Answer
0.13 m
Work Step by Step
By combining equations 10.5, 10.16 we can get,
$\frac{2\pi}{T}=\sqrt {\frac{g}{L}}$ ; Solving for the length L, we get,
$L=\frac{T^{2}g}{4\pi^{2}}$
Difference between the final and initial = $L_{2}-L_{1}$
$L_{2}-L_{1}=\frac{T_{2}^{2}g}{4\pi^{2}}-\frac{T_{1}^{2}g}{4\pi^{2}}=\frac{(T_{2}^{2}-T_{1}^{2})g}{4\pi^{2}}$
$L_{2}-L_{1}=[(1.45\space s)^{2}-(1.25\space s)^{2}]\frac{9.8\space m/s^{2}}{4\pi^{2}}\approx0.13\space m$
