area of aperture \(=f^{2},\) area \(\propto \frac{1}{\text { exposure time }}\)

New aperture \(=1.4 \mathrm{f} \quad\) area \(=1.96 \mathrm{f}^{2}\)

Ratio of areas \(=1.96 .\) since area of aperture has increased thus exposure time will decrease in same proportion, i.e.,

new time \(=\frac{1}{(60 / 196)}=\frac{1}{31}\) \(sec.\)