Solution:

It has only one stable state. When a triggering pulse of less than $1 / 3 \mathrm{vcc}$ is applied to trigger pin the monostable.

The multivibrator will change its stale from low to high. The external capacitor connected to pin 6 will start charging exponentially through the RA resistor.

The external capacitor connected to pin 6 will start charging exponentially through the RA resistor.

When the voltage across the capacitor reaches $[2 / 3 \vee c]$, comparator (1), 0/p will become high which will make the multivibrator low,

This means a monostable multivibrator will generate a small pulse whose width will be given by,

TON $=1.1$ RA.C

$ \begin{aligned}\\ \text { Design 1 Ton } &=10 \mathrm{sec}[\text { given }] \\\\ \text { ToN } &=1.1 R A \cdot C \\\\ \text { Let } C &=0.1 \mathrm{msec} \\\\ \therefore 10 \mathrm{sec} &=(1.1) \mathrm{RA}(0.1 \mathrm{msec}) \\\\ \therefore R A &=90.9 \mathrm{k} \Omega\\ \end{aligned}\\ $