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thanx alti'll tell u a very easy method for solvin these typo qs,
remember, a path difference of lambda(one wavelength) = a phase difference of 2pi radians
from here we can form an equation , path diff/phase differnce=lamda/2pi radians
now find the wavelength from tha data given,
and substitute the values in the equation 2 get the phase difference!
hope that helpsss
Thanx a bunch gala) first find the wavelength = speed/frequency
= 340/500 = 0.68m
Formula for path difference (you need to know this) = 2(pie) x path difference (which is 0.17 in this case) / wavelength(0.68)
= 2(pie) x 0.17 / 0.68 = 1/2(pie)
ok as it involves hieght we r gona take kinetic energy and potential energy too..How do we do this? do we use V^2=U^2 + 2as ???View attachment 12904
wow you really listed most of the estimates.. thanks dude u rock..
can u tell me some more? like the approximate diameter of a uranium atom or young modulus of a metal or i think its called the range of alpha particles.... and any other like these that come across ur mind.. thanks bud..
Q 27 its D... it can either be A or D...it cant be A cz lines cant be parallel there
yeah, how do we do this??whats the answer?
Is it C??
is the ans D for 27?
Q no 27 from 1 anti node to node will be lamda/4=x (x is da distance)
oh yeap D.. how did u get it? I thought 2x since 0.5lambda=x.. but not rightis the ans D for 27?
Why are we assuming that the initial and final velocity are 0?? It dusn't say it is dropped from rest?its C
--> for first case when the sphere is dropped
u^2 = 0 + 2as (let s be total height of the fall)
i.e. u = (2as)^1/2.........................(i)
-->2nd case for bounce
0 = v^2 - 2a(s/2) (since the height is half of total height this time)
or, as = v^2
i.e. v = (as)^1/2
now,
v\u = (as/ 2as)^ 1/2
= (1/2)^1/2
Lambda=4x, as distance btwn 1 node and antinode =lambda/4oh yeap D.. how did u get it? I thought 2x since 0.5lambda=x.. but not right
ohh okay thank u so muchLambda=4x, as distance btwn 1 node and antinode =lambda/4
thankyou so much that certainly cleared it up..youngs modulus of materials are like to the power 10 or 11 (eg: 10^11 or 10^10)
you can find out the diameter of uranium atom by the formula of volume
4/3 pie (r^3)
mass of uranium is 238u
where u=1.67x10^-27
so the mass is around 10^-25
the range of alpha particles is like a few cms like 5-6cm in air n can be stopped by a piece of paper/cardboard
it travels with 10% speed of light, has greatest ionization effect
and is the ans D for q,34???ohh okay thank u so much
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