Answer
(a) increase
(b) $1.710mm$
(c) $1.955s$ before; $1.957s$ after
Work Step by Step
(a) We know that $T=2\pi \sqrt{\frac{L}{g}}$ and $\Delta L=\alpha L_{\circ}\Delta T$. Theses two equations show that the length of the pendulum increases as the temperature increases and time period is directly proportional to the length; hence the period of the pendulum increases as well.
(b) We know that
$\Delta L=\alpha L_{\circ}\Delta T$
We plug in the known values to obtain:
$\Delta L=(12.0\times 10^{-6}K^{-1})(0.9500m)(150C^{\circ})$
$\implies \Delta L=0.00171m=1.710mm$
(c) We can find the time period of the pendulum before the temperature increases
$T_{before}=2\pi \sqrt{\frac{L}{g}}$
We plug in the known values to obtain:
$T_{before}=2\pi\sqrt{\frac{0.9500m}{9.81m/s^2}}$
$\implies T_{before}=1.955s$
and the period of the pendulum after the temperature increases
$T_{after}=2\pi \sqrt{\frac{L}{g}}$
We plug in the known values to obtain:
$T_{after}=2\pi \sqrt{\frac{0.95171m}{9.810m/s^2}}$
$\implies T_{after}=1.957s$
