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16 Answers
14.) the 4th one..
two things to remember:
I) the sign of B will change.
II) the graph will follow the eqn. B= μ0i/2Πr [edited]
(a rectangular hyperbola)
1SKY PLZ MAKE ME CLEAR ABOUT DA ANSWER :) PF PROBLEM 4 or .. coz that option was dere.. but i cudnt understand how it happened .. ar why will the graph follkow E = kQ/r^2?.. plz explain clearly :)
106in Q28.. nothing else is given about θ2?? i think somethings missing.. some info. has to be given abt Q..
1nope :Odat was what da sum was given i have checked [1] plz try den 86
1oops sorry [3] B is at the vertical pin point.. and i think this is undergoing ANGULAR SHM isnt it?.. plz solve
3never saw such a post ... always forget to check general phy . section [2][2][2][2][2][2]..
okie any wayz ..... that sq. plate y have u scribbled on the right part of it???????? is that a hinge ????
1Question: 86
For small angular displacement θ
The total energy of the system can be written as
T=1/2 k (bθ)2+1/2 I ω2
where I is the moment of inertia about B.
Since the energy is conserved.
dT/dθ=0
or
kb2θ dθ/dt + I ω dω/dt=0
since ω=dθ/dt
we get
ω(kb2θ+I dω/dt)=0
or
d2θ/dt2=-kb2θ/I
Thus ang. acceleration is proportional to ang. displacement
or
ω2=kb2/I
or Time period T=2Î √I/kb2