IIT-JEE 2010 - Career Point

8
Solution
Set # 7
Physics Challenging Problems
Questions were Published in November Issue
1. As the resistances of voltmeters in upper branch are
R, R/2, R/4 ......................
the equivalent circuit is as shown below
3. From current division formula we can conclude that
current in upper and lower branch are in the ratio of
1 : 2.
a
R
R/2 R/4
V
Lower Branch
upper Branch
...................
the resistance of upper branch is
= R + R/2 + R/4 + ............. up to infinite
⎛ 1 1 ⎞
= R ⎜1
+ + + ..... ⎟
⎝ 2 4 ⎠
⎛ 1 ⎞
= R ⎜ ⎟ = 2R
⎝1−1/
2 ⎠
further the equivalent circuit is
a
R
V
upper branch
Lower Branch
the resistance of voltmeter V should be 2R so that
current in upper and lower branch is same.
2. Entire upper branch is having the resistance of 2R
and voltmeter V 1 is having the resistance of R so we
can conclude that equivalent resistance of all the
voltmeters in upper branch except V 1 is R and the
upper branch is as follows:
a V 1 V 2 V 3
.....up to infinite b
a
i
R
C
V 1 =X V 2 =Y
As reading of voltmeter V 1 is X = i.R
sum of the readings of voltmeters is Y = i.R
Except V 1 in upper branch
So,
X = Y
R
b
b
b
4.
a
i
2i
R
C
R′ = R
voltmeter V
Reading of voltmeter V 1 is i.R
Reading of voltmeter V is (2i.)R
So V = 2V 1
a.
x.
A
b.
d x.
l = length of rod = b – a
l.
B
charge on element of length d x is d q
d q = λd x as λ = 3x
d q = 3xd x
Equivalent current due to element of length d x
ω
d i = ω.d q =
2 (3xd x) π
ω
Total equivalent current i =
∫di
=
∫
(3xd x )
2 π
3ω
=
2 π
⎡
2
x ⎤
⎢ ⎥
⎣ 2
⎦
b
a
3ω
=
2 π
⎛ b
⎜
⎝
3 ω
= (b 2 – a 2 )
4π
Option A is correct
(B) Equivalent current
2
R
− a
2
2
⎞
⎟
⎠
b
a
b
3 ω
= . 2 2 π
(b2 – a 2 )
XtraEdge for IIT-JEE 20 DECEMBER 2009