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Q30- d sin theta=n lambda
Q37 - its not D as no current will pass if both switch are open. Then its not even A as both are closed so current will be I. In B S2 is open and S1 is closed so current just flows in series so current is still I. In C current flows through parallel circuit as S1 is open and S2 closed and current is not equal in parallel so it will not be I
Q37 - its not D as no current will pass if both switch are open. Then its not even A as both are closed so current will be I. In B S2 is open and S1 is closed so current just flows in series so current is still I. In C current flows through parallel circuit as S1 is open and S2 closed and current is not equal in parallel so it will not be I
hahah no problem
Q30- d sin theta=n lambda
sin theta=1 as its normal to the grating. then d=1/N. N=300 per mm means 300000 per m. so 1/300000=(690*10^-9)*n
n will be around 4. then multiply by 2 and u get 9
you always multiply by 2 to find the maxima. i think it was because n is only for one side or something...i dont really remember the reason
but just remember 2 do it everytime you always multiply by 2 to find the maxima. i think it was because n is only for one side or something...i dont really remember the reason
no diameter gets halved but the radius gets decreasd by 1/4th then. Try it using a value for diameter and find the area. Then repeat using half value of diameter. Area will be 1/4thand for 27..if the diameters being halved shouldnt it be 1/2?
sorry can you do 36 as well..last question
you have to split it into 2 parts. 1st take the positive part. I=2. then find P. P=I^2R=2*2*100=400Wand for 27..if the diameters being halved shouldnt it be 1/2?
sorry can you do 36 as well..last question
ok for 15 take f=kvhttp://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_12.pdf
Can any one help me with the Question No:7,15,22...
Please anyone
link pls
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