HC Verma Solutions for Class 11 Physics Chapter 1 – Introduction to Physics
HC Verma Solutions for Class 11 Physics Chapter 1 – Introduction to Physics

HC Verma Physics books are the most preferred books among students of CBSE schools. Students can be found referring to the chapters as well as practice questions at the end of each of these chapters, in the books. Students follow these textbooks religiously since quite a few questions in these also appear in exams.

HC Verma Solutions for Class 11 Physics Chapter 1 – Introduction to Physics

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Page No 8:

Question 1:

The metre is defined as the distance travelled by light in

1299, 792, 458second. Why didn’t people choose some easier number such as

1300, 000, 000second? Why not 1 second?

Answer:

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, 458s
Hence, the metre is defined as the distance travelled by light in

1299, 792, 458s.
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.

Question 2:

What are the dimensions of:
(a) volume of a cube of edge a,
(b) volume of a sphere of radius a,
(c) the ratio of the volume of a cube of edge a to the volume of a sphere of radius a?

Answer:

(a) Volume of a cube of edge a,

V=a×a×ai.e., [V] = L

×L

×L = L3

(b) Volume of a sphere of radius a,

V=43π(a)3i.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.

Page No 9:

Question 3:

Suppose you are told that the linear size of everything in the universe has been doubled overnight. Can you test this statement by measuring sizes with a metre stick? Can you test it by using the fact that the speed of light is a universal constant and has not changed? What will happen if all the clocks in the universe also start running at half the speed?

Answer:

The validity of this statement cannot be tested by measuring sizes with a metre stick, because the size of the metre stick has also gotdoubled 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.

Question 4:

If all the terms in an equation have same units, is it necessary that they have same dimensions? If all the terms in an equation have same dimensions, is it necessary that they have same units?

Answer:

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

Question 5:

If two quantities have same dimensions, do they represent same physical content?

Answer:

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.

Question 6:

It is desirable that the standards of units be easily available, invariable, indestructible and easily reproducible. If we use foot of a person as a standard unit of length, which of the above features are present and which are not?

Answer:

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.

Question 7:

Suggest a way to measure:
(a) the thickness of a sheet of paper,
(b) the distance between the sun and the moon.

Answer:

(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.

Question 1:

Which of the following sets cannot enter into the list of fundamental quantities in any system of units?
(a) length, mass and velocity,
(b) length, time and velocity,
(c) mass, time and velocity,
(d) length, time and mass.

Answer:

(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.

Question 2:

A physical quantity is measured and the result is expressed as nu where u is the unit used and n is the numerical value. If the result is expressed in various units then
(a)

n ∝ size of u(b)

n ∝ u2(c)

n ∝ u(d)

n ∝ 1u.

Answer:

(d)

n ∝ 1uThe 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.

Question 3:

Suppose a quantity x can be dimensionally represented in terms of M, L and T, that is,

(x)=Ma Lb Tc. The quantity mass
(a) can always be dimensionally represented in terms of L, T and x,
(b) can never be dimensionally represented in terms of L, T and x,
(c) may be represented in terms of L, T and x if a = 0,
(d) may be represented in terms of L, T and x if

a≠0.

Answer:

(d) may be represented in terms of L, T and x if

a≠0If 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.

Question 4:

A dimensionless quantity
(a) never has a unit,
(b) always has a unit,
(c) may have a unit,
(d) does not exist.

Answer:

(a) may have a unit

Dimensionless quantities may have units.

Question 5:

A unitless quantity
(a) never has a non-zero dimension,
(b) always has a non-zero dimension,
(c) may have a non-zero dimension,
(d) does not exist.

Answer:

(a) never has a non-zero dimension

A unitless quantity never has a non-zero dimension.

Question 6:

∫dx2ax-x2=an sin-1xa-1.

The value of n is
(a) 0
(b) −1
(c) 1
(d) none of these.
You may use dimensional analysis to solve the problem.

Answer:

(a) 0

[ax] = [x2]
⇒ [a] = [x]    …(1)

Dimension of LHS = Dimension of RHS

⇒dxx2=an⇒LL=an       …(2) [L0]=[an]n=0

Question 1:

The dimensions ML−1 T−2 may correspond to
(a) work done by a force
(b) linear momentum
(c) pressure
(d) energy per unit volume.

Answer:

(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.

Question 2:

Choose the correct statements(s):
(a) A dimensionally correct equation may be correct.
(b) A dimensionally correct equation may be incorrect.
(c) A dimensionally incorrect equation may be correct.
(d) A dimensionally incorrect equation may be incorrect.

Answer:

(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.

Question 3:

Choose the correct statements(s):
(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.
(d) The dimension of a derived quantity is never zero in any base quantity.

Answer:

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=1137where 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.

Question 1:

Find the dimensions of
(a) linear momentum,
(b) frequency and
(c) pressure.

Answer:

(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-2Dimesion of area=L2∴ Dimesion of pressure=MLT-2L2=ML-1T-2

Question 2:

Find the dimensions of
(a) angular speed ω,
(b) angular acceleration α,
(c) torque

τand
(d) moment of interia I.
Some of the equations involving these quantities are

ω=θ2-θ1t2-t1, α=ω2-ω1t2-t1, τ=F.r and I=mr2.
The symbols have standard meanings.

Answer:

(a) Dimensions of angular speed,

ω=θt=M0L0T-1(b) Angular acceleration,

α=ωtHere, ω = [M0L0T−1] and t = [T]
So, dimensions of angular acceleration = [M0L0T−2]

(c) Torque, τ =Frsin

θ
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]

Page No 10:

Question 3:

Find the dimensions of
(a) electric field E.
(b) magnetic field B and
(c) magnetic permeability

μ0.
The relevant equation are

F=qE, F=qvB, and B=μ0I2 π a;where F is force, q is charge, v is speed, I is current, and a is distance.

Answer:

(a) Electric field is defined as electric force per unit charge.
i.e.,

E=Fq

Also, F=MLT-2 and q=ATSo, 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

Question 4:

Find the dimensions of
(a) electric dipole moment p and
(b) magnetic dipole moment M.
The defining equations are p = q.d and M = IA;
where d is distance, A is area, q is charge and I is current.

Answer:

(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]

Question 5:

Find the dimensions of Planck’s constant h from the equation E = hv where E is the energy and v is the frequency.

Answer:

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

Question 6:

Find the dimensions of
(a) the specific heat capacity c,
(b) the coefficient of linear expansion α and
(c) the gas constant R.
Some of the equations involving these quantities are

Q=mcT2-T1, lt=l01+αT2-T1and

PV=nRT.

Answer:

(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∆TSo,

α=LLT=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

Question 7:

Taking force, length and time to be the fundamental quantities, find the dimensions of
(a) density,
(b) pressure,
(c) momentum and
(d) energy.

Answer:

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

Question 8:

Suppose the acceleration due to gravity at a place is 10 m/s2. Find its value if cm/(minute)2.

Answer:

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

Question 9:

The average speed of a snail is

0·020miles/ hour and that of a leopard is 70 miles/ hour. Convert these speeds in SI units.

Answer:

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/sFor the leopard, average speed = 70 mi/h =

70×1.6×10003600m/sec=31 m/s

Question 10:

The height of mercury column in a barometer in a Calcutta laboratory was recorded to be 75 cm. Calculate this pressure in SI and CGS units using the following data : Specific gravity of mercury =

13·6, Density of

water=103 kg/m3, g= 9·8 m/s2at Calcutta. Pressure

=hρgin usual symbols.

Answer:

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 dyne104 cm2=10×105 dyne/cm2

Question 11:

Express the power of a 100 watt bulb in CGS unit.

Answer:

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

Question 12:

The normal duration of I.Sc. Physics practical period in Indian colleges is 100 minutes. Express this period in microcenturies. 1 microcentury = 10−6 × 100 years. How many microcenturies did you sleep yesterday?

Answer:

1 microcentury = 10−6 × 100 years = 10−4 × 365 × 24 × 60 minutes

1 min=110-4×365×24×60 microcentury⇒100 min=110-4×365×24×60×100=105365×144=1.9 microcenturiesSuppose, I slept x minutes yesterday.
∴ x min = 0.019x microcenturies

Question 13:

The surface tension of water is 72 dyne/cm. Convert it in SI unit.

Answer:

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

Question 14:

The kinetic energy K of a rotating body depends on its moment of inertia I and its angular speed ω. Assuming the relation to be

K=kIaωBwhere k is a dimensionless constant, find a and b. Moment of inertia of a sphere about its diameter is

25Mr2

Answer:

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

Question 15:

Theory of relativity reveals that mass can be converted into energy. The energy E so obtained is proportional to certain powers of mass m and the speed c of light. Guess a relation among the quantities using the method of dimensions.

Answer:

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 [LT−1]b

Equating the dimensions of both sides, we get:
[ML2T−2] = [M]a [LT−1]b
⇒ a = 1, b = 2
∴ E = kmc2

Question 16:

Let I = current through a conductor, R = its resistance and V = potential difference across its ends. According to Ohm’s law, product of two of these quantities equals the third. Obtain Ohm’s law from dimensional analysis. Dimensional formulae for R and V are

ML2I-2T-3and

ML2T-3I-1respectively.

Answer:

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

Question 17:

The frequency of vibration of a string depends on the length L between the nodes, the tension F in the string and its mass per unit length m. Guess the expression for its frequency from dimensional analysis.

Answer:

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

Question 18:

Test if the following equations are dimensionally correct:
(a)

h=2S cosθρrg,(b)

v=Pρ,(c)

V=π P r4t8 η l(d)

v=12 πmglI;where h = height, S = surface tension,

ρ= density, P = pressure, V = volume,

η=coefficient of viscosity, v = frequency and I = moment of interia.

Answer:

(a)

h=2S cos θρrgHeight, [h] = [L]
Surface Tension,

S=FL=MLT-2L=MT-2Density,

ρ=MI=ML-3T0Radius, [r] = [L], [g]= [LT−2]
Now,

2Scos θρrg=MT-2ML-3T0 L LT-2=M0L1T0=LSince the dimensions of both sides are the same, the equation is dimensionally correct.

(b)

ν=PρVelcocity, [ν] = [LT−1]
Pressure,

P=FA=ML-1T-2Density,

ρ=MV=ML-3T0Now,

Pρ=ML-1T-2ML-312=L2T-21/2=LT-1Since the dimensions of both sides of the equation are the same, the equation is dimensionally correct.

(c)

V=π P r4t8 η lVolume, [V] = [L3]
Pressure,

P=FA=ML-1T-2[r]= [L] and [t] = [T]
Coefficient of viscosity,
η=F6πrv=MLT-2LLT-1=ML-1T-1Now,

πP r4t8ηl=ML-1T-2 L4 TML-1T-1 L=L3Since the dimensions of both sides of the equation are the same, the equation is dimensionally correct.

(d)

ν=12πmglIFrequency, ν = [T−1]

mglI=M LT-2 LML2⇒ML2T-2ML212=T-1Since the dimensions of both sides of the equation are the same, the equation is dimensionally correct.

Question 19:

Let x and a stand for distance. Is

∫dxa2-x2=1asin-1axdimensionally correct?

Answer:

Dimension of the left side of the equation=

∫dxa2-x2=∫LL2-L2=L0Dimension of the right side of the equation =

1a sin-1 ax=L-1So,

∫dxa2-x2≠1a sin-1 axSince the dimensions on both sides are not the same, the equation is dimensionally incorrect.

Chapterwise HC Verma Solutions Class 11 Physics :

About the Author – HC Verma

HC Verma, the author of many popular and well-renowned Physics books, was born on 8 April 1952. Passing out from one of the most prestigious colleges in the country, IIT Kanpur, he worked as an experimental physicist in the Department of Nuclear Physics.

His most famous works which he is known for include the two-volume Concepts of Physics. He also worked for the social upliftment of economically weaker children through his organization named Shiksha Sopan. He is also the recipient of the Padma Shri, which is considered India’s fourth-highest civilian award. He received the same because of his contribution and valuable work in the field of Physics. 

HC Verma Solutions