Messages posted by johri_anshuman

In the fig. the total charge on both the spheres in Q. The spheres are solid and charge is uniformly distributed on both spheres. Find the electric field at the point at distance (+-) r/2 from origin (i.e. centre of the smaller sphere).

I m not living in Delhi. How can I get admission in DCE on the basis of AIEEE rank???

helloz dude..i m a reso passout this year...u get ARR (all reso rank) for every CT, PT, MT and JPT... u hv to jus keep on improving it... so wats d need to know others progress???

yes one can easily differentiate btw E & M fields..keep a charged particle at rest in d field... if it comes into motion, it is electric field... if not... its magnetic field... :)

I got IIT-JEE AIR 4999 and AIEEE AIR 5141.

What r d possible options i can get???

Please refer me which one is d best to opt.

IIT JEE 2009 result aaj kitne baje aayega???

10th class marksheet and hallticket...

have a look at this http://www.jee.iitb.ac.in/aggregate.htm

actual cut off was 172 for general category......

yes...

a molecule can be optically active even if it has an axis of symmetry...

for a molecule to be optically active it should NOT have a plane of symmetry or a center of symmetry....

equation of the form y = A sin(wt -kx) is for plane waves...

y = (A/x) sin(wt - kx) is for spherical waves

y = (A/(root x)) sin (wt - kx) is for spherical waves...

simply plot the v-t graph and calculate area under the graph... distance cannot be -ve so take the modulus...

its coming out to be 8 m...

d option is correct...

let the acceleration of block 2m = A and that of m = a...

now simply use the result T*a = constant...

in the FBD make the tensions...

A * 2T = a* T

A = a/2

now the problem can be easily be solved...

Refer to the fig...

The field is in +i direction so v_0x will remain un affected...

so v(x) at any later time = v_0x and trajectory (1) will be a straight line...

Now for the trajectory (2) due to the y component

$\omega$ = eB/m

as the path followed will be circular so the y component along y will vary simple harmonically...

so using the equation of SHM v=A $\omega$ cos $\omega$ t

v_y = v_0ycos $\omega$ t

For trajectory (3) due to z component $\omega$ = eB/m

again as the motion due to v_0z is circular so the z component of velocity varies simple harmonically...

v_z=v_0zcos( $\omega$ t + $\Pi$ /2) { Clearly from the figure the phase difference due to the trajectories (2) and (3) is pi/2 }

=>v_z= - v_0z sin $\omega$ t

so the vector equation at any later time will be

v = v₀_x $\hat{i}$ + v_0ycos $\omega$ t $\hat{j}$ - v_0z sin $\omega$ t $\hat{k}$

it was juz ok.... could have done a lot better...

attempted 83... confident of atleast 70 to be correct...

The limit does not exist...
the function is defined only for natural numbers...

pls post up a better diagram... its not visible clearly... and there is no magnetic field in the fig...

anchitsaini has given a perfect solution...
thanks dude...

A ring whose linear charge density varies as

=₀cos@ where @ is a shown in the fig. Find the electric field (with direction) at point P. P is at distance x from the ring of radius R.

Solid Sphere

Pootential = i think potential of a particular point is not defined (not sure abt this)

Field = 0 --- Conducting solid sphere=> total charge remains on the surface. Using gauss law enclosed charge =0 inside the sphere

= kQ/r^2 for r>=R (i.e. outside the sphere)

Capacitance=0 -- coz a solid conduction sphere will never store charge

Hollow Shpere

Potential = kQ/R

Field = 0 for r = kQ/r^2 for r>=R

Capacitance= 4piER

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