝ Mr−3 n = – 3
8. Baxis = 0
4
µπ×
3
2M
d= B
Bequator = 0
4
µπ×
3
M
d =
1
20
3
2M
4 d
µ × π =
1
2Baxis
But, Baxis = B
∴ Bequator = B
2
9. Ba = 0
4
µπ
× 3
2M
r = 10–7 ×
2 3
2 4
(10 10 )−
×
×
∴ Ba = 8 × 10–4 T.
10. Be = 0
4
µπ
.3
M
r= 10–7 ×
3
0.2
(0.08) = 3.9 × 10–5 T
11. Torque on a bar magnet in earth’s magnetic
field (BH) is HMB sin ,τ = θ τ will be maximum
if sin θ = maximum i.e., θ = 90o. 12. τ = MB sin θ ∴ τ = (mL) B sin θ ∴ 25 × 10−6 = (m × 5 × 10−2) × 5 × 10−2 × sin 30 ∴ m = 2 × 10−2 A-m
13. From τ = M B sin θ, M = B sin
τ
θ
∴ M = 24.5 10
0.25 sin30
−×× °
= 0.36 JT–1
15. Angle between geographic meridian and
magnetic meridian = Declination i.e., θ = 58′ Also, angle of dip (δ) is given by,
tan δ = V
H
B 0.3
B 3= ×
4
4
10
0.3 10
−
−×=
1
3
∴ δ = 30°
16. tan δ = V
H
B
B
tan δ′ = V
H
B
B′ = V
H
B
B cos45°
tan δ′ = tan tan30
cos45 cos45
δ °=° °
= 2
3= 0.8164
∴ δ′ = 39°14′. 17. When the magnet is placed with its N pole
towards north of earth, neutral points are obtained on the equatorial line and B2 = BH i.e.,
02 2 3/ 2
M
4 (d )
µ
π + l = BH
∴ BH = 7
2 2 3/ 2
10 0.4
(0.1 0.05 )
− ×+
= 2.86 × 10−5 T
∴ BH = 0.286 G 19. According to Curie law, for a paramagnetic
material,
χ ∝ T
1 ∴ 2
1
χχ
= 1
2
T
T
∴ 3
χ
χ =
2
273 27
T
+ =
2
300
T
∴ T2 = 300 × 3 = 900 K ∴ T2 = 900 − 273 = 627 °C
Magnetism 04
2
Physics Vol-II (Med. and Engg.)
2
20. For a temporary magnet, the hysteresis loop should be long and narrow.
21. I = M
V =
6
3
4 2 1.25 10−× × ×
= 3 × 105 A-m2/m3
∴ I = 3 × 105 A/m 22. M = Niπr2 = 300 × 15 × π × (7 × 10−2)2
∴ M = 69.27 Am2
23. The point P lies on axial line of magnet N1S1.
∴ B1 = 0 13
1
2M
4 (O P)
µ⋅
π =
7
3
10 2 12.5
(0.05)
− × ×
∴ B1 = 0.02 T (along PX) The point P lies on the equatorial line of the
magnet N2S2.
∴ B2 = 0 23
2
M
4 (O P)
µ⋅
π =
7
3
10 12.5
(0.05)
− ×
∴ B2 = 0.01 T (along PY) The resultant magnetic field at point P,
B = 2 21 2B B+ = 2 2(0.02) (0.01)+
∴ B = 2.236 × 10−2 T
24. →
τ = M→
× B→
= MB sin θ At θ = 90°, τmax = MB = 10−4 Nm At θ = 30°, τ = MB sin 30°
∴ τ = 10−4 × 1
2=
1
2× 10−4 Nm
25. µ = B 1.6
H 1000= = 1.6 × 10−3 Tm A−1
µr = 3
70
1.6 10
4 10
−
−
µ ×=
µ π× ≈ 1.3 × 103
χ = µr − 1 = 1.3 × 103 − 1 = 1.299 × 103 ≈ 1.3 × 103
5 cm 5 cm
Y1
B2 B
θ
S1 O1 N1B1P O2
X
S2
N2
1
Chapter 05 : Electromagnetic Induction and
Alternating Current
2. The resistance of copper loop is less than that
of aluminium loop, hence induced current will
be more in the copper loop as compared to that
of the aluminium loop.
3. Charge induced, q = d
R
− φ
= (0 NBA) NBA
R R
− −=
= 450 0.02 100 10
2
−× × ×
∴ q = 5 × 10−3
C.
5. e = MdI
dt
For e = M, dI
dt= 1
6. N1 = 50 turns/cm = 5000 turns/metre = 5000 l
turns in l metre
N2 = 200
A = 4 cm2 = 4 × 10
−4 m
2
M = 0 1 2N N Aµ
l
= 7 44 10 5000 200 4 10− −π× × × × ×l
l
∴ M = 5.024 × 10−4
H
7. XC = 1
2 Cπν
XC ∝ 1
ν
C 2
C 1
(X )
(X ) = 1
2
ν
ν =
50
200 =
1
4
(XC)2 = 4
4 = 1 Ω
8. XL = ωL = 2πνL
L = LX
2πν =
50 7
2 22 50
×
× × = 0.16 H
9. I = RV 36
R 90= = 0.4 A
As V2 = 2 2
R LV V+
∴ VL = 2 2
RV V− = 2 2120 36− = 114.5 V
As VL = I(XL)
∴ XL = LV 114.5286.2
I 0.4= = Ω
As XL = 2πνL
∴ L = LX 286.2
2 2(3.14)60=
πν= 0.76 H
10. The resonant frequency is given by,
ν0 = 1
2 LCπ=
6 8
1
2 100 10 4 10− −×π × × ×
= 6
1
2 2 10−π× ×
∴ ν0 = 610
4π=
25
π× 10
4 Hz
11. Z = ( )22
L CR X X R+ − =
Hence I = 100
10 = 10 A.
12. E = 2 2
R C LV (V V )+ −
= 2 2(40) (80 40)+ −
∴ E = 1600 1600+ = 2(1600) = 40 2 V.
13. ν0 = 1
2 LCπ =
( )6
1
2 5 80 10−
π × ×
= 50
2π =
25
π Hz
14. E = E0 sin (ωt + φ)
Erms = 0E
2=
200
2
Power, P = Erms Irms cosφ
∴ Irms = rms
P
E cos× φ =
1000 2
200 cos60× °= 10 2 A
15. E0 = NABω and I0 = 0E
R
∴ I0 = NAB
R
ω =
1000 2 0.2 60
6000
× × × = 4 A
18. As M = 0 1 2N N Aµ
l
∴ M becomes 4 times.
19. tan φ = LX
R =
2 fL
R
π=
400 (15/16 )
300
π× π =
5
4
∴ φ = tan–1
(5/4)
20. P
S
N
N = P
S
E
E ⇒
S
75
N =
120
2400 =
1
20
∴ NS = 75 × 20 = 1500.
Electromagnetic Induction and Alternating Current 05
1
Chapter 06 : Electromagnetic Waves
1. uE = 2
0
1E
2ε
= 121
8.85 102
−× × × (170)2
∴ uE = 1.28 × 10−7
J/m3
3. Here, By = 10
–6 sin[7 × 10
11 t + 400 πx]
The Y-component of the magnetic field is
given by,
By = B0 sin 2πt
T
+ λ
x
Comparing the given equation with the above
equation,
2π
λ = 400 π
∴ λ = 2
400 = 5 × 10
–3 m
4. λ = c
ν =
8
6
3 10
3.4 10
××
= 88.24 m
8. The pressure exerted is given as,
Pressure (P) = I
c =
8
1.2
3 10× =
8
2
N0.4 10
m
−×
10. Suppose the charge on the capacitor at time t
is Q, the electric field between the plates of
the capacitor is 0
QE .
A=ε
The flux through the area considered is
φE = 0
Q.4A
Aε =
0
4Q
ε
∴ The displacement current,
ID = E0
d
dt
φε = 0
0
4 dQ
dt
ε
ε = 4I.
11. B0 = 0E
c =
8
90
3 10× = 3 × 10
–7 T
12. Charge oscillating sinusoidally is given by
Q = q0 sin ωt
Displacement current,
ID = dq
dt = q0 ω cos ωt
(ID)max = q0 ω = q0 × 2πν
= 2 × 10–6
× 2 × 3.14 × 8 × 105
≈ 10 A
14. E = Bc = 40 × 10−6
× 3 × 108 = 12,000 V/m.
15. v = r r
c
µ ε
= 83 10
1.42 2.31
×
×= 1.65 × 10
8 m/s
Electromagnetic Waves 06
1
Chapter 07 : Ray Optics
1. All others give the images which are either
diminished or of same size. 4. Brightness depends upon aperture or size
while image formation is independent of size
of the mirror.
5. µ = c
v =
100
70 = 1.43
6. m = f
f u− ∴ – 4 =
f
f ( 12)− − ∴ f = –9.6 cm
R = 2 × f = – 19.2 cm
7. 1
v +
1
u =
1
f
∴ 1
v =
1
f –
1
u =
1
18−–
1
27− =
1
54
−
∴ v = – 54 cm
Negative sign of v indicates the screen should
be placed in front of the concave mirror as it is
a real image.
8. Q m = v
u− also
1 1 1
f v u= + ⇒
u u1
f v= +
⇒ u u
1v f
− = −v f
u f u
−⇒ =
− so
fm
f u=
−
9. µg = c
v ∴ v =
g
c
µ =
83 10
1.5
× = 2 × 10
8 ms
–1
v = Thicknessof glassplate(d)
Timerequired to travel through it (t)
∴ t = d
v=
3
8
2 10
2 10
−×
× = 10
–11 s
10. µ = c
v =
8
8
3 10
1.5 10
×
× = 2
ic = sin–1
1
µ = sin
–1 1
2
ic = 30°
11. Lens formula 1
v –
1
u =
1
f
u is always negative, v is positive.
12. m = f
f u+ ∴ – 2 =
1
31
u3+
∴ u = – 0.5 m
−ve sign follows from the sign conventions.
13. f = 1
P =
1
25 = 0.04 m = 4 cm
M.P. = D
1f
+
= 25
14
+
= 7.25
14. M = – o
e
f
f ∴ – 10 = – o
e
f
f
∴ fo = 10 fe and fo + fe = 44
fe = 4 cm ∴ fo = 40 cm.
15. 2
v
µ – 1
u
µ = 2 1
R
µ −µ
∴ 1
v –
1.5
3− =
1 1.5
5
−
− ∴ v = – 2.5 cm
16. 1
f = (µ – 1)
1 2
1 1
R R
−
∴ 1
20+ = (1.55 – 1)
1 1
R R
− −
∴ R = 40 × 0.55 = 22.0 cm
17. wµg = 1.54
1.33, ic = sin
–1
1.33
1.54
= 59° 42′ 18. P = 15.5 – 5.5 = 10 D
∴ f = 1
P = 0.10 m
19. δ = A (µ – 1), δb = A (µb – 1), δr = A (µr – 1)
∴ D2 = A (1.525 – 1)
∴ D1 = A (1.520 – 1)
⇒ D2 > D1
20. M = – o
e
f
f Negative sign for inverted image.
∴ – 5 = – o
e
f
f
∴ fo = 5 fe and fo + fe = 24
∴ fo = 20 cm, fe = 4 cm 21. For the person to see nearby objects clearly
(at 25 cm), the image should be formed at
75 cm by the corrective lens. The image will
then act as an object and the final image will
be formed on retina. The corrective lens is
convex lens.
u = – 25 cm, v = – 75 cm
1
v –
1
u =
1
f ∴
1
75− –
1
25− =
1
f
∴ f = 37.5 cm = 0.375 m
P = 1
f (m) =
1
0.375 = + 2.67 D
22. 1
f = P = (µ – 1)
1 2
1 1
R R
−
∴ P = (1.5 – 1) 1 1
R R
− − +
∴ P = – 0.5 × 2
R = −0.5 ×
2
0.3 = –
10
3 D
Ray Optics07
1
Chapter 8 : Wave Optics
1. As medium changes there is no change in
frequency whereas wavelength in medium
become 1
µ times in air.
3. The ray near the base has to travel larger
distance through prism hence it is delayed than
the ray near the apex which has to travel very
less distance through prism, hence emerges
first.
7. cg = a
g
c
µ=
83 10
1.5
× = 2 × 10
8 m/s
t = g
d
c =
8
8
4 10
2 10
×
× = 2 s
8. w
g
c
c = wµg =
g
w
µ
µ
∴ cw = g
w
µ
µ × cg =
( )
( )
3 / 2
4 / 3× 2 × 10
8
∴ cw = 8910 m/s
4×
9. Angle made with surface = 30°
∴ i = 90° − 30° = 60°
1.5 = sini
sin r
∴ sin r = sin i
1.5 =
sin60
1.5
° = 0.5773
∴ r = 35°17′
Ratio of the width
2
1
W
W=
cos r
cos i =
cos 35 17
cos 60
′°
° =
0.8163
0.5 = 1.633
10. ∆λ = 706 − 656 = 50 nm = 50 × 10
−9 m
As v
c
∆λ=
λ
∴ v = c∆λ
×λ
= 9
8
9
50 103 10
656 10
−
−
×× ×
×
= 2.3 × 107 ms
−1
Wave Optics08
1
Chapter 09 : Interference of light
1. The phase difference of 2π corresponds to
path difference of λ. Hence the phase difference of δ corresponding to path difference x will be
2π
λ =
x
δ or
2 xπ
λ = δ
2. 93 λ = 0.0465 mm = 465 × 10–7 m
∴ λ = 746510
93−× = 5 × 10–7 m = 5000 Å
3. 1
2
I
I =
2122
a
a =
25
4
∴ 1
2
a
a =
5
2
max
min
I
I=
21 2
21 2
(a a )
(a a )
+
− =
( )
( )
2
2
5 2
5 2
+
− =
49
9 4. Fringes overlap so that it looks like uniform
illumination. 6. Different points in filament emit light
independently and have no fixed phase relationship. Hence, interference is not possible.
8. β = β′
∴ D
d
λ =
D
2d
′λ
∴ D′ = 2D 9. yb = yr ∴ (n + 1)λb = nλr ∴ (n +1) × 5 × 10–5 = n × 7.5 × 10–5 ∴ 5n + 5 = 7.5 n ∴ 2.5 n = 5 ∴ n = 2
10. β = D
d
λ
β ∝ D ∴ Decrease in D by 25% will bring 25%
decrease in β.
11. yn = n D
d
λ and ny′ =
n D
d
′ ′λ
As they coincide yn = ny′
∴ n
n′ =
′λ
λ =
900
750 =
6
5
5y′ = y6 = n D
d
λ =
9
3
6(750 10 )2
(2 10 )
−
−
×
×= 4.5 mm
12. For incoherent waves, Imax = nI0
∴ n = max
0
I
I =
32
2 = 16
Interference of Light 09
1
Chapter 10 : Diffraction and Polarisation
of light .11.3
1. 2
1
x
x= 2
1
λ
λ =
3600
5400
∴ x2 = x12
1
λ λ
= (0.3) × 3600
5400 = 0.2 mm
6. dθ = 1′ =
o1
60
= 180 60
π ×
radian
If d is the actual distance between the pillars
and D is distance between the pillars and the
person then,
d
D = dθ
∴ d = D dθ = 15000 × 180 60
π×
≈ 4.36 m
7. According to Brewster’s law,
µ = tan ip
∴ ip = tan–1
(µ)
8. dmin = 10
3
1.22 x 1.22 6600 10 5
D 3 10
−
−
λ × × ×=
×
= 2.01 × 10–3
m
9. R.P. = 7
D 4.2
1.22 1.22 6 10−=
λ × ×
= 6710
1.22× = 5.74 × 10
6
10. a sin θ = λ ∴ a = sin
λθ
= 75 10
sin1
−×°
= 75 10
0.0175
−×
∴ a = 0.028 mm 11. Depending upon amount of diffraction
resolution of an optical instrument is observed. 12. Intensity of light transmitted by the first
polariser is half of the intensity of unpolarised
light = 18 Wm–2
13. The width of central maxima = 2 D
d
λ
= 7
3
2
2 2.1 5 101.4 10 m 1.4 mm
0.15 10
−−
−
× × ×= × =
×
14. R.P. of telescope ∝ (diameter of aperture) 16. According to Brewster’s law,
µ = 1.57 = tan ip
∴ ip = 57.5°, r = 90° – 57.5° = 32.5° 17. Sound waves cannot be polarised because they
are longitudinal. Light waves can be polarised
because they are transverse.
18. Using law of Malus,
Intensity of light transmitted from 1st polaroid,
I1 = A2 cos
2 θ = A
2 cos
2 40°
∴ Intensity of emergent light,
I2 = I1 cos2 40°
= A2 cos
2 40 cos
2 40
= 0.34 A2
Diffraction and Polarization of Light 10
1
Chapter 11 : Dual Nature of Matter and
Radiation
1. Light is packets of energy called as quantas
which is photon. 4. Stopping potential is independent of intensity.
6. E = φ + K.Emax or hc
λ = φ + eV0
∴ λ = 0
hc
eVφ+
7. Einstein’s photoelectric equation is
21mv
2= hv − φ
φ = hν – 21mv
2
9. Given
max
1
hcK.E. = −φ
λ ….(i)
and max
2
hc2K.E. = − φ
λ ….(ii)
Dividing equation (ii) by equation (i), we get
2
1
h c
2h c
− φ λ =
− φ λ
which gives 2 1
1 2
hc(2 )φ = λ −λ
λ λ
10. λ = h
p=
h
mv.
11.
12. λ0 = hc
φ =
( ) ( )34 8
19
6.6 10 3 10
4.125 1.6 10
−
−
× × ×
× ×
= 3000 × 10−10 m = 3000 Å
13. λ ∝ 1
v ⇒ If v = 0, λ = ∞
14. K.E. = eV = 1.602 × 10–19 × 100 = 1.602 × 10–17 J 15. E = 180 eV = 180 × 1.6 × 10−19 = 2.88 × 10−17 J
Now λ = h
2mE
∴ λ = 34
131 17 2
6.6 10
2 9 10 2.88 10
−
− −
×
× × × ×
= 34
24
6.6 10
7.2 10
−
−
××
= 0.916 × 10−10 m = 0.9 Å
I
50°
V = 54V
φ 0
Dual Nature of Matter and
Radiation 11
1
Chapter 12 : Atoms and Nuclei
1. Energy released = E4 − E1
= −2
13.6
4−
2
13.6
1
−
= 12.75 eV 3. In each α-emission, the mass number
decreases by 4 and atomic number decreases
by 2. In each beta emission, the mass number
remains unchanged, but atomic number
increases by 1. 4. T= 50.8 days
∴ λ = 0.693
T=
0.693
50.8 = 1.36 ×10
−2 per day
N = 40% of N0 = 0.4 N0
As, 0N
N= e
λt
∴ ln 0N
N
= λt
∴ t =
0Nn
N
λ
l
= ( )
2
n 2.5
1.36 10−×
l=
2
0.916
1.36×10−
= 67.37 days
6. Energy required 2 2
13.6 13.6
n 10= = = 0.136 eV
7. Shortest wavelength comes from n1 = ∞ to
n2 = 1 and longest wavelength comes from
n1 = 6 to n2 = 5 in the given case. Hence,
2 2
min
1 1 1R
1
= − λ ∞ = R
∴ 2 2
max
1 1 1 36 25 11R R R
5 6 25 36 900
− = − = = λ ×
∴ max
min
λλ
=
900
11R
R
× =
900
11
8. E3 = − 13.6
9= − 1.51 eV
E4 = −13.6
16= − 0.85 eV
∴ E4 − E3 = 0.66 eV 10. Reaction (C) is incorrect because the mass
number is not conserved.
12. R = Ro 1
3A = 1.1 × 10−15
× 1
3(16)
= 2.77 × 10−15
m 14. For the ionization of second He electron, He
+
will act as hydrogen like atom.
Hence ionization potential = Z2 × 13.6
= (2)2 × 13.6
= 54.4 eV
16. 0
21115 115
48 50Cd Sn
β − →
17. nα = A A 200 168 32
84 4 4
′− −= = =
n β = (2nα − Z + Z′) = (16 − 90 + 80)
= 16 − 10 = 6
18. Wave number 1
ν =λ
= 8
1
5896 10−× = 16961 per cm 19. Two months = 2 half-lives. The activity of the
sample will become 2
1
2, i.e., one-fourth in
2 months. 20. As charge number is fixed (= 92), therefore,
number of protons and electrons is same. As
atomic weight is greater by 3, therefore, 238
92 U
contains 3 more neutrons. 21. Penetration power of γ is 100 times of β, while
that of β is 100 times of α. 22. For Lyman series,
Lyman 2 2
max
c 1 1 3RcRc
(1) (2) 4
ν = = − = λ
For Balmer series,
Balmer 2 2
max
c 1 1 5RcRc
(2) (3) 36
ν = = − = λ
∴ Lyman
Balmer
27
5
ν=
ν
23. For ground state, n = 1
For first excited state, n = 2
As rn ∝ n2
∴ radius becomes 4 times. 24. E = mc
2 = (1 × 10
–3)(3 × 10
8)
2 = 9 × 10
13 J
25. nα = A A
4
′−=
232 2086
4
−=
nβ = (2nα – Z + Z′) = (12 − 90 + 82) = 4 27. Nuclear fusion takes place in stars which
results in joining of nuclei accompanied by
release of tremendous amount of energy.
28. As is known, P.E. = −2 K.E. or P.E.
K.E. = −2
Atoms and Nuclei12
1
Chapter 13 : Electronic Devices
1. Donor atoms occupy the energy levels between conduction and valence band placed towards conduction band.
4. For intrinsic semiconductor , ne = nh 5. Both the impurities possess 3 valence
electrons. 6. In case of half-wave rectifier, we get output
current only for positive input as diode is forward biased.
10.
15. Id.c. = 61
π= 19.4 mA
∴ d.c. power output = Id.c.2 × RL
= (19.4 × 10−3)2 × 800 = 0.3 watt 16. P = VI = 9.1 × 40 = 364 mW 17. Vo = Vz = 8 volt 18. The breakdown field = 106 V/m, width of
depletion region = 2.5 × 10–6 m ∴ Vbreakdown = E × d = 106 × 2.5 × 10−6 ∴ Vbreakdown = 2.5 V 19. VBB = IBRB + VBE
∴ RB = BB BE
B
V V
I
− = 6
5.5 0.5
10 10−
−
× = 500 kΩ
20. Av = L
i
R
r
β and Ap =
2L
i
R
r
β = βAv
∴ p
v
A
A = β = 62
21. IE = IC + IB = 25 + 1 = 26 mA
∴ α = C
E
I
I =
25
26
22. I = 3
150 50
5 10
−
× = 20 mA
IL= 5 mA ….[Given] ∴ Zener current IZ = 15 mA
23. IB = i
i
V
R =
0.01
1000 = 0.01 × 10−3 A = 0.01 mA
∴ IC = βIB = 50 × 0.01 = 0.5 mA = 500 µA
24. α = 0.95
∴ β = 1
α
−α=
0.95
1 0.95− = 19
Also, β = C
B
I
I
∆
∆
∴ ∆IC = β(∆IB) = 19 × 0.4 = 7.6 mA 25. The energy of emission,
E = hν = hc
λ =
34 8
10
6.62 10 3 10
5890 10
−
−
× × ×
×
= 3.37 × 10−19 J
= 19
19
3.37 10
1.6 10
−
−
×
× = 2.11 eV
∴ For, λ = 5890 Å, E = 2.11 eV ∴ The condition for emission of electrons is, hν > Eg. But here, hν < Eg [Eg = 3eV] ∴ For emission of electrons, λ < 5890 Å is a must.
B
E C
+
−
−
+Forward Biased
Reversed Biased
Electronic Devices13
1
Chapter 12 : Atoms and Nuclei
1. ν = 1
2 LCπ =
9 6
1
2 2 10 45 10− −π × × ×
= 710
6π = 0.53 MHz
2. Maximum distance covered by space wave
communication
= 6 42Rh 2 6.4 10 100 12.8 10 m= × × × = ×
< 4 × 104 m
= 40 km
Since the distance between receiver and
transmitter is 80 km, the wave can be only
propagated by a sky wave.
∴ νc = ( )1/2
max9 N
= ( )1/ 212
9 10
∴ νc = 9 MHz 6. fSB = fc ± fm = 1800 ± 0.8 = 1800.8 kHz and
1799.2 kHz
7. c c
2Lν= =
λ L
2
λ = Q
83 10
2 0.3
×=
× = 500 MHz.
8. Time = 3
2
8
4000 101.33 10 s
3 10
−×= ×
× ≈ 13 ms
10. Number of stations
= B.W.
2 Highest modulating frequency×
= 240,000
2 12000×
= 10 11. d = 2hR
= 62 250 6.4 10× × ×
= 56.568 × 103m ≈ 56.6 km
12. d = 2Rh
Population covered
= πd2 × population density
= π(2Rh) × ρ
= 22
7× 2 × 6400 × 0.1 × 4000
= 1.6 × 107
13. Since, total power
2
a
PtPcm
12
= +
Here, ma = 0.6
∴ Pc = 2
8 8
(1 0.18)0.61
2
= ++
= 8
6.8 kW1.18
=
Communication Systems14