If the length of the pendulum in pendulum clock increases

$\mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}}}$

$\frac{\Delta \mathrm{T}}{\mathrm{T}}=\frac{1}{2} \frac{\Delta \ell}{\ell}$

$\Delta \mathrm{T}=\frac{1}{2} \times \frac{0.1}{100} \times 24 \times 3600$

$\Delta \mathrm{T}=43.2$

Ans. 3