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In an aluminium sheet there is a hole diameter 1 m and is horizontally mounted on a stand. On this hole an iron sphere of diameter 1.004 m is resting. Initial temperature of this system is 25^(@)C. Find at what temperature, the iron sphere will fall down through the hole in sheet. The coefficients of linear expansion for aluminium and iron are 2.4xx10^(-5) and 1.2xx10^(-5) respectively.

Answer» <html><body><p></p>Solution :A value of coefficient of <a href="https://interviewquestions.tuteehub.com/tag/linear-1074458" style="font-weight:bold;" target="_blank" title="Click to know more about LINEAR">LINEAR</a> expansion for aluminium is more than that for <a href="https://interviewquestions.tuteehub.com/tag/iron-1051344" style="font-weight:bold;" target="_blank" title="Click to know more about IRON">IRON</a>, it expands faster than iron. So at some higher temperature when diameter of hole will exactlky become equal to that of iron sphere, the sphere will pass throughthe hole. Let it happen at some higher temperature T. Thus we have at this temperature T, <br/>` ("diameter of hole")_(Al) = ("diameter of sphere")_("iron")` <br/> `1[1+alpha_(Al) (T-<a href="https://interviewquestions.tuteehub.com/tag/25-296426" style="font-weight:bold;" target="_blank" title="Click to know more about 25">25</a>)]=1.004[1+alpha_("iron") (T-25)]` <br/> `alpha_(Al)(T-25)=<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>.004+1.004 alpha_("iron") (T-25)` <br/> `or T=((0.004)/(alpha_(Al) -1.004alpha_("iron"))+25)""^(@)C or T=(0.004)/(1xx2.4xx10^(-5)-1.004xx1.2xx10^(-5))+25 or T =359.7^(0)C`</body></html>


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