a
Here mass of the liquid, \(m = m _2- m _1=17.3-10.1=7.2\,gm\)

So the relative error, \(\frac{\Delta m }{ m }=\pm\left[\frac{\Delta m _1}{ m _1}+\frac{\Delta m _2}{ m _2}\right]\) or \(\Delta m =\pm\left[\frac{\Delta m _1}{ m _1}+\frac{\Delta m _2}{ m _2}\right] \times m =\pm[0.1 / 10.1+0.1 / 17.3] \times 7.2=\pm 0.1\,gm\) Thus mass of liquid with possible accuracy \(=(7.2 \pm 0.1)\,gm\)