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SOLUTIONS TO CONCEPTS
CHAPTER – 1
1
...
M0L0 T 2
[M0L0T–2]
t
T
–2
2 –2
c) Torque = F r = [MLT ] [L] = [ML T ]
2
2
2 0
d) Moment of inertia = Mr = [M] [L ] = [ML T ]
2
MLT
a) Electric field E = F/q =
[MLT 3I1 ]
[IT ]
b) Angular acceleration =
3
...
c) Magnetic permeability 0 =
4
...
h=
6
...
02 mile/hr
0
...
6 1000
–1
Converting to S
...
units,
m/sec [1 mile = 1
...
0089 ms
3600
c) Gas constant = R =
7
...
9
...
6 1000
= 31 m/s
3600
1
...
Height h = 75 cm, Density of mercury = 13600 kg/m , g = 9
...
G
...
Units, P = 10 × 10 dyne/cm
2
11
...
I
...
G
...
Unit = 10 erg/sec
12
...
9 microcentury
13
...
I
...
072 N/m
14
...
Let energy E M C where M = Mass, C = speed of light
a
b
E = KM C (K = proportionality constant)
Dimension of left side
2 –2
E = [ML T ]
Dimension of right side
a
a
b
–1 b
M = [M] , [C] = [LT ]
2 –2
a
–1 b
[ML T ] = [M] [LT ]
a = 1; b = 2
So, the relation is E = KMC
2
2 –3 –2
16
...
Frequency f = KL F M M = Mass/unit length, L = length, F = tension (force)
–1
Dimension of f = [T ]
Dimension of right side,
a
a
b
–2 b
c
–1 c
L = [L ], F = [MLT ] , M = [ML ]
–1
a
–2 b
–1 c
[T ] = K[L] [MLT ] [ML ]
0 0 –1
M L T = KM
b+c
a+b–c
L
–2b
T
Equating the dimensions of both sides,
b+c=0
…(1)
–c + a + b = 0
…(2)
–2b = –1
…(3)
Solving the equations we get,
a = –1, b = 1/2 and c = –1/2
–1 1/2
So, frequency f = KL F M
–1/2
=
K 1/ 2 1/ 2 K
F
F M
L
L
M
1
...
a) h =
2SCos
rg
LHS = [L]
MLT 2
[MT 2 ]
L
Surface tension = S = F/I =
–3 0
Density = = M/V = [ML T ]
–2
Radius = r = [L], g = [LT ]
RHS =
2Scos
[MT 2 ]
[M0L1T0 ] [L]
3 0
rg
[ML T ][L][LT 2 ]
LHS = RHS
So, the relation is correct
b) v =
p
where v = velocity
–1
LHS = Dimension of v = [LT ]
–1 –2
Dimension of p = F/A = [ML T ]
–3
Dimension of = m/V = [ML ]
p
[ML1T 2 ]
[L2T 2 ]1/ 2 = [LT 1 ]
[ML3 ]
RHS =
So, the relation is correct
...
d) v =
1
(mgl / I)
2
–1
LHS = dimension of v = [T ]
RHS =
[M][LT 2 ][L]
(mgl / I) =
2
[ML ]
–1
= [T ]
LHS = RHS
So, the relation is correct
...
Dimension of the left side =
Dimension of the right side =
So, the dimension of
dx
2
(a x )
L
2
2
(a x )
≠
0
2
(L L )
1 1 a
–1
sin = [L ]
a
x
dx
2
2
1 1 a
sin
a
x
So, the equation is dimensionally incorrect
...
3
= [L ]
Chapter-I
20
...
4