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# HC Verma Class XII Physics Chapter 1– Introduction to Physics

Exercise : Solution of Questions on page Number : 08

The speed of light in vacuum is 299,792,458 m/s.
Then time taken by light to cover a distance of 1 metre in vacuum = 1299, 792, 458 s
Hence, the metre is defined as the distance travelled by light in 1299, 792, 458 s.
As 300,000,000 m/s is an approximate speed of light in vacuum, it cannot be used to define the metre.
The distance travelled by light in one second is 299,792,458 m. This is a large quantity and cannot be used as a base unit. So, the metre is not defined in terms of second.

(a) Volume of a cube of edge a, V = a×a×a
i.e., [V] = L × L × L = L3
(b) Volume of a sphere of radius a, V = 43π(a)3
i.e., [V] = L × L × L = L3
(c) The ratio of the volume of the cube to the volume of the sphere is a dimensionless quantity.

Exercise : Solution of Questions on page Numbe: 09

The validity of this statement cannot be tested by measuring sizes with a metre stick, because the size of the metre stick has also got doubled overnight.
Yes, it can be verified by using the fact that speed of light is a universal constant and has not changed.
If the linear size of everything in the universe is doubled and all the clocks in the universe starts running at half the speed, then we cannot test the validity of this statement by any method.

Yes, if all the terms in an equation have the same units, it is necessary that they have the same dimension.
No, if all the terms in an equation have the same dimensions, it is not necessary that they have the same unit. It is because two quantities with different units can have the same dimension, but two quantities with different dimensions cannot have the same unit. For example angular frequency and frequency, both have the dimensions [T-1] but units of angular frequency is rad/s and frequency is Hertz.Another example is energy per unit volume and pressure.Both have the dimensions of [ML-1T-2] but units of pressure is N/m2 and that of energy per unit volume is J/m3

No, even if two quantities have the same dimensions, they may represent different physical contents.
Example: Torque and energy have the same dimension, but they represent different physical contents.

If we use the foot of a person as a standard unit of length,features that will not be present are variability, destructibility and reproducible nature and the feature that will be present is the availability of the foot of a person to measure any length.

(a) The thickness of a sheet of paper can roughly be determined by measuring the height of a stack of paper.
Example: Let us consider a stack of 100 sheets of paper. We will use a ruler to measure its height. In order to determine the thickness of a sheet of paper, we will divide the height of the stack with the number of sheets (i.e., 100).
(b) The distance between the Sun and the Moon can be measured by using Pythagoras theorem when the Earth makes an angle of 90∘ with the Sun and the Moon. We already know the distances from the Sun to the Earth and from the Earth to the Moon. However, these distances keep on changing due to the revolution of the Moon around the Earth and the revolution of the Earth around the Sun.

(b) length, time and velocity
We define length and time separately as it is not possible to define velocity without using these quantities. This means that one fundamental quantity depends on the other. So, these quantities cannot be listed as fundamental quantities in any system of units.

(d) n ∝ 1u
The larger the unit used to express the physical quantity, the lesser will be the numerical value.
Example:
1 kg of sugar can be expressed as 1000 g or 10000 mg of sugar.
Here, g (gram) is the larger quantity as compared to mg (milligram), but the numerical value used with gram is lesser than the numerical value used with milligram.

(d) may be represented in terms of L, T and x if a≠0
If a = 0, then we cannot represent mass dimensionally in terms of L, T and x, otherwise it can be represented in terms of L, T and x.

(a) may have a unit
Dimensionless quantities may have units.

(a) never has a non-zero dimension
A unitless quantity never has a non-zero dimension.

(a) 0
[ax] = [x2]
⇒ [a] = [x] …(1)
Dimension of LHS = Dimension of RHS
⇒  dxx2 = an
⇒  LL = an …(2)
[L0] = [an]n = 0

(c) pressure
(d) energy per unit volume
[Work done] = [ML2T−2]
[Linear momentum] = [MLT−1]
[Pressure] = [ML−1 T−2]
[Energy per unit volume] = [ML−1 T−2]
From the above, we can see that pressure and energy per unit volume have the same dimension, i.e., ML−1 T−2.

(a) A dimensionally correct equation may be correct.
(b) A dimensionally correct equation may be incorrect.
(d) A dimensionally incorrect equation may be incorrect.
It is not possible that a dimensionally incorrect equation is correct. All the other situations are possible.

The statements which are correct are:
(a) All quantities may be represented dimensionally in terms of the base quantities.
(b) A base quantity cannot be represented dimensionally in terms of the rest of the base quantities.
(c) The dimensions of a base quantity in other base quantities is always zero.
Statement (d) is not correct because A derived quantity can exist which is dimensionless for example fine structure constant which is given by
α = 2πe2hc = 1137  where e is the electric charge and c is the speed of light and h is Planks constant.α is a derived quantity and is dimensionless.

(a) Linear momentum = mv
Here, [m] = [M] and [v] = [LT−1]
∴ Dimension of linear momentum, [mv] = [MLT−1]
(b) Frequency = 1Time
∴ Dimension of frequence = 1T=[M0L0T-1]
(c) Pressure = ForceArea
Dimesion of force = MLT – 2 Dimesion of area = L2
∴ Dimesion of pressure = MLT – 2L2 = ML – 1T – 2

(a) Dimensions of angular speed, ω = θt = M0L0T – 1
(b) Angular acceleration, α=ωt
Here, ω = [M0L0T−1] and t = [T]
So, dimensions of angular acceleration = [M0L0T−2]
(c) Torque, τ =Frsinθ
Here, F = [MLT−2] and r = [L]
So, dimensions of torque = [ML2T−2]
(d) Moment of inertia = mr2
Here, m = [M] and r2 = [L2]
So, dimensions of moment of inertia = [ML2T0]

Exercise : Solution of Questions on page Number : 10

(a) Electric field is defined as electric force per unit charge.
i.e., E = Fq
Also, F = MLT-2 and q = AT So, dimension of electric field, [E] = MLT-3A-1
(b) Magnetic field, B=Fqv
Here, F = MLT-2, q = AT and v = LT-1 So, dimension of magnetic field, [B] = MLT-2AT LT-1=ML0T-2A-1
(c) Magnetic permeability, μ0 = B×2πrI
Here, B=MT-2A-1 and r=LSo, dimension of magnetic permeability, [μ0]=MT-2A-1 ×LA=MLT-2A-2

(a) Electric dipole moment, P = q.(2d)
Here, [q] = [AT] and d = [L]
∴ Dimension of electric dipole moment = [LTA]
(b) Magnetic dipole moment, M = IA
Here, A = [L2]
∴ Dimension of magnetic dipole moment = [L2A]

E = hv, where E is the energy and v is the frequency
Here, E=ML2T-2 and v=T-1So, h=Ev=ML2T-2T-1=ML2T-1

(a) Specific heat capacity, C=Qm∆T
Q=ML2T-2 and T=TSo, C=ML2T-2M T=L2T-3
(b) Coefficient of linear expansion,α=L1-L0L0∆T
So, α = LL T = T – 1
(c) Gas  constant, R=PVnT
Here, P = ML – 1T – 2, [n] = [mol] = [N], [T] = [T] and V = L3So, R = ML – 1T – 2 L3N T = ML2T – 3N – 1

Let F be the dimension of force.
(a) Density=mv=force/accelerationvolume
[Acceleration] = LT – 2 and [volume] = L3
∴ [Density] = F/LT – 2L3 = FL4T – 2 = FL – 4T2
(b) Pressure=forcearea
Area = L2
∴ Pressure = FL2 = FL – 2
(c) Momentum = mv = (force/acceleration) × velocity
Acceleration=LT-2 and velocity = LT – 1
∴ Momentum = FLT – 2 × LT – 1 = FT
(d) Energy=12mv2=forceacceleration×velocity2
∴ Energy = FLT – 2 × LT – 12 = FL

Acceleration due to gravity, g = 10 m/s2
∴ g = 10 m/s2 = 10  × 100 cm × 1160min2
∴ g = 1000 × 3600 cm/min2 = 36 × 105 cm/min2

1 mi = 1.6 km
1 km = 1000 m
For the snail, average speed = 0.02 mi/h = 0.020 × 1.6 × 10003600 = 0.0089 m/s
For the leopard, average speed = 70 mi/h = 70 × 1.6 × 10003 600m/sec = 31 m/s

Height, h = 75 cm = 0.75 m
Density of mercury = 13600 kg/m3
g = 9.8 m/s2
In SI units, pressure = hρg = 0.75 × 13600 × 9.8 = 10 × 104 N/m2 (approximately)
In CGS units, pressure = 10 × 104 N/m2 = 10 × 104 × 105 dyne 104 cm2 = 10 × 105 dyne/cm2

In SI unit, watt = joule/s
In CGS unit, 1 joule = 107 erg
So, 100 watt = 100 joule/s = 100 × 107 erg 1 s = 109 erg/s

1 microcentury = 10−6 × 100 years = 10−4 × 365 × 24 × 60 minutes
1 min = 110 –  4 × 365 × 24 × 60 micro century
⇒ 100 min = 110 – 4 × 365 × 24 × 60 × 100 = 105365 × 144 = 1.9 micro centuries
Suppose, I slept x minutes yesterday.
∴ x min = 0.019x microcenturies

1 dyne = 10−5 N
1 cm = 10−2 m
∴ 72 dyne/cm = 72 × 10 – 5 N10 – 2 m = 72 × 10 – 3 N/m = 0.072 N/m

Kinetic energy of a rotating body is K = kI aωb.
Dimensions of the quantities are [K] = [ML2T−2], [I] = [ML2] and [ω] = [T−1].
Now, dimension of the right side are [I]a = [ML2]a and [ω]b = [T−1]b.
According to the principal of homogeneity of dimension, we have:
[ML2T−2] = [ML2]a [T−1]b
Equating the dimensions of both sides, we get:
2 = 2a
⇒  a = 1
And,
−2 = −b
⇒  b = 2

According to the theory of relativity, E α macb
⇒ E = kmacb, where k = proportionality constant
Dimension of the left side, [E] = [ML2T − 2]
Dimension of the right side, [macb] =  [M]a [L T − 1]b
Equating the dimensions of both sides, we get:
[ML2T−2] = [M]a [LT−1]b
⇒ a = 1, b = 2
∴ E = kmc2

Dimensional formula of resistance, [R] = [ML2A − 2T − 3] …(1)
Dimensional formula of potential difference, [V] = [ML2A − 1T − 3] …(2)
Dimensional formula of current, I = [A]
Dividing (2) by (1), we get:
VR = ML2A – 1T – 3ML2A – 2T – 3 = A
⇒ V = IR

Frequency, f ∝ LaFbmc
f = kLaFbmc …(1)
Dimension of [f] = [T−1]
Dimension of the right side components:
[L] = [L]
[F] = [MLT−2]
[m] = [ML−1]
Writing equation (1) in dimensional form, we get:
[T−1] = [L]a [MLT−2]b [ML−1]c
[M0L0T−1] = [Mb + c La + b − c T−2b]
Equating the dimensions of both sides, we get:
b + c = 0 ….(i)
a + b − c = 0 ….(ii)
−2b = −1 ….(iii)
Solving equations (i), (ii) and (iii), we get:
a  = -1, b = 12 and c = -12
∴ Frequency, f = kL − 1 F1/2m−1/2 = kLF1/2 m-12 = kL Fm
⇒ f = kL Fm

(a) h=2S cos θρrg
Height, [h] = [L]
Surface Tension, S=FL=MLT-2L=MT-2
Density, ρ=MI=ML-3T0
Radius, [r] = [L], [g]= [LT−2]
Now, 2Scos θρrg = MT – 2ML – 3T0 L LT-2 = M0L1T0 = L
Since the dimensions of both sides are the same, the equation is dimensionally correct.
(b) ν=Pρ
Velcocity, [ν] = [LT−1]
Pressure, P=FA=ML-1T-2
Density, ρ=MV=ML-3T0
Now, Pρ=ML-1T-2ML-312=L2T-21/2=LT-1
Since the dimensions of both sides of the equation are the same, the equation is dimensionally correct.
(c) V=π P r4t8 η l
Volume, [V] = [L3]
Pressure, P = FA = ML – 1T-2
[r]= [L] and [t] = [T]
Coefficient of viscosity,
η = F6πrv = MLT – 2LLT – 1 = ML – 1T – 1
Now, πP r4t8ηl = ML – 1T – 2 L4 TML – 1T – 1 L = L3
Since the dimensions of both sides of the equation are the same, the equation is dimensionally correct.
(d) ν = 12πmglI
Frequency, ν = [T−1]
mgl I = M LT – 2 LML2
⇒ML2T-2ML212=T-1
Since the dimensions of both sides of the equation are the same, the equation is dimensionally correct.