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KCET 2022 Physics Solved Paper
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© examsnet.com
Question : 3
Total: 60
A tiny spherical oil drop carrying a net charge
q
is balanced in still air, with a vertical uniform electric field of strength
81
7
π
×
10
5
V
∕
m
. When the field is switched off, the drops is observed to fall with terminal velocity
2
×
10
−
3
ms
−
1
. Here
g
=
9.8
m
∕
s
2
, Viscosity of air is
1.8
×
10
−
5
Ns
∕
m
2
and the density of oil is
900
kg
m
−
3
. The magnitude of '
q
′
is
1.6
×
10
−
19
C
3.2
×
10
−
19
C
0.8
×
10
−
19
C
8
×
10
−
19
C
Validate
Solution:
Here,
E
=
81
π
7
×
10
5
Vm
−
1
v
=
2
×
10
−
3
ms
−
1
η
=
1.8
×
10
5
Nsm
−
2
ρ
=
900
kgm
−
3
When the electric field is switched off, let the drop falls with terminal velocity
v
, then
v
=
2
r
2
(
ρ
−
σ
)
g
9
η
or
r
=
[
9
v
η
2
(
ρ
−
σ
)
g
]
1
2
∴
q
=
1
E
×
4
3
π
ρ
g
[
9
v
η
2
(
ρ
−
σ
)
g
]
=
7
81
π
×
10
5
×
4
3
×
π
×
900
×
9.8
×
[
9
×
8
×
10
−
5
×
2
×
10
−
3
2
×
900
×
9.8
]
3
2
On solving we get,
q
=
8
×
10
−
19
C
© examsnet.com
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