Category Archives: HC VERMA

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Newton second law and pendulum

A pendulum is hanged in the car.
Car starts to move by constant acceleration
How to measure acceleration by angle the pendulum?

Asked Afshin



If the pendulum makes an angle θ with the vertical, then a = g tan θ

This is based on the idea of pseudo force experienced by a body in an accelerated frame of reference.

When a body is in an accelerated frame of reference, it feels as if it is acted upon by a force equal to its mass multiplied by the acceleration of the system. Please have a look at the diagram below.


Please feel free to ask further doubts on the topic via comments

YDSE (Young’s Double Slit Experiment)

In the question we have a simple YDSE setup , there’s just the addition of a glass slab of thickness t and ,refractive index u, in front of the top slit ( S1). Now we submerge the entire setup into water (given refractive index u1). We now have to find the locations of the points, which are the points of constructive interference( max intensity).

I considered 3 approaches, first-

I calculated the optical path difference in water = t(u/u1 -1)- yd/D. where y is the reference coordinate of the maxima( any general position) from the original central maxima, d is the slit separation , and D is the distance from screen. For constructive interference path difference = n*wavelength( in water). Now wavelength in water= “wavelength in air”/u1 Therefore we get our answer.

Approach two:

I found out the path difference in water and multiplied it by u1 , to get the path difference in air , and I got the same result when I equated this path difference to- n*wavelength(in air).

Approach three- I found the path difference in air – t(u-1)-yd/D and divided this by u1, to get path difference in water, and equated this to n* wavelength (water). This time around , I did not get the answer.

Can anyone explain the difference between the second and third approaches. And also please do tell me, if I have committed some error.

Posted by Garvit Sharma

Score More in Physics without fear – A Physics teacher’s advice

When I interacted with students of class 11 and 12, I came to know that Physics is one of the subjects they FEAR most during exam time. Though it seems difficult to score high marks, it is not actually so. Here are some suggestions you can easily follow to score more in Physics without much strain.

“Nobody plans to fail, but plans to fail”

Yes, A well planned study habit is essential to score more.

If you analyse the pattern of exam in class XI and XII (CBSE) the most of the marks are distributed among 3 marks questions and 5 marks questions which emphasizes the knowledge of concepts. SO, if you find it difficult to pass, then the following plan of action will easily work out.

  • Identify all portions from which 5 marks questions can be asked and practice them well.
  • Identify all portions from which three marks questions can be asked and practice them well.
  • Learn all Laws, definitions, principles
  • Practice all diagrams.
  • Practice all solved problems given in text

These links may be helpful



Just try this routine and see whether you are scoring more or not.

For those already got pass marks can also try the above to better plus

  • Solve all exercises and additional exercises from NCERT Textbook.
  • Get a collection of all one marks questions and Two marks questions preferably from previous ten years’ question papers and find answers to them.

Those who want to excel further and look forward to entrance examinations, IIT, MBBS etc the following links to books will help.


Buy Physics Books, H.C Verma and more at lowest price!

Force – Impulse – Momentum

Hi All! I’m having mega problems with one part of a physics assignment. Looking for any help.
The details are as follows:

The diagram shows the normal force on Christine’s feet vs. time, as recorded by a force plate while she stands still initially (until point B), then jumps off the plate. (This trial is separate from the one in the previous problem. The graph is over-simplified and idealised, compared to reality.) When her feet leave the plate, the normal force is zero.

1)What is the magnitude of the (upward) impulse generated by the normal force of Christine during the time interval of her jump off the plate?

2)What is the magnitude of the downward impulse due to gravity during this interval?

3)What is the net impulse which propels her upwards when she jumps off the plate? (Recall, the net force on her is the normal force minus the force of gravity.)

4)What is her change in speed upwards for this process?

The graph has NORMAL FORCE (N) on the y-axis and TIME (s) on the X axis.
The line is at a constant 550 N until point B (1.75 seconds) at which time it shoot up vertically to 1550 N at a time of 1.95 seconds. It peaks at this time and position then drops down to 0 N at 2.15 seconds.

Thanks in advance for any guidance that can be provided!


Equilibrium – Numerical Problem

SOLID HEMISPHERE OF RADII R EACH, are placed in contact with each other with each other with their flat faces on a rough horizontal surface. A sphere of mass m and radius R is placed symmetrically on top of them. The normal reaction between the top sphere and any hemisphere assuming the system to be in state equilibrium is

Understanding Projectile Motion

IN A PROJECTILE MOTION WE TAKE MOTION IN A HORIZONTAL DIRECTION AS g=0 while in vertical motion as g=-9.8 meter / second square why?

Asks Atanu PAUL

English: motion of the projectile

Projectile Motion is an example for motion in two dimension. It can be studied easily by resolving into two components – The vertical motion is affected by gravity and the horizontal motion is not affected by gravity.

Therefore we take g=0 when we consider the horizontal component of motion as the projectile is not accelerated in the horizontal direction. The horizontal component of velocity remains constant through out its motion.

The following presentation will help you understand the concept better.

A PowerPoint Presentation on projectile_motion

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