Answer:
[tex]d_{steel} = 25.026 cm[/tex]
Explanation:
Given data:
diameter of cylinder 25 cm
[tex]T_1 = 20[/tex] degree celcius
[tex]T_2 = 150[/tex] degree celcius
from the information given in the question we have
[tex]\delta_{brass} + 25 + \delta_{steel} + d_{steel}[/tex]
[tex]25\times \alpha \times \delta T + 25 = d_{steel} \times \alpha\times \delta T + d_[steel}[/tex]
[tex]25(1+ \alpha_{brass} \times \delta T) = d_s(1+ \alpha \times \delta T)[/tex]
SOLVING FOR[/tex] d_{steel}[/tex]
[tex]d_{steel} = \frac{25(1+\alpha_{brass} \times \delta t}{1+ \alpha_{steel}\times \delta T}[/tex]
[tex]d_{steel} = \frac{25(1+ 3\times 10^{-5} (C^0)^{-1} \times (150-20)}{1+ 1.2\times 10^{-5} (C^0)^{-1} \times (150-20)}[/tex]
[tex]d_{steel} = 25.026 cm[/tex]
