IITJEE Solved problem

IITJEE solved problems are two among those problems send to us by visitors of our blog as their querries. You can also send us your querries to us to get them solved.

Problem 1. Two masses m1 and m2 are connected by a spring of spring constant k and are placed on a frictionless horizontal surface. Initially the spring is stretched through a distance x0 when the system is released from rest. Find the distance moved by the messes before they again come to rest.

Solution.

Consider blocks plus spring a system and if no external forces acts on the sysyem then centre of mass of system will remain at rest. Mean position of two SHM's would be the unstretched position.If m1 moves towards right through a distance x1 and m2 moves towards left through a distance x2before spring acquires natural length then
x1+ x2= x0 .....................(1)
where x1 and x2 would be the amplitudes of blocks m1 and m2 resp. Since centre of mass of system will remain same so,
m1x1=m2x2
thus from 1
x1=m2x0/(m1 + m2)
x2=m1x0/(m1 + m2)
to get back to rest position masses m1 and m2 would have to travel distances x1 and x2 resp. , so total distance travelled by m1 before comming to rest is
2m2x0/(m1 + m2)
and total distance travelled by m2 before comming to rest is,
2m1x0/(m1 + m2)

Problem 2. Suppose the surface charge density over a sphere of radius R depends on a polar angle θ as σ=σ0cosθ, where σ0 is a positive constant. Find the electric field strength vector inside the given sphere.
Solution.
You can visualize such a charge distribution as a result of a small relative shift of two uniformly charged balls of radius R, whose charges are equal in magnitude but opposite in sign. In such a representation inner charges cancel out each other and only charges on the surface survives having charge density σ=σ0cosθ. Charge density is maximum along +Z and -Z where the distance between the surfaces is maximum.


Electric field due to positively charged sphere E+r+/3ε0
Electric field due to positively charged sphere E-=-ρr-/3ε0
Electric field at P
E = E++E- =ρ(r+-r-)/3ε0
or,
Ea/3ε0=ρakˆ/3ε0
now,
ρa = charge per unit area =σ0
hence,
E0kˆ/3ε0

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