If the length of the pendulum in pendulum clock increases by 0.1%, then the error in time per day is:

We begin with the basic formula for the time-period of a simple pendulum:

$$T = 2\pi \sqrt{\frac{L}{g}}$$

Here $$T$$ is the time-period, $$L$$ is the length of the pendulum and $$g$$ is the acceleration due to gravity.

Now the length of the pendulum is increased by $$0.1\%$$. In fractional form, an increase of $$0.1\%$$ means

$$\frac{\Delta L}{L} = 0.1\% = \frac{0.1}{100} = 0.001.$$

We want to find the corresponding fractional change in the time-period. For small changes we use the differentiation (or error-propagation) form of the above formula. Taking logarithms and differentiating, we have

$$\ln T = \ln(2\pi) + \tfrac12\ln L - \tfrac12\ln g,$$

so differentiating:

$$\frac{\Delta T}{T} = \tfrac12\,\frac{\Delta L}{L} - \tfrac12\,\frac{\Delta g}{g}.$$

Because $$g$$ is constant for the clock, $$\Delta g = 0$$. Therefore

$$\frac{\Delta T}{T} = \tfrac12\,\frac{\Delta L}{L}.$$

Substituting the given fractional change in length,

$$\frac{\Delta T}{T} = \tfrac12 \times 0.001 = 0.0005.$$

To express this as a percentage:

$$0.0005 = 0.05\%.$$

This positive value means the time-period increases, so each swing now takes longer than it should; the clock will run slow.

Next we convert this fractional error in one period to the total error accumulated in one day. A day has

$$24 \text{ hours} \times 60 \text{ min/hour} \times 60 \text{ s/min} = 86400 \text{ s}.$$

The same fraction $$0.0005$$ of this total time will be lost (the clock lags) in one day:

$$\Delta t_{\text{day}} = 0.0005 \times 86400 \text{ s}.$$

Carrying out the multiplication,

$$\Delta t_{\text{day}} = 0.0005 \times 86400 = 43.2 \text{ s}.$$

The clock therefore loses $$43.2$$ seconds per day.

Hence, the correct answer is Option A.