સળિયાની લંબાઈમાં તાપમાનના વધારા સાથે થતો ફેરફાર

$ \Delta l = \alpha l \Delta \theta$

$\therefore$ $\Delta l / l = \alpha \Delta \theta$

$\Delta { {l}}\,\, = \,\,\alpha { {l}}\,\,\Delta \theta \,\,\,\,\therefore \,\,\frac{{\Delta { {l}}}}{{{l}}}\,\, = \,\,\alpha \Delta \theta $

ગજિયા લોલક્નો આવર્તકાળ $T\,\, = \,\,2\pi \sqrt {\frac{{{l}}}{g}} \,\,\,$

$\,\therefore \,\,T\, \propto \,{{{l}}^{\frac{1}{2}}}\,\,\therefore \,\,T\,\, = \,\,k{{{l}}^{\frac{1}{2}}}\,\,\,\,\therefore \Delta T\,\, = \,\,k\left( {\frac{1}{2}} \right){{{l}}^{ – \frac{1}{2}}}\,\,\Delta {{l}}$

$\therefore \,\,\,\,\frac{{\Delta T}}{T}\,\, = \,\,\frac{1}{2}\,\frac{{\Delta {{l}}}}{{ {l}}}\,\,\,\, = \,\,\frac{1}{2}\,(\alpha \Delta \theta )\,\, = \,\,\frac{1}{2}\,(2\,\, \times \,\,{10^{ – 6}})\,(10)$

$\therefore \,\,\frac{{\Delta T}}{T}\,\% \,\, = \,\,1\,\, \times \,\,{10^{ – 5}}\,\, \times \,\,100\,\,\, = \,\,1\,\, \times \,\,{10^{ – 3}}\,\% $