A blacksmith fixes iron ring on the rim of the wooden wheel of a bullock cart. the diameter of the rim and the iron ring are 5.24 m and 5 .; at 27 °C (Normal temperature). For that the ring is to be heated so as to fit the rim of the wheel ( Iron becomes nearly white when it is heated at temperature 1 100 °C approximately). Why the blacksmith needs to do that ? According to you what may be the possible temperature range to what the ring is to be heated ??
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Answer:
Given L
1
=5.434m;L
2
=5.443m;
T
1
=37
0
C,T
2
=?;α=1.20×10
−1
K
−1
∴α=
L
1
(T
2
−T
1
)
(L
2
−L
1
)
∴T
2
−T
1
=
αL
1
(L
2
−L
1
)
=
1.20×10
−5
×5.434
5.443−5.434
or
T
2
−37=
1.20×5.434
0.009×10
5
=
1.2×5.434
900
=138.02
∴T
2
=138.02+37=175.02
T
2
=175.02
0
C
Explanation:
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