Gandhi Nagar, Hyderabad, India - 500020.
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Telugu Mother Tongue (Native)
English Proficient
Hindi Basic
jntu kakinada 2014
Bachelor of Technology (B.Tech.)
1, New Bakaram, Usyatam Residency, Gandhi Nagar, Kavadiguda
-meri gold appartment
Gandhi Nagar, Hyderabad, India - 500020
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Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in BTech Tuition
2
BTech Branch
Other Engineering topics, BTech 1st Year Engineering
Experience in School or College
2 years in vignan college
Type of class
Crash Course, Regular Classes
Class strength catered to
One on one/ Private Tutions
Taught in School or College
Yes
BTech 1st Year subjects
Engineering Mathematics (M1), Advanced Mathematics (M2)
Teaching Experience in detail in BTech Tuition
I have a teaching experience of 2 years in vignan engineering college. I have a very good track record in private tutoring in engineering mathematics with 100 percent pass percentage. Parents of the students concerned like srinivas sir, and many others has congratulated me for my efforts on their children who earlier has math phobia now being passed in mathematics that too in graduate level and there after children are confident enough to tackle any kind of mathematics. My teaching style mostly relies on removing the fear of student and allow themfall in love with mathematics. Real time examples of every topics makes them understand the concept and there Building a confidence in the student in such a way that they are going change the shape of India with their knowledge.
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 12 Tuition
4
Board
CBSE, ISC/ICSE, State
ISC/ICSE Subjects taught
Mathematics
CBSE Subjects taught
Mathematics
Taught in School or College
Yes
State Syllabus Subjects taught
Mathematics
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 11 Tuition
4
Board
CBSE, ISC/ICSE, State
ISC/ICSE Subjects taught
Mathematics
CBSE Subjects taught
Mathematics
Experience in School or College
i have taught for students in vignan college for 2 years.
Taught in School or College
Yes
State Syllabus Subjects taught
Mathematics
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 10 Tuition
4
Board
ICSE, CBSE
CBSE Subjects taught
Science, Mathematics
ICSE Subjects taught
Mathematics, Physics
Taught in School or College
Yes
5 out of 5 1 review
Umar Wahab
"Siddardha sir has a lot of experience in teaching. He is a very good tutor with some exceptional teaching skills. "
1. Which BTech branches do you tutor for?
Other Engineering topics and BTech 1st Year Engineering
2. Do you have any prior teaching experience?
Yes
3. Which classes do you teach?
I teach BTech Tuition, Class 10 Tuition, Class 11 Tuition and Class 12 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 2 years.
Answered on 23/10/2019 Learn CBSE/Class 12/Science/Physics/Unit 1-Electrostatics/ELECTROSTATIC POTENTIAL AND CAPACITANCE/NCERT Solutions/Exercise 2
we know that 1/2 CV2 is the energy stored in a capactor with capacitance C when connected across the voltage source V.
the property of capacitor is it acquires charge when it is connected to voltage source.
after this was this connected it acts as a energy source untill the charge in it is dissipated or until an equal energy source is countered to it.
so in case 1: when it is connected to the 200V source
energy acquired by the 600pF capacitor is 1.2 * 10-5
that means the energy 1.2 *10-5 is shared between two capacitors in equal amount
that means energy lost is half of the intial total energy and that is equal to energy gained by the second capacitor = 0.6 *10-5 joules
Answered on 23/10/2019 Learn CBSE/Class 9
you can do it by grouping two digits together form right like in this case 1369. now start division
step 1: choose the number such that its square is close to our left end pair (here 13)
3)1369(3
09
04 69
step 2 : Now multiplie the divisor with 2 (3 *2 =6)
choose another number (here 7), to place on the right side of the multiplied divisor (here 6) such that both the digits combined and multiplied with the chosen number will give either the exact value (here 496) or close to our required value.
67)496(7
496
0
so our square root of 1369 is our first digit 3 and our second digit 7 =37
or
for quick calculations follow this method
we know the square of 30 =900
and the square of 40 =1600
so the square of some number between 30 to 40 will be 1369
Now check the middle number, i.e. 35
sq of 35 =1225
that means our required number lies between 35 to 40
Now go for unit digit. Our unit digit is 9, which is given by 3 or 7 only when squared.
From 35 to 40 only 37 square will be 9 as a unit digit.
Our answer is 37.
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in BTech Tuition
2
BTech Branch
Other Engineering topics, BTech 1st Year Engineering
Experience in School or College
2 years in vignan college
Type of class
Crash Course, Regular Classes
Class strength catered to
One on one/ Private Tutions
Taught in School or College
Yes
BTech 1st Year subjects
Engineering Mathematics (M1), Advanced Mathematics (M2)
Teaching Experience in detail in BTech Tuition
I have a teaching experience of 2 years in vignan engineering college. I have a very good track record in private tutoring in engineering mathematics with 100 percent pass percentage. Parents of the students concerned like srinivas sir, and many others has congratulated me for my efforts on their children who earlier has math phobia now being passed in mathematics that too in graduate level and there after children are confident enough to tackle any kind of mathematics. My teaching style mostly relies on removing the fear of student and allow themfall in love with mathematics. Real time examples of every topics makes them understand the concept and there Building a confidence in the student in such a way that they are going change the shape of India with their knowledge.
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 12 Tuition
4
Board
CBSE, ISC/ICSE, State
ISC/ICSE Subjects taught
Mathematics
CBSE Subjects taught
Mathematics
Taught in School or College
Yes
State Syllabus Subjects taught
Mathematics
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 11 Tuition
4
Board
CBSE, ISC/ICSE, State
ISC/ICSE Subjects taught
Mathematics
CBSE Subjects taught
Mathematics
Experience in School or College
i have taught for students in vignan college for 2 years.
Taught in School or College
Yes
State Syllabus Subjects taught
Mathematics
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 10 Tuition
4
Board
ICSE, CBSE
CBSE Subjects taught
Science, Mathematics
ICSE Subjects taught
Mathematics, Physics
Taught in School or College
Yes
5 out of 5 1 review
Umar Wahab
"Siddardha sir has a lot of experience in teaching. He is a very good tutor with some exceptional teaching skills. "
Answered on 23/10/2019 Learn CBSE/Class 12/Science/Physics/Unit 1-Electrostatics/ELECTROSTATIC POTENTIAL AND CAPACITANCE/NCERT Solutions/Exercise 2
we know that 1/2 CV2 is the energy stored in a capactor with capacitance C when connected across the voltage source V.
the property of capacitor is it acquires charge when it is connected to voltage source.
after this was this connected it acts as a energy source untill the charge in it is dissipated or until an equal energy source is countered to it.
so in case 1: when it is connected to the 200V source
energy acquired by the 600pF capacitor is 1.2 * 10-5
that means the energy 1.2 *10-5 is shared between two capacitors in equal amount
that means energy lost is half of the intial total energy and that is equal to energy gained by the second capacitor = 0.6 *10-5 joules
Answered on 23/10/2019 Learn CBSE/Class 9
you can do it by grouping two digits together form right like in this case 1369. now start division
step 1: choose the number such that its square is close to our left end pair (here 13)
3)1369(3
09
04 69
step 2 : Now multiplie the divisor with 2 (3 *2 =6)
choose another number (here 7), to place on the right side of the multiplied divisor (here 6) such that both the digits combined and multiplied with the chosen number will give either the exact value (here 496) or close to our required value.
67)496(7
496
0
so our square root of 1369 is our first digit 3 and our second digit 7 =37
or
for quick calculations follow this method
we know the square of 30 =900
and the square of 40 =1600
so the square of some number between 30 to 40 will be 1369
Now check the middle number, i.e. 35
sq of 35 =1225
that means our required number lies between 35 to 40
Now go for unit digit. Our unit digit is 9, which is given by 3 or 7 only when squared.
From 35 to 40 only 37 square will be 9 as a unit digit.
Our answer is 37.
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