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Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEEPage No 8       

Ques 1: The metre is defined as the distance travelled by light in  Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE second. Why didn't people choose some easier number such as Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE second? Why not 1 second?
Ans: 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 Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE
Hence, the metre is defined as the distance travelled by light in Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE
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.

Ques 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?

Ans: (a) Volume of a cube of edge a, V=a x a x a
i.e., [V] = L x L x L = L3
(b) Volume of a sphere of radius a, Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

i.e., [V] = L x L x L = L3 

(c) The ratio of the volume of the cube to the volume of the sphere is a dimensionless quantity.
 

Page No 9       

Ques 1: 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? 
Ans: 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.

Ques 2: 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? 
Ans: Yes, if all the terms in an equation have the same units, it is necessary that they also have the same dimensions. However, if all the terms in an equation have the same dimensions, it is not necessary that they have the same units. This 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 of [T-1], but the unit of angular frequency is rad/s, while frequency is measured in Hertz.
  • Energy per unit volume and pressure both share the dimensions of [ML-1T-2], but the unit of pressure is N/m2, whereas the unit of energy per unit volume is J/m3.


Ques 3: If two quantities have same dimensions, do they represent same physical content? 
Ans: No, even if two quantities have the same dimensions, they may represent different physical quantities. For example, torque and energy share the same dimensions but represent different physical concepts.

Ques 4: 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? 
Ans: If we use the foot of a person as a standard unit of length, several features will be affected:

  • Variability: The foot of a person can vary in size from one individual to another, making it an inconsistent standard.
  • Destructibility: The foot can be damaged or altered, leading to inaccuracies in measurement.
  • Reproducibility: It is challenging to reproduce a specific person's foot consistently for measuring purposes.
  • Availability: The foot is readily available for use as a measuring tool.

In summary, while the foot is available for measurement, it lacks the qualities of invariance, indestructibility, and reproducibility.


Ques 5: Suggest a way to measure:
(a) the thickness of a sheet of paper,
(b) the distance between the sun and the moon. 

Ans: (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.

Ques 6: 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. 

Ans: (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.

Ques 7: 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) Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE
(b) Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE
(c) Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE
(d) Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE
Ans:  (d) Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

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.

Ques 8: Suppose a quantity x can be dimensionally represented in terms of M, L and T, that is, [x] = MaLbT. 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.

Ans: (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. 

Ques 9: A dimensionless quantity
(a) never has a unit,
(b) always has a unit,
(c) may have a unit,
(d) does not exist.

Ans: (c) may have a unit
Dimensionless quantities may have units.

Ques 10: 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. 

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

Ques 11: Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

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

Ans: (a) 0

[ax] = [x2]

⇒ [a] =[x]  ..(1)

Dimension of LHS = Dimension of RHS 

Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

n=0

Ques 12: 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. 

Ans: (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.

Ques 13: 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. 

Ans: (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.

Ques 14: 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. 

Ans: 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 Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE 

where e is the electric charge and c is the speed of light and h is Planks constant. 

α is the derived quantity and is dimensionless.

Ques 15: Find the dimensions of
(a) linear momentum,
(b) frequency and
(c) pressure. 

Ans: (a) Linear momentum = mv
Here, [m] = [M] and [v] = [LT−1]
∴ Dimension of linear momentum, [mv] = [MLT−1]
(b) Frequency = Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

∴ Dimension of frequence = Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

(c) Pressure = Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

Dimension of force = [MLT-2]

Dimension of area = [L2]

∴ Dimension of Pressure = Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

Ques 16: Find the dimensions of
 (a) angular speed ω,
(b) angular acceleration α,
 (c) torque τ and
(d) moment of interia I.
 Some of the equations involving these quantities are 

Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

The symbols have standard meanings. 
Ans: (a) Dimensions of angular speed, Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

(b) Angular acceleration,  Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

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


Page No 10         

Ques 1: Find the dimensions of
(a) electric field E.
(b) magnetic field B and
(c) magnetic permeabilityIntroduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

The relevant equation are

Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

where F is force, q is charge, v is speed, I is current, and a is distance. 
Ans: (a) Electric field is defined as electric force per unit charge.
 i.e.,
Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

Also, [F] = [MLT-2] and [q] = [AT]

So, dimension of elelctric field, [E] = [MLT-3A-1]

(b) Magnetic field,  Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

Here , Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

So, dimension of magnetic field, Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

(c) Magnetic permeability, Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

Here, [B] = [MT-2A-1] and [r] = [L]

So, dimension of magnetic permeability, Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

Ques 2: Find the dimensions of
(a) electric dipole moment p and
(b) magnetic dipole moment M.
The defining equations are p = q.d and MIA;
where d is distance, A is area, q is charge and I is current. 

Ans: (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]

Ques 3: Find the dimensions of Planck's constant h from the equation E = hv where E is the energy and v is the frequency. 
Ans: E = hv, where E is the energy and v is the frequency 

Here, [E]=[ML2T−2] and [v]=[T−1]

So, Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

Ques 4: 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=mc(T2−T1), lt=l0[1+α(T2−T1)] and PV=nRT.
Ans: (a) Specific heat capacity, Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

[Q]=[ML2T−2] and [T]=[K]

So, Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

(b) Coefficient of linear expansion, Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

So, Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

(c) Gas constant, Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

Here, Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

So, Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

Ques 5: Taking force, length and time to be the fundamental quantities, find the dimensions of
(a) density,
(b) pressure,
(c) momentum and
(d) energy. 

Ans: Let F be the dimension of force.

(a) Density = Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

(b) Pressure= Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

[Area]=[L2]

∴ [Pressure]  Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

(c) Momentum = mv = (force/acceleration) × velocity 

Acceleration = Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

∴ [Momentum] Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

(d) Energy = Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

∴ [Energy] = Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

Ques 6: Suppose the acceleration due to gravity at a place is 10 m/s2. Find its value if cm/(minute)2
Ans: Acceleration due to gravity, g = 10 m/s2

g = 10 m/s2 = 10 × 100 cm Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

g = 1000 × 3600 cm/min2 = 36 × 105 cm/min2

Ques 7: The average speed of a snail is 0⋅020 miles/ hour and that of a leopard is 70 miles/ hour. Convert these speeds in SI units.
Ans: 1 mi = 1.6 km
1 km = 1000 m
For the snail, average speed = 0.02 mi/h Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE= 0.0089 m/s

For the leopard, average speed = 70 mi/h = Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

Ques 8: 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/s2 at Calcutta. Pressure =hρg in usual symbols.
Ans: 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/m2Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

Ques 9: Express the power of a 100 watt bulb in CGS unit. 
Ans: In SI unit, watt = joule/s
In CGS unit, 1 joule = 107 erg

So, 100 watt = 100 joule/s Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

Ques 10: The normal duration of I.Sc. Physics practical period in Indian colleges is 100 minutes. Express this period in microcenturies. 1 microcentury = 106 × 100 years. How many microcenturies did you sleep yesterday? 
Ans: 1 microcentury = 10−6 × 100 years = 10−4 × 365 × 24 × 60 minutes 

1 min = Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

Suppose, I slept x minutes yesterday.
x min = 0.019x microcenturies

Ques 11: The surface tension of water is 72 dyne/cm. Convert it in SI unit. 
Ans: 1 dyne = 10−5 N
1 cm = 10−2 m
∴72 dyne/cm Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

Ques 12: 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=kIaωB where k is a dimensionless constant, find a and b. Moment of inertia of a sphere about its diameter is 2/5Mr2 
Ans: 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

Ques 13: 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. 
Ans: 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

Ques 14: 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−3 and ML2T−3I−1 respectively.
Ans: Dimensional formula of resistance, [R] = [ML2A−2T−3] ...(1)
Dimensional formula of potential difference, [V] = [ML2A−1T−3] ...(2)
Dimensional formula of current,  = [A]
Dividing (2) by (1), we get: 

Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

⇒ V = IR

Ques 15: 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. 
Ans: 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 + bc T−2b]
Equating the dimensions of both sides, we get:
bc = 0               ....(i)
a + bc = 0      ....(ii)
−2b = −1              ....(iii)
Solving equations (i), (ii) and (iii), we get: 

Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

Ques 16: Test if the following equations are dimensionally correct: 

Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

where h = height, S = surface tension, ρ= density, P = pressure, V = volume, η= coefficient of viscosity, v = frequency and I = moment of interia. 
Ans:
Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

Height, [h] = [L]
Surface Tension, Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

Density, Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

Radius, [r] = [L], [g]= [LT−2

Now, Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

Since the dimensions of both sides are the same, the equation is dimensionally correct. 

(b) Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

Velcocity, [ν] = [LT−1]

Pressure, PIntroduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE

Density , Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE
Now, Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE
Since the dimensions of both sides of the equation are the same, the equation is dimensionally correct.
Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE
Volume, [V] = [L3]
Pressure, P
Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE
[r]= [L] and [t] = [T]
Coefficient of viscosity,
Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE
Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE
Since the dimensions of both sides of the equation are the same, the equation is dimensionally correct.
Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE
Frequency, ν = [T−1]
Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE
Since the dimensions of both sides of the equation are the same, the equation is dimensionally correct.

Ques 17: Let x and a stand for distance. Is Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEEdimensionally correct?
Ans: Dimension of the left side of the equation Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE
Dimension of the right side of the equation Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE
Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE
Since the dimensions on both sides are not the same, the equation is dimensionally incorrect. 

The document Introduction to Physics - H.C Verma, Exercises Solved | HC Verma Solutions - JEE is a part of the JEE Course HC Verma Solutions.
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FAQs on Introduction to Physics - H.C Verma, Exercises Solved - HC Verma Solutions - JEE

1. What are the key topics covered in H.C. Verma's "Introduction to Physics"?
Ans. H.C. Verma's "Introduction to Physics" covers fundamental topics in physics that are essential for students preparing for competitive exams like JEE. Key topics include mechanics, thermodynamics, waves, optics, electromagnetism, and modern physics. Each section is designed to build a strong conceptual understanding and problem-solving skills.
2. How can I effectively use H.C. Verma's book to prepare for the JEE exam?
Ans. To effectively use H.C. Verma's book for JEE preparation, start by thoroughly reading the theory presented in each chapter. After understanding the concepts, attempt the solved examples to see their practical application. Finally, practice the exercises at the end of each chapter, focusing on both solved and unsolved problems to enhance problem-solving skills and speed.
3. Are the exercises in H.C. Verma's book sufficient for JEE preparation?
Ans. Yes, the exercises in H.C. Verma's "Introduction to Physics" are considered sufficient for JEE preparation. They cover a wide range of difficulty levels and concepts, providing a comprehensive practice platform. However, it is also advisable to refer to additional resources and previous years' JEE papers for a broader range of questions and exam patterns.
4. What are some effective strategies for solving problems in H.C. Verma's physics exercises?
Ans. Effective strategies for solving problems in H.C. Verma's exercises include: 1. Understanding the underlying concepts before attempting problems. 2. Drawing diagrams to visualize the problem. 3. Breaking down complex problems into simpler parts. 4. Reviewing similar solved examples to guide your approach. 5. Practicing regularly to improve speed and accuracy.
5. How does H.C. Verma's book differ from other physics textbooks for JEE?
Ans. H.C. Verma's "Introduction to Physics" is known for its clear explanations, logical flow of concepts, and emphasis on understanding over rote memorization. Unlike some other textbooks, it provides a balanced mix of theory and practical problems, encouraging students to think critically and apply concepts, which is crucial for success in the JEE exam.
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