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Manipal Entrance Test 2015 Physics Paper
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© examsnet.com
Question : 13
Total: 50
In a nuclear reactor, undergoes fission liberating
200
MeV
of energy. The reactor has a
10
%
efficiency and produces 1000 MW power. If the reactor is to function for
10
yr
, then find the total mass of uranium required.
36.5
×
10
3
kg
36
×
10
2
kg
39.5
×
10
3
kg
38.2
×
10
3
kg
Validate
Solution:
Total energy produced by the reactor in time t = 10yr.
E
=
1000
×
10
6
×
10
×
3.15
×
10
7
J
=
3.15
×
10
7
J
∵ Efficiency
=
output energy
input energy
Input energy caused by fission
=
output energy
Efficiency
=
3.15
×
10
17
(
10
100
)
=
3.15
×
10
18
J
Energy produced by one fission of
235
U
= 200 MeV
=
200
×
1.6
×
10
−
13
J
=
3.2
×
10
−
11
J
Therefore, number of fissions required
=
total energy
energy per fission
=
3.15
×
10
18
3.2
×
10
−
11
=
9.8
×
10
28
Hence, mass of uranium required is given by
m
=
N
N
a
×
235
=
9.8
×
10
28
6.02
×
10
26
=
38.2
×
10
3
kg
© examsnet.com
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