Question

Steel cylinder of diameter exactly 1 cm at 30 degree celsius is to be fitted into a hole in a steel plate the diameter of the holy 0.999 70 cm at 30 degree Celsius to what temperature should be plate be heated given Alpha Steel equals to 1.1 into 10 to the power minus 5 degree Celsius to the power minus one​

Answer 1

Answer:

Change in diameter with change in temperature is given by l

0

αΔθ.

Hence, Δθ=

αl

0

Δl

=

1.1×10

−5

×0.9997

0.0003

=27.3

∘

C

Hence final temperature=57.3

∘

C

Explanation:

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Answer 2

Answer:

57.3°C

Explanation:

for exactly fitting a cylinder into hole the diameter of the cylinder and hole should be equal.But in this case ,the diameter is different.We have to increase the diameter

∆d is increase in diameter

∆d=1-0.99970

∆d=0.0003

According to thermal expansion

∆d= d×alpha×∆T

here,∆d=0.0003cm

alpha=1.1×10^-5 C^-1

∆T=∆d/d×alpha

=0.0003/1×1.1×10^-5

=27.3 °C

to increase the diameter we have to heat plate at

T+∆T

30+27.3

57.3°C