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If the length of the pendulum in pendulum clock increases by 0.1%, then the error in time per day is:
Explanation
Time period, $T = 2\pi\sqrt{\ell/g} \therefore \Delta T/T = (1/2)(\Delta\ell/\ell)$ Error in time period in one day $\Rightarrow \Delta T = (1/2) \times (0.1/100) \times 24 \times 3600 = (1/2) \times (0.1/100) \times 86400$ $\Rightarrow \Delta T = 43.2s$