Focal length of a convex lens of refractive index 1.5 is 2 cm. Focal length of lens when immersed in a liquid of refractive index 1.25 will be 

5 cm               

$\text{Focal length of lens in air is }\frac{1}{f_a} = (\mu_g - 1)(\frac{1}{R_1} - \frac{1}{R_2})$

$\text{Focal length of lens in liquid is }\frac{1}{f_l} = (\frac{\mu_g}{\mu_l} - 1)(\frac{1}{R_1} - \frac{1}{R_2})$

$\frac{f_l}{f_a} = \frac{\mu_g - 1}{\frac{\mu_g}{\mu_l} - 1} = \frac{1.5 - 1}{\frac{1.5}{1.25} - 1} = \frac{0.5}{0.2} = 2.5$

$\Rightarrow f_l = 5cm$