Answer
The fractional change in the period is $0.0002$
Work Step by Step
We can find the increase in length of the steel rod when it is heated:
$\Delta L = \alpha~\Delta T~L$
$\Delta L = (12\times 10^{-6}~K^{-1})(-25~K)~(2.5000~m)$
$\Delta L = -7.50~\times 10^{-4}~m$
We can find ratio of the new length to the original length:
$\frac{2.5000~m-7.50~\times 10^{-4}~m}{2.5000~m} = 0.9997$
Let $T$ be the original period of the pendulum. We can find an expression for the new period of the pendulum:
$T' = 2\pi\sqrt{\frac{I}{mg(L'/2)}}$
$T' = 2\pi\sqrt{\frac{\frac{1}{3}mL'^2}{mg(L'/2)}}$
$T' = 2\pi\sqrt{\frac{2L'}{3g}}$
$T' = 2\pi\sqrt{\frac{2(0.9997~L)}{3g}}$
$T' = \sqrt{0.9997}\times 2\pi\sqrt{\frac{2L}{3g}}$
$T' = \sqrt{0.9997}\times T$
$T' = 0.9998\times T$
The fractional change in the period is $1-0.9998$ which is $0.0002$
