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Tardigrade
Question
Physics
The resistance of a wire is 5 Ω at 50 ° C and 6 Ω at 100 ° C. The resistance of the wire at 0 ° C will be
Q. The resistance of a wire is
5Ω
at
50
∘
C
and
6
Ω
at
100
∘
C
. The resistance of the wire at
0
∘
C
will be
588
141
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A
3Ω
B
2Ω
C
1Ω
D
4Ω
Solution:
R
t
​
=
R
0
​
(
1
+
α
t
)
Where
R
t
​
=
resistance of a wire at
t
∘
C
,
R
0
​
=
resistance of a wire at
0
∘
C
,
α
=
temperature coefficient of resistance.
∴
R
50
​
=
R
0
​
[
1
+
α
50
]
and
R
100
​
=
R
0
​
[
1
+
α
100
]
or
R
50
​
−
R
0
​
=
R
0
​
α
(
50
)
R
100
​
−
R
0
​
=
R
0
​
α
(
100
)
Dividing (i) by (ii), we get
6
−
R
0
​
5
−
R
0
​
​
=
2
1
​
or
10
−
2
R
0
​
=
6
−
R
0
​
or
R
0
​
=
4Ω