The period of oscillation of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{L}{g}}$ , where L is the length of the pendulum and g is the acceleration due to gravity. The length is measured using a meter scale which has $2000$ divisions. If the measured value L is $50\, cm,$ the accuracy in the determination of g is $1.1\%$ and the time taken for 100 oscillations is 100 seconds, what should be the resolution of the clock (in milliseconds) .