If the length of the pendulum in pendulum clock increases by 0 . 1...

$T = 2\pi\sqrt{\dfrac{l}{g}}$

$T ∝ l\dfrac{1}{2}$

$\dfrac{\DeltaT}{T} = \dfrac{1}{2}\dfrac{\Deltal}{l}$

$\dfrac{\DeltaT}{24 \times 3600} = \dfrac{1}{2} \times \dfrac{0 . 1}{100}$

$\DeltaT = \dfrac{1}{2} \times \dfrac{0 . 1}{100} \times 24 \times 3600$

$\DeltaT = 43 . 2\text{s}$