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Kartikey Sharma

Shahdara, Delhi, India - 110032

Kartikey Sharma Class 11 Tuition trainer in Delhi

Kartikey Sharma

Tutor

Shahdara, Delhi, India - 110032.

1 Student taught

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Overview

Kartikey Sharma describes himself as Tutor. He conducts classes in Autocad, Class 10 Tuition and Class 11 Tuition. Kartikey is located in Shahdara, Delhi. Kartikey takes at students Home, Regular Classes- at his Home and Online Classes- via online medium. He has 3 years of teaching experience . Kartikey has completed Diploma in civil engineering from C.R.R. institute of technology, Kanjhawala, Delhi in 2017. He is well versed in Hindi and English.

Languages Spoken

Hindi

English

Education

C.R.R. institute of technology, Kanjhawala, Delhi 2017

Diploma in civil engineering

Address

Shahdara, Delhi, India - 110032

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Class 11 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 11 Tuition

3

Board

State, CBSE

CBSE Subjects taught

Chemistry, English, Accountancy, Physics, Mathematics, Business Studies, Computer Science, Biology

Taught in School or College

No

State Syllabus Subjects taught

Chemistry, English, Accountancy, Biology, Mathematics, Physics, Business Studies

Teaching Experience in detail in Class 11 Tuition

A teacher has to know how to listen. Because his first role is listening to others, and assisting them. I know a nice number of teachers who know how to listen to their students, their families, other colleagues, admins, and everybody else. The teacher will listen to the kids for example, in an attentive way, realizing and learning what’s going on with their lives at the school – with prudence anyway. At the beginning of the school year he learns their names quite nicely, because he gets interested in the persons he has in charge. He doesn’t confine his job to teaching and disappearing through the door when his “block” has been delivered to his students. He’s prone to take in charge his students’ issues, problems and joys alike. He teaches with high standards of professionalism and with affection, with a clever and insightful affection – with prudence anyway.

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FAQs

1. Which school boards of Class 12 do you teach for?

State, CBSE

2. Have you ever taught in any School or College?

No

3. Which classes do you teach?

I teach Autocad, Class 10 Tuition, Class 11 Tuition, Class 12 Tuition, Class 6 Tuition, Class 7 Tuition, Class 8 Tuition, Class 9 Tuition, Class I-V Tuition and Engineering Diploma Tuition Classes.

4. Do you provide a demo class?

Yes, I provide a free demo class.

5. How many years of experience do you have?

I have been teaching for 3 years.

Answers by Kartikey Sharma (4)

Answered on 18/09/2018 CBSE/Class 11/Science/Physics/Unit 4-Motion of System of Particles/Chapter 7-System of Particles and Rotational Motion

Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same... ...more

Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time?

Let m and r be the respective masses of the hollow cylinder and the solid sphere.The moment of inertia of the hollow cylinder about its standard axis, I1 = mr2The moment of inertia of the solid sphere about an axis passing through its centre, I2 = (2/5)mr2We have the relation:τ = IαWhere,α... ...more

Let m and r be the respective masses of the hollow cylinder and the solid sphere.
The moment of inertia of the hollow cylinder about its standard axis, I1 = mr2
The moment of inertia of the solid sphere about an axis passing through its centre, I2 = (2/5)mr2
We have the relation:
τ = Iα
Where,
α = Angular acceleration
τ = Torque
I = Moment of inertia
For the hollow cylinder, τ1 = I1α1

 

For the solid sphere, τn = Inαn

As an equal torque is applied to both the bodies, τ= τ2
∴ α/ α1  =  I1  /  I2  =  mr2  /  (2/5)mr2
α2 > α1     …(i)
Now, using the relation:
ω = ω0 + αt
Where,
ω0 = Initial angular velocity
t = Time of rotation
ω = Final angular velocity
For equal ω0 and t, we have:
ω ∝ α … (ii)
From equations (i) and (ii), we can write:
ω2> ω1
Hence, the angular velocity of the solid sphere will be greater than that of the hollow cylinder.

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Answered on 18/09/2018 CBSE/Class 11/Science/Physics/Unit 4-Motion of System of Particles/Chapter 7-System of Particles and Rotational Motion

Two particles, each of mass m and speed v, travel in opposite directions along parallel lines separated... ...more

Two particles, each of mass m and speed v, travel in opposite directions along parallel lines separated by a distance d. Show that the vector angular momentum of the two particle system the same whatever be the point about which the angular momentum is taken.

Let at a certain instant two particles be at points P and Q, as shown in the following figure. Angular momentum of the system about point P: Lp = mv × 0 + mv × d = mvd …(i)Angular momentum of the system about point Q:LQ = mv × d + mv × 0 = mvd ….(ii) Consider... ...more

Let at a certain instant two particles be at points P and Q, as shown in the following figure.

Angular momentum of the system about point P:

Lp = mv × 0 + mv × d  =  mvd   …(i)
Angular momentum of the system about point Q:
LQ = mv × d + mv × 0   =  mvd   ….(ii)

 

Consider a point R, which is at a distance yfrom point Q, i.e.,
QR = y
∴ PR = d – y
Angular momentum of the system about point R:
LR = mv × (d – y) + mv × y
mvd – mvy + mvy
mvd  ….(iii)
Comparing equations (i)(ii), and (iii), we get:
LP = LQ = LR    …(iv)

We infer from equation (iv) that the angular momentum of a system does not depend on the point about which it is taken.

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Answered on 18/09/2018 CBSE/Class 11/Science/Physics/Unit 4-Motion of System of Particles/Chapter 7-System of Particles and Rotational Motion

A cylinder of mass 10 kg and radius 15 cm is rolling perfectly on a plane of inclination 30°. The... ...more

A cylinder of mass 10 kg and radius 15 cm is rolling perfectly on a plane of inclination 30°. The coefficient of static friction us = 0.25.
(a) How much is the force of friction acting on the cylinder?
(b) What is the work done against friction during rolling?
(c) If the inclination O of the plane is increased, at what value of 8 does the cylinder begin to skid, and not roll perfectly?

Mass of the cylinder, m = 10 kgRadius of the cylinder, r = 15 cm = 0.15 mCo-efficient of kinetic friction, µk = 0.25Angle of inclination, θ = 30°Moment of inertia of a solid cylinder about its geometric axis, I = (1/2)mr2The various forces acting on the cylinder are shown in the following... ...more

Mass of the cylinder, m = 10 kg
Radius of the cylinder, r = 15 cm = 0.15 m
Co-efficient of kinetic friction, µ= 0.25
Angle of inclination, θ = 30°
Moment of inertia of a solid cylinder about its geometric axis, I = (1/2)mr2
The various forces acting on the cylinder are shown in the following figure:

 
The acceleration of the cylinder is given as:

a = mg Sinθ / [m + (I/r2) ]
mg Sinθ / [m + {(1/2)mr2/ r2} ]
= (2/3) g Sin 30°
= (2/3) × 9.8 × 0.5  =  3.27 ms-2


(a) Using Newton’s second law of motion, we can write net force as:
fnet = ma
mg Sin 30° – f = ma
f = mg Sin 30° – ma
= 10 × 9.8 × 0.5 – 10 × 3.27
49 – 32.7 = 16.3 N

(b) During rolling, the instantaneous point of contact with the plane comes to rest. Hence, the work done against frictional force is zero.

(c) For rolling without skid, we have the relation:
μ = (1/3) tan θ
tan θ = 3μ = 3 × 0.25
∴ θ = tan-1 (0.75) = 36.87°.

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Answered on 18/09/2018 CBSE/Class 11/Science/Physics/Unit 4-Motion of System of Particles/Chapter 7-System of Particles and Rotational Motion

Prove the result that the velocity v of translation of a rolling body (like a ring, disc, cylinder or... ...more

Prove the result that the velocity v of translation of a rolling body (like a ring, disc, cylinder or sphere) at the bottom of an inclined plane of a height h is given by,v2=2gh/(1+k2/R2) using dynamical consideration (i.e., by consideration of forces and torques). Note k is the radius of gyration of the body about its symmetry axis, and R is the radius of the body. The body starts from rest at the top of the plane.

A body rolling on an inclined plane of height h,is shown in the following figure: m = Mass of the body R = Radius of the bodyK = Radius of gyration of the bodyv = Translational velocity of the bodyh =Height of the inclined planeg = Acceleration due to gravityTotal energy at the top of the plane,... ...more

   

A body rolling on an inclined plane of height h,is shown in the following figure:

 

m = Mass of the body

= Radius of the body
K = Radius of gyration of the body
= Translational velocity of the body
=Height of the inclined plane
g = Acceleration due to gravity
Total energy at the top of the plane, E­1= mgh
Total energy at the bottom of the plane, Eb = KErot + KEtrans
= (1/2) I ω2 + (1/2) mv2
But I = mk2 and ω = v / R
∴ Eb = (1/2)(mk2)(v2/R2) + (1/2)mv2
= (1/2)mv2 (1 + k2 / R2)
From the law of conservation of energy, we have:
ET = Eb
mgh = (1/2)mv2 (1 + k2 / R2)
∴ v = 2gh / (1 + k2 / R2)
Hence, the given result is proved.

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Class 11 Tuition 4.0

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 11 Tuition

3

Board

State, CBSE

CBSE Subjects taught

Chemistry, English, Accountancy, Physics, Mathematics, Business Studies, Computer Science, Biology

Taught in School or College

No

State Syllabus Subjects taught

Chemistry, English, Accountancy, Biology, Mathematics, Physics, Business Studies

Teaching Experience in detail in Class 11 Tuition

A teacher has to know how to listen. Because his first role is listening to others, and assisting them. I know a nice number of teachers who know how to listen to their students, their families, other colleagues, admins, and everybody else. The teacher will listen to the kids for example, in an attentive way, realizing and learning what’s going on with their lives at the school – with prudence anyway. At the beginning of the school year he learns their names quite nicely, because he gets interested in the persons he has in charge. He doesn’t confine his job to teaching and disappearing through the door when his “block” has been delivered to his students. He’s prone to take in charge his students’ issues, problems and joys alike. He teaches with high standards of professionalism and with affection, with a clever and insightful affection – with prudence anyway.

Class 12 Tuition 4.0

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 12 Tuition

3

Board

State, CBSE

CBSE Subjects taught

Chemistry, English, Accountancy, Physics, Mathematics, Business Studies, Computer Science, Biology

Taught in School or College

No

State Syllabus Subjects taught

Chemistry, English, Accountancy, Biology, Mathematics, Physics, Business Studies

Teaching Experience in detail in Class 12 Tuition

A teacher has to know how to listen. Because his first role is listening to others, and assisting them. I know a nice number of teachers who know how to listen to their students, their families, other colleagues, admins, and everybody else. The teacher will listen to the kids for example, in an attentive way, realizing and learning what’s going on with their lives at the school – with prudence anyway. At the beginning of the school year he learns their names quite nicely, because he gets interested in the persons he has in charge. He doesn’t confine his job to teaching and disappearing through the door when his “block” has been delivered to his students. He’s prone to take in charge his students’ issues, problems and joys alike. He teaches with high standards of professionalism and with affection, with a clever and insightful affection – with prudence anyway.

Engineering Diploma Tuition 4.0

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Engineering Diploma Branch

Civil Engineering Diploma

Civil Engineering Diploma Subject

Highway Engineering, Irrigation Engineering, Professional Practices, Applied Mathematics (CE and ME Group), Building Services And Entrepreneurship Development, Computer Aided Drawing, Development of Life Skills, Surveying

Type of class

Crash Course, Regular Classes

Class strength catered to

One on one/ Private Tutions, Group Classes

Taught in School or College

No

Class 6 Tuition 4.0

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 6 Tuition

2

Board

CBSE

CBSE Subjects taught

English, Computers, Mathematics, Hindi, EVS, Social Science, Science

Taught in School or College

No

Class 7 Tuition 4.0

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 7 Tuition

2

Board

CBSE

CBSE Subjects taught

English, Computers, Mathematics, Hindi, EVS, Social Science, Science

Taught in School or College

No

Class 8 Tuition 4.0

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 8 Tuition

2

Board

CBSE

CBSE Subjects taught

English, Computers, Mathematics, Hindi, EVS, Social Science, Science

Taught in School or College

No

Class 9 Tuition 4.0

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

CBSE

CBSE Subjects taught

English, Hindi, Mathematics, Social science, Science

Taught in School or College

No

Class 10 Tuition 4.0

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

CBSE

CBSE Subjects taught

English, Hindi, Mathematics, Social science, Science

Taught in School or College

No

Class I-V Tuition 4.0

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class I-V Tuition

1

Fees

₹ 200 per hour

Board

CBSE

CBSE Subjects taught

Hindi, English, Computers, Science, Mathematics, EVS

Taught in School or College

No

Spoken English classes 4.0

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Spoken English classes

3

Age groups catered to

10 yrs to 25 yrs, Below 10 yrs, Above 25 yrs

Lived or Worked in English Speaking Country

No

Awards and Recognition

No

Certification

None

Demo Class Provided

Yes

Profession

Tutor

Language of instruction offered

Hindi to English, English to English

Curriculum Expertise

ICSE/ISC, CBSE, State, International

Citizen of English Speaking Country

No

Class strength catered to

One on one/ Private Tutions, Group Classes

Teaching done in

Vocabulary, English Grammar, Basic Spoken English

Teaching at

Home

Teaching Experience in detail in Spoken English classes

I started my teaching career in 2015 working for a few privately-working language people. I taught a number of hours in each place, so spent most of my day on the metro getting from one side of the city to the other. Getting your foot in the door was how it worked back then, and to be honest, little has changed there regarding working in big cities. Before going solo I explored my city and worked Nationally as an english to hindi dubbing script writer. I did this to perfect my language & related skills. It worked in as far as learning Skills, but being a student has its age limitations, and knowing that I wanted to travel and learn more languages led me to going on my own. My other work experiences involved tutoring school and college students for different subjects in both the languages English/Hindi as convenient. I manage groups of people from different backgrounds, learning styles and cultures who have to work together constructively over short, intense periods of time. This can get sticky, so a sense of calm and humour is essential. Listening and providing a pastoral ear is also key, as well as being 100% open and objective to what may come up. When training I might be observing lessons, helping people in their planning, marking essays and even teaching myself as well as answering emails from previous trainees with their eclectic questions.

MS Word Training 4.0

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in MS Word Training

2

MS Office Software Training 4.0

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in MS Office Software Training

2

Autocad classes 4.0

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Autocad classes

2

Soft Skills Training Classes 3.5

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

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Answers by Kartikey Sharma (4)

Answered on 18/09/2018 CBSE/Class 11/Science/Physics/Unit 4-Motion of System of Particles/Chapter 7-System of Particles and Rotational Motion

Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same... ...more

Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time?

Let m and r be the respective masses of the hollow cylinder and the solid sphere.The moment of inertia of the hollow cylinder about its standard axis, I1 = mr2The moment of inertia of the solid sphere about an axis passing through its centre, I2 = (2/5)mr2We have the relation:τ = IαWhere,α... ...more

Let m and r be the respective masses of the hollow cylinder and the solid sphere.
The moment of inertia of the hollow cylinder about its standard axis, I1 = mr2
The moment of inertia of the solid sphere about an axis passing through its centre, I2 = (2/5)mr2
We have the relation:
τ = Iα
Where,
α = Angular acceleration
τ = Torque
I = Moment of inertia
For the hollow cylinder, τ1 = I1α1

 

For the solid sphere, τn = Inαn

As an equal torque is applied to both the bodies, τ= τ2
∴ α/ α1  =  I1  /  I2  =  mr2  /  (2/5)mr2
α2 > α1     …(i)
Now, using the relation:
ω = ω0 + αt
Where,
ω0 = Initial angular velocity
t = Time of rotation
ω = Final angular velocity
For equal ω0 and t, we have:
ω ∝ α … (ii)
From equations (i) and (ii), we can write:
ω2> ω1
Hence, the angular velocity of the solid sphere will be greater than that of the hollow cylinder.

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Answered on 18/09/2018 CBSE/Class 11/Science/Physics/Unit 4-Motion of System of Particles/Chapter 7-System of Particles and Rotational Motion

Two particles, each of mass m and speed v, travel in opposite directions along parallel lines separated... ...more

Two particles, each of mass m and speed v, travel in opposite directions along parallel lines separated by a distance d. Show that the vector angular momentum of the two particle system the same whatever be the point about which the angular momentum is taken.

Let at a certain instant two particles be at points P and Q, as shown in the following figure. Angular momentum of the system about point P: Lp = mv × 0 + mv × d = mvd …(i)Angular momentum of the system about point Q:LQ = mv × d + mv × 0 = mvd ….(ii) Consider... ...more

Let at a certain instant two particles be at points P and Q, as shown in the following figure.

Angular momentum of the system about point P:

Lp = mv × 0 + mv × d  =  mvd   …(i)
Angular momentum of the system about point Q:
LQ = mv × d + mv × 0   =  mvd   ….(ii)

 

Consider a point R, which is at a distance yfrom point Q, i.e.,
QR = y
∴ PR = d – y
Angular momentum of the system about point R:
LR = mv × (d – y) + mv × y
mvd – mvy + mvy
mvd  ….(iii)
Comparing equations (i)(ii), and (iii), we get:
LP = LQ = LR    …(iv)

We infer from equation (iv) that the angular momentum of a system does not depend on the point about which it is taken.

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Answered on 18/09/2018 CBSE/Class 11/Science/Physics/Unit 4-Motion of System of Particles/Chapter 7-System of Particles and Rotational Motion

A cylinder of mass 10 kg and radius 15 cm is rolling perfectly on a plane of inclination 30°. The... ...more

A cylinder of mass 10 kg and radius 15 cm is rolling perfectly on a plane of inclination 30°. The coefficient of static friction us = 0.25.
(a) How much is the force of friction acting on the cylinder?
(b) What is the work done against friction during rolling?
(c) If the inclination O of the plane is increased, at what value of 8 does the cylinder begin to skid, and not roll perfectly?

Mass of the cylinder, m = 10 kgRadius of the cylinder, r = 15 cm = 0.15 mCo-efficient of kinetic friction, µk = 0.25Angle of inclination, θ = 30°Moment of inertia of a solid cylinder about its geometric axis, I = (1/2)mr2The various forces acting on the cylinder are shown in the following... ...more

Mass of the cylinder, m = 10 kg
Radius of the cylinder, r = 15 cm = 0.15 m
Co-efficient of kinetic friction, µ= 0.25
Angle of inclination, θ = 30°
Moment of inertia of a solid cylinder about its geometric axis, I = (1/2)mr2
The various forces acting on the cylinder are shown in the following figure:

 
The acceleration of the cylinder is given as:

a = mg Sinθ / [m + (I/r2) ]
mg Sinθ / [m + {(1/2)mr2/ r2} ]
= (2/3) g Sin 30°
= (2/3) × 9.8 × 0.5  =  3.27 ms-2


(a) Using Newton’s second law of motion, we can write net force as:
fnet = ma
mg Sin 30° – f = ma
f = mg Sin 30° – ma
= 10 × 9.8 × 0.5 – 10 × 3.27
49 – 32.7 = 16.3 N

(b) During rolling, the instantaneous point of contact with the plane comes to rest. Hence, the work done against frictional force is zero.

(c) For rolling without skid, we have the relation:
μ = (1/3) tan θ
tan θ = 3μ = 3 × 0.25
∴ θ = tan-1 (0.75) = 36.87°.

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Answered on 18/09/2018 CBSE/Class 11/Science/Physics/Unit 4-Motion of System of Particles/Chapter 7-System of Particles and Rotational Motion

Prove the result that the velocity v of translation of a rolling body (like a ring, disc, cylinder or... ...more

Prove the result that the velocity v of translation of a rolling body (like a ring, disc, cylinder or sphere) at the bottom of an inclined plane of a height h is given by,v2=2gh/(1+k2/R2) using dynamical consideration (i.e., by consideration of forces and torques). Note k is the radius of gyration of the body about its symmetry axis, and R is the radius of the body. The body starts from rest at the top of the plane.

A body rolling on an inclined plane of height h,is shown in the following figure: m = Mass of the body R = Radius of the bodyK = Radius of gyration of the bodyv = Translational velocity of the bodyh =Height of the inclined planeg = Acceleration due to gravityTotal energy at the top of the plane,... ...more

   

A body rolling on an inclined plane of height h,is shown in the following figure:

 

m = Mass of the body

= Radius of the body
K = Radius of gyration of the body
= Translational velocity of the body
=Height of the inclined plane
g = Acceleration due to gravity
Total energy at the top of the plane, E­1= mgh
Total energy at the bottom of the plane, Eb = KErot + KEtrans
= (1/2) I ω2 + (1/2) mv2
But I = mk2 and ω = v / R
∴ Eb = (1/2)(mk2)(v2/R2) + (1/2)mv2
= (1/2)mv2 (1 + k2 / R2)
From the law of conservation of energy, we have:
ET = Eb
mgh = (1/2)mv2 (1 + k2 / R2)
∴ v = 2gh / (1 + k2 / R2)
Hence, the given result is proved.

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Kartikey Sharma describes himself as Tutor. He conducts classes in Autocad, Class 10 Tuition and Class 11 Tuition. Kartikey is located in Shahdara, Delhi. Kartikey takes at students Home, Regular Classes- at his Home and Online Classes- via online medium. He has 3 years of teaching experience . Kartikey has completed Diploma in civil engineering from C.R.R. institute of technology, Kanjhawala, Delhi in 2017. He is well versed in Hindi and English.

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