#question. Sir, is it possible for random shaped closed body to have 4pie starad. Solid angle. Or is it necessary to have a spherical surface for 4pie solid angle. ?
here the light source was kept directly on the axis perpendicular to the center of circular hole. how do we calculate the light power if light source is not lying on the axis??
sir in electricity please add videos of how to solve symmetric circuits( i mean folding circuit into half..) etc also if possible explain about y delta transformation....
SIR IF THERE IS A ROTATING AND TRANSLATING ROD IN FRONY OF AN INFIINITE LONG CURRENT CARRING WIREHW TO FIND EMF ----IS IT LIKE THIS LET THE LISTANCE FRM IT BE R AND TREATING IT AS CONSTANT VELOCITY PERPENDICULR TO WIRE IS Vcm+XW .. EMF=INTEGRATION XWPROJECTION IN DIRECTION OF VCM+Vcm x DX xB
+Arabinda Biswas Pl post the queries of concerned topic under the concerned videos only or in the INTERACT tab of www.physicsgalaxy.com... you are disturbing other viewers by posting such questions in random videos... be careful next time...
Your lectures are truly helpful and well presented, but please correct me if I am wrong here Horizontal distance from light source to cardboard is (R^2 - a^2)^1/2. Then Cos(Theta) = [ R^2 - a^2)^1/2] / R
thanks
Thanks a lot 🙏
Which topic is this like ??in Physics which chpt
this video(topic) saved a lost person in physics 5 years after its released. Thank you very much!
i think u came from the checklist video
Who?
This video(topic) saved a lost person in physics 8 year after it released. Thank you very much!
Animations dope❤
thank you very much sir this cleared absolutely everything.
Thank you ❤️
Wow
Good morning sir
Thank you sir
How can we derive omega = 2 phi (1 _ cos theta ) kindly derive it. Thank you.
I am in IIT yet i still watch sir's videos to revise such topics ! Thank you sir for providing us with such valuable content 🙏🙏
Which iit brother? I'm in class 12 now...
@@prajinshankarkarthik8013 IIT Hyd
so amazing 🔥🔥🔥🔥
Again thanks a lot for this noble cause.......
Thanks sir!
Thnx a lot sir for providing us videos even in ur busy schedule.
ji
#question.
Sir, is it possible for random shaped closed body to have 4pie starad. Solid angle. Or is it necessary to have a spherical surface for 4pie solid angle. ?
Spherical 3D surface in space.
No @jatinbhatt
If we find solid angle of cubical room's ceiling (upper inner part)it will be 360 only.
I mean to say it's not imp. To be a spherical body
here the light source was kept directly on the axis perpendicular to the center of circular hole. how do we calculate the light power if light source is not lying on the axis??
sir in that problem how cos theta = r/root of r sq +a sq ?
base/hypotenuse bro
…basic trigo
@@aashreykumar9886 adj/hyp
A living legend
sir in electricity please add videos of how to solve symmetric circuits( i mean folding circuit into half..) etc also if possible explain about y delta transformation....
+vishnu bhaskar see all adv illustrations to understand the symmetry ckts... I will use the method in Booster classes of electric circuits...
@@physicsgalaxyworld sir I just want to thank you 🙏🏻 for your work
SIR IF THERE IS A ROTATING AND TRANSLATING ROD IN FRONY OF AN INFIINITE LONG CURRENT CARRING WIREHW TO FIND EMF ----IS IT LIKE THIS LET THE LISTANCE FRM IT BE R AND TREATING IT AS CONSTANT VELOCITY PERPENDICULR TO WIRE IS Vcm+XW .. EMF=INTEGRATION XWPROJECTION IN DIRECTION OF VCM+Vcm x DX xB
+Arabinda Biswas Pl post the queries of concerned topic under the concerned videos only or in the INTERACT tab of www.physicsgalaxy.com... you are disturbing other viewers by posting such questions in random videos... be careful next time...
sir forgive me
Sir how are you writing as if theta=pi/2 then omega=2pi and also of theta=pi then omega=4pi...?
Please elaborate a bit.
Thankyou very much.
In the given expression of solid angle when you put Θ=π/2 then on solving it gives ω=2π and same for Θ=π we get ω=4π
Thank you very very much sir...I purchased your book- physics galaxy and it was excellent really useful. These videos on solid angle are excellent!
how you find the value of omega ?
Your lectures are truly helpful and well presented, but please correct me if I am wrong here
Horizontal distance from light source to cardboard is (R^2 - a^2)^1/2. Then Cos(Theta) = [ R^2 - a^2)^1/2] / R
No horizontal distance of center of hole to source is specified as r... pl watch the video carefully again...
You are correct. My oversight.
Just saved my ass prepping for an oral exam tomorrow! Thanks!!
Is there an oral exam in this pandemic😂😂
Thankyou sir