The correct options are
A time taken by particle to cross region -I for
1st time is
π√2ml5qE B time taken by particle to cross region -I for
2nd time is
π√2ml13qE C Angular speed of
1st revolution in magnetic field is
√5qE2mlUsing work-energy theorem till string become taut
Work done by
qE = change in K.E
⇒qEl=12mv2−0 ⇒v=√2qElm This speed
v is vertically upwards.
When string becomes taut an impulse is received and speed (tangential) of the bob becomes
vcos30∘=√3qEl2m Further using work energy theorem
Work done by
qE = Change in K.E,
qEl2=12mv′2−12m(vcos30∘)2 qEl2=12mv′2−12m3qEl2m qEl2=12mv′2−123qEl2 mv′2=5qEl2 speed of the bob as it enters magnetic field
v′=√5qEl2m For the first half revolution in the magnetic field particle will move with same constant speed
v′ ∴ time period to cross region
I for first time =
πlv′=π√2ml5qE After
1st half revolution in magnetic field, particle enters the region of electric field. Due to work done by electric field there is change in KE.
Let
v1 be velocity of particle as it enters region-
I for the second time.
Work done by
qE = Change in K.E, from lowermost to uppermost point.
qE.2l=mv212−m25qEl2m v1=√13qEl2m time period to cross region
I for second time =
πlv1=π√2ml13qE Angular speed of
1st half revolution in region
I=
v′l=√5qE2ml Angular speed of
2nd half revolution in region =
v1l=√13qE2ml