Physics Video Lecture
Connecting Students in 221 Countries

Video Lectures

High School/JEE/NEET/IPhO Physics | 17-19 Yrs

Basics of Solid Angle
Duration: 5.39 Min
 
Add to my sequences
Bookmark This Video
Duration: 4.13 Min
 
Add to my sequences
Bookmark This Video
 
Add to my sequences
Bookmark This Video
 
Add to my sequences
Bookmark This Video
  1.  
  2. Sir, if we take limits from 0 to theta, won't that only give us half the surface area of the spherical surface?, i.e. S/2?? shouldn't we take limits from -theta to theta??
    7 months ago by

    Post your answer here

  3.  
  4. why we can't derive a relation between theta and omega by using simple cone (without top sphereical surface.
    11 months ago by

    Post your answer here

  5.  
  6. is this derivation important for JEE ?
    5 years ago by Utkarsh

    Post your answer here

  7.  
  8. sir, the area of this curved surface(part of sphere) is obtained by assuming the width of ring as Rd(phi) which I understand. but is the area of flat circular surface which is called base of cone is (pi/2)(R^2)(1-cos(2(phi))) which is obtained by taking width of ring as dx where x=Rsin(phi) and so dx=Rcos(phi)d(phi) then area of flat circular surface S=integration 0 to (phi) of 2(pi)xdx which is S=integration 0 to (phi) of 2(pi)(Rsin(phi))(Rcos(phi)d(phi)) and if all this is correct then equating it with S=(omega)(R^2) we get (omega)=((pi)/2)(1-cos(2(phi))) which contradicts the original result. kindly elaborate.
    7 years ago by bruce wayne

    Post your answer here