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9 | 3939-3942 | 5 cm |
|
ue (25/6) cm = 4 17 cm The
separation between the objective and the eye-piece should be (7 |
9 | 3940-3943 | |
|
ue (25/6) cm = 4 17 cm The
separation between the objective and the eye-piece should be (7 5 +
4 |
9 | 3941-3944 | 17 cm The
separation between the objective and the eye-piece should be (7 5 +
4 17) cm = 11 |
9 | 3942-3945 | The
separation between the objective and the eye-piece should be (7 5 +
4 17) cm = 11 67 cm |
9 | 3943-3946 | 5 +
4 17) cm = 11 67 cm Further the object should be placed 1 |
9 | 3944-3947 | 17) cm = 11 67 cm Further the object should be placed 1 5 cm from
the objective to obtain the desired magnification |
9 | 3945-3948 | 67 cm Further the object should be placed 1 5 cm from
the objective to obtain the desired magnification 9 |
9 | 3946-3949 | Further the object should be placed 1 5 cm from
the objective to obtain the desired magnification 9 27
(a)
m = ( fO/fe) = 28
(b)
m = f
f
f
e
O
O
1
+25
= 33 |
9 | 3947-3950 | 5 cm from
the objective to obtain the desired magnification 9 27
(a)
m = ( fO/fe) = 28
(b)
m = f
f
f
e
O
O
1
+25
= 33 6
Rationalised 2023-24
350
Physics
9 |
9 | 3948-3951 | 9 27
(a)
m = ( fO/fe) = 28
(b)
m = f
f
f
e
O
O
1
+25
= 33 6
Rationalised 2023-24
350
Physics
9 28
(a)
fO + fe = 145 cm
(b)
Angle subtended by the tower = (100/3000) = (1/30) rad |
9 | 3949-3952 | 27
(a)
m = ( fO/fe) = 28
(b)
m = f
f
f
e
O
O
1
+25
= 33 6
Rationalised 2023-24
350
Physics
9 28
(a)
fO + fe = 145 cm
(b)
Angle subtended by the tower = (100/3000) = (1/30) rad Angle subtended by the image produced by the objective
=
O
140
h
h
f
=
Equating the two, h = 4 |
9 | 3950-3953 | 6
Rationalised 2023-24
350
Physics
9 28
(a)
fO + fe = 145 cm
(b)
Angle subtended by the tower = (100/3000) = (1/30) rad Angle subtended by the image produced by the objective
=
O
140
h
h
f
=
Equating the two, h = 4 7 cm |
9 | 3951-3954 | 28
(a)
fO + fe = 145 cm
(b)
Angle subtended by the tower = (100/3000) = (1/30) rad Angle subtended by the image produced by the objective
=
O
140
h
h
f
=
Equating the two, h = 4 7 cm (c)
Magnification (magnitude) of the eye-piece = 6 |
9 | 3952-3955 | Angle subtended by the image produced by the objective
=
O
140
h
h
f
=
Equating the two, h = 4 7 cm (c)
Magnification (magnitude) of the eye-piece = 6 Height of the
final image (magnitude) = 28 cm |
9 | 3953-3956 | 7 cm (c)
Magnification (magnitude) of the eye-piece = 6 Height of the
final image (magnitude) = 28 cm 9 |
9 | 3954-3957 | (c)
Magnification (magnitude) of the eye-piece = 6 Height of the
final image (magnitude) = 28 cm 9 29
The image formed by the larger (concave) mirror acts as virtual object
for the smaller (convex) mirror |
9 | 3955-3958 | Height of the
final image (magnitude) = 28 cm 9 29
The image formed by the larger (concave) mirror acts as virtual object
for the smaller (convex) mirror Parallel rays coming from the object
at infinity will focus at a distance of 110 mm from the larger mirror |
9 | 3956-3959 | 9 29
The image formed by the larger (concave) mirror acts as virtual object
for the smaller (convex) mirror Parallel rays coming from the object
at infinity will focus at a distance of 110 mm from the larger mirror The distance of virtual object for the smaller mirror = (110 –20) =
90 mm |
9 | 3957-3960 | 29
The image formed by the larger (concave) mirror acts as virtual object
for the smaller (convex) mirror Parallel rays coming from the object
at infinity will focus at a distance of 110 mm from the larger mirror The distance of virtual object for the smaller mirror = (110 –20) =
90 mm The focal length of smaller mirror is 70 mm |
9 | 3958-3961 | Parallel rays coming from the object
at infinity will focus at a distance of 110 mm from the larger mirror The distance of virtual object for the smaller mirror = (110 –20) =
90 mm The focal length of smaller mirror is 70 mm Using the mirror
formula, image is formed at 315 mm from the smaller mirror |
9 | 3959-3962 | The distance of virtual object for the smaller mirror = (110 –20) =
90 mm The focal length of smaller mirror is 70 mm Using the mirror
formula, image is formed at 315 mm from the smaller mirror 9 |
9 | 3960-3963 | The focal length of smaller mirror is 70 mm Using the mirror
formula, image is formed at 315 mm from the smaller mirror 9 30
The reflected rays get deflected by twice the angle of rotation of the
mirror |
9 | 3961-3964 | Using the mirror
formula, image is formed at 315 mm from the smaller mirror 9 30
The reflected rays get deflected by twice the angle of rotation of the
mirror Therefore, d/1 |
9 | 3962-3965 | 9 30
The reflected rays get deflected by twice the angle of rotation of the
mirror Therefore, d/1 5 = tan 7° |
9 | 3963-3966 | 30
The reflected rays get deflected by twice the angle of rotation of the
mirror Therefore, d/1 5 = tan 7° Hence d = 18 |
9 | 3964-3967 | Therefore, d/1 5 = tan 7° Hence d = 18 4 cm |
9 | 3965-3968 | 5 = tan 7° Hence d = 18 4 cm 9 |
9 | 3966-3969 | Hence d = 18 4 cm 9 31
n = 1 |
9 | 3967-3970 | 4 cm 9 31
n = 1 33
CHAPTER 10
10 |
9 | 3968-3971 | 9 31
n = 1 33
CHAPTER 10
10 1
(a)
Reflected light: (wavelength, frequency, speed same as incident
light)
l = 589 nm, n = 5 |
9 | 3969-3972 | 31
n = 1 33
CHAPTER 10
10 1
(a)
Reflected light: (wavelength, frequency, speed same as incident
light)
l = 589 nm, n = 5 09 ´ 1014 Hz, c = 3 |
9 | 3970-3973 | 33
CHAPTER 10
10 1
(a)
Reflected light: (wavelength, frequency, speed same as incident
light)
l = 589 nm, n = 5 09 ´ 1014 Hz, c = 3 00 ´ 108 m s–1
(b)
Refracted light: (frequency same as the incident frequency)
n = 5 |
9 | 3971-3974 | 1
(a)
Reflected light: (wavelength, frequency, speed same as incident
light)
l = 589 nm, n = 5 09 ´ 1014 Hz, c = 3 00 ´ 108 m s–1
(b)
Refracted light: (frequency same as the incident frequency)
n = 5 09 ´ 1014Hz
v = (c/n) = 2 |
9 | 3972-3975 | 09 ´ 1014 Hz, c = 3 00 ´ 108 m s–1
(b)
Refracted light: (frequency same as the incident frequency)
n = 5 09 ´ 1014Hz
v = (c/n) = 2 26 × 108 m s–1, l = (v/n) = 444 nm
10 |
9 | 3973-3976 | 00 ´ 108 m s–1
(b)
Refracted light: (frequency same as the incident frequency)
n = 5 09 ´ 1014Hz
v = (c/n) = 2 26 × 108 m s–1, l = (v/n) = 444 nm
10 2
(a)
Spherical
(b)
Plane
(c)
Plane (a small area on the surface of a large sphere is nearly
planar) |
9 | 3974-3977 | 09 ´ 1014Hz
v = (c/n) = 2 26 × 108 m s–1, l = (v/n) = 444 nm
10 2
(a)
Spherical
(b)
Plane
(c)
Plane (a small area on the surface of a large sphere is nearly
planar) 10 |
9 | 3975-3978 | 26 × 108 m s–1, l = (v/n) = 444 nm
10 2
(a)
Spherical
(b)
Plane
(c)
Plane (a small area on the surface of a large sphere is nearly
planar) 10 3
(a)
2 |
9 | 3976-3979 | 2
(a)
Spherical
(b)
Plane
(c)
Plane (a small area on the surface of a large sphere is nearly
planar) 10 3
(a)
2 0 × 108 m s–1
(b)
No |
9 | 3977-3980 | 10 3
(a)
2 0 × 108 m s–1
(b)
No The refractive index, and hence the speed of light in a
medium, depends on wavelength |
9 | 3978-3981 | 3
(a)
2 0 × 108 m s–1
(b)
No The refractive index, and hence the speed of light in a
medium, depends on wavelength [When no particular
wavelength or colour of light is specified, we may take the given
refractive index to refer to yellow colour |
9 | 3979-3982 | 0 × 108 m s–1
(b)
No The refractive index, and hence the speed of light in a
medium, depends on wavelength [When no particular
wavelength or colour of light is specified, we may take the given
refractive index to refer to yellow colour ] Now we know violet
colour deviates more than red in a glass prism, i |
9 | 3980-3983 | The refractive index, and hence the speed of light in a
medium, depends on wavelength [When no particular
wavelength or colour of light is specified, we may take the given
refractive index to refer to yellow colour ] Now we know violet
colour deviates more than red in a glass prism, i e |
9 | 3981-3984 | [When no particular
wavelength or colour of light is specified, we may take the given
refractive index to refer to yellow colour ] Now we know violet
colour deviates more than red in a glass prism, i e nv > nr |
9 | 3982-3985 | ] Now we know violet
colour deviates more than red in a glass prism, i e nv > nr Therefore, the violet component of white light travels slower than
the red component |
9 | 3983-3986 | e nv > nr Therefore, the violet component of white light travels slower than
the red component 10 |
9 | 3984-3987 | nv > nr Therefore, the violet component of white light travels slower than
the red component 10 4
1 2 10
0 28 10
4 14 |
9 | 3985-3988 | Therefore, the violet component of white light travels slower than
the red component 10 4
1 2 10
0 28 10
4 14 – 2
– 3
m = 600 nm
10 |
9 | 3986-3989 | 10 4
1 2 10
0 28 10
4 14 – 2
– 3
m = 600 nm
10 5
K/4
10 |
9 | 3987-3990 | 4
1 2 10
0 28 10
4 14 – 2
– 3
m = 600 nm
10 5
K/4
10 6
(a) 1 |
9 | 3988-3991 | – 2
– 3
m = 600 nm
10 5
K/4
10 6
(a) 1 17 mm
(b) 1 |
9 | 3989-3992 | 5
K/4
10 6
(a) 1 17 mm
(b) 1 56 mm
10 |
9 | 3990-3993 | 6
(a) 1 17 mm
(b) 1 56 mm
10 7
0 |
9 | 3991-3994 | 17 mm
(b) 1 56 mm
10 7
0 15°
10 |
9 | 3992-3995 | 56 mm
10 7
0 15°
10 8
tan–1(1 |
9 | 3993-3996 | 7
0 15°
10 8
tan–1(1 5) ~ 56 |
9 | 3994-3997 | 15°
10 8
tan–1(1 5) ~ 56 3o
Rationalised 2023-24
351
Answers
10 |
9 | 3995-3998 | 8
tan–1(1 5) ~ 56 3o
Rationalised 2023-24
351
Answers
10 9
5000 Å, 6 × 1014 Hz; 45°
10 |
9 | 3996-3999 | 5) ~ 56 3o
Rationalised 2023-24
351
Answers
10 9
5000 Å, 6 × 1014 Hz; 45°
10 10 40 m
CHAPTER 11
11 |
9 | 3997-4000 | 3o
Rationalised 2023-24
351
Answers
10 9
5000 Å, 6 × 1014 Hz; 45°
10 10 40 m
CHAPTER 11
11 1
(a)
7 |
9 | 3998-4001 | 9
5000 Å, 6 × 1014 Hz; 45°
10 10 40 m
CHAPTER 11
11 1
(a)
7 24 × 1018 Hz (b) 0 |
9 | 3999-4002 | 10 40 m
CHAPTER 11
11 1
(a)
7 24 × 1018 Hz (b) 0 041 nm
11 |
9 | 4000-4003 | 1
(a)
7 24 × 1018 Hz (b) 0 041 nm
11 2
(a)
0 |
9 | 4001-4004 | 24 × 1018 Hz (b) 0 041 nm
11 2
(a)
0 34 eV = 0 |
9 | 4002-4005 | 041 nm
11 2
(a)
0 34 eV = 0 54 × 10–19J (b) 0 |
9 | 4003-4006 | 2
(a)
0 34 eV = 0 54 × 10–19J (b) 0 34 V (c) 344 km/s
11 |
9 | 4004-4007 | 34 eV = 0 54 × 10–19J (b) 0 34 V (c) 344 km/s
11 3
1 |
9 | 4005-4008 | 54 × 10–19J (b) 0 34 V (c) 344 km/s
11 3
1 5 eV = 2 |
9 | 4006-4009 | 34 V (c) 344 km/s
11 3
1 5 eV = 2 4 × 10–19 J
11 |
9 | 4007-4010 | 3
1 5 eV = 2 4 × 10–19 J
11 4
(a)
3 |
9 | 4008-4011 | 5 eV = 2 4 × 10–19 J
11 4
(a)
3 14 × 10–19J, 1 |
9 | 4009-4012 | 4 × 10–19 J
11 4
(a)
3 14 × 10–19J, 1 05 × 10–27 kg m/s (b) 3 × 1016 photons/s
(c) 0 |
9 | 4010-4013 | 4
(a)
3 14 × 10–19J, 1 05 × 10–27 kg m/s (b) 3 × 1016 photons/s
(c) 0 63 m/s
11 |
9 | 4011-4014 | 14 × 10–19J, 1 05 × 10–27 kg m/s (b) 3 × 1016 photons/s
(c) 0 63 m/s
11 5
6 |
9 | 4012-4015 | 05 × 10–27 kg m/s (b) 3 × 1016 photons/s
(c) 0 63 m/s
11 5
6 59 × 10–34 J s
11 |
9 | 4013-4016 | 63 m/s
11 5
6 59 × 10–34 J s
11 6
2 |
9 | 4014-4017 | 5
6 59 × 10–34 J s
11 6
2 0 V
11 |
9 | 4015-4018 | 59 × 10–34 J s
11 6
2 0 V
11 7
No, because n < no
11 |
9 | 4016-4019 | 6
2 0 V
11 7
No, because n < no
11 8
4 |
9 | 4017-4020 | 0 V
11 7
No, because n < no
11 8
4 73 × 1014 Hz
11 |
9 | 4018-4021 | 7
No, because n < no
11 8
4 73 × 1014 Hz
11 9
2 |
9 | 4019-4022 | 8
4 73 × 1014 Hz
11 9
2 16 eV = 3 |
9 | 4020-4023 | 73 × 1014 Hz
11 9
2 16 eV = 3 46 × 10–19J
11 |
9 | 4021-4024 | 9
2 16 eV = 3 46 × 10–19J
11 10 (a)
1 |
9 | 4022-4025 | 16 eV = 3 46 × 10–19J
11 10 (a)
1 7 × 10–35 m (b) 1 |
9 | 4023-4026 | 46 × 10–19J
11 10 (a)
1 7 × 10–35 m (b) 1 1 × 10–32 m (c) 3 |
9 | 4024-4027 | 10 (a)
1 7 × 10–35 m (b) 1 1 × 10–32 m (c) 3 0 × 10–23 m
11 |
9 | 4025-4028 | 7 × 10–35 m (b) 1 1 × 10–32 m (c) 3 0 × 10–23 m
11 11 l = h/p = h/(hn/c) = c/n
CHAPTER 12
12 |
9 | 4026-4029 | 1 × 10–32 m (c) 3 0 × 10–23 m
11 11 l = h/p = h/(hn/c) = c/n
CHAPTER 12
12 1
(a) No different from
(b) Thomson’s model; Rutherford’s model
(c) Rutherford’s model
(d) Thomson’s model; Rutherford’s model
(e) Both the models
12 |
9 | 4027-4030 | 0 × 10–23 m
11 11 l = h/p = h/(hn/c) = c/n
CHAPTER 12
12 1
(a) No different from
(b) Thomson’s model; Rutherford’s model
(c) Rutherford’s model
(d) Thomson’s model; Rutherford’s model
(e) Both the models
12 2
The nucleus of a hydrogen atom is a proton |
9 | 4028-4031 | 11 l = h/p = h/(hn/c) = c/n
CHAPTER 12
12 1
(a) No different from
(b) Thomson’s model; Rutherford’s model
(c) Rutherford’s model
(d) Thomson’s model; Rutherford’s model
(e) Both the models
12 2
The nucleus of a hydrogen atom is a proton The mass of it is
1 |
9 | 4029-4032 | 1
(a) No different from
(b) Thomson’s model; Rutherford’s model
(c) Rutherford’s model
(d) Thomson’s model; Rutherford’s model
(e) Both the models
12 2
The nucleus of a hydrogen atom is a proton The mass of it is
1 67 × 10–27 kg, whereas the mass of an incident a-particle is
6 |
9 | 4030-4033 | 2
The nucleus of a hydrogen atom is a proton The mass of it is
1 67 × 10–27 kg, whereas the mass of an incident a-particle is
6 64 × 10–27 kg |
9 | 4031-4034 | The mass of it is
1 67 × 10–27 kg, whereas the mass of an incident a-particle is
6 64 × 10–27 kg Because the scattering particle is more massive than
the target nuclei (proton), the a-particle won’t bounce back in even
in a head-on collision |
9 | 4032-4035 | 67 × 10–27 kg, whereas the mass of an incident a-particle is
6 64 × 10–27 kg Because the scattering particle is more massive than
the target nuclei (proton), the a-particle won’t bounce back in even
in a head-on collision It is similar to a football colliding with a tenis
ball at rest |
9 | 4033-4036 | 64 × 10–27 kg Because the scattering particle is more massive than
the target nuclei (proton), the a-particle won’t bounce back in even
in a head-on collision It is similar to a football colliding with a tenis
ball at rest Thus, there would be no large-angle scattering |
9 | 4034-4037 | Because the scattering particle is more massive than
the target nuclei (proton), the a-particle won’t bounce back in even
in a head-on collision It is similar to a football colliding with a tenis
ball at rest Thus, there would be no large-angle scattering 12 |
9 | 4035-4038 | It is similar to a football colliding with a tenis
ball at rest Thus, there would be no large-angle scattering 12 3
5 |
9 | 4036-4039 | Thus, there would be no large-angle scattering 12 3
5 6 ´ 1014 Hz
12 |
9 | 4037-4040 | 12 3
5 6 ´ 1014 Hz
12 4
13 |
9 | 4038-4041 | 3
5 6 ´ 1014 Hz
12 4
13 6 eV; –27 |