Marathahalli Ramesh Nagar, Bangalore, India - 560017.
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English
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IIT 2002
Bachelor of Technology (B.Tech.)
Marathahalli Ramesh Nagar, Bangalore, India - 560017
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Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Teaching Experience in detail in Quantitative Aptitude Coaching
Will be able to tech only if material is provided. Can take up topics randomly if required.
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 9 Tuition
2
Board
ICSE
ICSE Subjects taught
Mathematics, Chemistry
Experience in School or College
Private Coaching Institute
Taught in School or College
Yes
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 10 Tuition
2
Board
ICSE
ICSE Subjects taught
Mathematics, Chemistry
Taught in School or College
No
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Engineering Entrance Coaching classes
3
Engineering Entrance Exams
IIT JEE Coaching Classes
IITJEE Coaching
IIT JEE Foundation Course, IIT JEE Crash Course
Type of class
Crash Course, Regular Classes
IIT-JEE Subjects
Physics , Maths
Teaching Experience in detail in Engineering Entrance Coaching classes
Good subject matter expertise in all relevant topics of the syllabus.
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 11 Tuition
2
Board
ISC/ICSE
ISC/ICSE Subjects taught
Physics, Mathematics
Taught in School or College
No
Teaching Experience in detail in Class 11 Tuition
Good Subject matter knowledge when handling problems in the past.
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 12 Tuition
2
Board
ISC/ICSE
ISC/ICSE Subjects taught
Physics, Mathematics
Taught in School or College
No
Teaching Experience in detail in Class 12 Tuition
Good Subject matter knowledge when handling problems in the past.
1. Which classes do you teach?
I teach Class 10 Tuition, Class 11 Tuition, Class 12 Tuition, Class 9 Tuition, Engineering Entrance Coaching, Mathematics Tuition and Quantitative Aptitude Classes.
2. Do you provide a demo class?
Yes, I provide a free demo class.
3. How many years of experience do you have?
I have been teaching for less than a year.
Answered on 22/09/2018 Learn CBSE/Class 11/Science/Physics/Unit 6-Gravitation/Chapter 8-Gravitation
Answered on 22/09/2018 Learn CBSE/Class 11/Science/Physics/Unit 6-Gravitation/Chapter 8-Gravitation
The initial energy consisting of kinetic and potential would be -GMm/R + 0.5 mv², where R is radius of Mars, M is mass of Mars, and 'm' is mass of the body
Considering that energy loss is there to the tune of 20%, the energy remaining that would be conserved is 0.8*(-GMm/R + 0.5 mv²).
At the furthest point, where the rocket turns back, kinetic energy is 0 and potential energy would be -GMm/(R+h) where h is the height in question.
Answered on 22/09/2018 Learn CBSE/Class 11/Science/Physics/Unit 6-Gravitation/Chapter 8-Gravitation
The initial energy consisting of kinetic and potential would be -GMm/R + 0.5 mv², where R is radius of Mars, M is mass of Mars, and 'm' is mass of the body
Considering that energy loss is there to the tune of 20%, the energy remaining that would be conserved is 0.8*(-GMm/R + 0.5 mv²).
At the furthest point, where the rocket turns back, kinetic energy is 0 and potential energy would be -GMm/(R+h) where h is the height in question.
Answered on 20/09/2018 Learn CBSE/Class 11/Science/Physics/Unit 6-Gravitation/Chapter 8-Gravitation
Total energy of a moving body (of mass m)from earth's surface is given by the sum of potential and kinetic energy viz -(GMm/r)+0.5mv²
To have enough vel so as to reach space, where the potential energy becomes negligible and velocity of body reaches 0, implies that if we use the above equation and equate it to 0, we will get the escape vel.
Thus we see that escape vel should not be dependent on the mass of the body, but depends on vertical vel i.e direction of projextion, height from earth as well as location of earth(since earth is not completely spherical).
Answered on 15/09/2018 Learn CBSE/Class 11/Science/Physics/Unit 2-Kinematics/Chapter 4-Motion in a Plane
Taking wind velocity direction (45 degrees) into account, the velocity component along east and north direction would be 72 cos 45 or 36√2.
If the boat goes at 51 towards north, the relative wind velocity would be 51-36√2 towards south and 36√2 towards east.
Taking the above into account, the flag would be showing a eastern direction as the north-south relative component (51-36√2) comes close to 0.
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Teaching Experience in detail in Quantitative Aptitude Coaching
Will be able to tech only if material is provided. Can take up topics randomly if required.
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 9 Tuition
2
Board
ICSE
ICSE Subjects taught
Mathematics, Chemistry
Experience in School or College
Private Coaching Institute
Taught in School or College
Yes
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 10 Tuition
2
Board
ICSE
ICSE Subjects taught
Mathematics, Chemistry
Taught in School or College
No
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Engineering Entrance Coaching classes
3
Engineering Entrance Exams
IIT JEE Coaching Classes
IITJEE Coaching
IIT JEE Foundation Course, IIT JEE Crash Course
Type of class
Crash Course, Regular Classes
IIT-JEE Subjects
Physics , Maths
Teaching Experience in detail in Engineering Entrance Coaching classes
Good subject matter expertise in all relevant topics of the syllabus.
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 11 Tuition
2
Board
ISC/ICSE
ISC/ICSE Subjects taught
Physics, Mathematics
Taught in School or College
No
Teaching Experience in detail in Class 11 Tuition
Good Subject matter knowledge when handling problems in the past.
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 12 Tuition
2
Board
ISC/ICSE
ISC/ICSE Subjects taught
Physics, Mathematics
Taught in School or College
No
Teaching Experience in detail in Class 12 Tuition
Good Subject matter knowledge when handling problems in the past.
Answered on 22/09/2018 Learn CBSE/Class 11/Science/Physics/Unit 6-Gravitation/Chapter 8-Gravitation
Answered on 22/09/2018 Learn CBSE/Class 11/Science/Physics/Unit 6-Gravitation/Chapter 8-Gravitation
The initial energy consisting of kinetic and potential would be -GMm/R + 0.5 mv², where R is radius of Mars, M is mass of Mars, and 'm' is mass of the body
Considering that energy loss is there to the tune of 20%, the energy remaining that would be conserved is 0.8*(-GMm/R + 0.5 mv²).
At the furthest point, where the rocket turns back, kinetic energy is 0 and potential energy would be -GMm/(R+h) where h is the height in question.
Answered on 22/09/2018 Learn CBSE/Class 11/Science/Physics/Unit 6-Gravitation/Chapter 8-Gravitation
The initial energy consisting of kinetic and potential would be -GMm/R + 0.5 mv², where R is radius of Mars, M is mass of Mars, and 'm' is mass of the body
Considering that energy loss is there to the tune of 20%, the energy remaining that would be conserved is 0.8*(-GMm/R + 0.5 mv²).
At the furthest point, where the rocket turns back, kinetic energy is 0 and potential energy would be -GMm/(R+h) where h is the height in question.
Answered on 20/09/2018 Learn CBSE/Class 11/Science/Physics/Unit 6-Gravitation/Chapter 8-Gravitation
Total energy of a moving body (of mass m)from earth's surface is given by the sum of potential and kinetic energy viz -(GMm/r)+0.5mv²
To have enough vel so as to reach space, where the potential energy becomes negligible and velocity of body reaches 0, implies that if we use the above equation and equate it to 0, we will get the escape vel.
Thus we see that escape vel should not be dependent on the mass of the body, but depends on vertical vel i.e direction of projextion, height from earth as well as location of earth(since earth is not completely spherical).
Answered on 15/09/2018 Learn CBSE/Class 11/Science/Physics/Unit 2-Kinematics/Chapter 4-Motion in a Plane
Taking wind velocity direction (45 degrees) into account, the velocity component along east and north direction would be 72 cos 45 or 36√2.
If the boat goes at 51 towards north, the relative wind velocity would be 51-36√2 towards south and 36√2 towards east.
Taking the above into account, the flag would be showing a eastern direction as the north-south relative component (51-36√2) comes close to 0.
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