Teacher’s Manual
TEXTBOOK SOLUTIONS
Exercise
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1.1 ...................................................1
2.1 ...................................................2
3.1...................................................2
3.2...................................................4
4.1 ...................................................5
4.2 ...................................................5
4.3 ...................................................6
4.4...................................................6
5.1...................................................7
5.2...................................................9
5.3 .................................................10
6.1 .................................................11
6.2 .................................................11
6.3 .................................................12
7.1 .................................................12
7.2 .................................................13
7.4 .................................................14
7.5 .................................................15
8.1 .................................................17
8.2 .................................................17
8.3 .................................................18
9.1 .................................................19
9.2 .................................................20
9.3 .................................................20
10.1 .................................................22
10.2 .................................................23
10.3 .................................................23
10.4 .................................................24
10.5 .................................................24
10.6 .................................................25
10.7 .................................................26
10.8 .................................................28
11.1 .................................................28
11.2 .................................................29
11.3 .................................................30
11.4 .................................................31
11.5 .................................................31
12.1 .................................................32
12.2 .................................................32
12.3 .................................................33
12.4 .................................................33
13.1 .................................................34
13.2 .................................................35
13.3 .................................................35
Exercise
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14.1 .................................................36
14.2 .................................................36
15.1 .................................................36
15.2 .................................................37
16.1 .................................................39
16.2 .................................................40
16.3 .................................................40
17.1 .................................................41
17.2 .................................................42
18.1 .................................................42
18.2 .................................................43
18.3 .................................................44
19.1 .................................................44
19.2 .................................................45
20.1 .................................................47
20.2 .................................................48
20.3 .................................................49
21.1 .................................................50
22.1 .................................................51
23.1 .................................................51
23.2 .................................................53
23.3 .................................................54
24.1 .................................................54
24.2 .................................................55
27.1 .................................................56
27.2 .................................................56
28.1 .................................................57
28.2 .................................................57
28.3 .................................................59
28.4 .................................................59
28.5 .................................................60
28.6 .................................................60
29.1 .................................................61
29.2 .................................................62
30.1 .................................................64
30.2 .................................................64
30.3 .................................................65
31.1 .................................................66
32.1 .................................................67
32.2 .................................................68
32.3 .................................................68
32.4 .................................................69
33.1 .................................................69
1
Teacher’s Manual
TEXTBOOK SOLUTIONS
–1
Exercise 1.1
(vi) 100 km h = 100 000 m h
Q1 Area = length × length ⇒ Unit of area
2
= (m)(m) = m = the square metre
–1
–1
000
------------------- m s = 27.78 m s
= 100
60 × 60
–9
Q2 P = mv ⇒ Unit of p = (Unit of m)(Unit of v)
–1
–1
= (kg)(m s ) = kg m s = the kilogram metre
per second
–1
–2
Unit of v
ms
- = ------------- = m s
Q3 Unit of a = ---------------------
Q5
–6
–5
9
(ix) 5 Gm = 5 × 10 m
Q10 W = Fs ⇒ 1 Joule = 1 N m
–3
Mass
kg m , density = ------------------
Volume
–3
kg
- = kg m
⇒ Unit of density = -----3
m
F
of Force
-------------------------------P = --- ⇒ Unit of P = Unit
A
Unit of Area
Newton
- = Newton per square
= --------------------------------Metre squared
=Nm
(viii) 10 µW = 10 × 10 W = 1 × 10 W
–2
= the metre per second squared
Q4
(vii) 5 nN = 5 × 10 N
Q9 F = ma ⇒ Unit of force = (kg)(m s )
–2
⇒ 1 N = 1 kg m s
s
Unit of t
–1
–2
2
Q11 1 J = 1 N m = 1(kg m s )(m) = 1 kg m s
–1
2
–2
–1
2
–2
Q12 1 W = 1 J s = (1 kg m s )s = 1 kg m s
–3
metre
–2
Q6 (a) 10 000 (i.e. 100 × 100 )
(b) 1 000 000 (i.e. 100 × 100 × 100 )
(c) 1000
Q7
2
(i) 5 cm = 5 × 10
–4
m
2
–4
m = 4 × 10
m
3
2
(ii) 40 cm = 40 × 10
3
(iii) 1 cm = 1 × 10
–6
2
–6
3
–3
m
2
3
(iv) 456 cm = 456 × 10 m
–4
3
= 4.56 × 10 m
9
–6
(v) 1 000 000 000 = ( 1 × 10 ) × 10
3
3
3
= 10 m = 1000 m
Q8
3
5
(i) 105 km = 105 × 10 m = 1.05 × 10 m
(ii) 57 mm = 57 × 10
(iii) 6.67 × 10
– 11
–3
m = 5.7 × 10
–2
m
– 11
–2
cm = ( 6.67 × 10 ) (× 10 )
– 13
= 6.67 × 10 m
27
27
–3
(iv) 6 × 10 grams = ( 6 × 10 ) (10 ) kg
24
= 6 × 10 kg
–3
–3
kg
g
9 × 10 kg
9 × 10
---------------------------- = -------------------- ------(v) 9 --------=
3
3
–6
3
cm
cm
1 × 10
m
3
= 9 × 10 kg m
–3
1
Real World Physics
Exercise 2.1
Exercise 3.1
8
3.8 × 10
--------------------- = ---------------------- = 1.27 s
Q1 t = Distance
8
Speed
Q1 u = 30
3 × 10
Q2 Incident ray, reflected ray, normal, angle of
incidence, angle of reflection.
Q5 Answer = 1 m (see diagram)
⇒
⇒
⇒
1m
v = 50
f=?
1
1
1
--- + --- = --u
v
f
1
1
1
------ + ------ = --30
50
f
5+3
1
------------ = --150
f
150
f = --------- = 18.75
8
focal length = 18.75 cm
2m
Q2 u = 20
v = 30
Image virtual
⇒
16
8
3.97 × 10
--------------------- = --------------------------- = 1.32 × 10 s
Q6 t = Distance
8
Speed
3 × 10
⇒
= 4.2 years
Q7 From the diagram (i) 50° (ii) 50° (iii) 40°
30˚
Q3 u = 15
40˚
20˚
50˚
– 1--- = --1v
⇒ f = 60 cm
f = 10
1
1
1
--- + --- = --u
v
f
1
1
1
------ + --- = -----15
v
10
1
1
1
--- = ------ _ -----v
10
15
3–2
1
--- = -----------v
30
1
1
--- = -----v
30
10˚
30˚
⇒
1
1
1
------ – ------ = --20
30
f
3–2
1
------------ = --60
f
1
1
--- = -----f
60
1
--u
v=?
⇒ v = 30 cm
Image is real (v + )
------ = 2
Magnification = --v- = 30
Q9 From the diagram, length = 0.9 m
u
Image is twice the size of object
⇒ its height is 4 cm
0.3 m
0.6m
0.3 m
0.6 m
1.2m
0.6 m
2
15
1.8 m
f
Teacher’s Manual
Q4 u = 10
f = 20
1
1
1
--- + --- = --u
v
f
1
1
1
------ + --- = -----10
v
20
1
1
1
--- = ------ – -----v
20
10
v=?
⇒
v = –20 cm
Q7 Real image: --v- = 2 ⇒ v = 2u
u
1
1
1
--- + --- = --u
v
f
1
1
1
--- + ------ = -----u
2u
50
2+1
1
------------ = -----2u
50
⇒
∴ Image is 20 cm from mirror. It is virtual.
2u = 150
------ = 2
m = --v- = 20
Virtual image:
u
10
v
--u
Q5 f = 40
Magnification = 4 = --vu
⇒ v = 4u.
Find u.
1
1
1
--- + --- = --u
v
f
1
1
1
--- + ------ = --u
4u
f
4+1
1
------------ = --4u
f
5
1
4u
------ = --- ⇒ f = -----4u
f
5
5 × 40
⇒
u = --------------4
= 2 and
1
--u
u = 75 cm
– 1--- = --1v
v = 2u
1
1
1
--- – ------ = -----u
2u
50
2–1
1
------------ = -----2u
50
⇒
2u = 50
u = 25 cm
-----i.e. 40 = 4u
5
Q8 Image upright
1
--u
– 1--- = --1v
f
u = 50 cm
Q6
1
(i) Real image: 1--- + 1--- = ----u
v
20
v
--- = 3 ⇒
u
1
1
1
--- + ------ = -----u
3u
20
3+1
1
------------ = -----3u
20
4
1
------ = -----3u
20
v = 3u
3u = 80
⇒
1
1
1
--- – ------ = --------u
3u
100
3–1
1
------------ = --------3u
100
= 3 and
⇒ Virtual image
Magnification = 3 = --vu
⇒ v = 3u
⇒ 200 = 3u
--------- = 66⅔ cm
u = 200
3
Q9 Image is real and at the focus since light from a
distant object arrives as parallel light.
Q10 Magnification = 1--u = 26⅔ cm
⇒
(ii) Virtual image
v
--u
f
1
--u
– 1--- = --1-
v = +3u
1
1
1
--- – ------ = -----20
u
3u
3–1
1
------------ = -----3u
20
2
1
------ = -----3u
20
v
f
⇒
⇒
⇒
3
v
1
--- = --⇒
u
3
1
1
1
--- + --- = --u
v
f
1
1
1
------ + --- = -----3v
v
40
4
1
------ = -----3v
40
--------v = 160
3
---------
u = 3v = 3  160
3 
u = 3v
= 160 cm
3u = 40
------ = 13⅓ cm
u = 40
3
3
Real World Physics
Exercise 3.2
Q1 1--- – 1--- = – --1u
v
Q5 v = 4 cm
hi = 4, ho = 6
m = --v-
f
u
v
--u
u = 10, f = 12, v = ?
m=
1
-----10
Image is virtual and
5.45 cm behind mirror.
1
– 1--- = – ----v
12
1
1
1
------ + ------ = --10
12
v
6+5
1
------------ = --⇒ 1--60
v
v
60
v = ------ = 5.45 cm
11
=
5.45
---------10
= .545
-----= 11
60
Q2 u = 30, f = 12, v = ?
1
1
1
--- – --- = – --u
v
f
1
1
1
------ – --- = – -----30
v
12
1
1
1
------ + ------ = --30
12
v
Q6
1
--u
---------- = 0.29
Magnification = --v- = 8.57
Q3 f = 10, Mag. =
1
1
1
--- – --- = – --u
v
f
1
1
1
------ – --- = – -----4v
v
10
1–4
1
------------ = – -----4v
10
Q4
1
--u
–
1
--v
=
– --1f
v = 10
∴
=
1
1
- = – -----– -----u
20
( --3- )
30
1
--4
+3
-----4v
⇒ 4v = u
-----= +1
u = 4v ⇒
10
30 = 4v
u = 30 cm
m=
1
--u
–
v
--u
2
--5
1
--v
of image
--------------------------------------= Height
Height of object
= --vu
= – --1f
1
1
1
------ – ------ = – --25
10
f
2–5
1
------------ = – --50
f
–3
1
------ = – --50
f
------ = 16⅔
⇒ f = 50
3
1
3
1
--- – --- = – -----u
u
20
–2
1
------ = – -----⇒
u
20
u = 40 cm
Q7 hi = ½ho
h
-----i
ho
v = 7.5 cm
m=
2u = 5v
2u = 5(10)
u = 25
cm
3
Virtual
u
ho
u
6
u
4
4
--- = --- ⇒ u = 6 cm
6
u
1
1
1
--- – --- = – --u
v
f
1
1
1
--- – --- = – --6
4
f
2–3
1
------------ = – --12
f
1
1
– ------ = – --⇒ f = 12
12
f
m = --v- = 1--u
3
1
1
1
--- – --- = – ------ Find u.
u
v
20
v = u---
⇒ v = 8.57 cm
v
--u
h
m = -----i = --v- ⇒ 4--- = --v-
= 1--- = --v-
⇒ u = 2v
1
1
1
------ – --- = – -----2v
v
40
1–2
1
------------ = – -----2v
40
1
1
- = -----+ ----2v
40
⇒ v = 20
u
2
∴ u = 40 cm
Q8 m = 1--- = --v1
--u
cm
–
2
1
--v
u
= – --1f
v = u--2
1
--u
1
1
- = – --– -----u
f
( --2- )
– 1--- = – --1u
f
∴
f=u
∴ Object must be a distance in front of
mirror equal to its focal length.
4
Teacher’s Manual
Exercise 4.1
Exercise 4.2
Q1
an g
Sin 25°
.4226
- = ------------- = 1.496989 = 1.5
= ---------------------
Q1
Q2
an d
60°
------------------ = 2.42
= Sin
Real depth
- = an w
Q2 ------------------------------------
Q3
an w
Sin 16.4°
.2823
Sin 21°
=
Sin 30°
-----------------Sin r
Sin 30°
-----------------1.33
= 1.33 ⇒ Sin r =
= .3759
an g
Real depth
5
- = ---------- = 1.5
= -----------------------------------Apparent depth
Apparent depth
10
------------------------------------∴
Apparent depth
⇒
3.33
= 1.33
10
- = 7.52 m
Apparent depth = --------1.33
⇒ r = 22.1°
1
1
2
- = ------- = --= ------
Q4
an g
= 1.5
Q5
gnw
1
1
- = ---------- = 0.88
= -------
Q6
Sin 35°
gna = -----------------an g
gna
Sin 69°
1
1
------- = --------------Sin 35 °
gn a
------------------Sin 69 °
Sin i
- = dn a
Q7 ----------------Sin 15°
⇒ Sin i =
1.5
3
1.13
wng
=
ang
Sin 15°
-----------------2.42
Real depth
Real depth
4
- = anw⇒ ------------------------- = --Q3 -----------------------------------Apparent depth
⇒ Real depth =
0.8
4 × 0.8
---------------3
3
= 1.07 m
Q4 Real depth = 6
Appears as cube ⇒ Apparent depth = 5
6
∴ Refractive index = --- = 1.2
5
69°
------------------ = 1.63
= Sin
Sin 35°
Sin i
1
------------------ = ---------Sin 15°
2.42
0.2588
---------------- = 0.10695
2.42
∴
=
–1
i = Sin (.10695) = 6.1395, i.e. i = 6.1°
Sin r
- = 2.42 ⇒ Sin r = 0.8277
Q8 ----------------Sin 20°
⇒ r = 55.9°
Q5 Let x = the real depth = thickness of block, i.e.
Real depth = x, Apparent depth = x – 3.33
an g
Real depth
x
- ⇒ 1.5 = ------------------= -----------------------------------Apparent depth
x – 3.33
1.5x – 1.5 ( 3.33 ) = x
1.5x – x = ( 1.5 ) ( 3.33 )
0.5x = 4.995
------------- = 9.99 cm
x = 4.995
0.5
5
Real World Physics
Exercise 4.3
Q1 n =
Exercise 4.4
c air
------------------c medium
⇒ cmediun =
c air
-------n
Sin C
8
3 × 10
----------------1.5
=
8
= 2 × 10 ms
–1
8
c air
–1
8
3 × 10
- = ----------------- = 1.24 × 10 ms
Q2 cd = -------
n
1
1
- = ------------------- = 1.56
Q1 n = -------------
2.42
Sin 40°
1
1
- = ------------------------ = 2.40
Q2 n = ------------Sin C
Sin 24.6°
1
Q3 n = ------------Sin C
8
c air
3
3 × 10
- = ----------------- = --- = 1.5
Q3 n = ------8
cm
c
cm
2
2 × 10
8
3 × 10
air
- = ------------------------- = 2.4
Q4 n = ------8
Q4
⇒
1
Sin C = 1--- = ------
⇒
⇒
Sin C = 0.8333
C = 56.4°
1.2
1
- = 0.7518797
Sin C = 1--- = --------n
1.25 × 10
1.33
⇒
Q5
n
C = 48.75°
1
- ⇒
Sin C = 1--- = --------n
1.66
C = 37.04°
1
1
1
- = ------------------- = ---------------- = 1.56
Q6 n = ------------Sin C
Sin 40°
0.6428
Real depth
12
- ⇒ 1.56 = ------------------------------------n = -----------------------------------Apparent depth
Apparent depth
⇒ Apparent depth = 7.69 cm
Q7 n = 1.33
⇒
tan C
i.e.
1
-------------Sin C
= --r4
= 1.33 ⇒ C = 48.75°
⇒
r = 4 tan( 48.75° )
radius = 4.56 m
Air
r
C
4m
C
Water
------- ⇒ C = 48.99 °
Q8 Tan C = 2.3
2
1
1
- = ------------------------- = 1.33
n = -----------Sin C
6
Sin 48.99°
Teacher’s Manual
Q9 The angle of incidence of the ray on the glass
air surface is 45° . In going from glass to air the
angle of incidence must be greater than the
critical angle if total internal reflection is to
occur, i.e. for total internal reflection 45° > C
i.e. C < 45°
⇒ Sin C < Sin 45° ⇒ Sin C < 0.707
1
1
∴ --------------- > ------------Sin
C
Exercise 5.1
Q1 1--- + 1--- = --1u
v
⇒
f
1
-----40
1
1
- = --+ ----25
f
⇒ f = 15.38 cm, real.
Q2 1--- + 1--- = --1u
v
⇒
f
1
--------100
1
1
1
1
- ⇒ --- = ------ – --------+ 1--- = ----v
30
v
30
100
100 – 33
- ⇒ v = 42.86 cm
= -----------------------( 30 ) ( 100 )
0.707
1
- = refractive index of the glass
But -------------Sin C
1
- = 1.414
and ------------
h
-----i
ho
= --v-
hi
---4
------------= 42.86
u
100
0.707
∴ Refractive index must be greater than 1.414
( 42.86 )
- = 1.71 cm
hi = 4-------------------100
Image is 1.71 cm high, real, and 42.86 cm
from lens.
Q3 1--- + 1--- = --1u
=
v
2–3
-----------60
=–
1
-----20
⇒
f
1
-----60
1
1
1
1
- ⇒ --- = ------ – -----+ 1--- = ----v
30
v
30
20
⇒ v = – 60
⇒ Image is virtual and 60 cm from lens.
h
-----i
ho
------ (2) = 6 cm
= --v- ⇒ hi =  60

u
20
Image is 6 cm high.
v
Q4 --u- = 2, f = 50
Real image
Virtual image
1
1
1
--- + --- = --u
v
f
1
1
1
--- + ------ = -----u
2u
50
2+1
1
------------ = -----2u
50
1
1
1
--- – --- = --u
v
f
1
1
1
--- – ------ = -----u
2u
50
1
1
------ = -----2u
50
⇒
⇒
2u = 150
u = 75 cm
u = 25 cm
7
Real World Physics
1
--u
Q5 f = 20
+ 1--- = --1v
Let f be the focal length of lens.
f
Image is twice the height of object
⇒ m = 2 = --v⇒ v = 2u
u
Real image
Virtual image
Then for postion (i) we have 1--- + 1--- = --1-
(1)
also u + v = 80 ⇒ u = 80 – v
(2)
1
1
1
--- + --- = --u
v
f
1
1
1
--- + ------ = -----u
2u
20
1
2+1
------------ = -----2u
20
1
3
------ = -----20
2u
In postion (ii)
⇒
u
1
1
1
--- – --- = --u
v
f
1
1
1
--- – ------ = -----u
2u
20
2–1
1
------------ = -----2u
20
1
1
------ = -----2u
20
2u = 60
u = 30 cm
⇒
1
--u
u = 10 cm
m=
∴
=
0.08
+ = --1f
1
1
---------------- + -----10
0.2667
v
--u
1
--u
3u = 0.08v
= --1-
1
--------4
-----)
( 15
1
--f
( v – 16 ) ( 96 – v )
( v – 16 ) ( 96 – v )
2
1
1
1
48 + 32
80
- = --- ⇒ --- = ---------------------- = -----------+ ----48
f
( 32 ) ( 48 )
f
u
f=
1
--4
1536
v
u
m
Q9 (i)
(ii)
Position
1
80 cm
16 cm
Position
2
32
4
2
48 – 16
32
2
- = ------ = --Postion (ii) m = --v- = -----------------
Q8
8
v – 16
6
3
------ = --- = --Postion (i) m = --v- = 48
f
i.e. f = 25 cm
80 cm
80 – v + 16
⇒ f = 19.2cm
15
1
1
------ + --- = --4
4
f
15 + 1
1
--------------- = --- ⇒
4
f
u
v
32v = 1536 ⇒ v = 48
u = 80 – v ⇒ u = 32
1
-----32
+ 1--- = --14
1
1
- + -------------+ 1--- = --------------------------
2
15
4
-----15
∴ This becomes
96v – ( 16 ) ( 96 ) – v + 16v = 80v – v
v
m = --v- = 15 ⇒ v = 15u ⇒ u = -----
⇒u=
From (2) u = 80 – v
( v – 16 ) ( 96 – v ) = v ( 80 – v )
Q7 Convex lens, since a concave lens does not
form a real image.
=
v – 16
= 3.85
f = 0.2597 m
f = 26.0 cm
+
u + 16
80
= --------------------------------------
u = 0.2667
v=4
v
80
---------------------v ( 80 – v )
u
(3)
f
v – 16 + 96 – v
= --------------------------------------
f
1
--v
v – 16
v + 80 – v
----------------------v ( 80 – v )
1
--f
1
--u
1
1
- = --+ -------------
f
3.75 + 0.1 = --1-
⇒
Object distance = u + 16
Image distance = v – 16
1
1
- + -------------+ 1--- = --------------
1
-------------80 – v
1
--v
0.08 )10
u = (-------------------3
f
From (1) and (3)
3
- , i.e.
Q6 Here v = 10 and magnification = --------3
---------0.08
1
--------------u + 16
∴
v
32 + 16
48
3
u and v are interchangable.
∴ Distance between lenses = 13.3 – 8.0
= 5.3 cm
1
--u
+ 1--- = --1v
f
⇒
1
---------13.3
⇒ f = 4.995 cm
1
1
- = --+ -----8.0
f
Teacher’s Manual
Exercise 5.2
Q1 Concave lens formula is ⇒ 1--- – 1--- = – --1u
v
Q4 Mag. = 1--- = --v- . Let focal length be f.
f
2
u = 2v ⇒ v = u---
u = 30, f = 20
1
-----30
– 1--v
1
– --v
1
– --v
–
=
=
=
u
2
1
--v
1
= – ----20
1
1
- – -----– ----20 30
–3–2
---------------60
–5
-----60
⇒
1
--v
=
5
-----60
1
--u
1
--u
60 = 5v
-----v = 60
5
–
–
1
--v
1
--u
--2
=
=
– --1f
1
– --f
1
2
1
--- – --- = – --u
u
f
1–2
1
------------ = – --u
f
v = 12 cm
Concave lens image is always virtual.
⇒ –
1
--u
= – --1- ⇒ u = f
f
i.e. the distance from the object to the lens is
equal to the focal length of the lens.
Q2 f = 20 ; u = 20 ; v = ?
Concave lens ⇒
1
1
1
--- – --- = – --u
v
f
1
1
1
------ – --- = – -----20
v
20
1
1
1
– --- = – ------ – -----v
20 20
1
2
– --- = – -----v
20
Mag. = --v-
u
10
1
= ------ = --20
2
Height of image
⇒ --------------------------------------Height of object
2v = 20
v = 10 cm
=
1
--2
∴ Height of image
= 1--- Height of object =  1--- (5) = 2.5 cm
2
2
Image is virtual; 10 cm from lens, of height
2.5 cm.
Q3 m = 1--- ⇒ --v- = 1--3
u
1
1
1
--- – --- = – --u
v
f
1
1
1
------ – --- = – -----3v
v
60
1–3
1
------------ = – -----3v
60
–2
–1
------ = -----3v
60
3
⇒ u = 3v
f = 60
⇒ 3v = 120
⇒ v = 40 cm
u = 3v
⇒ u = 120 cm
9
Real World Physics
Exercise 5.3
Q1
Q2
Q3
Q4
–1
1
- = +2.5 m
(i) p = --1- = ------
f
0.4
1
1
- = –1.67
(ii) p = --- = -----f
0.6
1
- m = 8.33 cm
f = --1- = ----p
12
1
- = 40 m
f = --1- = -----------p
0.025
m
Q9 Let f 1 = focal length of convex lens.
1
1
1
--- = ----- + ----f1
f2
f
1
1
1
----- = ------ + -----f1
40
40
–1
1
1
1
--- = ----- + ----f1
f2
f
–1
1
1
------ = ------ – ----40
20
f2
1
1
1
----- = ------ + -----f2
20
40
–1
1
- = 0.0625 m = 6.25 cm
(ii) f = --1- = -----
Q5
16
(i) p = p 1 + p 2 = – 4 – 8 = – 12 m
–1
1
- = 0.0833 m = 8.33 cm
(ii) f = --1- = ----p
Its sign is + ∴ convex.
Q7 For convex lens f = 10 cm
1
------0.1
= 10 m
–1
For concave lens f = 15 cm
⇒ p2 =
–1
---------0.15
= –6.67
p = p 1 + p 2 = 10 – 6.67 = 3.33 m
–1
⇒ Focal length of combination
1
- = 0.3 m = +30 cm
= --------3.33
Sign + ⇒ combination is convex.
ho = 2 cm,
1
--u
u = 20, v = ?, f = 30
+ 1--- = --1v
f
1
1
------ + 1--- = ----- ⇒ v = – 60 cm
20 v 30
⇒ Image is virtual.
h
-----i
ho
------ ( 2 ) = 6 cm
= --v- ⇒ h i =  60
20
u
1
1
1
- = ----- + ------ ⇒ f 1 = 30 cm
Q8 ----20
10
f1
f 2 = 13.33 cm
12
Q6 p = p 1 + p 2 = – 0.02 + 0.05 = 0.03 m
⇒ p1 =
60
1
-----40
1
1
= ----– ----f1
40
⇒ f 1 = 20 cm
Q10 Let f 2 = focal length of concave lens.
(i) p = p 1 + p 2 = 6 + 10 = 16 m
p
⇒
–1
Teacher’s Manual
Exercise 6.1
Q1
Exercise 6.2
–3
--------------------------------Q1 Average velocity = Displacement
–2
(i) 20 × 10 s = 2 × 10 s
(ii) 500 ms = 500 × 10
–3
Time
= 0.5 s
(iii) 4000 ms = 4000 × 10
–3
=
= 6.67 m s
–1
South East
Q2 s = ut = ( 30 ) ( 10 ) = 300 m South
s=4s
(iv) 1 µs = 1 × 10 s
–6
Q3
(i) s = ut = 10 × 1 = 10 m
(v) 50 µs = 50 × 10 s = 5 × 10 s
(ii) s = ut = 10 × 10 = 100 m
(vi) ½ hr = (½)(60)(60) s = 1800 s
(iii) s = ut = ( 10 ) ( t ) = 10t metres
(vii) Two days = (2)(24)(60)(60) s = 172 800 s
(iv) s = ut = ( 10 ) ( 2 × 10 ) = 2 × 10 m
–6
–5
6
Q2 1 × 10 s =
(i) 2 m North
–5
A
(ii) 2 m South
6
1 × 10
--------------------------------( 24 ) ( 60 ) ( 60 )
–6
Q4
(x) One year = (365)(24)(60)(60)
= 31 536 000 s
days = 11.57 days
2
(iii) 2 m West
Q3 no working out reqd.
Q4
2000 m South East
--------------------------------------------5 × 60 s
(iv) 2 m East
θ
B
(i) 6.25 km = 6.25 × 10 m
AC
2
2
2
= 2 + 2 ⇒ AC =
C
2
3
8 = 2.83
–2
(ii) 1 cm = 1 × 10 m
(iii) 20 mm = 20 × 10
tan θ = 2--- = 1 ⇒ θ = 45°
–3
2
–2
= 2 × 10 m
(v) 2.83 m W 45° N (North West)
–9
(iv) 4 nm = 4 × 10 m
(vi) 2.83 m E 45° S (South East)
--------------------Q5 (a) Average speed = Distance
Q5 Distance = perimeter of semi-circle = ½( 2 π r )
= π r = 5 π = 15.71 m
Time
2000
- = 8.33 ms
= ------------------
–1
( 4 ) ( 60 )
Displacement of B from A = 10 m East
–1
--------- = 10 ms
(b) Average speed = 100
10
15.71
--------------------- = ------------(i) Average speed = Distance
Time
= 1.571 m s
Distance
200 000
- = ------------------- = 8000 s
Q6 Time = ----------------------------------
Q7 Time =
Average speed
Distance
----------------------------------Average speed
=
25
2 000 000
-----------------------200
10
(ii) Average velocity
= 10 000 s
–1
10
--------------------------------- = -----= Displacement
Time
10
–1
= 1 m s East
Q8 Distance travelled = Circumference of circle
= 2 π r = (2)( π )(30)
Speed =
Q9
Distance
--------------------Time
=
( 2 ) ( π ) ( 30 )
--------------------------( 90 )
=
2π
-----3
–1
= 2.09 m s
(i) s = ut = ( 2 ) ( 25 ) = 50 m
(ii) s = ut = ( 2 ) (½)(60)(60) = 3600 m
–3
(iii) s = ut = ( 2 ) ( 1 × 10 ) = 2 × 10
(iv) s = ut = 2t metres
–3
m
Q6 If the speed of the car is changing it means the
car would travel 30 m in the next second if its
speed stopped changing at that instant. If the
speed of the car is constant then it will travel
30 m in the next second.
Q7 No; Yes, an object moving in a circle at a
steady speed. The velocity is changing since
the direction of motion is changing but the
speed is constant.
11
Real World Physics
Q8 no working out reqd.
Q9 Magnitude of overall displacement
=
2
2
3 +4 = 5m
Exercise 7.1
–2
v–u
30 – 10
20
2
- = ------------------ = ------ = 6 --- m s North
Q1 a = ----------
Q2
Direction of overall displacement is East θ °
North where Tan θ = 4--- ⇒ θ = 53.13 °
Q3
Average velocity =
Q4
=
5m
--------5
=1ms
–1
3
Displacement
--------------------------------Time
E 53.13 ° N
t
3
3
3
–2
v–u
10 – 0
a = ----------- = --------------- = 5 m s West
t
2
–2
–2
25 – 40
------------------ = –1.5 m s = 1.5 m s in
10
direction to the initial motion.
(i) Velocity gained = Acceleration × Time
–1
= (3)(1) = 3 m s
(ii) (3)(4) = 12 m s
–1
(iii) (3)(13.5) = 40.5 m s
Exercise 6.3
(iv) 3t m s
–1
–1
Q1 No calculations required.
–2
v–u
10 – 0
- = --------------- = 1.667 m s
Q5 a = ----------
Q2 No calculations required.
Q6 a =
Q3 No calculations required.
12
opposite
t
v–u
----------t
=
–2
6
9 West – 5 East
--------------------------------------2.5
= 5.6 m s West
– ( –5 )
= 9------------------2.5
Teacher’s Manual
Exercise 7.2
Q1 u = 10, a = 2, t = 12, v = ?, s = ?
v = u + at
= 10 + 2 ( 12 )
= 34 m s
s = ut + 1--2- at
2
= ( 10 ) ( 12 ) +
–1
Q7 u = 20, a = – 3, v = 0, t = ?
v = u + at
0 = 20 – 3 ( t )
2
1
--- ( 2 ) ( 12 )
2
------ = 6.667 s
t = 20
= 120 + 144
= 264 m
3
Q8
(i) u = 120, a = 10, t = 12, v = ?
Q2 u = 14, v = 30, t = 20, a = ?, s = ?
s = ut + 1--2- at
v = u + at
30 = 14 + a ( 20 )
16 = 20a
s = ( 14 ) ( 20 ) + 1--2- ( 0.8 ) ( 20 )
2
v = 240 m s
s = 280 + 160
s = 440 m
-----a = 16
20
= 0.8 m s
v = u + at
v = 120 + ( 10 ) ( 12 )
2
v = u + at
v = 120 – 10 ( 12 )
Q3 u = 2, v = 12, s = 50, a = ?, t = ?
2
2
2
2
12 = 2 + 2a ( 50 )
144 – 4 = 1000
a = 1.4 m s
v = u + at
12 = 2 + 1.4t
v = 0ms
Q9
10
- = 7.14 s
t = ------
=
=
=
=
u + at
60 + a4
4a
–2
– 10 m s
s = ut + 1--2- at
–1
v = 25 m s in original direction
of motion
2
s = ( 60 ) ( 4 ) + 1--2- ( – 10 ) ( 4 )
2
(ii) u = 5, a = – 4, v = ?, t = 5
s = 160 m
2
3000 = 6 ( 60 ) + 1--2- a ( 60 )
2
3000 – 360 = 1800a
a = 1.467 m s
(i) u = 5, a = 4, v = ?, t = 5
v = u + at
v = 5 + 4(5)
Q5 u = 6, t = 60, s = 3000, a = ?, v = ?
s = ut + 1--2- at
–1
1.4
–2
Q4 u = 60, v = 20, t = 4, a = ?, s = ?
v
20
– 40
a
in original direction
of motion.
(ii) u = 120, a = – 10, t = 12, v = ?
–2
v = u + 2as
–1
v = u + at
v = 6 + ( 1.467 ) ( 60 )
v = 94.02 m s
v = u + at
v = 5 + ( –4 ) ( 5 )
v = – 15 m s
–1
–1
i.e. 15 m s in
opposite direction of
original motion v.
–1
–2
Q6 u = 30, v = 10, s = 200, a = ?
2
2
v = u + 2as
2
100 = 30 + 2a ( 200 )
a = –2 m s
–2
13
Real World Physics
Q10 Distance S 1 of rear of bike from P after t
seconds.
Exercise 7.4
Q1 a =
v–u
----------t
=
S 1 = 12t
Distance S 2 of front of car from P after t
seconds.
2
s = ut + 1--2- at
S2 = ( 0 )( t ) +
2
1
--- ( 2 )t
2
= t
2
2
2
t – 12t = 0, i.e. t ( t – 12 ) = 0
⇒ t = 0 or t = 12
⇒ Car meets bike after 12 seconds
Distance of bike from P at this instant
= S 1 = ( 12 ) ( 12 )
= 144 m from P.
14
= 445.3 cm s
–2
= 4.453 m s
–2
–2
–1
4 × 10
- = 0.3 m s
Q2 u = -------------------
v=
Car meets bike when S 2 = S 1 i.e. t = 12t
4.3   1.45
 --------- – --------- 0.04  0.04
-------------------------------------1
8  ------
 50
0.1333
–2
4 × 10
-------------------0.0167
= 2.4 m s
–1
2
2
2
2
–u
( 2.4 ) – ( 0.3 )
= ----------------------------------s = 1.4 ∴ a = v----------------
2s
2 ( 1.4 )
= 2.025 m s
–2
Teacher’s Manual
Exercise 7.5
Q1 u = 0, s = 60, a = 9.8, t = ?, v = ?
2
2
2
2
v =
1176
(ii) Alternative Solution
2
2
v = u + 2as
2
2
v = 80 + 2 ( – 9.8 ) ( 96 )
–1
v = ± 67.22 m s
v = u + at
67.22 = 80 – 9.8t
t = 1.30 s
– 67.22 = 80 – 9.8t
t = 15.02 s
v = u + 2as
v = 0 + 2 ( 9.8 ) ( 60 )
v = 34.29 m s
v = u + at
34.29 = 9.8t
t = 3.5 s
–1
Q2 u = 200, a = – 9.8, v = 0, s = ?, t = ?
2
2
v = u + 2as
v = u + at
0 = 200 – 9.8t
t = 20.41 s
2
0 = ( 200 ) + 2 ( – 9.8 ) s
2
s =
200
----------19.6
= 2040.8 m
Q6
(i) u = ?, s = 100, t = 2, a = – 9.8
s = ut + 1--2- at
100 = u ( 2 ) + --12- ( – 9.8 ) ( 2 )
Q3 u = 0, a = 9.8, t = 3, s = ?
s = ut + 1--2- at
2
100 = 2u – 19.6
⇒ u = 59.8 m s
2
s = ( 0 ) ( 3 ) + 1--2- 9.8 ( 3 )
2
–1
2
(ii) u = 59.8, v = 0, s = ?, a = – 9.8
s = 44.1 m
Q4 u = 0, v = 22, s = 30, a = ?, t = ?
2
2
v = u + 2as
2
2
2
v = u + 2as
2
22 = 0 + 2a ( 30 )
a = 8.067 m s
–2
v = u + at
22 = 0 + 8.067t
t = 2.73 s
0 = ( 59.8 ) + 2 ( – 9.8 )s
s = 182.45 m
(iii) First find the two times at which it is
100 m above ground.
u = 59.8, a = – 9.8, s = 100, t = ?
Q5
(i) u = 80, v = 0, a = – 9.8, s = ?
2
2
v = u + at
2
80 + 2 ( – 9.8 ) ( s ) 0 = 80 – 9.8t
80
2
t = -----80
9.8
---------- = 326.5 m
19.6
t = 8.16 s
96, t = ?
2
ut + 1--2- at
v = u + 2as
0 =
s =
(ii)
s =
s =
96 = 80t – 1--2- ( 9.8 )t
2
s = ut + 1--2- at
2
100 = ( 59.8 )t – ( 1--2- )9.8t
2
2
4.9t – 59.8t + 100 = 0
2
59.8 ± ( 59.8 ) – 4 ( 4.9 ) ( 100 )
t = -------------------------------------------------------------------------( 2 ) ( 4.9 )
t = 10.204 s or t = 2
Time difference 10.204 –2 = 8.204 s
2
4.9t – 80t + 96 = 0
2
80 ± 80 – 4 ( 4.9 ) ( 96 )
t = ---------------------------------------------------------
( 2 ) ( 4.9 )
⇒ t = 15.02 or 1.304 s
15
Real World Physics
Q7
(i) u = 24, v = 0, s = ?, a = – 9.8
2
2
v = u + 2as
Q8 On Earth:
u = ?, s = 2, a = – 9.8, v = 0
2
2
0 = u + 2 ( – 9.8 ) ( 2 ) ⇒ u =
⇒ Total height above ground:
= 29.39 + 16
= 45.39 m
u = 6.26099 m s
On Moon:
u = 6.26099 =
2
t 1 = 2.449 s
t2 =
s = ?, v = 0
2
If body 1 moves for t seconds, body 2 moves
for t – 10 seconds. Distance of body 2 from A:
S 2 = ( 0 ) ( t – 10 ) + 1--2- 2 ( t – 10 )
2
1
--- at 2
2
2
Bodies meet when: S 1 = S 2
2
40t = ( t – 10 )
2
2
40t = t – 20t + 100 ⇒ t – 60t + 100 = 0
2
2
60 ± 60 – 4 ( 1 ) ( 100 )
t = -------------------------------------------------------2
60 ± 56.57
= ------------------------2
t = 58.28 s or t = 1.715 s
t must be greater than 10 ∴ t = 58.28 s
Total time = t 1 + t 2
= 2.449 + 3.044
= 5.493 s
(ii) Alternative solution
u = 24, a = – 9.8, t = ?, s = – 16
It takes the second body 58.28 – 10 = 48.28 s
to catch up.
Distance from A = S 1 = ut
2
1
--- at
2
– 16 = 24t – 1--2- 9.8t
– 9.8
----------,
6
⇒ s = 12 m
( 2 ) ( 45.39 )
-------------------------- = 3.044 s
9.8
s = ut +
39.2, a =
9.8
- )s
0 = 39.2 + 2 ( –--------6
Time t 2 from greatest height to ground:
u = 0, a = 9.8, s = 45.39, t 2 = ?
45.39 = 0 + 1--2- 9.8t 2
–1
Q9 First body: after t seconds of motion it will be
a distance S 1 from A where S 1 = 40t
o = 24 – 9.8t 1
s = ut 2 +
39.2
v = u + 2as
(ii) Time t 1 from point of projection to
greatest height:
u = 24, v = 0, a = – 9.8, t 1 = ?
v = u + at 1
2
v = u + 2as
2
0 = 24 – 2 ( 9.8 )s
s = 29.39 m
2
= ( 40 ) ( 58.28 )
= 2331.2 m
2
4.9t – 24t – 16 = 0
[Solve: t = 5.49 s or a negative solution]
(iii) From greatest height to ground:
u = 0, a = 9.8, t = 3.044, v = ?
v = u + at
v = 0 + ( 9.8 ) ( 3.044 )
v = 29.83 m s
–1
Q10
u = u, a = – 9.8, t = 6, v = – 8
v = u + at
– 8 = u + ( – 9.8 ) ( 6 )
u = 50.8 m s
s = ut + 1--2- at
–1
2
2
s = ( 50.8 ) ( 6 ) – ( 1--2- ) ( 9.8 ) ( 6 ) = 128.4 m
Height above ground = 128.4 + 40
= 168.4 m
16
Teacher’s Manual
Exercise 8.1
Exercise 8.2
2
2
2
Q3 (iv) R = 3 + 4 = 25 ⇒ R = 5 N
tan θ = 3--4- ⇒ θ = 36.87°
(v)
2
2
2
R = 2 +2 = 8⇒R =
tan θ =
2
2
--2
2
= 1 ⇒ θ = 45°
2
(vi) R = 5 + 12 = 169
8 = 2.83 N
Q1 (a) R 1 = 2 2 + 2 2 = 2.83
∴ Overall resultant = 2.83 +5 = 7.83 N in
direction of 5 N force
2
2
(b) R 1 = 3 + 3 = 4.24 N
Resultant = 6 + 4.24 = 10.24 N in
direction of 6 N force.
2
⇒R =
tan θ =
169 = 13
5
-----12
⇒ θ = 22.62°
2
(c) R 1 = 10 + 10 = 14.14 N
R = 14.14 – 10 = 4.14 N
at 45° to each of the perpendicular 10 N
forces.
2
2
(d) R 1 = 4 + 4 = 5.66
R = 5.66 – 2 = 3.66 N (in opposite
direction to 2 N force.
17
Real World Physics
Exercise 8.3
Q1 Horizontal component = 200 Cos 70 ° = 68.40 N
Vertical component = 200 Sin 70 ° = 187.94 N
Q2
(i) Horizontal force on cart = Horizontal
component = 3500 Cos 25 ° = 3172.1 N
Q10
(i) Displacement: Comp along ox = 2 Cos 30 °
= 1.732 m
Comp along oy = 2 Sin 30 = 1.0 m
y
(ii) Vertical force on cart = Vertical
component = 3500 Sin 25 ° = 1479.2 N
2
30˚
Q3 Horizontal component = 100 Cos 60 ° = 50
Vertical component = 100 Sin 60 ° = 86.6
x
O
Q4 Horizontal component = 200 Sin 40 ° = 128.6
Vertical component = 200 Cos 40 ° = 153.2
Q5 Resolve velocity into components parallel to
and perpendicular to bank.
Parallel comp = 6 Cos 40 ° = 4.596 m s
–1
Perpendicular comp = 6 Sin 40 ° = 3.857 m s
(ii) Velocity: Comp along ox = –5 Sin 30
–1
= –2.5 m s
–1
Comp along oy = 5 Cos 30 = 4.33 m s
y
–1
5
30˚
Time to cross lake =
=
Width of lake
--------------------------------Perp comp
2000
------------- = 518.5
3.857
60˚
30˚
s
2m
P
30˚
x
O
Distance travelled parallel to side in this time
= (Parallel comp) × Time
= (4.596)(518.5) = 2383 m
Distance from start= ( 2000 ) 2 + ( 2383 ) 2
= 3111.1 m
(iii) Acceleration:
Comp along ox = –3 Cos 30
–2
= –2.60 m s
–2
Comp along oy = –3 Sin 30 = –1.5 m s
Q6 Comp parallel to roof = 40 Sin 30 ° = 20 N
Comp perp to roof = 40 Cos 30 ° = 34.64 N
Q7 Horizontal component = 4 Cos 30 ° = 3.46 m s
–2
Vertical component = 4 Sin 30 ° = 2 m s
–1
y
Q8 Parallel component = 30 Cos 20 ° = 28.19 N
Perpendicular component = 30 Sin 20 ° = 10.26 N
Q9 Parallel component = W Cos θ
Perpendicular component = W Sin θ
18
3
30˚
P
30˚
O
x
Teacher’s Manual
Exercise 9.1
Q1 F = ma = ( 20 ) ( 5 ) = 100 N
(ii) s = ut + 1--2- at 2
–2
F
4
- = ------ = 0.4 m s
Q2 a = ---
s = 0 + ( 1--2- ) ( 4 ) ( 20 )
Q3 F = ma = ( 100 ) ( 2 ) = 200 N
s = 800 m
m
2
10
------------ = 1333.3 kg
Q4 m = F--- = 4000
u = 80, v = 0, a = ?, t = 0.1
Q5 u = 6, F = 40, m = 10, t = 12
v = u + at
0 = 80 + a ( 0.1 )
a
3
–2
F
40
- = ------ = 4 m s
(i) a = ---
m
10
a =
(ii) v = u + at
v = 6 + ( 4 ) ( 12 )
v = 54 m s
–2
F
40 – 10
- = ------------------ = 1.5 m s
Q9 (a) a = ---
(b) a =
(c) a =
2
= 72 + 288
= 360 m
m
F
---m
F
---m
=
=
20
–2
100 – 20
--------------------- = 4 m s
20
–2
600 – 600
------------------------ = 0 m s
20
v = u + at
v = 20 + ( 1.5 ) ( 2 ) = 23 m s
v = 20 + ( 4 ) ( 2 ) = 28 m s
Q6 u = 2, v = 10, t = 4
v = 20 + 0 = 20 m s
v = u + at
10 = 2 + a4
a = 2ms
–2
5
2
= ( 6 ) ( 12 ) + 1--2- ( 4 ) ( 12 )
= 800 m s
F = ma = ( 800 ) ( 500 ) = 4 × 10 N
–1
1
(iii) s = ut + --2- at
– 80
--------0.1
–1
–1
–1
–2
F
500 – 400
- = ------------------------ = 16.67 m s
Q10 a = ---
m
–2
F = ma = ( 20 ) ( 2 ) = 40 N
Q11 Car 100 km h
(i) w = ( 1 ) ( 9.8 ) = 9.8 N
–3
(ii) w = ( 1 × 10 ) ( 9.8 ) = 9.8 × 10 N
(iii) w = ( 105 ) ( 9.8 ) = 1029 N
–1
–1
× 1000
--------------------------- = 27.78 m s
= 100
60 × 60
u = 27.78, v = 0, s = 100, a = ?
v 2 = u 2 + 2as
0 = ( 27.78 ) 2 + 2 ( a ) ( 100 )
Q7 w = mg
–3
6
27.78 ) 2
a = (-------------------200
a = 3.859 N
(iv) w = ( m )9.8 = 9.8 m newtons
Q8 F = ma
------------ = 500 kg
m = F--- = 2000
a
4
u = 0, t = 20, a = 4, v = ?
F = ma = (1200)(3.859) = 4630 N = the force
needed to stop the car in 100 m
Force available = 2000 N
Answer = No
Force = 4630 N
(i) v = u + at
v = 0 + ( 4 ) ( 20 )
v = 80 m s
–1
19
Real World Physics
Exercise 9.2
Exercise 9.3
Q1 Driving force = 600 N, because if car is moving
at uniform speed, resultant force is zero.
–2
F
1000 – 600
- = --------------------------- = 0.5517 m s
a = ---
m
725
u = 0, v = 100 km h
a = 0.55, t = ?
–1
Q1 P = mv = (800)(20) = 16 000 kg m s
Q2 (a) P = mv = (1200)(30) = 36 000 kg m s
(b) P = mv = (1200)(0) = 0 kg m s
–1
= 27.778 m s ,
–1
–1
–1
–1
× 1000
--------------------------- = 27.78 m s
(c) 100 km/hr = 100
60 × 60
v = u + at
27.778 = 0 + 0.5517t ⇒ t = 50.35 s
v–u
Q2 Deceleration of bullet = ----------
P = mv = (1200)(27.78)
–1
= 33 333.3 kg m s
–1
–1
(d) 500 cm s = 5 m s
–1
P = mv = (1200)(5) = 6000 kg m s
t
4
–2
50 – 200
- = 3 × 10 ms
= --------------------
0.005
4
Force = ma = ( 0.002 ) ( 3 × 10 ) = 60 N
–1
000
1
---------------- = 33 --- m s
Q3 P = mv ⇒ v = ---P- = 40
m
Q3
(i) Object moving at constant speed
⇒ a = 0 ⇒ T = 19 600 N
(ii) Accelerating down F = ma
⇒ 19 600 – T = ( 2000 ) ( 2 )
T = 15 600 N
Q4 Reading on balance = Tension
Net force = Mass × Acceleration
v=
P
---m
=
40 000
---------------2
= 20 000 m s
(iii) 10g – T = 10 ( 2 ) ⇒ T = 78 N
(iv) 10g – T = 10 ( 0 ) ⇒ T = 98 N
(v) 10g – T = 10 ( 9.8 ) ⇒ T = 0 N
Q5 Reading on scales = Normal reaction N
Net force = Mass × Acceleration
(i) N – 980 = 100(0) ⇒ N = 980 N
(ii) N – 980 = 100(0) ⇒ N = 980 N
(iii) 980 – N = 100(0) ⇒ N = 980 N
(iv) N – 980 = 100(3) ⇒ N = 1280 N
(v) 980 – N = 100(3) ⇒ N = 680 N
(vi) 980 – N = 100(9.8) ⇒ N = 0 N
Q6 Force = Rate of change of momentum
---------- = 45 N
= 2.25
0.05
20
3
–1
Q4 Total momentum = (20)(40) + (50)(–20)
= –200
–1
= 200 kg m s in the direction in which the
50 kg mass moves.
Q5 Momentum before = Momentum after
000
---------------( 800 × 20 ) + (1500)(0) = 2300v ⇒ v = 16
2300
(i) T – 98 = 10 ( 0 ) ⇒ T = 98 N
(ii) T – 98 = 10 ( 2 ) ⇒ T = 118 N
1200
= 6.96 m s
–1
Q6 P before = P after ⇒ (6000)(10) + (2000)(2)
–1
= 8000v ⇒ v = 8 m s in original direction
of motion.
Q7 P before = P after ⇒ (6000)(10) + (2000)(–20)
= 8000v
–1
60 000 – 4000 = 8000v ⇒ v = 7 m s in
original direction of motion.
Q8 P before = P after
–3
0 = ( 10 × 10 ) ( 400 ) + 2v ⇒ v = –2 m s
–1
i.e. gun recoils backwards at 2 m s
–1
Q9 P before = P after
(100)(10) + (60)(–15) = 100v + (60)(8)
⇒ 1000 – 900 = 100v + 480 ⇒ 100v = –380
–1
⇒ v = –3.8 m s i.e. 100 kg block moves with
–1
3.8 m s in opposite direction to its initial
velocity.
Teacher’s Manual
Q10 P.C.M.
( m ) ( 0.6 ) + ( 3m ) ( 0 ) = ( m ) ( 0.2 ) + ( 3m )x
Q13
30
50 m s–1
0.6 = 0.2 + 3x
–1
3x = 0.4 ⇒ x = 0.133 m s
B
A
m
0.6ms–1
A
m
3m
20 m s–1
0.2 ms–1
3m
θ
4,000 kg
x
x
5,000 kg
After
(i) 0 = 500v + 2 × 500 ⇒ v = –2 m s
–1
i.e. recoil velocity is 2 m s
–1
(ii) For gun: u = 2, v = 0, a = ?, s = 0.25
v 2 = u 2 + 2as
0 = 2 2 + 2(a)(0.25)
4
- = –8 m s
a = – ------
–2
0.5
F = ma (500)(8) = 4000 N
–3
Q12 P before = P after ⇒ ( 12 × 10 ) ( 200 ) +
–3
( 10 ) ( 0 ) = 10v + ( 12 × 10 ) ( 50 )
2.4 + 0 = 10v + 0.6 ⇒ 10v = 1.8
–1
⇒ v = 0.18 m s
Force exerted on block
=
v
B
Before
Q11
y
1,000 kg
(1000)(50) = 5000x ⇒ x = 10
(4000)(20) = 5000y ⇒ y = 16
v=
10 2 + 16 2 = 18.87 m s
–1
------ ⇒ θ = 32°
Tan θ = --x = 10
y
16
–1
Velocity of wreck = 18.87 m s at 32° with
direction of lorry’s initial velocity.
Q14 Apply P.C.M. in direction of emission of mass
P before = P after
–1
0 = (10)(2000) + (5000)(–x) ⇒ x = 4 m s
2,000 m s–1
10 kg
5,000 kg
40 m s–1
40
θ
v
Change in momentum of block
-------------------------------------------------------------------------Time taken
x
–1
( 10 ) ( 0.18 ) – ( 10 ) ( 0 )
- = 900 N
= -------------------------------------------------0.002
Force exerted on bullet
in momentum of bullet
--------------------------------------------------------------------------= Change
Time taken
Thus rocket recoils at 4 m s . It still keeps its
–1
40 m s :
Resultant velocity: v =
–1
= 40.1995 m s
–1
–1 ( 4 )
x
- = Tan ------- ⇒ θ = 5.7°
θ = Tan  ----
40
–3
–3
( 12 × 10 ) ( 50 ) – ( 12 × 10 ) ( 200 )
= --------------------------------------------------------------------------------------= – 900 N
0.002
∴ Forces have the same magnitudes QED
40 2 + 4 2
40
–1
Velocity of rocket = 40.1995 m s at 5.7°
with original direction of motion.
Q15 (0.08)(5) + 0 = 0.280v
–1
v = 1.43 m s
Change in momentum for 80 g mass
= (final momentum – initial momentum)
–1
= (0.08)(1.43 – 5) = – 0.286 kg m s
Change in momentum for 200 g mass
–1
= (.2)(1.43) = 0.286 kg m s
in momentum
0.286
----------------------------------------------------- = ------------- = 2.86 N
Force = Change
Time
0.1
21
Real World Physics
Exercise 10.1
Q1
200
- kg = 0.2 kg
(i) 200 g = -----------
–3
4
---- = ------------- = 333.33 kg m
Q4 ρ = m
1000
Q5 ρ =
(ii) 4 g = 0.004 kg
5
5
× 10
(iii) 2 × 10 g = 2----------------= 200 kg
1000
(iv) 24 mg = 24 × 10
Q6 ρ =
–3
24 × 10
- kg
g = ----------------------
–3
=
–5
3
(i) 1 cm = 1 × 10
–6
m
= 11 180 kg m
ρ
4
------------------------4
1.05 × 10
= 3.8 × 10
3
= 1.2 × 10
–6
m
–4
m
m
3
3 cm = 3 × 10
3
–6
m
3
3
∴ m = ρv
–6
(iii) 4 litres = 4000 cm = 4000 × 10 m
–3
3
= 4 × 10 m
6
3
–4
3
3
–6
3
(iv) 2 × 10 cm = ( 2 × 10 ) ( 1 × 10 ) = 2 m
4
–6
= ( 1.05 × 10 ) ( 3 × 10 )
3
⇒ m = 0.0315 kg
Q7 m = ρ V = ( 1.36 × 10 ) ( 1 × 10 ) = 0.0136 kg
4
Q3
2
(i) 1cm = 1 × 10
–4
m
2
2
(ii) 220 cm = 220 × 10
= 2.2 × 10
2
(iii) 4 mm = 4 × 10
4
2
–6
–4
–2
m
(iv) 3 × 10 cm = 3 m
22
–3
3
(ii) 120 cm = 120 × 10
6
=
0.012
4
1.8 × 10
---------------------1.61
---∴ V= m
1000
= 2.4 × 10 kg
Q2
V
m
---V
m
---V
2
2
m
m
2
2
–6
–3
m
( 20 × 10 )
---- ⇒ V = ---- = ---------------------------Q8 ρ = m
3
V
ρ
7.3 × 10
= 2.74 × 10
Volume = Area × Thickness
–6
–6
2.74 × 10 = A × 2 × 10
2
⇒ A = 1.37 m
–6
m
3
Teacher’s Manual
Exercise 10.2
Exercise 10.3
F
× 9.8
980
---------------------- = --------- = 1960 Pa
Q1 P = --= 100
F
--------- = 20 Pa
Q1 P = --= 100
Q2 P =
Q3
A
F
--A
A
5
=
60
----------------------–4
25 × 10
0.5
Q2 P = ρgh = (1000)(9.8)(33) = 323 400 Pa
= 24 000 Pa
(i) Area of side A
–2
–2
–3
2
= ( 3 × 10 ) ( 5 × 10 ) = 1.5 × 10 m
(ii) Area of side B
–2
–2
–3
2
= ( 3 × 10 ) ( 9 × 10 ) = 2.7 × 10 m
(iii) Area of side C
–2
–2
–3
2
= ( 5 × 10 ) ( 9 × 10 ) = 4.5 × 10 m
Force = Weight of block = (4)(9.8) = 39.2 N
4
F
39.2
(i) P = --= -----------------------= 2.613 × 10 Pa
–3
A
0.5
1.5 × 10
39.2
(ii) P = -----------------------= 1.452 × 10 Pa
–3
4
2.7 × 10
39.2
(iii) P = -----------------------= 8.711 × 10 Pa
–3
3
Q3
(i) P = ρgh = ( 10 ) ( 9.8 ) ( 0.2 ) = 1960 Pa
3
(ii) P = ρgh = ( 13.6 × 10 ) ( 9.8 ) ( 0.2 )
= 26 656 Pa
3
Q4
(i) P = ρgh = (1000)(9.8)(1) = 9800 Pa
(ii) P = ρgh = (1000)(9.8)(1.1) = 10 780 Pa
F
(iii) P = --⇒ F = PA = (9800)(0.2)(0.3)
A
= 588 N
F
(iv) P = --⇒ F = PA = (10 780)(0.2)(0.3)
A
= 646.8 N
Resultant upward force on block
= 646.8 – 588 = 58.8 N
Weight of block = 70 N ⇒ it will sink.
4.5 × 10
F
Q4 P = --∴ F = P A = (400)(0.06) = 24 N
A
5
–4
Q5 F = P A = ( 1 × 10 ) ( 621 × 10 ) = 6210 N
2
Q6 F = P A = ( 500 ) ( π ( 0.1 ) ) = 15.7 N
F
Q7 P = --⇒ F = PA = (556)(2.2)(2.2) = 2691 N
A
= weight of oil
m
W
----- ∴ ρ = ---- = ------W = mg ⇒ m = W
g
=
2691
--------------------------( 9.8 ) ( 2.2 ) 3
V
= 25.79 kg m
gV
–3
F
32 ) ( 9.8 )
- = 62 388.7 Pa
Q8 P = --= (---------------------2
A
π ( 0.04 )
23
Real World Physics
Exercise 10.4
Exercise 10.5
5
Q1 P 1 = 1 × 10 , V 1 = 3 m
3
Gravity solutions
– 11
Gm 1 m 2
( 6.7 × 10 ) ( 1 ) ( 1 )
-----------------------------------------------Q1 F = -----------------=
2
5
P 2 = 3 × 10 , V 2 = ?
d
P1 V 1 = P2 V 2
V2 =
P1 V 1
------------P2
V2 = 1 m
Q2
= 6.7 × 10
5
× 10 × 3
= 1-------------------------5
d
(i) P 1 V 1 = P 2 V 2
5
–6
–6
5
–6
( 1 × 10 ) ( 40 × 10 )
- = 25 000 Pa
P 2 = -------------------------------------------------–6
160 × 10
(ii) P 1 V 1 = P 2 V 2
d
–6
80 × 10
–6
–6
5
–6
( 1 × 10 ) ( 40 × 10 )
P 2 = -------------------------------------------------–6
1 × 10
– 11
24
22
Gm 1 m 2
( 6.7 × 10 ) ( 6 × 10 ) ( 7 × 10 )
---------------------------------------------------------------------------------Q5 F = -----------------=
2
2
(i) P 1 V 1 = P 2 V 2
5
–6
5
( 1 × 10 ) ( 700 × 10 ) = ( 2 × 10 )V 2
5
–6
( 1 × 10 ) ( 700 × 10 )
V 2 = ----------------------------------------------------5
2 × 10
3
m = 350 cm
5
3
–6
5
(ii) ( 1 × 10 ) ( 700 × 10 ) = ( 7 × 10 )V 2
5
–6
( 1 × 10 ) ( 700 × 10 )
V 2 = ----------------------------------------------------5
= 1 × 10
–4
7 × 10
3
m = 100 cm
3
5
–6
( 1 × 10 ) ( 700 × 10 )
(iii) V 2 = ----------------------------------------------------4
= 1.4 × 10
5 × 10
–3
3
m = 1400 cm
3
Q5 20 Pa m since at a fixed temperature PV is a
constant by Boyle’s Law.
24
20
N
– 11
24
30
Gm 1 m 2
( 6.7 × 10 ) ( 6 × 10 ) ( 1.9 × 10 )
--------------------------------------------------------------------------------------Q6 F = -----------------=
2
2
11
( 1.5 × 10 )
= 3.39 × 10
6
No calculations required.
–4
8
( 3.8 × 10 )
d
d
= 4 000 000 Pa = 4 × 10 Pa
= 3.5 × 10
6
( 1.7 × 10 )
= 1.95 × 10
5
N
= 123.34 N
(iii) P 1 V 1 = P 2 V 2
( 1 × 10 ) ( 40 × 10 ) = ( 1 × 10 )P 2
–6
– 11
22
Gm 1 m 2
( 6.7 × 10 ) ( 76 ) ( 7 × 10 )
-------------------------------------------------------------------Q4 F = -----------------=
2
2
–6
5
–6
( 1 × 10 ) ( 40 × 10 )
- = 50 000 Pa
P 2 = -------------------------------------------------–6
2
= 1.508 × 10
d
5
3
6
( 6.4 × 10 )
– 11
Gm 1 m 2
( 6.7 × 10 ) ( 90 ) ( 1000 )
-----------------------------------------------------------Q3 F = -----------------=
2
2
( 1 × 10 ) ( 40 × 10 ) = ( 80 × 10 )P 2
Q4
N
= 745.9 N
( 1 × 10 ) ( 40 × 10 ) = ( 160 × 10 )P 2
Q3
– 11
– 11
24
Gm 1 m 2
( 6.7 × 10 ) ( 76 ) ( 6 × 10 )
-------------------------------------------------------------------Q2 F = -----------------=
2
2
3 × 10
3
1
22
N
Teacher’s Manual
Exercise 10.6
– 11
27
( 6.7 × 10 ) ( 1.9 × 10 )
= ------------------------------------------------------------2
--------Q1 g = GM
2
7
( 7 × 10 )
R
= 25.98 m s
= 26 m s
–2
–2
Q7 Value on surface = 9.8
Half value on surface = 4.9
Let d = distance from centre of Earth to object.
2
GM
--------- ⇒ d = --------(i) 4.9 = GM
2
4.9
d
W = mg = (90)(26) = 2340 N
– 11
24
( 6.7 × 10 ) ( 6 × 10 )
= --------------------------------------------------------
– 11
22
( 6.7 × 10 ) ( 1.9 × 10 )
= ------------------------------------------------------------2
--------Q2 g = GM
2
4.9
⇒ d = 9.058 × 10 m
6
6
( 1.7 × 10 )
R
= 1.62 m s
–2
6
Radius of Earth = 6.4 × 10
∴ Height above surface =
6
6
6
9.058 × 10 – 6.4 × 10 = 2.7 × 10 m
W = mg = (60)(1.62) = 97.2 N
– 11
30
( 6.7 × 10 ) ( 1.9 × 10 )
= ------------------------------------------------------------2
GM
-s
Q3 gs = ----------2
6
i.e. h = 2.7 × 10 m
8
( 6.956 × 10 )
Rs
= 263 m s
g
9.8
GM
- = ------- = 0.98 = --------(ii) Acceleration = ----2
–2
10
Q4 Weight = force of gravity
GM
-------------- ⇒ g = --------⇒ mg = GMm
2
2
R
0.98
d
0.98
= 2.03 × 10
Q5 If g d = acceleration at distance d, then weight
of object of mass m = mg d = grav force of
attraction.
d
7
2
6
( 9.8 ) ( 6.4 × 10 )
gR
--------- ⇒ M = --------- = --------------------------------------Q8 g = GM
2
– 11
G
R
6.7 × 10
= 5.99 × 10
R
( R + h)
by previous question
– 11
24
–2
( 6.7 × 10 ) ( 6 × 10 )
- = 9.51 m s
= ----------------------------------------------------------2
6
6
24
kg
GM
--------- ⇒ 9.8 = --------Q9 g = GM
2
2
GM
∴ g h = --------------------2-
Q6 g =
7
Height = d – R = 2.03 × 10 – 6.4 × 10
7
= 1.39 × 10
At height h above surface of planet of radius R,
distance from centre d = R + h
GM
-------------------- ,
2
(R + h)
d
– 11
24
GM
( ( 6.7 × 10 ) ( 6 × 10 ) )
⇒ d = --------- = --------------------------------------------------------------
R
GMm
GM
- ⇒ g d = --------∴ mg d = ------------2
2
10
R
Distance from centre of Earth is 3R.
–2
GM
1 GM
1
- = ---  --------- = --- (9.8) = 1.089 m s
g h = ------------2
 2
9
( 3R )
9
R
Q10 No calculation required.
3
( 6.4 × 10 + 100 × 10 )
GM
GM
m
- and g e = -----------eQ11 g m = -----------2
2
Rm
g
Re
M
R
2
1 
⇒ -----m- =  --------m  ------e- = ( 0.04 )  --------ge
M e Rm
0.37
2
∴ g m = g e (1.081) = (9.8)(0.292) = 2.86 m s
–2
25
Real World Physics
Q12 Let P be the point where resultant force = 0
Suppose P is a distance y from the Earth and a
distance x from the moon:
Exercise 10.7
Q1
(i) M = Fd = ( 10 ) ( 0.4 ) = 4 N m
GM m
GM m
(ii) M = Fd = ( 10 ) ( 0 ) = 0 N m
y
x
(iii) M = Fd = ( 10 ) ( 0.9 ) = 9 N m
e
m
- = ----------------⇒ ---------------2
2
Where m = mass of object placed at P
2
M
Mm
Mm
M
x
- = -------⇒ ------2e- = -------⇒ ---⇒ --x = --------m
2
2
y
x
Me
y
Also x + y = 3.8 × 10
y
y = 3.8 × 10 – x
8
x = --------m ( 3.8 × 10 – x )
1
x = ----- ( 3.8 × 10 8 – x )
81
8
⇒ 9x = 3.8 × 10 – x
⇒ 10x = 3.8 × 10
(i) – ( 10 ) ( 0.4 ) – ( 1 ) ( 0.5 ) – ( 12 ) ( 0.7 )
+ ( 28 ) ( 0.5 ) = +1.1
8
8
M
Me
Me
Q2
8
7
x = 3.8 × 10 m
Resultant gravity force is zero when
7
3.8 × 10 m from Moon.
(ii) – ( 12 ) ( 0.2 ) + ( 10 ) ( 0.1 ) + 5 ( 0.5 ) = +1.1
+ ( 5 ) ( 0.7 ) + ( 10 ) ( 0.3 ) + 1 ( 0.2 )
– ( 28 ) ( 0.2 ) = + 1.1
(iii) (a) ( 12 ) ( 0.4 ) + ( 60 ) ( 0.2 ) = X ( 0.4 )
⇒ X = 42 N
(b) ( 30 ) ( X ) = ( 10 ) ( 0.25 )
⇒ X = 0.0833 m = 8.33 cm
(c) ( 20 ) ( 0.35 ) + ( 40 ) ( 0.15 )
= ( 10 ) ( 0.15 ) + 30 ( X )
⇒ X = 0.3833 m = 38.33 cm
(d) ( 20 ) ( 0.4 ) + ( 40 ) ( 0.2 )
= ( 60 ) ( 0.1 ) + ( 10 ) ( 0.3 ) + X ( 0.2 )
⇒ X = 35
Q4 Weight W acts vertically down through 50 cm
mark.
Take moments about 45 cm mark:
W ( 0.05 ) = ( 60 ) ( 0.35 ) ⇒ W = 420 N
Q5 Weight of stick = ( 20 + 40 ) – ( 10 + 20 + 28 )
⇒W=2N
Moments about A:
Clockwise:
(10)(10) +(20)(30) + (2)(50) + (28)(100)
= 3600 N cm
Anti-clockwise:
(20)(20) +(40)(80) = 3600 N cm
∴ Sum of moments = 3600 – 3600 = 0
Moments about B:
Clockwise:
(20)(10) +(2)(20) + (28)(70) = 2200 N cm
Anti-clockwise:
(10)(20) +(40)(50) = 2200 N cm
∴ Sum of moments = 0
26
Teacher’s Manual
Moments about C:
Clockwise:
(20)(30) + (28)(50) = 2000 N cm
Anti-clockwise:
(40)(30) +(20)(20) + (10)(40) = 2000 N cm
∴ Sum of moments = 0
Q7
Y
Boy
(100)(9.8)
= 980 N
X(7) = (980)(5) ⇒ X = 700 N
∴ Man exerts 700 N and boy ( 980 – 100 )
= 280 N
x
Man
Q6 Let R 1 = force at 10 cm and R 2 = force at
70 cm.
Let W = weight of beam. It acts at 50 cm (i.e.
in middle). Take moments about 10 cm mark:
R1 + R2 = W ⇒ R1 =
X
Man
Take moments about boy:
Moments about D:
Clockwise:
(40)(20) + (20)(80) = 2400 N cm
Anti-clockwise:
(10)(90) +(20)(70) + (2)(50) = 2400 N cm
∴ Sum of moments = 0
( R 2 ) ( 0.6 ) = W ( 0.4 ) ⇒ R 2 =
10 m
3m
Boy
Man supports three times weight of boy
⇒
2
--- W
3
Man exerts 3--- ( 980 ) N = 735 N
Boy exerts
1
--- W
3
4
1
--- ( 980 ) N
4
= 245 N
Take moments about centre:
(735)(X) = (245) (5) ⇒ X = 1.667 m
Q8
8m
x
Man
8–x
200 x 9.8
= 1,960 N
Boy
Man exerts 4--- of 1960 = 1568 N
Boy exerts
5
1
--5
of 1960 = 392 N
Moments about weight: (1568)x = (392)(8 – x)
⇒ 1568x + 392x = 3136 ⇒ x = 1.6 m
27
Real World Physics
Exercise 10.8
Exercise 11.1
Q1 F ( 0.05 ) = ( 300 ) ( 1.5 ) ⇒ F = 9000 N
Q1 W = Fs = ( 10 ) ( 30 ) = 300 J
Q2 T = Fd = ( 40 ) ( 0.5 ) = 20 N m
˙ 000 J
Q2 W = Fs = ( 400 ) ( 30 ) = 12
Q3 T = Fd ⇒ 85 = F ( 0.4 ) ⇒ F = 212.5 N
20 000
----- = ---------------- = 58.82 m
Q3 W = Fs ⇒ s = W
Q4 T = Fd ⇒ 600 = F ( 0.1 ) ⇒ F = 6000 N
Q4 W = mg = ( 100 ) ( 9.8 ) = 980 N
˙ 800 J
W = Fs = ( 980 ) ( 60 ) = 58
F
340
Q5 W = Fs = ( 20 ) ( 9.8 ) ( 1 ) = 196 J
Q6 W = Fs = ( 60 ) ( 9.8 ) ( 8 ) = 4704 J
Q7 u = 0, v = 30, t = 10, a = ?, s = ?
Find a:
v = u + at
30 = 0 + a ( 10 )
⇒a = 3ms
Find s:
–2
s = ut + 1--2- at
2
2
s = 1--2- ( 3 ) ( 10 )
s = 150 m
F = ma ⇒ F = ( 800 ) ( 3 ) = 2400 N
W = Fs = ( 2400 ) ( 150 ) = 360 000 J
Q8 u = 0
–1
–1
–1
80 000
- m s = 22.22 m s
v = 80 km h = --------------------( 60 ) ( 60 )
t = 20, a = ?, s = ?
Find a:
–2
v = u + at ⇒ 22.22 = a(20) ⇒ a = 1.11 m s
Find F:
F = ma ⇒ F = (1000)(1.11) = 1110 N
Find s:
2
s = ut + 1--2- at ⇒ s = 1--2- ( 1.11 ) ( 20 ) 2 ⇒ s = 222 m
Find W:
W = Fs = (1110)(222) = 246 420 J
The answer is 246 914 J if you retain all
decimal places throughout the calculation.
Q9 Let h = Vertical height
h
- ⇒ h = 10 Sin 40 °
Sin 40 ° = ----10
W = Fs = (105)(9.8) 10 Sin 40 ° = 6614.28 J
If ladder was vertical:
W = Fs = (105)(9.8)(10) = 10 290 J
28
Teacher’s Manual
Exercise 11.2
2
2
Q1 E k = 1--- mv =  1--- (20) ( 12 ) = 1440 J
2
2
Q8 Work done on bullet = loss in E k
–3
2
= ( 0.5 ) ( 4 × 10 ) ( 200 ) = 80 J
2
Q2 E k = 1--- (1000) ( 28 ) = 392 000 J
Q3 E k =
2
1
--2
) ( 1.5 × 10 )
– 16
J
( 9.1 × 10
= 1.024 × 10
80
----- = ------- = 160 N
W = Fs ⇒ F = W
7 2
– 31
s
Work done = Fs = (160)(0.25)
= 40 J = loss in E k
2E k
2
Q4 E k = 1--- mv ⇒ m = -------2
2
Remaining E k = 80 J – 40 J = 40 J
v
=
2
Q5 E k = 1--- mv ⇒ v
2
( 2 ) ( 160 000 )
-------------------------------2
( 20 )
= 800 kg
2E
m
= --------k-
Q6
(i)
2
( 4 × 10 ) ( 400 ) = 320 J
Ek
2
1
= 1--- m ( 2v ) 2 = 4 ( --2- mv )
2
⇒ E k has quadrupled.
2
100 000  2
000)  --------------------( 60 ) ( 60 )
2
2
2
m 1 ( 4v 2 ) = m 2 v 2
2
(iii)
mv 2
2
2
Q10 1--- m 1 v 1 = 1--- m 2 v 2
2
(ii) 1--- ( 800 ) ( 28 ) = 313 600 J
1
--- (10
2
2
1
--2
Double speed to 2v
200
–3
–1
∴ 40 = 1--- ( 4 × 10 –3 )v 2 ⇒ v = 141.4 m s
Q9 E k =
–1
( 2 ) ( 4000 )
= ------------------------ = 6.32 m s
1
--2
0.5
m
1
- ⇒ m1 : m2 = 1 : 16
∴ ------1 = ----m2
16
6
= 3.9 × 10 J
–3
150 000  2
- = 26 J
(iv) 1--- ( 30 × 10 )  --------------------
2
Q7
( 60 ) ( 60 )
2
(i) 1--- (20)(3) = 90 J
2
2
(ii) 1--- (20)(30) = 9000 J
2
(iii) Change in E k = 9000 – 90 = 8910 J
(iv) Work done = change in E k
= Force × distance
∴ 8910 = (12) s ⇒ s = 742.5 m
(v) From above: 8910 J
29
Real World Physics
Exercise 11.3
Q8 (10)(30) + 0 = 25v ⇒ v = 12 m s
Q1 E p = mgh = ( 20 ) ( 9.8 ) ( 600 )
= 117 600 J
–1
2
Initial E k = 1--- ( 10 ) ( 30 ) = 4500 J
Q2 E p = mgh ∴ 4000 = ( 4 ) ( 9.8 )h
∴ 4000 = 39.2h
Final E k =
2
2
1
--- ( 25 )12
2
= 1800 J
Loss in E k = 2700 J
------------ = 102 m
h = 4000
39.2
Q9
Q3 E p = mgh = ( 3 ) ( 9.8 ) ( 60 )
= E K = 1764 J
2m
2
–1
1764 = 1--- ( 3 )v ⇒ v = 34.3 m s
2
100 m s–1
Q4 Loss in kinetic energy = gain in potential
energy.
1
--2
Conservation of momentum:
1000
Q5 E p = mgh = ( 100 ) ( 9.8 ) ( 50 ) = 49 000 J
E p at 30 m = ( 100 ) ( 9.8 ) ( 30 ) = 29 400 J
Kinetic Energy = Loss in potential energy
= 19 600 J
Q6 Loss in E p = Gain in E k
2
m ( 9.8 ) ( 0.5 ) = 1--- mv
2
( 2 ) ( 9.8 ) ( 0.5 ) = 3.13 m s
–1
Q7 Loss in Potential Energy = Gain in Kinetic
energy
mg(2 – 2 Cos 40 °) = 1--- mv 2
2
30
u
6 
- ( 100 ) = 5.006 u ⇒ u = 0.1199 m s
⇒  ----------
h = 510.2 m
v=
5 kg
2
( m ) ( 100 ) = m ( 9.8 ) ( h )
v =
6 grams
4g ( 1 – Cos 40° ) = 3.028 m s
–1
Gain in E p = Loss in E k
( 5.006 ) ( 9.8 )h = 1--- ( 5.006 ) ( 0.1199 ) 2
2
⇒ h = 0.073 cm
–1
Teacher’s Manual
Exercise 11.4
Exercise 11.5
P
4000
O
- × 100 = ------------ × 100 = 80%
Q1 % efficiency = -----PI
5000
600 000
----- = ------------------- = 50 kw
Q1 P = W
t
12
mgh
( 2000 ) ( 9.8 ) ( 20 )
-------------- = ----------- = ---------------------------------------Q2 Power out = Work
E
18 000
- = 10 W
Q2 P = --= --------------------t
Time
( 30 ) ( 60 )
60
% efficiency =
7
2 × 10
----- = ------------------------------ = 2778 W
Q4 P = W
t
( 2 ) ( 60 ) ( 60 )
6
t
39 200
× 100 = ---------------× 100
60 000
Q3 25% of input power = 130 kW
7
Q6 E = Pt = ( 130 × 10 ) (5)(60) = 3.9 × 10 J
( 50 ) ( 9.8 ) ( 50 )
----- = ---------------------------------- = 2042 W
Q7 P = W
PO
------PI
= 65.3%
Q5 E = Pt = (60)(5)(60)(60) = 1.08 × 10 J
3
10
= 39 200 W
E
000
---------------- = 1000 W
Q3 P = --= 60
t
t
75% of power becomes heat ∴ amount
--------- 75 kW = 390 kW
converted to heat =  130

25
12
( 50 ) ( 9.8 ) ( 2 ) ( 25 )
----- = ------------------------------------------- = 408 W
Q8 P = W
t
60
( 40 ) ( 9.8 ) ( 50 ) ( 0.16 )
----- = -------------------------------------------------- = 78.4 W
Q9 P = W
t
40
6
1 × 10
----- = -------------------- = 12.987 s
Q10 t = W
3
P
77 × 10
Q11 W = Pt = (1000)(60)(60) = 3.6 MJ
31
Real World Physics
Exercise 12.1
Exercise 12.2
π
- rad
Q1 180 ° = π rad ⇒ 1 ° = -------180
Q2
Q3
10 π
π
- rad = ------ rad
⇒ 10 ° = -------180
18
48 π
- = 0.878 rad
48 ° = -------180
180°
π
--- rad = ----------- = 36 °
5
5
180
π
--- rad = --------- = 25.7 °
7
7
180
1 rad = --------- = 57.296 °
π
4.2 ) ( 180 )
- = 240.64 °
4.2 rad = (------------------------π
θ = -s ⇒ s = r θ
r
(i) s = (3)(1) = 3 m
Q1
Q5 θ =
v
-r
θ
--t
(i) ω =
( 2 ) ( 60 )
------------= 0.333
3
⇒
= 0.111 rad s
–1
π
--2
- = 14.15 s
t = ---θ- = -----------w
0.111
π
 --3- 
- = 9.43 s
(ii) t = ---θ- =  ------------
ω
 0.111
Q2 r = 10 m, ω = 4 rad s
–1
v = ω r = ( 4 ) ( 10 ) = 40 m s
–1
------------ = 25 s
v = s- ⇒ t = -s- = 1000
t
Q3 ω =
Q4
v
v
-r
=
3
------0.4
40
= 7.5 rad s
–1
(i) v = ω r = ( 6 ) ( 0 ) = 0 m s
–1
(ii) v = ω r = ( 6 ) ( 0.3 ) = 1.8 m s
(iii) s = (3)  π--- = 2.36 m
Q4 θ =
t
(ii) ω =
(ii) s = (3)( π ) = 9.42 m
(iv)
–1
40
- = 0.333 m s
(i) v = s- = ------------------
4
(
123 ) π
- = 6.44 m
s = (3)  ---------------180 
s
s
3
- ⇒ r = --- = ------- = 0.857 cm
θ
r
3.5
6
s
- ⇒ s = r θ = ( 93 × 10 ) ( 0.00932 )
r
–1
–1
( 400 ) ( 2 π )
- rad s
Q5 ω = ------------------------
60
–1
( 400 ) ( 2 π )
- (2) = 83.78 m s
v = ω r =  -----------------------
60
Q6
R
5
--------- = 0.384 m
Q6 s = r θ = ( 2.2 )  10π

E
R
= 8.67 × 10 miles
53˚
180
2π
(i) ω = θ--- = -------------------------------t
( 24 ) ( 60 ) ( 60 )
= 7.27 × 10 –5 rad s –1
–5
6
(ii) v = ω R = ( 7.27 × 10 ) × ( 6.4 × 10 )
= 465 m s
–1
(iii) v = ω R = ω ( R E Cos 53 )
= ( 7.27 × 10 –5 ) ( 6.4 × 10 6 ) Cos 53
–1
= 280 m s
32
Teacher’s Manual
Exercise 12.3
Exercise 12.4
2
2
– 11
24
2
( 6.7 × 10 ) ( 6 × 10 )
--------- = ----------------------------------------------------------Q1 v = GM
6
6
1200 ) ( 25 )
mv
- = 3750 N
Q1 F = --------= (----------------------------r
200
2
R
( 6.4 × 10 ) + ( 50 × 10 )
2
Q2 F = m ω r = ( 4 ) ( 4.5 ) ( 0.2 ) = 16.2 N
2
Q3
(i) a =
v
----r
2
=
( 12 )
------------1
= 144 m s
⇒ v = 2669.8 m s
–2
F = ma = ( 5 ) ( 144 ) = 720 N
2
2
–2
-------- = 480 m s
(ii) a = v----- =  12
r
0.3 
–1
6
6
( 2 ) ( π ) [ ( 6.4 × 10 ) + ( 50 × 10 ) ]
---------- = -------------------------------------------------------------------------------T = 2πR
v
2669.8
= 132733.41 s = 36.87 hours
F = ma = ( 5 ) ( 480 ) = 2400 N
Q4
2 3
2
R
--------------- ⇒ R =
Q2 T = 4π
–1
4
- = 10 rad s
(i) ω = v-- = ------
(ii) a =
r
2
v
----r
=
0.4
2
4
------0.4
= 40 m s
GM
–2
2
3
T GM
---------------2
4π
30 min = (30)(60) s
(iii) F = ma = ( 10 ) ( 40 ) = 400 N
2
R =
3
– 11
24
[ ( 30 ) ( 60 ) ] ( 6.7 × 10 ) ( 6 × 10 )
--------------------------------------------------------------------------------------2
4π
100 000 2
2
mv
Q5 F = --------=
r
( 950 )  ----------------------
 ( 60 ) ( 60 )
-----------------------------------------100
Q6 v = r ω = ( 0.5 ) ( 20 ) = 10 m s
v2
v2
r
a
–1
( 500 2 )
Q7 a = ----- ⇒ r = ----- = -------------------- = 2834 m
v =
( 9 ) ( 9.8 )
Q8 Loss in E k = Gain in E p
2
2
1
1
--- m 5 – --- mv
2
2
 1--- ( 5 2 ) –  1---
 2
 2
6
= 3.2 × 10 m
= 7330 N
=
GM
--------R
– 11
24
( 6.7 × 10 ) ( 6 × 10 )
= -------------------------------------------------------6
( 3.2 × 10 )
8
1.2562 × 10 = 11 208 m s
3
2 3
2
11
2
2
π R
4 π ( 7.8 × 10 )
Q3 T = 4--------------⇒ T = ------------------------------------------------------------– 11
30
= mgh
GM
2
( 6.7 × 10
v = ( 9.8 ) ( 0.8 )
2
–1
12.5 – 7.84 = v----- ⇒ v = 3.05 m s
2
a=
v
----r
=
2
2
( 3.05 )
----------------0.8
= 11.63 m s
–1
–2
F = ma = ( 0.24 ) ( 11.63 ) = 2.79 N
) ( 2.0 × 10 )
= 1.3981 × 10
17
⇒ T = 3.739 × 10 s = 11.9 years
8
2
Q4 T ∝ R
2
TS
-------2
TE
3
3
3
RS
2
2 R
2
3
- ⇒ T S = T E  ------S  = T E ( 9.5 )
= -------3
 
RE
RE
TS = TE
3
( 9.5 ) = T E (29.28) years
Since period of Earth’s orbit = 1 year
33
Real World Physics
Q5
2
--------v = GM
Exercise 13.1
R
2
vs
-----2
ve
v s
 --- v e
=
2
Q1 F = ks ⇒ F = 4000 s
GM s
-----------Rs
-----------GM E
------------RE
–2
(i) s = 2 cm = 2 × 10 m
–2
∴ F = ( 4000 ) ( 2 × 10 ) = 80 N
R
Rs
(ii) 1000 = 4000 s ⇒ s = 0.25 m
M
ME
=  ------E  -------s-
Q2 F = ks ⇒ 8 = k ( 0.06 ) ⇒ k = 133.3
R E = R s and v s = 10v e
M
-------sME
⇒ ( 10 ) =
2
(i) F = ( 133.3 ) ( 0.02 ) = 2.67 N
⇒ M s = M E ( 100 )
24
= 6 × 10 × 100
26
= 6 × 10 kg
2 3
2
π R
⇒ R =
Q6 T = 4---------------
GM
2
3
T GM
--------------2
4π
2
– 11
24
( 86 400 ) ( 6.7 × 10 ) ( 6 × 10 )
-----------------------------------------------------------------------------2
4π
7
= 4.24 × 10 m
R =
3
– 11
24
2
GM
( 6.7 × 10 ) ( 6 × 10 )
--------- ⇒ v = --------- = --------------------------------------------------------v = GM
7
R
R
= 3079 m s
4.24 × 10
–1
–5
–1
2π
- = ( 7.27 × 10 ) rad s
Q7 ω = θ--- = --------------------------------
t
( 24 ) ( 60 ) ( 60 )
–5
6
(i) v = ω r = ( 7.27 × 10 ) ( 6.4 × 10 )
= 465 m s
2
–1
2
(ii) a = ω r = ( 7.27 × 10 –5 ) ( 6.4 × 10 6 )
–2
= 0.0338 m s
(0.03385 if you don’t round off ω )
34
(ii) 15 = ( 133.3 ) ( s ) ⇒ s = 11.25 cm
Teacher’s Manual
Exercise 13.2
Q1
Exercise 13.3
4
- = 0.2 s
(i) T = Time for one oscillation = ----20
1
--T
(ii) f =
=
1
------0.2
= 5 Hz
2
- = 2.84 s
Q1 T = 2 π --l- ⇒ T = 2π --------g
Q2 T =
g = 4π
------ = 0.4 s
Q2 T = 20
50
Q3
90
-----50
a = 4 ( 0.1 ) = 0.4 m s
a
a = 4s ⇒ s = ---
2
2 = ω ( 0.5 )
2
⇒ω = 4
= 1.8 s
2
l
-----2
T
2
–2
4 π ) ( 0.8 )
- = 9.75 m s
= (------------------------2
( 1.8 )
2
--------Q3 T = 2 π --l- ⇒ l = gT
2
a = 4s
2
a = ω s
9.81
∴l=
–2
g
2
( 9.8 ) ( 2 )
----------------------2
4π
4π
= 0.9929 m = 99.29 cm
4
=
Q4
0.5
------4
= 0.125 m
R
R
R
R
Q4 Force is max when accel is max and when
displacement is max, i.e. force is max at amplitude.
2
a= ω s
--------g = GM
2
2
R
2
1 = ω ( 0.1 ) ⇒ ω = 10
T = 2 π --l-
12
g
2
a= ω s
l
- =
T height = 2π ---------g
 ---- 16
5 = 10s max ⇒ s max = 0.5 m
16  2π --l-
g
= 4 (Period on surface) = 4(0.4) = 1.6 s
Q5 T = 2-----π- ⇒ ω = 2-----πω
R
R is four times bigger ⇒ g is 16 times smaller
F max
–2
60
- = ------ = 5 m s
a max = ----------m
1
g ∝ ----2
T
∴ω =
2π
------0.5
⇒ ω = 157.914
2
a = ω s ⇒ a = (157.914)(0.04)
–2
a = 6.3165 m s
2
F = ma = ( 4 ) ( 6.3165 ) = 25.27 N
Q6 a = ω s ⇒ 3 = ω ( 0.5 ) ⇒ ω =
2
2
6
2π
- = 2.565 s
T = 2-----π- = ------
ω
6
Q7 a = ω s ⇒ 2.5 = ω ( 0.14 ) ⇒ ω = 4.226
2
2
2π
- = 1.487 ⇒ f = 0.673 Hz
T = 2-----π- = ------------
ω
4.226
2π
- = 4.189
Q8 T = 2-----π- ⇒ ω = 2-----π- = ------
ω
T
2
1.5
2
a = ω s = ( 4.189 ) ( 0.25 ) = 4.39 m s
–2
35
Real World Physics
Exercise 14.1
Q1 T/K = t/ °C + 273.15
(i) = 0 + 273.15 = 273.15 K
Exercise 15.1
Q1 Q = C ∆θ = ( 1500 ) ( 80 – 10 ) = 105 000 J
1 000 000
---- = ------------------------ = 1667° C
Q2 Q = C ∆θ ⇒ ∆θ = Q
C
600
4000
Q
------- = ------------ =
10
∆θ
(ii) = –100 + 273.15 = 173.15 K
Q3 Q = C ∆θ ⇒ C =
(iii) = 20 + 273.15 = 293.15 K
Q4 Q = mc ∆θ = (2)(4180)(80 – 12) = 568 480 J
(iv) = 100 + 273.15 = 373.15 K
Q5 Q = mc ∆θ = (12)(390)(120 – 10) = 514 800 J
Q2 t/ °C = T/K –273.15
m ∆θ
( 1.6 ) ( 25 – 7 )
–1
= 385 J kg K
Q
Q7 Q = mc ∆θ ⇒ m = --------c ∆θ
(iii) = 373 – 273.15 = 99.85 °C
(iv) = 500 – 273.15 = 226.85 °C
–1
11 088
Q
Q6 Q = mc ∆θ ⇒ C = ----------= -------------------------------
(i) = 100 – 273.15 = –173.15 °C
(ii) = 273 – 273.15 = –0.15 °C
400 J K
–1
1 000 000
= --------------------------------------( 451 ) ( 100 – 20 )
= 12.32 kg
Q
4000
- = ----------------------------- = 2.39
Q8 Q = mc ∆θ ⇒ ∆θ = -----( 0.4 ) ( 4180 )
mc
∴ Find temperature of water
Exercise 14.2
Q1 No calculations required.
Q2 No calculations required.
= 20° C + 2.39° C = 22.39° C
= 20°
Q
Q9 Q = mc ∆θ ⇒ ∆θ = ------
For
For
mc
40 000
copper ∆θ 1 = ---------------------- = 51.3 ° C
( 2 ) ( 390 )
40 000
- = 22 ° C
aluminium ∆θ 2 = --------------------( 2 ) ( 910 )
Find temperature of copper
= 0 + 51.3 = 51.3 ° C
Find temperature of aluminium
= 20 + 22 ° C = 42 ° C
⇒ Copper reaches the higher temperature
Q10 Heat added = heat to copper + heat to water
= m c c c ∆θ ↑ + m w c w ∆θ ↑
= ( 0.08 ) ( 390 ) ( 60 – 18 )
+ ( 0.120 ) ) ( 4180 ) ( 60 – 18 )
= 1310.4 + 21067.2
= 22 377.6 J
Q11 Heat supplied Q = mc ∆θ
= (5)(4180)(90) = 1 881 000 J
supplied
1 881 000
--------------------------------------- = ------------------------ = 1881 s
Time = Energy
Power
36
1000
Teacher’s Manual
Exercise 15.2
Q12 Let θ = final temperature
Heat lost = Heat gained
( 0.03 ) ( 4180 ) ( 100 – θ )
=
( 0.1 ) ( 4180 ) ( θ – 20 )
3 ( 100 – θ ) = 10 ( θ – 20 )
300 – 3θ = 10 θ – 200
500 = 13 θ
⇒ θ = 38.5° C
Final Temp = 38.5° C
Q13 Heat lost by Copper = Heat gained by water +
Heat gained by Calorimeter
m c c c ∆ θ ↓ = m w c w ∆ θ ↑ + m cal c c ∆ θ ↑
( 0.12 ) ( 390 ) ( 100 – θ )
= ( 0.08 ) ( 4180 ) ( θ – 20 )
+ ( 0.06 ) ( 390 ) ( θ – 20 )
46.8 ( 100 – θ )
= 334.4 ( θ – 20 ) + 23.4 ( θ – 20 )
θ ( – 46.8 – 334.4 – 23.4 )
= ( – 46.8 ) ( 100 ) – 20 ( 334.4 ) – 20 ( 23.4 )
⇒ 404.6 θ = 11836
θ = 29.25° C
Q1 Q = ml
5
6
Q = ( 10 ) ( 3.3 × 10 ) = 3.3 × 10 J
5
Q2 Q = ml = ( 0.5 ) ( 3.3 × 10 ) = 1.65 × 10 5 J
= 165 kJ
6
1 × 10
---- = ---------------------- = 3.03 kg
Q3 Q = ml ⇒ m = Q
5
l
3.3 × 10
6
5
Q4 Q = ml = ( 0.4 ) ( 2.3 × 10 ) = 9.2 × 10 J
6
5
Q5 Q = ml = ( 0.08 ) ( 2.3 × 10 ) = 1.84 × 10 J
Q6 Heat needed = Heat to melt ice + Heat to raise
melted ice from 0° to 99°
= ml + mc ∆θ
5
= ( 3 ) ( 3.3 × 10 ) + ( 3 ) ( 4180 ) ( 99 )
= 2.23 MJ
Q7 Heat needed = Heat to melt ice + Heat to raise
water to 100° + Heat to vaporise water at
100°
= ml f + mc ∆θ ↑ + ml v
5
= ( 1 ) ( 3.3 × 10 ) + ( 1 ) ( 4180 ) ( 100 )
6
+ ( 1 ) ( 2.3 × 10 ) = 3.05 MJ
Q8 Q = mc ∆θ + ml
Q = ( 0.06 ) ( 4180 ) ( 100 – 15 )
6
+ ( 0.06 ) ( 2.3 × 10 )
Q = 21 318 + 138 000 = 159 318 J
Heat 
Heat to
Heat to bring melted
- =  ------------------- +  -------------------------------------------------
Q9  --------------- melt ice
 ice from 0° to 100° 
needed
= ml f + mc ∆θ ↑
5
= ( 0.06 ) ( 3.3 × 10 ) + ( 0.06 ) ( 4180 ) ( 100 )
= 44 880 J
Heat lost by steam = m s l v
6
∴ m s ( 2.3 × 10 ) = 44 880
∴ m s = 19.5 grams
37
Real World Physics
Q10 Let θ = final temp reached
Heat lost by water = Heat gained by ice
Heat lost by water = Heat to melt ice + Heat to
bring melted ice to final temp.
m w c w fall in temp = m ice l f + m ice c w rise
in temp
5
( 0.5 ) ( 4180 ) ( 80 – θ ) = ( 0.2 ) ( 3.3 × 10 )
+ ( 0.2 ) ( 4180 ) ( θ – 0 )
167 200 – 2090 θ = 66 000 + 836 θ
167 200 – 66 000 = (836 + 2090) θ
θ = 34.6° C
Q12 Heat lost by water + Heat lost by calorimeter
= Heat needed to melt ice at 0° C to water at
0° C + Heat needed to raise temperature of
melted ice from 0° C to 5.1° C
i.e. m w c w (fall in temp) + m c c c (fall in temp)
= MiceL + MiceCW(5.1)
⇒ (0.08)(4180)(25 – 5.1)
+ (0.05)(390)(25 – 5.1) = (0.02) L
+ (0.02)(4180)(5.1)
⇒ Latent heat of fusion of ice L
5
–1
3.3 × 10 J kg
3
Q11 Let θ = final temp reached
(i) Heat lost by steam = Heat gained by water
m s l v + m s c w (fall in temp) = m w c w (rise
in temp)
6
( 0.002 ) ( 2.3 × 10 )
+ ( 0.002 ) ( 4180 ) ( 100 – θ )
= ( 0.08 ) ( 4180 ) ( θ – 10 )
4600 + 8.36 ( 100 – θ )
4600 + 836 – 8.36 θ
4600 + 836 +3344
θ
= 334.4 ( θ – 10 )
= 334.4 θ –3344
= (334.4 + 8.36) θ
= 25.6° C
(ii) Heat lost by steam = Heat gained by
water + Heat gained by Cal
i.e. Heat lost by steam
= m w c w ∆θ ↑ + m c c c ∆θ ↑
⇒ 4600 + 8.36 ( 100 – θ )
= 334.4 ( θ – 10 )
+ 27.3 ( θ – 10 )
4600 + 836 – 8.36 θ
= 334.4 θ – 3344 + 27.3 θ – 273
4600 + 836 + 3344 + 273
= (334.4 + 8.36 + 27.3) θ
θ = 24.5° C
38
Q13 1 litre = 1000 cm
Q = mc ∆θ
Heat needed = Heat to bring kettle from
10° → 100° + Heat water 10° → 100°
= m al c al ( 90 ) + m w c w ( 90 )
= ( 0.5 ) ( 910 ) ( 90 ) + ( 1.7 ) ( 4180 ) ( 90 )
= 40 950 + 639 540
= 680 490 J
680 490
----------------- = ---------------------- = 272 s
Time taken = Energy
3
Power
2.5 × 10
= 4.54 min
6
6
------- ( 2.3 × 10 ) = 1.955 × 10 J
Q = ml =  1.7

2
6
1.955 × 10
- = 782 s = 13.03 min
Time = --------------------------3
2.5 × 10
–3
Q14 E p = mgh = ( 3.5 × 10 ) ( 9.8 ) ( 3000 ) = 102.9 J
Let mass melted = m
Q
102.9
- = ---------------------Q = ml f ⇒ m = --5
lf
= 3.1181 × 10
–4
3.3 × 10
kg = 0.31 grams
Residual mass of hailstones = 3.5 – 0.31
= 3.19 grams
Teacher’s Manual
Exercise 16.1
Q8 c = λ f = 12 × 40 = 480 m s
Q1 10 Hz = 10 cycles passing per second
1
- s = 0.1 s
Duration of 1 cycle = ----10
(i) f remains the same;
Q2 Wavelength = 2 × 2 m = 4 m
(ii) c = λ f , if f remains the same and c
doubles then λ doubles.
Q3 Amplitude = max. disp from undisturbed
postion = 1--- (2 m) = 1 m
Q9 15 min = 1--- period
2
Q4 c = λ f = (1.5)(5) = 7.5 m s
2
–1
∴ Period = 30 min = 30 × 60 sec = 1800 s
6
8
Q5 c = λ f = (3.125)( 96 × 10 ) = 3 × 10 m s
6
8
–1
–1
Q6 c = λ f = ( 6 ) ( 50 × 10 ) = 3 × 10 m s
(for station 1)
All radio waves travel at the same speed in air
(≈ vacuum)
∴ For station 2: c = λ f ⇒λ =
c
--f
Q7 λ = 5 × 10 ⇒ f =
c
--λ
λ = 1 × 10 –11 ⇒ f =
8
=
c
--λ
3 × 10
-------------------–9
5 × 10
⇒f=
1
--T
1
= ----------1800
–1
–1
× 1000
--------------------------- m s
v = 400 km hr = 400
60 × 60
= 111.11 m s
–1
c
111.11
- = 199 998 m
∴ c = λ f ⇒ λ = --- = --------------1
f
8
=
3 × 10
-------------------6
25 × 10
= 12 m
–9
–1
Hz
19
Hz
= 6 × 10
340
- = 850 Hz
Q10 In air c = λ f ⇒ f = --c- = ---------
In other medium λ
16
-)
( ----------1800
λ
0.40
c
--------- = 58.82 cm
= --- = 500
f
850
8
3 × 10
= --------------------– 11
1 × 10
= 3 × 10
Frequency range = 6 × 10
16
19
Hz
Hz – 3 × 10
39
Real World Physics
Exercise 16.2
Exercise 16.3
Q1 No calculations required.
fc
( 2000 ) ( 336 )
- = ------------------------------- = 2349.7 Hz
Q1 f ′ = ----------
Q2 No calculations required.
Q2 f ′ =
Q3 Distance from node to antinode = 5 m
∴ λ--- = 5 ⇒ λ = 20 m
Q3 f ′ =
4
c = fλ⇒ f=
Q4 c = f λ ⇒ λ =
c
--λ
c
--f
4
= ------ = 0.2 Hz
20
=
–2
60 × 10
----------------------6
c–u
fc
-----------c+u
fc
----------c–u
⇒ 340 –
336 – 50
( 2000 ) ( 336 )
= ------------------------------- = 1740.9
336 + 50
( 600 ) ( 340 )
⇒ 720 = --------------------------340 – u
600 ) ( 340 )
u = (--------------------------720
⇒ u = 56.67 m s
Hz
–1
–1
= 0.1 m
Distance between adjacent nodes = --λ- = 0.05 m
2
= 5 cm
–1
Q5 c = 340 m s , f = ?
8  λ--- = 2 m ⇒ λ = 4--- = 0.5 m
2
8
c
340
f = --- = --------- = 680 Hz
λ
0.5
Q4 f = 1000 Hz, u = 40 m s , c = 336 m s
fc
( 1000 ) ( 336 )
- = ------------------------------While approaching: f ′ = ---------( 336 – 40 )
c–u
= 1135.1 Hz
fc
( 1000 ) ( 336 )
- = ------------------------------While going away: f ′ = ----------( 336 + 40 )
c+u
= 893.6 Hz
Change in frequency = 1135.1 – 893.6
= 241.5 Hz
fc
( 2000 ) ( 336 )
- = ------------------------------Q5 Approaching f ′ = ---------c–u
( 336 – 20 )
= 2126.58 Hz
fc
( 2000 ) ( 336 )
- = ------------------------------Going away: f ′ = ----------c+u
( 336 + 20 )
= 1887.64 Hz
∴ Change in frequency
= 2126.58 – 1887.64 = 238.9 Hz
40
–1
Teacher’s Manual
Exercise 17.1
Q6 f = 4000 Hz
Highest note is heard
when at A
f ′ = 4200, c = 340,
u=?
B
1m
fc
(i) f ′ = ----------
For downward journey of stone:
u = 0, a = 9.8, t = 2, s = ?
c–u
⇒ 4200
2
s = ut + 1--- at
2
( 4000 ) ( 340 )
= -----------------------------( 340 – u )
s=
⇒ u = 16.19 m s
–1
(ii) Lowest note occurs at B: f ′ =
=
( 4000 ) ( 340 )
--------------------------------( 340 + 16.19 )
(iii) Time =
(iv) Time =
0.388
------------2
=
fc
f ′ = ----------
c–u
( f ) ( 340 )
= --------------------340 – u
⇒ (1000)(340 – u)
= 340f
= 0.388 s
–1
The problem becomes much more difficult if
we do not make this simplifying assumption.
u = 0, a = 9.8, s = ?, t 1 = ?
s = 4.9t 1
1
fc
f ′ = -----------
For sound travelling up:
Time for sound to travel up
800
s
- ⇒ s = 340t 2
= t 2 = --------
c+u
( f ) ( 340 )
= --------------------340 + u
(800)(340 + u)
= 340f
–1
= 888.9 Hz
( 336 – u )
( 200 ) ( 336 )
---------------------------( 208 )
⇒ u = 12.92 m s
⇒ s = 19.6 m
2
Train receding:
fc
( 200 ) ( 336 )
- ⇒ 208 = ---------------------------Q8 f ′ = ----------
⇒ 336 – u =
2
2
⇒ u = 37.78 m s
c–u
( 9.8 ) ( 2 )
2
s = 1--- 9.8t1
∴ 1000(340 – u) = 800(340 + u)
( 1000 ) ( 340 – 37.78 )
-------------------------------------------------340
 1---
 2
For stone falling down:
2π ( 1 )
-------------16.19
= 0.194 s
Q7 Train approaching:
∴f=
fc
-----------c+u
= 3818.2 Hz
Dis tan ce
----------------------Speed
1000
A
Q1 If we assume the time taken for the sound to
travel up from the bottom of the well is
negligible compared with time for stone to fall,
we have:
340
2
Now t 1 + t 2 = 2 ∴ t 2 = 2 – t 1
Equating 1 and 2:
2
= 340t 2
2
= 340 ( 2 – t 1 )
2
+ 340t 1 – 680 = 0
4.9t 1
4.9t 1
4.9t 1
Solving for t 1 :
2
– ( 340 ± ( 340 ) – 4 ( 4.9 ) ( – 680 ) )
2 ( 4.9 )
t 1 = ----------------------------------------------------------------------------------t 1 = 1.945 s
2
1 ⇒ s = 1--- ( 9.8 ) ( 1.945 )
2
= 18.5 m = depth of well
--------------------Q2 Speed = Distance
Time
⇒ Distance = (Speed) × (Time) = (340)(4)
= 1360 m
41
Real World Physics
Exercise 17.2
Exercise 18.1
1 T
1
200
- --- = -------------------- ---------- = 44.2 Hz
Q1 f = ----
Q2 f =
2l
µ
1
----2l
T
--µ
( 2 ) ( 0.8 ) 0.04
⇒ 4f l =
2 2
Q3 µ =
=
n
=
2
2
= 0.0625 kg m
–1
–6
λ = 1.269 × 10 m
–6
d Sin θ
( 2.5 × 10 ) ( Sin 40° )
- = ----------------------------------------------------Q2 λ = -----------------
n
1
–6
( 2 ) ( 0.8 ) 0.0625
–1
Mass
0.04
- = ---------- = 0.01 kg m
Q4 µ = ----------------
f=
( 2.8 × 10 ) ( Sin 65° )
----------------------------------------------------2
= 1.607 × 10 m
1 T
1
100
- --- = -------------------- ---------------- = 25 Hz
f = ---2l µ
m, n = 2, θ = 65, λ
–6
2 2
0.05
---------0.8
–6
d Sin θ
n λ = d Sin θ ⇒ λ = -----------------
T
--µ
⇒ T = 4 f l µ = (4)(500) (0.6) (0.02)
= 7200 N
Mass
----------------Length
Q1 d = 2.8 × 10
Length
4
1 T
1
400
----- --- = ---------------- ---------2l µ
( 2 ) ( 4 ) 0.01
–6
1
- mm = 0.005 mm = 5 × 10 m
Q3 d = --------
200
–6
1
- mm = 0.002 mm = 2 × 10 m
Q4 d = --------
500
= 25 Hz
–6
1
- mm = 0.00125 mm = 1.25 × 10 m
Q5 d = --------
800
Q5 f = ∝
d Sin θ
n λ = d Sin θ ⇒ λ = ------------------
T⇒f = k T
n
If tension is increased four times to 4T.
f new = k T new = k 4T = 2k T = 2 f
i.e frequency is doubled
Q6 f ∝
∴f=
260 
 -------- 40
d Sin θ
n λ = d Sin θ ⇒ λ = -----------------n
T
–6
260
---------40
( 2.5 × 10 ) Sin 31°
∴ λ = ------------------------------------------------- = 6.44 × 10
160 = 520 Hz
Q7
n=2
l
∴ f = 230 Hz
(ii) f ∝
⇒ fl = k
∴ k = (460)(0.6) = 276
276
∴ f = ---------
∴f=
42
l
276
--------1.5
= 184 Hz
0.3
θ
40
(i) f ∝ 1--- If length is doubled f is halved
–7
2
200
---------------------- = 581 Hz
(ii) When T is 200, f = 260
1
--l
m
400
260
---------40
(i) When T is 160, f =
Q7
–7
1
–6
1
- mm = 0.0025 mm = 2.5 × 10
Q6 d = -------m
T⇒f = k T
260 = k 40 ⇒ k =
–6
( 1.25 × 10 ) Sin 30°
∴ λ = ---------------------------------------------------- = 6.25 × 10
2.5
–6
1
- = 5 × 10 m
d = --------
200
------- = 0.12 ⇒
Tan θ = 0.3
λ =
2.5
d Sin θ
------------------n
Sin θ = 0.1191
–6
( 5 × 10 ) ( 0.1191 )
⇒ λ = ----------------------------------------------
2
= 2.978 × 10
–7
m
m
Teacher’s Manual
Q8
n=2
x
0.4 m
–6
1
- = 1.667 × 10 m
d = --------
–7
λ = 5 × 10 m
600
n λ = d Sin θ
–7
( 2 ) ( 5 × 10 )
- = 0.5999
⇒ Sin θ = n-----λ- = -------------------------------–6
⇒ θ = 36.86°
Tan θ =
x
------0.4
Q1 No calculations required.
1
Q2 d = ----------------= 2 × 10
5
θ
d
Exercise 18.2
1.667 × 10
⇒ x = ( 0.4 )Tan θ
5 × 10
–6
m
θ –θ
2
R
L
θ = -----------------
217.2 – 182.8
θ = -------------------------------2
⇒ θ = 17.2°
d Sin θ
n λ = d Sin θ ⇒ λ = -----------------n
= (0.4) Tan 36.86° ⇒ x = 0.2999 m
–6
1
- = 2.5 × 10 m
Q9 d = --------
400
–7
λ = 6.2 × 10 m
Sin θ
n = d------------------
n max
λ
–6
× 10
------------------------ = 4.03
= --d- = 2.5
–7
λ
6.2 × 10
Fourth order is highest.
–6
( 2 × 10 )Sin 17.2°
= -----------------------------------------------
1
λ = 5.914 × 10
6.2 × 10
m
Q3 163°30′ 182°45′ 200 217°15′ 243°30′
17.25
17.25
43.5
36.5°
–6
1
- mm = 2 × 10 m
d = --------
500
Both 1st order images give
–6
λ = d Sin θ = ( 2 × 10 ) Sin ( 17.25 )
–6
5 × 10
Q10 n max = --d- = -----------------------= 8.06
–7
λ
–7
= 5.931 × 10
–7
m
–6
( 2 × 10 ) Sin 36.5°
2nd order: λ = -------------------------------------------------2
Eight order is highest.
= 5.948 × 10
–7
m
–6
–7
( 2 × 10 ) Sin ( 43.5 )
- = 6.884 × 10 m
or λ = ----------------------------------------------------
2
Average value of correct λ = 5.94 × 10
6.884 × 10
is wrong.
–7
–7
m
m is different, therefore 243°30′
43
Real World Physics
Exercise 18.3
Exercise 19.1
–6
1
- mm = 0.0025 mm = 2.5 × 10
Q1 d = -------m
400
Q1 ε = ε r ε o = ( 2.2 ) ( 8.9 × 10
-----n λ = d Sin θ ⇒ Sin θ = nλ
= 1.958 × 10
d
–9
2 ) ( 710 × 10 )
For red light: Sin θ = (-------------------------------------–6
2.5 × 10
⇒ θ = 34.61 °
For blue light: Sin θ =
⇒ θ = 19.15 °
– 11
– 12
Fm
–1
– 11
4 × 10
ε
- = --------------------------- = 4.49
Q2 ε = ε r ε o ⇒ ε r = ---– 12
εo
8.9 × 10
1 Q1 Q2
(1)(3)
1
- -------------- = --------------------------------------- ---------------Q3 F = ----------2
– 12
2
–9
( 2 ) ( 410 × 10 )
--------------------------------------–6
2.5 × 10
∴ Angular separation
4 πε o
4 π ( 8.9 × 10
r
)
1
10
= 2.682 × 10 N
The force is the same on each
1 Q1 Q2
- -------------Q4 F = ----------2
4 πε o
= 34.61 – 19.15 = 15.46 °
r
–6
Q2
)
(i) Because of the bigger n factor, calculate
the angle between red and violet when
n = 1 and when n = 3 to see exactly.
(ii) Red has a longer λ , the bigger the λ ,
the bigger θ
d Sin θ
(iii) n λ = d Sin θ ⇒ λ = -----------------n
[n = 1, λ max = d;
n = 2, λ max = d--- ]
2
(iv) n max = --d- , bigger d, the more orders
λ
visible.
(v) In a grating, red is diffracted most, violet
least; in the prism it is the other way
around.
–6
1
( 3 × 10 ) ( 6 × 10 )
- -------------------------------------------------= -------------------------------------– 12
2
4 π ( 8.9 × 10
)
(4)
–2
= 1.0058 × 10 N
The force is attraction since the charges have
opposite sign.
1 Q1 Q2
- -------------Q5 F = ----------2
4 πε o
r
–6
4 π ( 8.9 × 10
–6
( 2 × 10 ) ( 2 × 10 )
-------------------------------------------------2
( 0.3 )
1
= -------------------------------------– 12
)
= 0.397 N towards the –2 µC charge
Q Q
1
1 2
- ------------Q6 F1 = -------2
4 πε
r
–6
4 π ( 7.2 × 10
= 5.526 × 10
)
–3
–6
( 2 × 10 ) ( 4 × 10 )
-------------------------------------------------2
( 0.4 )
1
= -------------------------------------– 10
N towards the –4 µC charge
– 19
– 19
1 ( 1.6 × 10 ) ( 1.6 × 10 )
- ---------------------------------------------------------------- = 14.306 N
Q7 F = ----------2
4 πε o
( 4 × 10
– 15
)
GM M
1 2
F = -------------------2
r
– 11
– 27
– 27
( 6.7 × 10 ) ( 1.67 × 10 ) ( 1.67 × 10 )
= -----------------------------------------------------------------------------------------------------2
( 4 × 10
= 1.17 × 10
44
– 35
N
– 15
)
Teacher’s Manual
Exercise 19.2
Q8 Let the size of each charge be Q
F=
1 Q1 Q2
------------ ------------4 πε o d 2
⇒ 0.2 =
2
Q
1
--------------------------------------- ----------------– 12
2
4 π ( 8.9 × 10 ) ( 0.02 )
⇒ Q = ( 0.2 ) ( 4 π ) ( 8.9 × 10
– 15
= 8.9472 × 10
2
– 12
) ( 0.02 )
2
–1
6
F
12
- = -------------------- = 3 × 10 N C
Q1 E = --–6
Q
4 × 10
3
–6
Q2 F = EQ = ( 3 × 10 ) ( 2 × 10 ) = 6 × 10
–3
N
–6
–9
F
F
7 × 10
- ⇒ Q = --- = -------------------- = 3.5 × 10 C
Q3 E = --3
Q
E
2 × 10
= 3.5 N C
⇒ Q = 8.9472 × 10
– 15
= 9.4589 × 10 C
–8
= 9.46 × 10 C
–8
–6
Q
4 × 10
- = ------------------------------------------------Q4 E = ---------------2
– 12
2
4 πε o r
4 π ( 8.9 × 10
1 Q1 Q2
- -------------Q9 F1 = -------2
4 πε
3
= 8.94 × 10 V m
x
4 πε ( 3x )
=
=
F1  1---
9
Q
Q5 E = ---------------2
4 πε o r
 1---
 9
–6
20 × 10
(i) ---------------------------------------------------------------------2
4 π ( 8.9 × 10
i.e. F2 is
 1---
 9
F1
–6
=
4 π ( 8.9 × 10
– 12
NC
–1
–3
) ( 1 × 10 )
= 1.7882 × 10
11
NC
–1
–6
20 × 10
(iii) --------------------------------------------------------------------2
–6
1 ( 8 × 10 ) ( 10 × 10 )
------------ ----------------------------------------------------2
4 πε o
( 0.1 )
4 π ( 8.9 × 10
– 12
–2
) ( 10 × 10 )
7
= 1.7882 × 10 N C
–6
( 0.1 )
–6
13
–6
8 × 10
= ------------------2
⇒ Q = 2 × 10
–3
) ( 0.1 × 10 )
20 × 10
(ii) -----------------------------------------------------------------2
–6
1 ( 10 × 10 )Q
------------ -------------------------------2
4 πε o
( 0.05 )
Q
----------------2
( 0.05 )
– 12
= 1.7882 × 10
Q10 Let the charge at C be Q coulombs
Force on 10 µC due to Q = Force on 10 µC due
to 8 µC .
⇒
–1
The direction is always from the 4 µ C charge
1 Q1 Q2
- -------------F2 = -------2
1 Q1 Q2
--------- ------------4 πε x 2
)(2)
–1
In each case the direction of E is towards the
charge.
C = + 2µC
Q6
Q
(i) E = ---------------2
4 πε o r
Its direction is away from the charge Q
Q
(ii) E = ---------------2
4 πε o r
Its direction is towards the charge Q
9
Q7 F = EQ = ( 3 × 10 ) ( 1.6 × 10
– 10
= 4.8 × 10 N
– 19
)
– 10
F
4.8 × 10
------------- = ---- = --------------------------Acceleration = Force
– 31
Mass
m
9 × 10
= 5.3333 × 10
20
ms
–2
45
Real World Physics
Q8
20 cm
+4µC
+ 2 µC
C
0.1 m
Q9 Electric field intensity at midpoint due
to +3 µC
0.1 m
–6
Q
3 × 10
- = -----------------------------------------------------= ---------------2
– 12
2
E at
4 πε o r
C due to +4 µC
4 π ( 8.9 × 10
5
=
–1
= 6.706 × 10 N C towards the –7 µC
charge
Q
----------------2
4 πε o r
Electric field intensity at midpoint due to
– 7 µC charge
–6
4 × 10
= ----------------------------------------------------– 12
2
4 π ( 8.9 × 10
) ( 0.2 )
) ( 0.1 )
–6
6
Q
7 × 10
- = -----------------------------------------------------= ---------------2
– 12
2
–1
= 3.5765 × 10 N C to the right
4 πε o r
Q
E at C due to + 2 µC = ---------------2
6
–1
Total field intensity =
5
6
6.706 × 10 + 1.565 × 10
–1
6
= 2.236 × 10 N C towards the – 7 µC charge
–6
2 × 10
= ----------------------------------------------------– 12
2
) ( 0.1 )
6
) ( 0.2 )
= 1.565 × 10 N C towards the – 7 µC charge
4 πε o r
4 π ( 8.9 × 10
4 π ( 8.9 × 10
–1
= 1.7883 × 10 N C to the left
6
Total field intensity at C
6
= 3.5765 × 10 – 1.7883 × 10
6
6
–1
= 1.7882 × 10 N C towards the +2 µC
charge
Force on +5 µC , F = E Q
6
–6
= ( 1.7882 × 10 ) ( 5 × 10 )
–6
F = EQ = ( 2.236 × 10 ) ( 2 × 10 )
= 4.47 N towards the – 7 µC charge
Q10
EB
+ 10 µC
EA
C
A
1–x
+5 µC
x
B
Let C be the point at which E = 0
Let x = distance from +5 µC at which E is zero
E at C due to +10 µC = E at C due to +5 µC
i.e. EA = EB
= 8.941 N towards the +2 µC charge
–6
1 10 × 10
------------ ----------------------4 πε o ( 1 – x ) 2
–6
1 5 × 10
- -------------------= ----------2
4 πε o
⇒
10
------------------2
(1 – x)
5
= ---2
⇒
(1 – x)
------------------10
2
2
x
= ----
x
5
2
( 1 – x ) = 2x
∴1–x =
2x
1 = ( 1 + 2 )x
⇒ x=
1
---------------1+ 2
or
x
2
∴ 1 – x = – 2x
1 = x – 2x
1 = x(1 – 2)
⇒ x=
x = 0.4142 m
1
---------------1– 2
x = – 2.4
Impossible
∴ E is zero at a distance of 0.4142 m from
the +5 µC charge
46
Teacher’s Manual
Q11
–12µC
EA
+ 5µC
A
B
0.3 m
EB
6
Q1 W = QV ⇒ V = W
----- = --- = 3 V
x
Q
To the right of the +5 µC the electric field
strength due to +5 µC ( E B ) is to the right and
electric field strength E A due to –12 µC is to
the left.
Let x = distance at which these two forces
cancel, i.e. E A = E B (in magnitude)
–6
12 × 10
-----------------------------------2
4 πε o ( 0.3 + x )
2
–6
=
5 × 10
---------------------2
4 πε o ( x )
⇒
12
-----------------------2
( 0.3 + x )
=
5
----2
x
2
12x = 5 ( 0.3 + x )
12x = 5 ( 0.3 + x )
⇒ x = 0.5463
or 12x = – 5 ( 0.3 + x )
⇒ x is negative; impossible
∴ E is zero 0.5463 m from the +5 µC charge
and not between the two charges
Exercise 20.1
2
–5
6 × 10
- = 10 V
Q2 W = QV ⇒ V = W
----- = ------------------–6
Q
6 × 10
Q3 W = QV = (4)(20) = 80 J
–6
Q4 W = QV = ( 8 × 10 ) (12) = 9.6 × 10
–5
J
– 16
4.8 × 10
- = 3000 V
Q5 W = QV ⇒ V = W
----- = -------------------------– 19
Q
Q6
1.6 × 10
(i) W = QV = ( 1.6 × 10
= 1.6 × 10
– 19
(ii) W = QV = ( 1.6 × 10
= 4.8 × 10
– 19
J
– 19
– 17
)(1)
) ( 300 )
J
Q7 Work = Force × Dis tan ce
E = Force on 1 C = 2 × 10 4 N
Work done in bringing 1 C from one plate to
the other.
W = (Force on 1 C)(Distance between plates)
4
= ( 2 × 10 ) (0.2) = 4000 J
Work done in bringing 1 C from one plate to
the other = p.d. between plates.
⇒ p.d. = 4000 V
47
Real World Physics
Q8 (i)
p.d. between plates = 400 V
⇒ 400 J done in bringing +1 C from one plate
to the other.
–2
W = Fs ⇒ 400 = (F)( 2 × 10 )
⇒ F = 20 000 N
(ii) E is the force on + 1 C ⇒
E = 20 000 N C –1
4
(iii) F = EQ = ( 2 × 10 )( 1.6 × 10
= 3.2 × 10
– 15
(iv) W = QV = ( 1.6 × 10
= 6.4 × 10
– 17
– 31
V
12
)
---- ⇒ Q = CV
Q4 C = Q
V
= ( 2 × 10
) ( 400 )
– 11
3
) ( 300 × 10 )
–6
= 6 × 10 C = 6 µC
J
2
) v = 6.4 × 10
– 17
–6
Q
6
4 × 10
- = ------------------- = 1.33 × 10 V
---- ⇒ V = --Q5 C = Q
C
12
V
3 × 10
–6
–7
5 × 10
---- = -------------------- = 4.167 × 10 F
Q7 C = Q
V
---- ⇒
Q8 C = Q
V
12
–6
Q = CV = ( 50 × 10 )(100)
–3
⇒ v = 1.19 × 10 m s
7
48
–6
–7
4 × 10
---- = -------------------- = 3.333 × 10 F
Q2 C = Q
= 8 × 10 C = 8 µC
2
( 9.1 × 10
1 × 10
–6
– 19
2
– 17
(vi) 1--- m v = 6.4 × 10
1
--2
V
V
(v) Since kinetic energy gained = potential
energy lost.
E k = 6.4 × 10 –17 J
⇒
–6
– 10
2 × 10
---- = -------------------- = 2 × 10 F
Q1 C = Q
4
– 12
---- ⇒ Q = CV = ( 8 × 10 )(1 000 000)
Q3 C = Q
N
– 19
Exercise 20.2
–1
= 5 × 10 C
Teacher’s Manual
Exercise 20.3
– 12
ε0 A
– 10
( 8.9 × 10 ) ( 0.02 )
- = ----------------------------------------------Q1 C = -------= 1.78 × 10 F
d
( 0.001 )
Q10 W = 1--- CV 2
– 12
–4
ε0 A
( 8.9 × 10 ) ( 150 × 10 )
- = --------------------------------------------------------------Q2 C = -------–3
d
⇒ 23 × 10
( 1 × 10 )
= 1.335 × 10
– 10
ε0
( 0.5 ) ( Q )
= -----------------------–6
2
–3
2.4 × 10
–4
( 8.9 × 10
)
Q11 W = 1--- CV 2 ⇒
( 0.01 ) = 1--- ( C ) ( 12 ) 2
2
8
= 1.1235 × 10 m
–4
Q12 P.d. across each is 40 V
---- ⇒ Q = CV
C = Q
( 2 × 10 )
= 4.45 × 10
– 11
V
– 11
– 10
F
–4
εA
( 7 ) ( 8.9 × 10 ) ( 25 × 10 )
- = -------------------------------------------------------------------Q5 C = ----–3
F
Q = C V = ( 1.5575 × 10
– 10
Q
-------------------–6
8 × 10
) ( 500 )
Q14
Q6 W = 1--- CV 2 = 1--- ( 6 × 10 –3 ) ( 200 ) 2 = 120 J
2
+
= 40
–4
C
1.067 × 10
= -----------------------------= 13.33 V
–6
Q
V 2 = ---C
–4
1.067 × 10
= -----------------------------= 26.67 V
–6
8 × 10
4 × 10
(i) P.d. across the resistor:
–3
3
V = I R = ( 0.2 × 10 ) ( 20 × 10 ) = 4 V
⇒ p.d. across capacitor = 40 – 4 = 36 V
–6
Q
4 × 10
---- ⇒ V = ---- = -------------------- = 2 V
Q7 C = Q
–6
C
C
Q
-------------------–6
4 × 10
Q
V 1 = ---C
–8
V
---and V = Q
–4
= 7.79 × 10 C
2
–4
⇒ Q(375 000) = 40 ⇒ Q = 1.067 × 10
( 1 × 10 )
⇒ C = 1.5575 × 10
–6
Q13 V 1 + V 2 = 40
⇒
– 10
–4
Q = ( 4 × 10 ) ( 40 ) = 1.6 × 10 C
) = 1.157 × 10
– 12
d
–6
Q = ( 8 × 10 ) ( 40 ) = 3.2 × 10 C
F
With perspex dielectric ε r = 2.6
C = (2.6)( 4.45 × 10
2
⇒ C = 1.39 × 10 F
2
– 12
–4
ε0 A
( 8.9 × 10 ) ( 100 × 10 )
- = --------------------------------------------------------------Q4 C = -------–3
d
1  Q 2
--- -----2 C 
⇒ Q = 3.32 × 10 C
F
–3
ε0 A
Cd
( 1 ) ( 1 × 10 )
- ⇒ A = ------- = --------------------------------Q3 C = -------– 12
d
⇒ W =
2
2 × 10
–6
---- ⇒ Q = CV = ( 60 × 10 ) ( 36 )
(ii) C = Q
V
W=
1
2
--- CV
2
=
1
–6
--- ( 2 × 10 ) ( 2 ) 2
2
= 4 × 10
–6
–3
= 2.16 × 10 C
J
(iii) Work done
–4
Q8 W = 1--- CV 2 = 1--- ( 4.6 × 10 –6 ) ( 8 ) 2 = 1.472 × 10 J
2
2
–6 2
0.5 ) ( 2 × 10 )
W = 1---  ------ = (--------------------------------------–6
2 C
Q9 W =
1
--- QV
2
=
1
–6
--- ( 7 × 10 ) ( 30 )
2
= 1.05 × 10
–4
J
= Energy stored
Q2
( 60 × 10 )
–8
= 3.33 × 10 J
49
Real World Physics
Exercise 21.1
Q1
–3
(i) 1 mA = 1 × 10 A
(ii) 0.05 mA = 0.05 × 10
Q6 Q = I t = (6)(4)(60)(60) = 86 400 C
–3
–5
= 5 × 10 A
(iii) 50 µA = 50 × 10 A = 5 × 10 A
–6
–5
(iv) 1000 mA = 1 A
(v) 0.2 µA = 0.2 × 10
Q2
–6
–7
= 2 × 10 A
Ch arg e on one electron
1
------------------------= 6.25 × 10 18 electrons
1.6 × 10 –19
Q8 Q = I t = (20)(6) = 120 C
1 C has 6.25 × 10 18 electrons
⇒ 120 C has (120)( 6.25 × 10 18 )
(i) 1 A = 1000 mA
3
5
(ii) 100 A = 100 × 10 mA = 10 mA
(iii) 0.025 A = 0.025 × 10
3
= 25 mA
(iv) 1 µA = 1 × 10 mA
–3
3
(v) 0.0006 A = 0.0006 × 10 mA = 0.6 mA
Q3
1 coulomb
Q7 Number of electrons = ---------------------------------------------------------
(i) 10 = 7 + x ⇒ x = 3 A
(ii) 2 + 3 = 1 + x ⇒ x = 4 A
(iii) x + 8 + 2 = 9 + 3 ⇒ x = 2 A
(iv) x + x + 4 = 2 + 3 + 5 ⇒ x = 3 A
= 7.5 × 10 20 electrons
Q9 Charge gone past
= (Nr. of electrons)(charge on one electron)
= ( 2 × 10 20 ) ( 1.6 × 10 –19 ) = 32 C
32
---- = ------ = 32 A
I = Q
t
1
36000
---- = --------------- = 7200 s
Q10 Q = I t ⇒ t = Q
I
5
Q11 Charge hitting screen in one hour
= Q = I t = ( 1 × 10 –3 )(60)(60) = 3.6 C
Ch arg e
Number of electrons = --------------------------------------------------------Ch arg e on one electron
Q4
(i) Q = I t = (3)(1) = 3 C
(ii) Q = I t = (3)(60) = 180 C
(iii) Q = I t = (30)(60)(60) = 10 800 C
Q5
10
---- = ------ = 1 A
(i) I = Q
(ii) I =
(iii) I =
50
t
Q
---t
Q
---t
=
=
10
1
-----60
10
-----1
= 0.01667 A
= 10 A
=
3.6
--------------------------1.6 × 10 –19
= 2.25 × 10 19 electrons
Q12 No calculations required.
Teacher’s Manual
Exercise 22.1
Exercise 23.1
------ = 5 Ω
Q1 R = V---- = 20
Q1 Total emf = Sum of indiviadual emfs
= 4 + 6 + 3 = 13 V
I
Q2 V = I R = (5)(12) = 60 V
Q2 (a) W = Q V = (1)(10) = 10 J
--------- = 2.3 A
Q3 I = V---- = 230
(b) W = Q V = (6)(10) = 60 J
(c) W = Q V = ( 1 ×
Q3 W = Q V ⇒ V =
W
----Q
10 –6 ) (10)
=
200
--------50
= 1×
Q4 R =
10 –5 J
= 4V
Q8
R =
∴
80
100
Q9 P = V I ⇒ I =
= --------- = 0.435 A
230
Q10 Total power = (50)(5) = 250 W
I
V
---I
=
3
24
-----2
= 12 Ω
Increase in resistance = 12 - 8 = 4 Ω
I
2
Q8 R = R 1 + R 2 = 2 + 3 = 5 Ω
------ = 6 A
I = V---- = 12
R
2
V = I R = (6)(5) = 30 V
------ = 10 Ω
Q9 R = V---- = 10
I
P
---V
∴
1
Resistance of bulb = 10 – 4 = 6 Ω
Q10 V 1 = I R 1 = (5)(2) = 10 V
P
250
- = --------- = 1.136 A
I = ---
V 2 = I R 2 = (5)(4) = 20 V
(i) P = V I = (10)(5) = 50 W
Energy = Power × time = (50)(5)(60)
= 15 000 J = 15 kJ
(ii) W = Q V = (1)(10) = 10 J
V 3 = I R 3 = (5)(8) = 40 V
V
Q11
= 3Ω
------ = 5 Ω
Q7 R = V---- = 10
(i) P = V I = (220)(4) = 880 W
(ii) W = P t = (880)(1)(60)(60)
= 316800 J = 3.168 MJ
=
10
12
-----4
------ = 8 Ω
Q6 R = V---- = 24
4000
----- = ------------ = 50 V
Q5 W = Q V ⇒ V = W
100
P
- = --------- = 0.435 A
Q6 P = V I ⇒ I = --V
230
Q7 P = V I = (12)(6.67) = 80 W
R
V
---I
Q5 V = I R = (5)(6) = 30 V
Q4 W = Q V = (20)(60) = 1200 J
Q
4
220
Total V = V 1 + V 2 + V 3 = 70 V
Q11 (a) --1- = 1--- + 1--- ⇒ R = 2 Ω
(b)
Q12 Total emf = 2 + 3 + 12 = 17 V
P = V I = (17)(2) = 34 W
(c)
Q13 V 1 = 6 ⇒ V 2 = 20 - 6 = 14 V
Power in bulb 2, P = V I = (3)(14) = 42 W
Energy = Power × Time = (42)(2)(60)(60)
= 302 400 J = 3.024 × 10 5 J
(d)
(e)
(f)
Q14 No calculations required.
Q15 Since both bulbs are identical and current is
everywhere the same in a series circuit, they
shine equally brightly.
R
1
--R
1
--R
1
--R
1
--R
1
-----Rp
=
=
=
=
=
4
4
1
1
--- + --- ⇒
4
1
1
1
1
------ + ------ + -----30 30 30
1
1
1
------ + ------ + -----30 60 90
1 1
--- + --- ⇒ R
8 8
1 1
--- + --- ⇒
2 4
R = 0.8 Ω
⇒ R = 10 Ω
⇒ R = 16.36 Ω
= 4Ω
R p = 1.33 Ω
R = 1.33 + 6 = 7.33 Ω
Q12
1
1
(i) --1- = ----- + ------ ⇒ R = 10 Ω
R
20
20
(ii) R = R 1 + R 2 = 20 + 20 = 40 Ω
(iii) --1- = 1--- + 1--- ⇒ R = 1.2 Ω
R
2
3
51
Real World Physics
1
1 1
- = --- + --- ⇒
Q13 ----Rp
2
3
R p = 1.2 Ω
Total resistance = 1.2 + 4 = 5.2 Ω
Current in 4 Ω resistor:
P.d. across 4 Ω resistor:
(ii) Current in 600 Ω resistor increases
⇒ p.d. across it increases,
⇒ the potential at A decreases.
V = I R = (3.846)(4) = 15.38 V
⇒ p.d. across parallel combination
= 20 – 15.38 = 4.615 V
Current in 2 Ω resistor:
(iii) p.d. across the parallel combination has
decreased,
⇒ current through 2 k Ω resistor
decreases.
20
- = 3.846 A
I = V---- = -----R
5.2
------------- = 2.31 A
I 1 = V---- = 4.615
R
2
Current in 3 Ω resistor:
------------- = 1.538 A
I 2 = V---- = 4.615
R
3
Q14 FIG (A) shows the circuit in a more intuitive
manner:
2000 Ω
400Ω
600Ω
FIG (A)
R = 400 Ω
(i) Resistance of parallel combination:
1
-----Rp
1
1
= -------- + ------------ ⇒ R p = 342.9 Ω
400
2400
Total resistance = 600 + 342.9
= 942.9 Ω
(ii) Current from battery,
10
- = 0.01061 A
I = V---- = -----------R
942.9
P.d. across 600 Ω resistor = I R
= (0.01061)(600) = 6.36 V
∴ p.d. across parallel combination
= 10 - 6.36 = 3.64 V
∴ Current in 2 k Ω resistor,
3.64
- = 0.0015 A
I = V---- = ----------R
52
(i) If R is reduced, total resistance of circuit
decreases, therefore current in the 600 Ω
resistor increases
2400
Teacher’s Manual
Exercise 23.2
Q1 A = π r 2 = π  d---
2
=
2
–2 2
0.22 × 10 
= π  --------------------------
2
3.80 × 10 –8 m 2
( 28.2 ) ( 3.8 × 10 –8 )
RA
ρ = ------- = --------------------------------------------- = 1.2 × 10 –6 Ω m
( 0.892 )
l
( 27.9 ) π ( 0.22 × 10 –3 ) 2
Rπd 2
- = -----------------------------------------------------Q2 ρ = -----------( 4 ) ( 0.856 )
4l
= 1.24 × 10 –6 Ω m
l
Q9 R ∝ --- and A is 4 times bigger
A
⇒ R is 4 times smaller
------- = 0.65 Ω
⇒ R = 2.6
4
If the resistance of the wire is to again be six
ohms (i.e. four times bigger), the length must
be four times longer.
 d
RA
Q10 ρ = ------
and A = π r 2 = π  --2-
l
RA
Q3 R = ρ-----l ⇒ l = ------
ρ
( 4 ) ( 0.16 × 10 –6 )
- =
= ---------------------------------------4.2 × 10 –7
RA
ρl
----- ⇒ l = ------ρ
A
( 12 ) π ( 0.25 × 10 –3 ) 2
= ------------------------------------------------1.7 × 10 –8
A
∴
l
Q4 R =
∴
l
1.524 m
ρ =
=
Rπ d2
-------------4l
Q11 Length halved ⇒ Resistance is halved to 5 Ω
= 138.6 m
Q5 No calculations required.
Rπd 2
( 400 ) π d 2
- ⇒ 1.2 × 10 –6 = ----------------------Q6 ρ = -----------( 4 ) ( 1.5 )
4l
⇒d =
⇒
d 2
( R )  π  --- 
  2 
----------------------------l
2
( 4 ) ( 1.5 ) ( 1.2 × 10 –6 )
------------------------------------------------( 400 ) ( π )
Cross-sectional area doubled ⇒ Resistance
is again halved to 2.5 Ω
Q12 Average diameter = 0.424 mm
( 6 ) π ( 0.424 × 10 –3 ) 2
Rπd 2
- = -------------------------------------------------ρ = -----------4l
( 4 ) ( 0.784 )
= 1.08 × 10 –6 Ω m
= 7.569 × 10 –5 m
Rπd 2
- ⇒ d =
Q7 ρ = ------------
=
4l
( 1.3 × 10 –6 ) ( 4 ) ( 1.3 )
-------------------------------------------------( 20 ) ( π )
4ρl
-------πR
= 3.28 × 10 –4 m
Q8 R ∝ l and l is 6 times bigger
⇒ R is 6 times bigger
⇒ R = (6)(4) = 24 Ω
R ∝ l and R is 10 times bigger
⇒ l is 10 times bigger
⇒ l = (10)(1.2) = 12 m
53
Real World Physics
Exercise 23.3
Q1
R
R
-----1- = -----3R2
R4
Q2
R
R
-----1- = -----3R2
R4
Exercise 24.1
⇒
4
15
--- = -----6
R4
⇒
R
10
------ = -----360
20
⇒ R 4 = 22.5 Ω
⇒ R 3 = 30 Ω
R
R
10R
4
Q3 -----1- = -----3- ⇒ -----------2- = ----- ⇒ R 4 = 0.4 Ω
R2
Q4
R4
R2
R
R2
R
R4
⇒
R
2R
--------1- = -----1X
R2
R4
R
R4
2R
X
(i) -----1- = -----3- becomes --------1- = -----3-
Q1 W = I 2 R t = ( 5 2 )(20)(3) = 1500 J
Q2 W = I 2 R t
= ( ( 12 × 10 –3 ) 2 )( 1 × 10 3 )( 4 × 60 )
= 34.56 J
Q3 W = I 2 R t = ( ( 0.5 ) 2 )(400)( 60 × 60 )
= 360 000 J
Q4
(i) P = I 2 R = ( 1 ) 2 (10) = 10 J
(ii) P = I 2 R = ( 2 ) 2 (10) = 40 J
⇒ X = 2 R2
(iii) P = I 2 R = ( 3 ) 2 (10) = 90 J
i.e. R 2 must be doubled
(ii) The left hand side of
R
R
-----1- = -----3R2
R4
(iv) P = I 2 R = ( 4 ) 2 (10) = 160 J
is doubled
so if R 4 is halved the right hand side of
the ratio is also doubled, and the bridge is
balanced. Thus half the value of R 4
Q5
(i) P = I 2 R = ( 2 ) 2 (1) = 4 J
(ii) P = I 2 R = ( 2 ) 2 (2) = 8 J
(iii) P = I 2 R = ( 2 ) 2 (10) = 40 J
(iv) P = I 2 R = ( 2 ) 2 (100) = 400 J
Q6 P = I 2 R = ( 6 ) 2 (40) = 1440 J
Energy = Power × time
= (1440)(60)(60) = 5 184 000 J
Q7 P = I 2 R ⇒ 100 = ( 0.5 ) 2 R
⇒ R =
100
---------0.25
⇒
Q8 P = I 2 R
⇒ R =
= 400 Ω
3 × 10 3
----------------100
3 × 10 3 = ( 10 ) 2 R
= 30 Ω
P
2 × 10 3
- = ----------------- = 8.7 A
Q9 (a) P = V I ⇒ I = ---
(b) P = I 2 R ⇒ R
V
P
= ---I2
=
230
2 × 10 3
----------------( 8.7 ) 2
= 26.42 Ω
W
100 × 10 6
----- ⇒ t = ----- = ----------------------(c) P = W
t
P
2000
= 50 000 s
--------- = 5.75 A
Q10 I = V---- = 230
R
40
W = I 2 R t = ( 5.57 ) 2 (40)(30) = 39 675 J
Q = I t = (5.75)(30) = 172.5 C
54
Teacher’s Manual
Exercise 24.2
Q11 P ∝ I 2 ∴ if the current doubles the power
is four times greater.
OR
Suppose
P 1 = R I 12
Now let current increase to 2 I 1
∴ New power P 2 = R ( 2I 1 ) 2 = 4 R I 12
i.e. P 2 = 4 P 1
Q12 Energy supplied to the water:
Q = m c ∆θ = (500)(4180(40)
= 8.36 × 10 7 J
= 80% of the electrical energy supplied
⇒ electrical energy supplied = 1.045 × 10 8 J
I2
W =
Rt
⇒ 1.045 × 10 8 = ( 20 ) 2 R(5)(60)(60)
⇒ R = 14.5 Ω
Q1 P = V I
⇒
⇒ I =
P
---V
------------ = 4.35 A
= 1000
230
The 13 A fuse is suitable.
Q2 Total power = (2)(500) + 1000 + 2000
= 5000 W
P
5000
- = ------------ = 21.73 A
I = --V
230
⇒ The 30 A fuse is suitable.
Q3 (Nr. of kW h) = (Power in kW)(Time in h)
= (2)(3) = 6 kW h
Q4 (Nr. of kW h) = (Power in kW)(Time in h)
------ ) = 0.05 kW h
= (0.075)( 40
60
Q13 Power P = V I = (230)(9) = 2070 W
Electrical energy needed:
Q = m c ∆θ = (2)(4180)(90) = 752 400 J
( 752400 )
------------------- = ---------------------- = 363.5 s
Time = Energy
Power
2070
Energy needed to boil off 3/4 of the water
= (Mass)(Specific latent heat of vaporisation)
= ml
= (0.75)(2)( 2.3 × 10 6 ) = 3.45 × 10 6 J
3.45 × 10 6
------------------- = ------------------------- = 1666.7 s
Time = Energy
Power
2070
55
Real World Physics
Exercise 27.1
Exercise 27.2
Q1 No calculations required.
Q2 F = BIL
2
- = 1.33 T
2 = B(3)(0.5) ⇒ B = ------------------( 3 ) ( 0.5 )
Q3 F = BIL = (2.5)(4)(2) = 20 N
⊥ to both B and wire
Q1 F = qvB = (2)(10)(2) = 40 N
–6
–3
Q2 F = qvB = ( 3 × 10 ) ( 200 ) ( 4 ) = 2.4 × 10 N
– 19
Q4 F = qvB ⇒ 2 × 10
⇒v=
F
4
4
- = ----------- = --- T
Q4 F = BIL ⇒ B = ---IL
3(1)
7
Q3 F = qvB = ( 1.6 × 10 ) ( 6 × 10 ) ( 4 )
–4
= 3.84 × 10 N
3
– 18
= ( 1.6 × 10
– 18
2 × 10
---------------------------------------– 19
( 2 ) ( 1.6 × 10 )
= 6.25 m s
– 19
)(v)(2)
–1
Q5 No calculations required.
Q5 No calculations required.
Q6 No calculations required.
Q6 Force on moving charge = centripetal force
2
mv
qvB = ---------
Q7 No calculations required.
r
Q8 F = BIL = (0.4)(10)(0.3) = 1.2 N;
M = Fd = ( 1.2 ) ( 0.08 ) = 0.096 N m;
Because ⊥ distance to axis decreases; yes;
(1.2)(0.16) = 0.192 N m
Q9 F = B ⊥ IL = (2)(Sin 30)(3)(3) = 9 N
zero degrees, i.e. when wire is parallel to
magnetic field
Q10 Parallel component = 3 Cos 60 = 1.5 T
⊥ component = 3 Sin 60 = 2.6 T
The ⊥ component causes the force on the wire
F = (3 Sin 60)(2.5)(0.5) = 3.25 N
perpendicular to wire and to the magnetic field.
( 1.6 × 10
– 19
7
) ( 2 × 10 ) ( 0.03 )
2
– 27
7
( 1.67 × 10 ) ( 2 × 10 )
= -----------------------------------------------------------
r
– 27
7 2
( 1.67 × 10 ) ( 2 × 10 )
------------------------------------------------------------------------– 19
7
( 1.67 × 10 ) ( 2 × 10 ) ( 0.03 )
r=
= 6.9583 m
2
– 31
2
mv
( 9.1 × 10 ) ( 2000 )
- = -------------------------------------------------------------------------Q7 r = --------– 19
–2
qvB
( 1.6 × 10
) ( 2000 ) ( 3 × 10 )
= 3.7916 × 10
–7
m
2
mv
--------Q8 --------= qvB ⇒ v = qBr
r
m
– 19
–3
–2
( 1.6 × 10 ) ( 2 × 10 ) ( 10 × 10 )
v = ------------------------------------------------------------------------------------– 31
( 9.1 × 10
7
= 3.52 × 10 m s
)
–1
2πr
--------------------- = --------- = T
Q9 Time for one resolution = Distance
2
mv
--------r
Speed
= Bev ⇒
v
-r
=
v
Be
-----m
2πm
∴ T = 2 π  -- = 2 π  ------ i.e. T = ----------v
Be
Be
r
Q10
m
(i) Number passing per second
= (0.1)(104) = 10.4 particles
–6
–5
(ii) (10.4)( 2 × 10 ) = 2.08 × 10 C
(iii) 2.08 × 10
–5
A
Q11 I = charge passing per second
12
– 19
---------- × ( 10 ) × ( 1.6 × 10 )
=  0.02

100
= 3.2 × 10
56
– 11
A
Teacher’s Manual
Exercise 28.1
Exercise 28.2
final Φ – initial Φ
5–1
- = ------------ = 2 V
Q1 E =  --------------------------------------------time


2
Q1 Φ = BA = ( 2 ) ( 0.3 ) = 0.6 Wb
2
Φ
------- = 0.8 m
Q2 Φ = B A ⇒ A = ----- = 0.4
B
final Φ – initial Φ
0 – 0.4
- = ---------------- = – 2 V
Q2 E =  --------------------------------------------time taken


0.2
0.5
–4
Q3 Φ = BA = ( 2 ) ( 100 × 10 ) = 0.02 Wb
2
4
2
2
–4
Total induced emf = ( 200 ) ( 2 ) = 400V
2
( 1m = 10 cm ⇒ 1cm = 10 m )
–2
Φ
2 × 10 
- = 1 T
Q4 Φ = B A ⇒ B = ---- =  ------------------------ 200 × 10 –4
A
–2
Φ
2
2 × 10
- = 6.667 m
Q5 Φ = B A ⇒ A = ---- = ------------------–
3
B
3 × 10
dΦ
2.4 – 0
Q3 E = N --------- = ( 600 )  ---------------- = 2400 V
 0.6 
dt
dΦ
4–2
Q4 E = N --------- = ( 200 )  ------------ = 1333.3 V
 0.3 
dt
E
---------------- = 333.3 A
I = --= 1333.3
R
2
A = πr ⇒ r =
A
--π
=
6.667
------------π
Q5 Change in flux = final Φ – initial Φ
–4
–4
= [ 2.4 ] [ 10 × 6 × 10 ] – [ 1.2 ] [ 10 × 6 × 10 ]
= 1.46 m
Q6 Change in Φ = final Φ – initial Φ
= 0.0072 Wb
= ( BA ) final – ( BA ) initial
–4
= 7.2 × 10
–4
= ( 2.4 ) ( 10 × 6 × 10 ) – ( 1.2 ) ( 10 × 6 × 10 )
= 0.0144 – 0.0072
–3
= 7.2 × 10 Wb
Q7 Φ = B ⊥ A = (4 Sin 30° )(0.2) = 0.4 Wb
Q8 Comp of B perpendicular to plane of coil
= B Sin 40 °
–3
= ( 2.5 × 10 ) ( Sin 40° )
Φ = B⊥ A
–3
–2 2
= ( 2.5 × 10 ) ( Sin 40° ) ( π ( 20 × 10 ) )
–4
= 2.02 × 10 Wb
4
–3
Wb
dΦ
Q6 E = N --------dt
( 6 ) ( 0.08 ) – ( 2 ) ( 0.08 )
= 100  ----------------------------------------------------- = 64 V


0.5
–4
Q7 Final Φ = BA = ( 2 ) ( 8 × 6 × 10 )
= 0.0096 Wb
Initial Φ = 0 Wb
–2
8 × 10 
Time taken for change =  ------------------ 3 
–2
= 2.667 × 10 s
final Φ – initial Φ
E =  -----------------------------------------------------


time
0.0096 – 0
= -----------------------------= 0.36 V
–2
2.667 × 10
Q8 Initial flux = 0, Final flux = BA
–4
= ( 4 ) ( 4 × 6 × 10 ) = 0.0096 T
⇒ 10 rotations per second
1
- s
1 rotation takes ----10
1
1
- = ------ s
Time to go from A to B =  1---  ----4 10
40
dΦ
0.0096 – 0
- = 76.8 V
E = N --------- = ( 200 )  -----------------------1


dt
-----40
57
Real World Physics
dΦ
final Φ – initial Φ
Q9 E = N --------- = N  --------------------------------------------time


dt
1
2 
- = ---- R
Q12 From Q11: IR = N  -------------------------t
t


B A–B A
2
(B – B )
– ( 2 × 10 –2 ) ( 3 × 10 –2 ) + 0
= 10 000  -----------------------------------------------------------------0.1


1
2
⇒ Q = NA ---------------------R
4
–2
–4
( 1 × 10 ) ( 2 × 10 ) ( 9 × 10 )
= ------------------------------------------------------------------------–1
–4
–5
500 ) ( 400 × 10 ) ( 0 – 1.8 × 10 )
= (-----------------------------------------------------------------------------------
10
1 × 10
= 18 × 10
–1
2
–2
–5
–5
( 5 × 10 ) ( 4 × 10 ) ( 1.8 × 10 )
= ------------------------------------------------------------------------------- = 3.6 × 10 C
= 1.8 V
10
Q10 Time to go from position 1 to position 2
= time for 1--- rev
4
=
1
 1--- ---- 4 40
sec
= 6.25 × 10
–3
s
final Φ – initial Φ
E = N  --------------------------------------------time


–2
( B ) ( 5 × 10 ) – 0
4 = 200  ----------------------------------------- 6.25 × 10 – 3 
–3
–1
4 ) ( 6.25 × 10 )
= 0.025 × 10 T
∴ B = (---------------------------------------–2
( 200 ) ( 5 × 10 )
= 2.5 × 10
–3
T
dΦ
---- ;
Q11 E = N --------- ; E = IR ; I = Q
t
dt
Let t be the time during which the flux density
changes from 0 to 0.25 T
Φ final – Φ initial
E = N  ----------------------------------------t


1
2 
IR = 1  -------------------------t


B A–B A
Q
---- R
A
= ---t ( B 1 – B 2 )
t
∴ QR = A ( B 1 – B 2 )
–3
( 4 × 10 ) ( 12 ) = A ( 0.25 – 0 )
–3
⇒ A = 192 × 10 m
⇒ A = 1.92 × 10
58
–1
Q
2
m
2
Teacher’s Manual
Exercise 28.3
Exercise 28.4
Q1 In one second: Work done = Electrical power
2
F × dist = I R
V
20
- = 14.14 V
Q1 V rms = ------o- = -----2
2
Q2 V o = V rms 2 = ( 20 ) 2 = 28.28 V
2
F ( 30 ) = ( 0.6 ) ( 20 ) ⇒ F = 0.24 N
Q2
–4
( 2 ) ( 5 × 12 × 10 ) – 0
- = 0.4 V
(i) E = ----------------------------------------------------–2
12 × 10 
 ---------------------

4
Q4 P = I rms V rms = ( 2 ) ( 110 ) = 220 W
2
Q5 P = I rms R
I = V---I =
Q3 V o = V rms 2 = ( 230 ) 2 = 325.3 V
2
⇒ 500 = I rms ( 20 )
R
0.4
------5
⇒ I rms = 5 A
I = 0.08 A
P = I rms V rms
2
(ii) Fd = I R
⇒ 500 = 5V rms
2
( F ) ( 4 ) = ( 0.08 ) ( 5 )
F = 8 × 10
–3
N
final Φ – initial Φ
Q3 E = N  --------------------------------------------time


2
BL – 0
= N  ------------------- = N ( BLv ) = NBLv
 L/v 
⇒ V rms = 100 V
V o = 100 2 = 141.42 V
Q6 I o = 3
3
- = 2.12 A
(i) I rms = -----2
2
(i) I =
E
--R
=
(ii)
NBL v
---------------R
(ii) In one second
Work done = Electrical power
2
3
P = I rms R =  ------- ( 200 ) = 900 W
 2
900
(iii) P = I rms V rms ⇒ V rms = --------2.12
= 424.53 V
2 2 2 2
N B L v 
2
- R
⇒ Fv = I R =  ------------------------2

R
(iv) V o = V rms 2 = 600.37 V
Q7 Power = V rms I rms
2
2 2
N B L v
⇒ F = -----------------------R
520
3
=  ---------  ------- = 780 W
 2   2
W
----- = 2000
P = ----t- ⇒ t = W
------------ = 2.56 s
780
P
–4
dΦ
400 ) ( 5 × 10 )
- = 200 V
Q8 E = N ------- = (-------------------------------------–3
dt
1 × 10
nett emf = 300 – 200 = 100 V
E
I = --R
100
I = --------200
I = 0.5 A
59
Real World Physics
Exercise 28.5
Exercise 28.6
V
Vo
Q2 I = V---- = 20
------ = 2 A
R
N
Ns
Q1 ------i = ------p-
Q1 No calculations required.
10
N s
⇒ V o = V i = V i  ----- N p
Q3 No calculations required.
Q4 No calculations required.
100
= 230  --------- = 46 V
 500
Q5 No calculations required.
N
Np
V
2000
- ⇒ V i = V o  ------p- = 4  ------------ = 80 V
Q2 ------i = ----- 100 
Ns
 N s
Vo
Q6 E = IR
–3
12 – (Induced emf) = ( 100 × 10 ) ( 40 )
⇒ Induced emf = 8 V
V i
3000
- N =  ------------ ( 60 ) = 818 turns
Q3 N p =  ---- V o s  220 
V
220 
- = 550 turns
Q4 N s = N p  -----o- = ( 10 000 )  ----------Vi
4000
V iI p = V oI s
( 4000 ) ( I p ) = ( 220 ) ( 5 ) ⇒ I p = 0.275 A
Output power = V o I s = ( 220 ) ( 5 ) = 1100 W
1100
- ) = 1222.22 W
Input power = 100 ( ----------90
∴ 4000I p = 1222.22 W ⇒ I p = 0.306 A
Q5 No calculations required.
60
Teacher’s Manual
Exercise 29.1
Q1
(i) E P lost, W = QV = eV
– 19
– 15
= ( 1.6 × 10 ) (8000) = 1.28 × 10 J
(ii) E K gained = E P lost = 1.28 × 10
– 15
J
2
– 31 2
(iii) E K = 1--- mv ⇒ (0.5) ( 9.1 × 10 )v
= 1.28 × 10
– 15
7
= 5.3 × 10 m s
6
(ii) 6 MeV = 6 × 10 eV
9
(iii) 2.5 GeV = 2.5 × 10 eV
1.6 × 10
1.28 × 10
---------------------------------------– 31
( 0.5 ) ( 9 × 10 )
⇒v=
(i) 3 keV = 3000 eV
1
- eV = 6.25 × 10
(iv) 1 J = -------------------------– 19
2
– 15
Q5
(v) 2 × 10
– 15
–1
2
Q2 1--- mv = eV
2
Q3
1
--2
2
1.6 × 10
(vi) 6.4 × 10
– 16
(vii) 5.6 × 10
– 19
– 16
6.4 × 10
- eV = 4000 eV
J = -------------------------– 19
1.6 × 10
– 19
5.6 × 10
- = 3.5 eV
J = -------------------------– 19
1.6 × 10
Q6 No calculations required.
mv = eV
2
--------- = 639.8 V
V = mv
2e
Q4
eV
– 15
2 × 10
- eV = 12 500 eV
J = -------------------------– 19
7
–1
v = 2eV
---------- = 5.93 × 10 m s
m
18
(i) 5 eV = 5 × 1.6 × 10
– 19
J = 8 × 10
(ii) 200 eV = 200 × 1.6 × 10
– 17
= 3.2 × 10 J
– 19
3
– 15
)J
) J
J
N
– 12
6
–1
F
2 × 10
- = ---------------------------------------- = 4.2 × 10 m s
v = -----– 19
qB
( 1.6 × 10
)(3)
Bqv
(vi) 4.2 eV = ( 4.2 ) ( 1.6 × 10
– 19
– 19
) J
) J
)
Q10 No calculations required.
11
Q11 ---e- = 1.76 × 10
m
– 19
J
Bq
– 31
7
–2
( 9.1 × 10 ) ( 5.6 × 10 )
- = 1.06 × 10 m
= ---------------------------------------------------------–2
– 19
( 3 × 10 ) ( 1.6 × 10
9
– 19
– 12
r
– 19
(v) 40 GeV = ( 40 × 10 ) ( 1.6 × 10
–9
= 6.4 × 10 J
= 6.72 × 10
= 1.41 × 10
J
6
) ( 2.1 × 10 ) ( 4.2 )
2
2
mv
mv
mv
- ⇒ r = ------Q9 --------= Bqv ⇒ r = ---------
(iv) 5 MeV = ( 5 × 10 ) ( 1.6 × 10
= 8 × 10
– 19
J
6
– 13
– 19
Q8 F = qvB
J
(iii) 40 keV = ( 40 × 10 ) ( 1.6 × 10
= 6.4 × 10
– 19
Q7 F = qvB = ( 1.6 × 10
1.6 × 10
--------------------------- = 1.76
m
– 19
1.6 × 10
⇒ m = -------------------------11
1.76 × 10
× 10
11
= 9.09 × 10
– 31
kg
61
Real World Physics
Exercise 29.2
Q1 Red light : E = hf = ( 6.6 × 10
– 34
= 2.64 × 10
2.64 × 10
– 19
14
) ( 4 × 10 )
Q6 2.2 eV = ( 2.2 ) ( 1.6 × 10
E
3.52 × 10
- = 5.33 × 10
E = hf ⇒ f = --= ----------------------------– 34
J
h
2.64 × 10
- eV = 1.65 eV
J = ----------------------------– 19
1.6 × 10
8
–7
3 × 10
- = 5.63 × 10 m
λ = --c- = -------------------------14
f
– 34
) ( 8 × 10 )
– 19
J
– 19
5.28 × 10
- eV = 3.3 eV
= ----------------------------– 19
1.6 × 10
8
14
3 × 10
- = 6 × 10 Hz
Q3 c = f λ ⇒ f = --c- = ------------------–7
λ
E = hf = ( 6.6 × 10
5 × 10
– 34
= 3.96 × 10
Q7 Number of Photons striking per second
= Number of electrons emitted per second
Current = Charge passing per second
= Charge on one electron × Number of
electrons passing per second
∴ 2 × 10
14
) ( 6 × 10 )
– 19
5.33 × 10
14
–6
= ( 1.6 × 10
13
2 × 10
- = 1.25 × 10 Photons
⇒ x = -------------------------– 19
J
1.6 × 10
–9
m
Q8 1 eV = 1.6 × 10
λ f = c ⇒ f = --cλ
– 34
8
hc
( 6.6 × 10 ) ( 3 × 10 )
E = hf = ------ = -----------------------------------------------------–9
λ
600 × 10
– 19
J
4 eV = 4 × 1.6 × 10
– 19
= 3.3 × 10
– 19
Q5 λ = 590 nm = 590 × 10
–9
J
6.6 × 10
J
= 9.70 × 10
m
Q9 (a) Φ = h f o = ( 6.6 × 10
c
--λ
– 34
= 1.32 × 10
14
14
– 19
J
(iii) Energy of source = 10 W = 10 joules per
second
Number of Photons emitted per second
Energy emitted per second
---------------------------------------------------------------Energy of one photon
= 2.98 × 10 19 Photons
15
) ( 2 × 10 )
J
= 8.25 eV
Hz
) ( 5.08 × 10 )
= 3.3528 × 10
Hz
1.6 × 10
= 5.08 × 10
– 34
– 18
14
– 18
1.32 × 10
(b) Work function in eV = ----------------------------– 19
590 × 10
(ii) E = hf = ( 6.6 × 10
– 19
6.4 × 10
---- = --------------------------Φ = h fo ⇒ fo = Φ
– 34
8
3 × 10
∴ frequency = ------------------------–9
62
= 6.4 × 10
– 19
h
=
)( x)
–6
Q4 λ = 600 nm = 600 × 10
(i) c = f λ ⇒ f =
– 19
=
10
-----------------------------------– 19
3.3528 × 10
Q10 Φ = 1.2 eV = 1.2 × 1.6 × 10
= 1.92 × 10
– 19
– 19
J
J
– 19
1.92 × 10
---- = -----------------------------Φ = h fo ⇒ fo = Φ
– 34
h
6.6 × 10
= 2.91 × 10
14
8
–6
3 × 10
- = 1.03 × 10 m
λ = --c- = -------------------------14
f
2.91 × 10
Hz
– 19
14
6.6 × 10
– 19
= 5.28 × 10
) J = 3.52 × 10
– 19
– 19
Q2 Blue light : E = hf = ( 6.6 × 10
– 19
J
Hz
Teacher’s Manual
Q11
f o = 8.8 × 10
14
Hz, f = 9.2 × 10
14
Q14 λ = 250 nm = 250 × 10
Hz
15
3 × 10
- = 1.2 × 10 Hz
f = --c- = ------------------------–9
λ
2
= 6.6 × 10
= 2.64 × 10
(ii) 2.64 × 10
14
– 20
14
( 9.2 × 10 – 8.8 × 10 )
– 20
250 × 10
Φ = 1.6 eV = ( 1.6 ) ( 1.6 × 10
J
= 2.56 × 10
– 20
J =
2.64 × 10
-----------------------------– 19
1.6 × 10
eV
Q12 λ = 350 nm = 350 × 10
–9
– 19
m
– 34
350 × 10
= 7.92 × 10
– 19
– 2.56 × 10
= 5.36 × 10
– 19
J
14
3 × 10
- = 8.57 × 10 Hz
c = λ f ⇒ f = --c- = ------------------------–9
= 3.2 × 10
– 19
J
15
= ( 6.6 × 10
– 19
) J
⇒ E K max
8
Max E K = 2 eV = 2 × 1.6 × 10
– 19
Max kinetic energy = hf – Φ
= 0.165 eV
λ
m
8
2
(i) 1--- mv max = hf – h f o = h(f – f o )
– 34
–9
J
J
1
--2
E K max = hf – Φ
( 9.1 × 10
– 31
) ( 1.2 × 10 ) – 2.56 × 10
– 19
2
)v = 5.36 × 10
– 19
⇒ v = 1.09 × 10 m s
6
– 19
– 34
3.2 × 10 = ( 6.6 × 10 ) ( 8.57 × 10
– 19
– 19
⇒ Φ = 5.66 × 10 – 3.2 × 10
= 2.46 × 10
– 19
– 14
) –Φ
Q15 E = hf = ( 6.6 × 10
– 34
= 6.072 × 10
J
–3
–3
Q13 0.2 mA = ( 0.2 × 10 ) A = ( 0.2 × 10 ) C s
Number of electrons in 1 C =
– 19
–1
6
) ( 92 × 10 )
– 26
joules
Number of Photons emitted per second
–1
6
2 × 10
- = 3.29 × 10
= -------------------------------– 26
1
--------------------------– 19
1.6 × 10
6.072 × 10
31
Photons
∴ Number of electrons leaving cathode
–3
0.2 × 10
- ⇒ Number of photons
= -------------------------– 19
1.6 × 10
–3
15
0.2 × 10
- = 1.25 × 10
= -------------------------– 19
1.6 × 10
Energy of one photon E = hf =
– 34
16
– 17
( 6.6 × 10 ) ( 2 × 10 ) = 1.32 × 10 J
∴ Energy per second =
 Energy of one ×  Number of Photons
 Photon  

per second
= ( 1.32 × 10
– 17
15
) ( 1.25 × 10 ) = 0.0165 J
= 1.65 × 10
–2
J
63
Real World Physics
Exercise 30.1
Exercise 30.2
Q1 No calculations required.
Q1 No calculations required.
Q2 No calculations required.
Q2 No calculations required.
Q3 No calculations required.
Q3
(i) 1--- remains
2
Q4 No calculations required.
(ii) 6 years = 2 half-lives ⇒ 1--- remains
Q5 No calculations required.
(iii) 9 years = 3 half-lives ⇒
Q6 Decrease in mass number = 238 – 230 = 8
⇒ 2 ∝ –particles emitted ⇒ atomic number
decreases by 4 (i.e. 2 × 2 ) to 88
To get back up to 90, 2 β -particles must be
emitted.
Q7 Decrease in mass number = 238 – 226 = 12
⇒ 3 ∝ –particles ⇒ decrease in atomic
number = 3(2) = 6 to 86 ∴ 2 β must be emitted
to get atomic number up to 88
Q8 No calculations required.
4
1
--8
remains
Q4 40 years = 4 half-lives
1
15
1
- = ------ remains and ------ has decayed.
⇒ ---4
16
2
Q5
16
5 ) ( 60 )
- = 15
(i) Number of half-lives = (-----------------( 20 )
(ii) 3 hours =
1
----9
2
( 3 ) ( 60 )
------------------20
half-lives = 9
1
remains undecayed, i.e. -------512
(iii) 40 mins = 2 half-lives ⇒
1
--4
remains
⇒ 3--- has decayed
4
(iv)
1
-----n
2
Q9 No calculations required.
Q10 No calculations required.
–3 –1
ln2
ln2
-------- ⇒ λ = -------- = --------- = 5.776 × 10 s
Q6 T --1- = ln2
2
λ
T1
--2
Q11 No calculations required.
Q12 4 ∝ –particles ⇒ Mass number decreases by
4 × 4 = 16 and atomic number decreases by
4×2 =8
3 β –particles ⇒ atomic number increases by
3 ∴ overall decrease in atomic number
=8–3=5
120
Q7 T 1--- = 5.5 minutes = (5.5)(60) seconds
2
–1
0.693
-------- = ----------------------- = 0.0021 s
λ = ln2
T1
--2
( 5.5 ) ( 60 )
= 2.1 × 10
–3
s
–1
ln2
- = 9.158 × 10
Q8 λ = ----------------------------------------------------------( 2.4 ) ( 365 ) ( 24 ) ( 60 ) ( 60 )
–9
s
–1
–4
ln2
-------- = ----------------- = 1.386 × 10 s
Q9 T 1--- = ln2
3
2
λ
5 × 10
ln2
-------- = -------------------- = 346.57 s = 5.776 min
Q10 T 1--- = ln2
–3
2
λ
2 × 10
0.693
-------- = ------------------------------ = 7200 s = 2 hours
Q11 T 1--- = ln2
–5
2
λ
9.627 × 10
6 hours = 3 half-lives ⇒ 1--- remains undecayed
8
64
Teacher’s Manual
Exercise 30.3
Q12 Activity = λ N
Q1 64 g = 6.02 × 10
3
3 × 10 = 8 × 10
⇒N=
3
3 × 10
-------------------–8
8 × 10
–8
6.02 × 10
1 g = --------------------------
N
= 3.75 × 10
10
64
23
2000 ) ( 6.02 × 10 )
= (------------------------------------------------64
2000 g
–3
= 1.6 × 10
15
13
= 1.88 × 10
Q2 218 g = 6.02 × 10
Bq
--2
( 6 × 10 )
= 1.73 × 10
λ
atoms
218
20
18
= 9.665 × 10
Q3 218 g has 6.02 × 10
Bq
15
23
atoms
atoms
23
–6
6.02 × 10 × 2.4 × 10
2.4 µg = ----------------------------------------------------------
ln2
-------- ⇒ λ = -------Q15 T --1- = ln2
2
atoms
–6
6.02 × 10 × 3.5 × 10
3.5 µg = ----------------------------------------------------------
ln2 
 ----------------- ( 4 ) ( 60 )
=
23
25
23
Q14 Activity = λ N
λ =
atoms
23
Q13 Activity = λ N = ( 8 × 10 ) ( 2 × 10 )
ln2
-------T1
23
T1
218
--2
= 6.628 × 10
ln2
λ = -----------------------------------------------------------------------18
15
atoms
–3 –1
0.693
------------- = ----------------------- = 3.73 × 10 s
λ = 0.693
( 3.1 ) ( 60 )
t1
( 8 × 10 ) ( 365 ) ( 24 ) ( 60 ) ( 60 )
--2
–3
= 2.747 × 10
15
Activity = λ N = ( 3.73 × 10 ) ( 6.628 × 10 )
– 27 –1
s
13
Activity = λ N = ( 2.747 × 10
= 1.7 × 10
– 10
– 27
) ( 6.2 × 10 )
alpha particles per second
10
--2
79 200
Activity = λ N
21
= λ ( 2.6 × 10 )
⇒ λ = 1.346 × 10
Q4 T 1 = 22 hrs = 79 200 s
–6 –1
0.693
- = 8.75 × 10 s
λ = ---------------
Q16 Activity = λ N
3.5 × 10
= 2.47 × 10 Bq
16
– 11
s
150 = 8.75 × 10
–1
23
T1
--2
=
ln2
-------λ
=
ln2
--------------------------------– 11
1.346 × 10
10
= 5.1496 × 10 s
–6
7
N ⇒ Ν = 1.74 × 10 atoms
6.02 × 10 atoms = 228 g ⇒ 1.714 × 10
– 15
atoms = 6.5 × 10 grams
7
= 1633 years
65
Real World Physics
Exercise 31.1
9
2
E
1.4 × 10
- = -------------------------------Q1 E = mc ⇒ m = ---2
2
( 3.00 × 10 )
= 1.56 × 10
– 6.646 × 10
–8
– 27
Mass lost = ( 3.344 × 10
Q6
8
c
kg
– 27
= 4.2 × 10
– 29
2
E = mc = ( 4.2 × 10
)(2)
kg
– 29
8 2
) ( 3.00 × 10 )
6
Q2 Mass lost = ( 4 × 10 ) ( 60 ) ( 60 )
2
= 3.78 × 10
8 2
6
E = mc = ( 4 × 10 ) ( 60 ) ( 60 ) × ( 3 × 10 )
= 1.296 × 10
27
J
Q7
59
27
1
60
27
Co + 0n →
– 12
J
Co + γ
Loss in Mass
Q3 Energy emitted in 1 year
= ( 9.7859 × 10
– 26
kg + 1.6749 × 10
– 27
kg )
23
= ( 3.6 × 10 ) ( 60 ) ( 60 ) ( 24 ) ( 365 )
= 1.1352 × 10
31
– ( 9.9520 × 10
J
14
E
1.1352 × 10
- = --------------------------------- = 1.2614 × 10 kg
m = ---2
2
2
– 10
– 27
= 3.2 × 10
Q5 Mass of reactants:
– 26
– 27
2.325210 × 10 + 6.646322 × 10 kg
– 26
= 2.9899 × 10
– 26
+ 1.672623 × 10
= 1.1349 × 10
– 27
kg
kg
– 30
kg
J
8
– 25
– 28
kg
)U = ( 6.68 × 10
2
Energy = E = mc = ( 1.2 × 10
– 12
– 27
)V
– 29
8 2
) ( 3 × 10 )
J
= Total initial energy
i.e. E α + E n = 1.08 × 10
8 2
– 12
– 25
2
– 27
2
⇒ 1--- ( 4.24 × 10 ) U + 1--- ( 6.68 × 10 ) V
2
J
= 1.08 × 10
= 709312.5 eV = 0.71 MeV
Energy available 7.68 – 0.71 = 6.97 MeV
17
-----18
2
– 12
Also V = 63.5U
Energy of α -particle = 7.68 MeV
Kinetic energy of proton =
6
) × 10 J
2
) ( 3 × 10 )
– 13
– 19
U
= 1.08 × 10
⇒ Energy taken in = E = mc
= ( 1.261 × 10
J
= 3.56 × 10
Gain in mass = 1.261 × 10
– 30
8 2
( 3 × 10 )
c
Q9 P.C.M. ⇒ ( 4.24 × 10
V
- = 63.5 : 1
⇒ ---
– 26
kg
– 11
2
E
3.2 × 10
- = --------------------------E = mc ⇒ m = ---2
2
kg
Mass of products
= 2.822706 × 10
– 12
– 11
– 29
) ( 3 × 10 )
Q8 200 MeV = 200 × ( 1.6 × 10
8 2
) ( 3.00 × 10 )
J
= 2.9898 × 10
– 29
= 1.251 × 10
8
Q4 E = mc = ( 8 × 10
of 6.97
= 6.58 MeV
66
E = mc = ( 1.39 × 10
( 3.00 × 10 )
= 7.2 × 10
) = 1.39 × 10
2
31
c
– 26
Solving simultaneously gives:
7
–1
5
–1
V = 1.78 × 10 m s
U = 2.81 × 10 m s
and
(see Exercise 32.1 Q1 for an alternative
method of solving a question like this.
Teacher’s Manual
Exercise 32.1
Q10
1
0n
1
Q1 Loss in Mass = 3.753 152 × 10
0
→ 1p + –1e
Loss in mass = 1.674 929 × 10
– ( 1.672 623 × 10
– 27
+ 9.109 390 × 10
i.e. Loss in mass = 1.395 × 10
2
E = mc = ( 1.395 × 10
= 1.26 × 10
(– 3.686 602 × 10
– 27
– 30
– 13
J
– 30
kg
– 31
)
= 8.678 × 10
E = mc
2
8 2
– 30
– 25
+ 6.646 322 × 10
– 27
)
kg
= ( 8.678 × 10
= 7.81 × 10
) ( 3 × 10 )
– 25
– 30
– 13
8 2
) ( 3 × 10 )
J = 4.88 MeV
M
v
--------- ⇒ --- = 55.5
Q2 M R u = M H v ⇒ --v- = --------R- = 222
MH
u
u
Ra
4
u
He
Rn
v
4
222
Ratio of kinetic energies
=
1
2
--- M R u
2
-----------------1
2
--- M H v
2
2
M
u 
u
- = --=  --------R- u----2- =  --v-  ---MH v
u  v 2
v
2
1
- ⇒ Kinetic energy of α -particle
= --------55.5
= 55.5 Kinetic energy of Radon
⇒ Kinetic energy of α -particle
Total energy
- × 55.5
= ----------------------------56.5
= 7.66 × 10
– 13
J
Q3 The β -particle is so light compared with a
Nitrogen nucleus that it gets virtually all of the
Kinetic energy (It is approximatley 28 000
times lighter)
2
E = mc = ( 2.77 × 10
= 2.493 × 10
β -particle
– 14
– 31
8 2
) ( 3 × 10 )
J = Kinetic energy of
67
Real World Physics
Exercise 32.2
Exercise 32.3
Q1 1 u = 1.66 × 10
E = mc
– 27
kg
Q1 Energy needed E = mc
2
= ( 2.5 × 10
= ( 1.66 × 10
= 1.49 × 10
– 27
– 10
8 2
– 28
2
8 2
) ( 3 × 10 )
– 11
) ( 3.00 × 10 )
= 2.25 × 10
J
⇒ Energy of each proton
– 10
1.49 × 10
- eV = 931 eV
= ----------------------------– 19
= 1.125 × 10
1.6 × 10
J
– 11
J
= 70 MeV
Q2 Decrease in Mass
2
= 238.050 784 – (234.043 593 + 4.002 603)
= ( 3 ) ( 2.5 × 10
= 0.004 588 u
= 6.75 × 10
E = mc
– 11
– 28
8 2
) ( 3 × 10 )
J
2
= (0.004588) ( 1.66 × 10
= 6.854 × 10
– 13
J
– 13
6.854 × 10
- eV
= -------------------------------– 19
1.6 × 10
= 4.28 MeV
68
Q2 E = mc
– 27
8 2
) ( 3 × 10 )
⇒ Energy of proton = 3.375 × 10
= 211 MeV
– 11
J
Teacher’s Manual
Exercise 32.4
Q1
(i) ud
(ii) ud
(iii) uud
(iv) udd
(v) us
(vi) us
(vii) uds
(viii) uus
(ix) dds
(x) dss
(xi) sss
Exercise 33.1
 + 2--- +  1--- = +1
 3  3
 – 2--- +  – 1--- = – 1
 3  3
 + 2--- +  + 2--- +  – 1---
 3  3  3
 + 2--- +  – 1--- +  – 1---
 3  3  3
 + 2--- +  + 1--- = +1
 3  3
 – 2--- +  – 1--- = –1
 3  3
 + 2--- +  – 1--- +  – 1---
 3  3  3
 + 2--- +  + 2--- +  – 1---
 3  3  3
 – 1--- +  – 1--- +  – 1---
 3  3  3
 – 1--- +  – 1--- +  – 1---
 3  3  3
 – 1--- +  – 1--- +  – 1---
 3  3  3
Q1 FSD = 15 mA
I 1 R1 = I 2 R2
= +1
(0.015)(5) = (0.985)(R)
=0
( 0.015 ) ( 5 )
- = 0.07614 Ω
⇒ R = ------------------------( 0.985 )
Q2 FSD = 10 mA = 0.01 A
V 1 + V 2 = 12
(0.01)(5) + (0.01)(R) = 12 ⇒ R = 1195 Ω
=0
= +1
= –1
= –1
= –1
Q3
(i) In parallel
FSD = 6 mA = 0.006 A
I 1 R1 = I 2 R2
⇒ (0.006)(10)
= (11.994)(R)
⇒ R = 0.005 Ω
(ii) In series
20 = (0.006)(10) + (0.006)R
⇒ R = 3323.3 Ω
Q4 Voltages in parallel are equal
⇒ (I)(4) = (20 – I)(0.02)
4I = 0.4 – 0.02I
4.02I = 0.4
0.4
- = 0.0995 A
I = ---------
Q5 I =
4.02
V
40
---- = -----------R
3005
= 0.01331 A
V 1 = I 1 R 1 = (0.01331)(5) = 0.0666 V
20
- = 0.001 A
Q6 I = V---- = --------------R
20 000
Need: 80 V across R
80
- = 80 000 Ω
V = IR ⇒ R = V---- = -----------I
0.001
∴ connect in series a resistor of 80 k Ω
69
Real World Physics
2
-------------
Q7 A = π r = π  d--- = π  0.085
2
2000 
2
2
–8
( 1.7 × 10 ) ( 18 )
R = ρ-----l = ---------------------------------------2
A
0.085
π  -------------
2000
R = 53.93 Ω
FSD = 2 mA = 0.002 A
(i) V = IR
V max = (0.002)(53.93) = 0.1079 V
(ii) 10 = I 1 R 1 + I 2 R 2
10 = (0.002)(53.93) + (0.002)R
R = 4946 Ω
70