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JEE Advanced 2020 Paper 2
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© examsnet.com
Question : 11
Total: 54
Two identical non-conducting solid spheres of same mass and charge are suspended in air from a common point by two non-conducting, massless strings of same length. At equilibrium, the angle between the strings is
α
.
The spheres are now immersed in a dielectric liquid of density
800
k
g
m
−
3
and dielectric constant
21
.
If the angle between the strings remains the same after the immersion, then
[JEE Adv 2020 P2]
electric force between the spheres remains unchanged
electric force between the spheres reduces
mass density of the spheres is 840 kg m
−
3
the tension in the strings holding the spheres remains unchanged
Validate
Solution:
Net force on a sphere will decrease, but force between spheres will be same.
When in air
tan
α
2
=
F
e
m
g
&
T
=
√
F
e
2
+
(
m
g
)
2
When immersed in liquid
tan
α
2
=
F
e
k
m
g
−
B
&
T
=
√
(
F
e
k
)
2
+
(
m
g
−
B
)
2
⇒
m
g
=
k
m
g
(
1
−
ρ
l
ρ
b
)
ρ
b
=
k
ρ
l
k
−
1
=
21
×
800
21
−
1
=
840
k
g
∕
m
3
© examsnet.com
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