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KEAM 2021 Physics and Chemistry Solved Paper
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© examsnet.com
Question : 29
Total: 120
10
18
fissions per second is required for producing power of
300
M
W
in a nuclear power station. To increase the power output to
360
M
W
the additional number of fissions required per second is
2
×
10
18
5
×
10
18
5
×
10
17
6
×
10
17
2
×
10
17
None of the above
Validate
Solution:
Given,
n
1
=
10
18
fissions per second
Power,
P
1
=
300
M
W
=
300
×
10
6
W
P
2
=
360
M
W
=
360
×
10
6
W
As, power
=
energy
time
number of fission
=
per second
(
n
)
×
energy per fission
time
If energy per fission and time is constant, then power
∝
n
⇒
P
1
P
2
=
n
1
n
2
⇒
n
2
=
n
1
×
P
2
P
1
=
10
18
×
360
×
10
6
300
×
10
6
=
1.2
×
10
18
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