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AIEEE 2011 Solved Paper
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© examsnet.com
Question : 49
Total: 91
Two identical charged spheres suspended from a common point by two massless strings of length
l
are initially a distance
d
(
d
<
<
1
)
apart because of their mutual repulsion. The charge begins to leak from both the spheres at a constant rate. As a result charges approach each other with a velocity
v
. Then as a function of distance
x
between them,
v
∝
x
−
1
y
∝
x
1
/
2
v
∝
x
v
∝
x
−
1
/
2
Validate
Solution:
At any instant
T
cos
θ
=
m
g
.
.
.
(
i
)
T
s
i
n
θ
=
F
e
.
.
.
(
i
i
)
⇒
s
i
n
θ
cos
θ
=
F
e
m
g
⇒
F
e
=
m
g
tan
θ
⇒
k
q
2
x
2
=
m
g
tan
θ
⇒
q
2
∝
x
2
tan
θ
s
i
n
θ
=
x
2
l
For
s
m
a
l
l
θ
,
s
i
n
θ
≈
tan
θ
∴
q
2
∝
x
3
⇒
q
d
q
d
t
∝
x
2
d
x
d
t
∴
d
q
d
t
=
const.
∴
q
∝
x
2
⋅
v
⇒
x
3
∕
2
α
x
2
⋅
v
[
.
as
q
2
∝
x
3
]
⇒
v
∝
x
−
1
∕
2
© examsnet.com
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