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I am a Chemical Engineer and Khan Academy Talent Hunt Finalist (2017)(please watch my winning video below from my Youtube channel "THE SCIENCE CUBE"). I specialize in teaching students of class 11 by giving them a very strong foundation. My teaching style includes -

1. Deep discussion on each lesson followed by numerical examples

2. I then give video lessons prepared by me for home revision along with solved video numerical examples. Many of my students depend a lot on these videos for exam preparation.

4. Students are then expected to work on numerical problems at home followed by a discussion of some important problems

5. Once a month, a 2-hour mock test is conducted followed by a discussion on the test.

I believe I cannot only provide a good education but also positive vibes for my students to build the required confidence.

1. Deep discussion on each lesson followed by numerical examples

2. I then give video lessons prepared by me for home revision along with solved video numerical examples. Many of my students depend a lot on these videos for exam preparation.

4. Students are then expected to work on numerical problems at home followed by a discussion of some important problems

5. Once a month, a 2-hour mock test is conducted followed by a discussion on the test.

I believe I cannot only provide a good education but also positive vibes for my students to build the required confidence.

No

English, Hindi

Bachelor of Technology (B.Tech.) from Kanpur in 1993.

Master of Business Administration (M.B.A.) from TAPMI in 1998.

Jayanagar 9th Block, Bangalore, India- 560069.

Class XI-XII Tuition (PUC)

Fees

₹ 500 per hour

Board

CBSE, ISC/ICSE

IB Subjects taught

ISC/ICSE Subjects taught

Physics

CBSE Subjects taught

Physics

IGCSE Subjects taught

Experience in School or College

Taught in School or College

No

State Syllabus Subjects taught

Class Location

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Class XI-XII Tuition (PUC)

2

Teaching Experience in detail in Class XI-XII Tuition (PUC)

I feel students learn best when they start liking the subject. For a student to like a subject, she needs to understand it very well. So my focus is largely on making students understand the depth of the topic being dealt with. This could include long drawn discussions, real life examples, improving understanding through working around numerical problems or watching some good Youtube videos (Khan Academy, The Science Cube etc.) Apart from this I believe a student gets better and then among the best through following regimen- 1. Mock tests. Once a month atleast. Practice is the key 2. Finding best way of solving problems - I encourage students to not leave a problem till they get conceptual clarity around the solution 3. I emphasize the need for proper nutrition and exercise. It plays big part in keeping one motivated. 4. Make best use of time of the day and students need to get creative about this. Mornings , when the brain is at peak performance, should be reserved to do mock tests or tackle difficult topics that you need to master. Afternoons should be used to do revision of something that they are familiar with. Evenings again can be used to work around subjects that you may be a little familiar 5. Students should not burn themselves out but indulge in something they like. Of-course in moderation 6. Students should get 7-8 hours of sleep every day. The productivity will be high if they do so. 7. Relax one day before the exam and get at least 8 hours of sleep

Rotation - Angular Position, Velocity...

Video5

Torque

Solving Pulley Problems

Static and Dynamic Friction

Analyzing Forces in an Elevato...

Understanding Impulse

The Rocket Equation

KE of Rotation

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4.9 from 14 reviews

4.9/514

M

- by Manish kumar

attended

posted on 23 May, 2018

"He is best teacher for my subject. I would recommend to all. "

Reply by Vishesh

Thanks, Manish! So much appreciate your words

S

- by Shaili Nigam

attended Class XI-XII Tuition (PUC)

posted on 29 Apr, 2018

"Vishesh exceeds expectations in the role of a mentor. He adapts to change easily, works well under pressure with a positive attitude. He is a listener and a people's person. The services provided by him. Wish him all the best. "

G

- by Gagandeep Singh Walia

attended Class XI-XII Tuition (PUC)

posted on 08 Apr, 2018

"Vishesh sir is excellent in teaching skills. His ability of going deep into a subject yet keeping it relevant to the topic is amazing. His explanation with relative, day-to-day examples always help to understand any topic so well that it is clear and will remain fresh for life. His mode of teaching is mostly discussion based and the student always is encouraged to ask questions that not only helps in understanding the subject better but also encourages the student to express himself and show his interest or disinterest without any hesitation thus leading to a right pace of learning and understanding.Sir has an ability to well understand when to be strict and when to be friendly. His this ability has always helped in finishing the syllabus in desired time frame. Vishesh sir is not just a great teacher but also has been a guide as he is a man of ethics, commitments and has orientation in owning the results. "

S

- by Sanchay

attended Class XI-XII Tuition (PUC)

posted on 08 Apr, 2018

"He is amazing teacher. His teaching style is very unique and innovative. I will recommend him to everyone who want to excel in physics. "

Have you attended the class of Vishesh? Write a Review

All you need to know about Motion, Velocity and Speed

Linear Motion also called rectilinear motion is a one-dimensional motion along a straight line, and can, therefore, be described mathematically using only one dimension. The linear motion can be of two...

"A projectile is fired at an angle of 30degree with the horizontal with velocity 10m/s. At what angle with the vertical should it be fired to get maximum range?" in CBSE/Class 11/Science/Physics, Tuition/Class XI-XII Tuition (PUC)

The angle is 45 degrees

Watch this video made by me to understand the concept better

https://www.youtube.com/watch?v=N4l0hJ9x4Tg&list=PLQgL7RKnNg_yrgYAhTNAEoshDNDHDBAWr&index=4&t=7s

1

| 0

"An elevator weighs 3000kg. What is its acceleration when the in the tension supporting cable is 33000N. Given that g = 9.8m/s2." in CBSE/Class 11/Science/Physics, Tuition/Class XI-XII Tuition (PUC)

Equate F (net) = ma

Then T - mg = ma (assume all vectors in + Y direction as positive)

33000 - 3000 X 9.8 = 3000 a

a = 1.2 m/s²

This essentially means that the net force on the elevator is making it move up with an acceleration of 1.2 m/s²

See my lecture notes for a better understanding

0

| 0

"Action and reaction forces do not balance each other. Why?" in CBSE/Class 11/Science/Physics, Tuition/Class XI-XII Tuition (PUC)

Let's say you push the wooden block, it is true the the block pushes you back with the same force. However, this does not mean that the force that you are exerting on the block has got cancelled. The wooden block will continue to to experience the push and so would you from the block being pushed. In a way they are 2 separate forces, each acting on separate object.

To understand the concept better imagine yourself pushing your friend and your friend pushing you back with the same force. Just because the forces are equal and opposite in magnitude, does not mean that you feel at ease. You feel the pain in your muscles because there is a force acting on your arms that stresses your muscles.

Scientifically put, you need to see the body being pushed in isolation. The force you are putting on the body is “tangible” and makes it move once the force of friction has been overcome.

This is the reason problems around Newton’s laws of motion are solved by using “free body diagrams”. This essentially requires you to label all forces that act on a body and then find the “Net force”, using vector algebra. This net force is then equated with the product of mass and the acceleration this net force is creating, to find the unknown in the equation.

1

| 0

"A lift is accelerated upward. Will the apparent weight of a person inside the lift increase, decrease or remain the same relative to its real weight? If the lift is going with uniform speed, then?" in CBSE/Class 11/Science/Physics, Tuition/Class XI-XII Tuition (PUC)

Put yourself on a weighing scales inside a stationery elevator. So your velocity here is zero meters per second. Let us examine what are the forces acting on your body. We know one of the force is “mg” which acts in the downward direction.The weighing scale will push you up with a certain force which is nothing but the Normal reaction and shows up as weight “W” on the weighing scale (We will take any vector pointing in the upward direction as positive and any vector acting in the downward direction as negative).

We write the forces acting on your body in accordance with Newton's second law of motion which says that the net force acting on you or on a body is equal to its mass into the acceleration induced or

F net = m X a

There are 2 forces acting on the body (1) mg acting in *downward direction *and we take it as a negative vector and (2) The normal reaction from the floor lift that is the weight or W (we take it as a positive vector since it points upwards)

So the net force is = W – mg and this should equal the product of resultant acceleration with mass. So the equation that will form is

F net = W - mg = ma

Or W = mg + ma

So if the lift is stationery, a = 0 and W = mg the way it should be. You simply see your weight in the scale

But if the lift starts accelerating up so that velocity increases resulting in positive acceleration “a”. We see that the entity “ma” in above equation adds to mg and *therefore the weight W increase.*

But if the lift is moving up with a uniform velocity, acceleration “a” will become zero (since acceleration happens only when velocity changes). In such a case again, W = mg or no change

Now imagine that you are reaching the top floor. The lift will start slowing down and this can happen only if a force acts on the lift in the opposite direction. Hence the acceleration “a” now is in downward direction hence negative. So the above equation will now become

W = mg + m X (-a) or

W = mg – ma

So you see when the lift is slowing down at the top, your weight will now reduce by “ma”.

1

| 0

Vishesh address

x Class XI-XII Tuition (PUC)

Fees

₹ 500 per hour

Board

CBSE, ISC/ICSE

IB Subjects taught

ISC/ICSE Subjects taught

Physics

CBSE Subjects taught

Physics

IGCSE Subjects taught

Experience in School or College

Taught in School or College

No

State Syllabus Subjects taught

Class Location

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Class XI-XII Tuition (PUC)

2

Teaching Experience in detail in Class XI-XII Tuition (PUC)

I feel students learn best when they start liking the subject. For a student to like a subject, she needs to understand it very well. So my focus is largely on making students understand the depth of the topic being dealt with. This could include long drawn discussions, real life examples, improving understanding through working around numerical problems or watching some good Youtube videos (Khan Academy, The Science Cube etc.) Apart from this I believe a student gets better and then among the best through following regimen- 1. Mock tests. Once a month atleast. Practice is the key 2. Finding best way of solving problems - I encourage students to not leave a problem till they get conceptual clarity around the solution 3. I emphasize the need for proper nutrition and exercise. It plays big part in keeping one motivated. 4. Make best use of time of the day and students need to get creative about this. Mornings , when the brain is at peak performance, should be reserved to do mock tests or tackle difficult topics that you need to master. Afternoons should be used to do revision of something that they are familiar with. Evenings again can be used to work around subjects that you may be a little familiar 5. Students should not burn themselves out but indulge in something they like. Of-course in moderation 6. Students should get 7-8 hours of sleep every day. The productivity will be high if they do so. 7. Relax one day before the exam and get at least 8 hours of sleep

CET Coaching classes

Subjects

Physics

"A projectile is fired at an angle of 30degree with the horizontal with velocity 10m/s. At what angle with the vertical should it be fired to get maximum range?" in CBSE/Class 11/Science/Physics, Tuition/Class XI-XII Tuition (PUC)

The angle is 45 degrees

Watch this video made by me to understand the concept better

https://www.youtube.com/watch?v=N4l0hJ9x4Tg&list=PLQgL7RKnNg_yrgYAhTNAEoshDNDHDBAWr&index=4&t=7s

1

| 0

"An elevator weighs 3000kg. What is its acceleration when the in the tension supporting cable is 33000N. Given that g = 9.8m/s2." in CBSE/Class 11/Science/Physics, Tuition/Class XI-XII Tuition (PUC)

Equate F (net) = ma

Then T - mg = ma (assume all vectors in + Y direction as positive)

33000 - 3000 X 9.8 = 3000 a

a = 1.2 m/s²

This essentially means that the net force on the elevator is making it move up with an acceleration of 1.2 m/s²

See my lecture notes for a better understanding

0

| 0

"Action and reaction forces do not balance each other. Why?" in CBSE/Class 11/Science/Physics, Tuition/Class XI-XII Tuition (PUC)

Let's say you push the wooden block, it is true the the block pushes you back with the same force. However, this does not mean that the force that you are exerting on the block has got cancelled. The wooden block will continue to to experience the push and so would you from the block being pushed. In a way they are 2 separate forces, each acting on separate object.

To understand the concept better imagine yourself pushing your friend and your friend pushing you back with the same force. Just because the forces are equal and opposite in magnitude, does not mean that you feel at ease. You feel the pain in your muscles because there is a force acting on your arms that stresses your muscles.

Scientifically put, you need to see the body being pushed in isolation. The force you are putting on the body is “tangible” and makes it move once the force of friction has been overcome.

This is the reason problems around Newton’s laws of motion are solved by using “free body diagrams”. This essentially requires you to label all forces that act on a body and then find the “Net force”, using vector algebra. This net force is then equated with the product of mass and the acceleration this net force is creating, to find the unknown in the equation.

1

| 0

"A lift is accelerated upward. Will the apparent weight of a person inside the lift increase, decrease or remain the same relative to its real weight? If the lift is going with uniform speed, then?" in CBSE/Class 11/Science/Physics, Tuition/Class XI-XII Tuition (PUC)

Put yourself on a weighing scales inside a stationery elevator. So your velocity here is zero meters per second. Let us examine what are the forces acting on your body. We know one of the force is “mg” which acts in the downward direction.The weighing scale will push you up with a certain force which is nothing but the Normal reaction and shows up as weight “W” on the weighing scale (We will take any vector pointing in the upward direction as positive and any vector acting in the downward direction as negative).

We write the forces acting on your body in accordance with Newton's second law of motion which says that the net force acting on you or on a body is equal to its mass into the acceleration induced or

F net = m X a

There are 2 forces acting on the body (1) mg acting in *downward direction *and we take it as a negative vector and (2) The normal reaction from the floor lift that is the weight or W (we take it as a positive vector since it points upwards)

So the net force is = W – mg and this should equal the product of resultant acceleration with mass. So the equation that will form is

F net = W - mg = ma

Or W = mg + ma

So if the lift is stationery, a = 0 and W = mg the way it should be. You simply see your weight in the scale

But if the lift starts accelerating up so that velocity increases resulting in positive acceleration “a”. We see that the entity “ma” in above equation adds to mg and *therefore the weight W increase.*

But if the lift is moving up with a uniform velocity, acceleration “a” will become zero (since acceleration happens only when velocity changes). In such a case again, W = mg or no change

Now imagine that you are reaching the top floor. The lift will start slowing down and this can happen only if a force acts on the lift in the opposite direction. Hence the acceleration “a” now is in downward direction hence negative. So the above equation will now become

W = mg + m X (-a) or

W = mg – ma

So you see when the lift is slowing down at the top, your weight will now reduce by “ma”.

1

| 0

Linear Motion also called rectilinear motion is a one-dimensional motion along a straight line, and can, therefore, be described mathematically using only one dimension. The linear motion can be of two...

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