\(\frac{\Delta Q }{\Delta t }=- K \left( T _{ avg }- T _0\right)\)
\((i)\) \(\frac{ ms \times 12}{2}=- K \left(\frac{98+86}{2}-22\right)\)
\(6=-\frac{ K }{ ms }\left[\frac{98+86}{2}-22\right]\)
\(6=-\frac{ K }{ ms }[70]..........(i)\)
\((ii)\) \(\frac{ ms \times 6}{\Delta t }=- K \left(\frac{75+69}{2}-22\right)\) \(\frac{6}{\Delta t }=-\frac{ K }{ ms }(50) \ldots \ldots \ldots \ldots\) (ii)
\((ii)\) \(\div\) \((i)\)
\(\frac{6}{\Delta t (6)}=\frac{50}{70}\)
\(\Delta t =\frac{7}{5}=1.4\,min\)
કારણ : બે પાતળા ધાબળા વચ્ચેનું હવાનું પડને લીધે જાડાઈ વધે છે.