Class 11-science H C VERMA Solutions Physics Chapter 1 - Introduction to Physics
Introduction to Physics Exercise 9
Solution 1
a) Linear momentum = mv
Dimension = [MLT-1]
b) Frequency = 1/T
Dimension = [T-1]
c) Pressure = Force/Area
Dimension = [MLT-2]/[L2] = [ML-1T-1]
Solution 2
a)Angular speed = radian/time = [M0L0T-1]
b)Angular acceleration = Angular speed/time = [M0L0T-2]
c)
Torque = 
= [M L2T-2]
d) Moment of Inertia = Mr2 = [ML2]
Introduction to Physics Exercise 10
Solution 3
a)Electric field= F/q= [MLT-2]/[IT] = [MLT-3I-1]
b) Magnetic field = F/qv = [MLT-2]/[IL] = [MT-2 I-1]
c) Magnetic Permeability= B×2πa/I = [MT-2 I-1][L]/[I]=[MLT-2I-2]
Solution 4
a) Electric dipole moment = qL = [IT]×[L]=[LIT]
b) Magnetic dipole moment = IA=[I]×[L2] = [IL2]
Solution 5
So, [h] = [ML2T-2]/[T-1] = [ML2T-1]
Solution 6
a) Specific heat constant c = Q/m(T2-T1) = [ML2T-2]/[M][θ] = [L2T-2θ-1]
b)
Coefficient of linear expansion α 
=[θ]
c) Gas constant = R = PV/nT = [ML-1T-2][L3]/[mol][θ] = [ML2T-2θ-1(mol)-1]
Solution 7
a)
Density = 
= [F/LT-2][L3] = [FL-4T-2]
b)
Pressure = 
= [F]\[L2] = [FL-2]
c)
Momentum = 
=[F]/[LT-2]×[LT-1]=
[FT]
d)
Energy = = 
×(velocity)2
= [F]/[LT-2]×[L2T-2] = [FL]
Solution 8
g = 10 metre/sec2 = 10×100×3600 cm/min2. = 36 × 105 cm/min2.
Solution 9
1 mile = 1.6km = 1600 m, and 1 hour = 3600 seconds.
Speed of snail = 0.02×1600/3600 metre per sec = 0.0089 m/sec.
Speed of leopard = 70×1600/3600 metre per sec = 31 m/sec.
Solution 10
H = 75cm, density of mercury = 13600kg/m3, g = 9.8ms-2
then, pressure = 105 N/m2.
In C.G.S. units, P = 10×105 = 106 dyne/cm2
Solution 11
1 watt = 1 Joule/sec = 107 erg/sec
So, 100 watt = 109 erg/sec.
Solution 12
1 microcentury = 10-4 years. = 10-4 ×365×24×60 min = 52.56 min.
So, 100 min = 100/52.26 microcentury = 1.9 microcentury
Solution 13
Surface tension of water is 72 dyne/cm
In S.I. unit, it is = 0.072 N/m
Solution 14
K = kIaωb, where k = constant and K = Kinetic energy
So, K = [ML2T-2]
Ia = [ML2]a and ωb = [T-1]b
So, [ML2T-2] = [MaL2aT-b]
So, a =1 and b = 2
Solution 15
Let us assume energy E α macb, or E = k macb, where k is a constant
Equating their dimensions ,
[ML2T-2] = [Ma][LbT-b]
From here, we see that a = 1 and b = 2
Solution 16
Dimension of R = [ML2I-2T-3]
Dimension of V = [ML2T-3I-1]
Now, R = [ML2I-2T-3] = [ML2I-1T-3]/[I] = V/[I] = V/I
or, V = IR.
Solution 17
Let, frequency ν = FaLbmc
or [T-1] = [MLT-2]a[Lb][Mc]
Equating the terms, we get -2a = -1, or a = ½, and c+a = 0, so c =-½
and a +b =0, so b = -½.
So, ν = F½L-½m-½
Solution 18
a) 
Here, h = [L], S = F/L = [MT-2], ρ = [ML-3], r = [L], g = [LT-2]
So,
=
[MT-2]//[ML-3L2T-2] = [L]
This relation is correct
b) Here, [v] = [LT-1], P = [ML-1T-2] and ρ = [ML-3]
Now,
P/ρ = [L2T-2], so, 
= [LT-1] = v.
This relation is correct
c)
V = 
V
= [L3], P = [ML-1T-2], t = [T], r = [L], 
=
[ML-1T-1],
Now,
=
[ML-1T-2][L4][T]/[MT-1] = [L3]
This relation is correct
d) v = [T-1], m = [M], g =[LT-2], I = [ML2]
Now,
mgl/I = [ML2T-2]/[ML2] = [T-2]
.Now, 
= [T-1]
Thus, this relation is also correct.
Solution 19
Dimension of the Integral = 
. But for 
, the dimension is [L-1]. So this expression is dimensionally incorrect.
