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A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0^(@)C. What is the change in the diameter of the hole when the sheet is heated to 227^(@)C ? Coefficient of linear expansion of copper =1.70xx10^(-5)K^(-1).

Answer» <html><body><p></p>Solution :But area of hole is `A_(1)` at <a href="https://interviewquestions.tuteehub.com/tag/temperature-11887" style="font-weight:bold;" target="_blank" title="Click to know more about TEMPERATURE">TEMPERATURE</a> <br/> `T_(1)=27+273=300K` <br/> `:.A_(1)=(pid_(1)^(2))/(4)=(pixx(4.24)^(2))/(4)` <br/> `:.A_(1)=4.4944pi" cm"^(2)` <br/> Area of hole becomes `A_(2)` at temperature `T_(2)=227+273=500K`. <br/> `:.A_(2)=A_(1)(1+beta Delta T)` <br/> `(pid_(2)^(2))/(4)=4.4944pi[1+2alpha(T_(2)-T_(1))]" where "beta=2alpha` <br/> `:.d_(2)^(2)=4xx4.4944[1+2xx1.7xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>)<a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a>(500-300)]` <br/> `:.d_(2)^(2)=4xx4.944[1+2xx1.7xx10^(-5)xx200]` <br/> `:.d_(2)^(2)=4xx4.944[1+0.0068]` <br/> `:.d_(2)^(2)=4xx4.944xx1.0068` <br/> `d_(2)^(2)=18.0998~~18` <br/> `:.d_(2)=sqrt(18.1)=4.2544" cm"` <br/> `:.` Change in diameter of hole `Deltad=d_(2)-d_(1)` <br/> `=4.2544-4.2400` <br/> `=0.0144" cm"` <br/> `=1.44xx10^(-2)" cm"`</body></html>


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