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Asked by dunnanagamani02 30th May 2021, 8:07 AM
Answered by Expert
Answer:
Qn.120 
 
NOR gate output is followed by two inverter because NAND gate is connected as inverter.
 
Hence equivalent circuit is NOR gate
 
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Qn.121
 
Magnetic field induction B at centre of loop of radius R is given as
 
begin mathsize 14px style B space equals space fraction numerator mu subscript o I over denominator 2 space R end fraction end style .................... (1)
where I is current in the loop and μo is the permeability of free space .
 
When helium nucleus that has charge 2e is moving with speed v in a circular loop ,
 
current I = ( 2e v ) = ( 2e ) × [ ( 2 π R ) / T ]  ....................(2)
 
where T = 2 s is period of revolution
 
If we substitute current I using eqn.(2) , we get magnetic field from eqn.(1) as
 
begin mathsize 14px style B space equals space fraction numerator mu subscript o over denominator 2 R end fraction cross times fraction numerator 2 e cross times left parenthesis 2 pi R right parenthesis over denominator T end fraction space equals space mu subscript o cross times 1.602 space cross times pi space cross times space 10 to the power of negative 19 end exponent space equals space 5.032 space cross times space 10 to the power of negative 19 end exponent space mu subscript o end style  T
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Qn.122
 
Magnetic field at centre of loop has two parts , (i) one part due to long wire carrying current I  and
(ii) second part due to current  I in circular loop
 
Magnetic field due to long wire , begin mathsize 14px style B subscript 1 space equals space fraction numerator mu subscript o I over denominator 2 pi space r end fraction end style
Magnetic field due to circular loop , begin mathsize 14px style B subscript 2 space equals space fraction numerator mu subscript o I over denominator 2 space r end fraction end style
Net magnetic field ,  begin mathsize 14px style B space equals space B subscript 1 space plus space B subscript 2 space equals space fraction numerator mu subscript o I over denominator 2 pi space r end fraction space plus space fraction numerator mu subscript o I over denominator 2 space r end fraction space equals space fraction numerator mu subscript o I over denominator 2 pi space r end fraction left parenthesis space pi space plus space 1 space right parenthesis end style
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Qn. 123 
 
Let charge on particle-X = q and charge on particle-Y = 2q 
 
Kinetic energies of particles after accelerated by same potential difference V is given as
 
begin mathsize 14px style 1 half m subscript x v subscript x superscript 2 space equals space q space V end style .................... (1)
begin mathsize 14px style 1 half m subscript y v subscript y superscript 2 space equals space 2 q space V end style  ....................(2)
From above equations (1) and (2) , we get   begin mathsize 14px style v subscript x over v subscript y equals square root of fraction numerator m subscript y over denominator 2 space m subscript x end fraction end root end style  ......................(3)
In magnetic field , magnetic force equals centripetal force . Hence we have 
 
begin mathsize 14px style m subscript x fraction numerator v subscript x superscript 2 over denominator R subscript 1 end fraction space equals space q space v subscript x space B end style ...........................(4)
begin mathsize 14px style m subscript y fraction numerator v subscript y superscript 2 over denominator R subscript 2 end fraction space equals space 2 q space v subscript y space B end style ........................ (5)
From above equations (4) and (5) we get begin mathsize 14px style v subscript x over v subscript y space equals space R subscript 1 over R subscript 2 cross times fraction numerator m subscript y over denominator 2 space m subscript x end fraction end style ....................(4)
By equating eqn.(3) and (4) , after simplification, we get begin mathsize 14px style m subscript x over m subscript y equals 1 half open parentheses R subscript 1 over R subscript 2 close parentheses squared end style
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Answered by Expert 30th May 2021, 2:49 PM
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