Tardigrade
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Tardigrade
Question
Physics
If temperature of an object is 140 ° F , then its temperature in centigrade is
Q. If temperature of an object is
140
∘
F
, then its temperature in centigrade is
2716
228
NTA Abhyas
NTA Abhyas 2020
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A
10
5
∘
C
0%
B
3
2
∘
C
0%
C
14
0
∘
C
25%
D
6
0
∘
C
75%
Solution:
Relation between
∘
C
and
∘
F
,
5
C
=
9
F
−
32
⇒
5
C
=
9
140
−
32
⇒
C
=
6
0
∘
C