March 2011 - Career Point

Physics Challenging Problems
Set # 11
This section is designed to give IIT JEE aspirants a thorough grinding & exposure to variety
of possible twists and turns of problems in physics that would be very helpful in facing IIT
JEE. Each and every problem is well thought of in order to strengthen the concepts and we
hope that this section would prove a rich resource for practicing challenging problems and
enhancing the preparation level of IIT JEE aspirants.
By : Dev Sharma
Solutions will be published in next issue
Director Academics, Jodhpur Branch
Passage # (Q. No. 1 to Q. No. 3 )
Light having photon energy hv is incident on a
metallic plate having work function φ to eject the
electrons. The most energietic electrons are then
allowed to enter in a region of uniform magnetic
field B as shown in figure. The electrons are
projected in x-z plane making an angle θ with x-
→
=
axis and magnetic field B B 0 i along x-axis.
Maximum pitch of halix described by an electron is
found to be P. Take mass of electron as m and
charge as q. Based on above information answer the
following questions.
y
z
1. The correct relation between P and B 0 is
(A) qPB 0 = 2πcosθ
2(hν − φ)
m
(B) qPB 0 = 2πcosθ
2(hν − φ)
m
(C) qPB 0 = 2π
2(hν − φ)
m
2πm
(D) P = ×
qB
hν − φ
0
2. Considering the instant of crossing origin at t = 0,
the z-coordinate of the location of electron as a
function of time t is
2m(hν − φ)
⎡ ⎛ qB0t
⎞⎤
(A) −
× sin θ⎢1
− cos⎜
⎟⎥ qB0
⎣ ⎝ m ⎠ ⎦
(B)
θ
B
e –
m(hν − φ)
⎡qB
t ⎤
× sin θ× sin
qB
⎢ ⎥
⎣ m ⎦
2
0
0
∧
x
(C)
(D)
−
2
0
m(hν − φ)
⎛ qB t ⎞
× sin θ× sin⎜
⎟
qB
⎝ m ⎠
0
m(hν − φ)
⎛ qB t ⎞
× sin⎜
⎟
qB ⎝ m ⎠
2 0
0
3. The plot between x-coordinate of the location of
electron as a function of time for different
frequencies v of the incident light, is
(A)
(C)
x
x
ν 2
(ν 1 > ν 2)
ν 1
ν 1
(ν 1 > ν 2)
ν 2
t
t
(B)
(D)
x
x
(ν 2 > ν 1)
ν 1 ν 2
ν 2
t
ν 1
(ν 1 > ν 2)
4. Two students were given a physics problem for
finding maximum extension of spring if blocks are
imparted velocities v 1 and v 2 when spring is
unstretched
v 1
m m v 2
1
2 1 2
By student A: m(v1 + v2
) = kx
2
2
1 2 1 2 1 2
By student B: mv1 + mv2
= kx
2 2 2
(A) Student A is correct, student B is wrong
(B) Student B is correct, student A is wrong
(C) Both are correct
(D) Both are wrong
5. Two particles having positive charges +Q and +2Q
are fixed at equal distance x from centre of an
conducting sphere having zero net charge and radius
r as shown. Initially the switch S is open. After the
switch S is closed, the net charge flowing out of
sphere is
t
XtraEdge for IIT-JEE 18 MARCH 2011