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  • Male, 56 Years
  • Activity Score100


Associated for 2 Years
  • Online
  • Qualification:
  • Experience:
    I have an experience of 25 years in teaching. I am dedicated to my profession and give my 100 percent to students what I have owing to get better result from them. I can teach Mathematics subject to students from class IX to cla... More [+]
  • Teaches:
    Physics, Mathematics, Chemistry, Algebra, Mechanical, Electronics, Electrical, Drawing, Spoken English, Public Speaking, IELTS, Effective Communication, Java and J2EE, C# (C Sharp), C, C++, .Net, MS Office, Computer for official job, Basic Computer, Advanced Excel
  • Board:
    ICSE, Local State Board
  • Areas:
  • Pincode:
Profile Details
Profile Details

Qualification :


Total Experience :

27 Years

I have an experience of 25 years in teaching. I am dedicated to my profession and give my 100 percent to students what I have owing to get better result from them. I can teach Mathematics subject to students from class IX to class XII.

Tutoring Option:

Home Tuition Only

Tutoring Approach:

I feel as a teacher it is my role to motivate students to overcome their inhibitions and shine, encourage them to sharpen their talents and groom them to face the world with confidence. I also strive to provide individual attention to each student, get to know them well, as I feel the learners should be more comfortable and confident about their teacher first.

Hourly Fees [INR]:


Class 9 - 10 Mathematics, Physics, Chemistry, Algebra, ICSE INR 150.00 /hour
Class 11 - 12 Mathematics, Physics, Chemistry, Algebra, Local State Board INR 250.00 /hour
English Speaking Public Speaking, IELTS, Spoken English, Effective Communication INR 350.00 /hour
Engineering Subjects Electrical, Electronics, Mechanical, Drawing INR 300.00 /hour
Programming Language .Net, Java and J2EE, C, C++, C# (C Sharp) INR 300.00 /hour
Basic Computer / Office Basic Computer, Computer for official job, MS Office, Advanced Excel INR 150.00 /hour
  • Answer:


  • Answer:

    C   is the correct answre as the larger the sphere the greater the charge assuming that both have same voltage when they were connected the voltage will continue to pass between them until it becomes equal on both sphere.


    (1/4pi *epsilon 0)*(Q1/R1) =(1/4pi *epsilon 0)*(Q2/R2)    



    since R2>R1 then the charge on the larger sphere will be more                   


  • Answer:

    The correct answer is B 

    According to newtons law the simple pendulum oscillation is based on its length not by its mass hence the oscillation would be contant

  • Answer:

    Answer B is correct

    The solution to this question is based on trignometry angle theta calculations, As it is given in the question the Length of the Adjecant side is 778.5m and opposite side that is total hight of the cliff is 450m, then the angle of the flight should be calculated as tan theta because the sides in consideration are Opposite and Adjacent hence it is Tan angle , the sides values  are given then  degrees of  the angle can be calculated by tan Inverse.that is

    Opp/Adj ---> 450 / 778.5 = 0.5780346  substituting this value in tan Table the angle is 30.029 almost 30 degrees.


  • Question: Why passwords are case sensitive

    Posted in: Computer Science | Date: 11/11/2017


    The passwords entering options are provided by the user is case sensitive as this feature is based on the W3c consortium guide lines , the main reason behind this is to give the user full control on password creation option yo make it somewaht difficult and safer 

    Note: Password are not only case sensitive but also provides special charecters selsction feature too.

  • Answer:

    The solution to this problem is given in details 

    1.The radius of bigger semicircle is 20 cm , beacause the small semi circle are half of the bigger semi circle as shown in the figure then:

    2. The small semi circle radius is 10 cm, there are total of 3 shaded small semi circles 

    3  Area (shaded portion) of single semi circle is (Pi * r *  r) / 2  --> ( 3.14159 * 10 * 10) / 2 = 157.0796 cm ^2

    4  Shaded portion of 3 small semi circle is 157.0796 * 3 =  471.2388 cm ^2

    5 Shaded portion of big semi circle - unshaded portion of small semi circle

    6 Area of big semi circle (3.1459 *20 *20) / 2 =628.318 - 157.0796 =  471.2389. 

    Total shaded area is 471.2389 + 471.2389 = 942.477 cm ^2


    Small semi circle perimeter is (2 * Pi * r) / 2 --->( 2 * 3.14159 * 10) / 2 = 31.4159 cm

    for three semi circle is 3 * 31.4159 = 94.2477 cm

    Big semi circle shaded portion perimeter is (2 * Pi * r) ---> (2 * 3.14159 * 20) = 125.6637 cm 

    Total shaded area perimeter is big semi crcle perimeter + three small semicircles--> 125.6637 + 94.2477= 219.9117 cm


  • Answer:

    The correct answer as per present international acceptance is seems to be A "parallax"but by close examing the major part in this method is afterall trgnometry

  • Question: Which one is a vector quantity?

    Posted in: Physics | Date: 18/11/2017


    The correct answer is D as the magnetic quantity has both magnitude and direction of thr flux

  • Question: A dimensionless quantity

    Posted in: Physics | Date: 18/11/2017


    A dimensionless quantity may have dimensionless unit Hence the answer to this question is D

  • Question: Physical quantities are  

    Posted in: Physics | Date: 18/11/2017


    The correct option is D

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