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H20...
in CH4 and BF3 the vector sum of individual dipole forces is zero ( cancel out)
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yes i got 1 that way kishan12.. but after that see that it is contained in the function of fractional part which is { }
so we get 0
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and if they are fractional part function then is the answer 0 ?
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superb krishnagopal... it looks easy now :)
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1 to power infinity is actually one...
but if its x tending to 1 but not = 1. then there are 2 cases
if x tends to 1 from left hand side then it will be a fraction and fraction to power infinity is 0.
if x tends to 1 from right hand side then it will be greater than 1 and power of infinity will make it equal to infinity.
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sin + cos
=  2 ( (1/ 2) sin + (1/  2) cos )
= 2 ( sin (  /4 + ) )
now as sin varies from -1 to +1
thus, value of above function varies from -2 to +2
*PLZ RATE*
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dR = d p l /A
= d krl/A
R = [a ] [b] krl / 2r (taking horizontal rings of increasing radius as element.. l is constant)
R = kl / 2  [a ][b ] r/r
= kl / 2 a[r]b
= kl / 2 (b - a)
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sorry this is the correct answer( i forgot to consider elements properly)
we can integrate resistance using formula pl/a
integrate krl/2 pi r taking rings as elements
we get k/2 pi [r] from a to b
= kl /2 pi (b-a)
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we have to find minimum value of what? tan square alpha or tan 2alpha? (jus asking the form of the terms)
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field is always in the outward direction of +ve charge.. as theres a cavity theres no charge inside so there wont be any field inside the cavity
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are those curly braces for fractional part function or have u just used them for clarity?
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message me on my nudgebook if u wanna know how to get this answer
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Ans) k r^2 / 2 pi ( r^2 - a^2 )
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at time almost equal to zero we can ignore the capacitance.
( similarly when time is infinity, we ignore the resistance placed in series after capacitance ... t = infinty means all current changes have taken place... all NCERT questions r supposed to be solved for t = infinity)
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