Power of a convex lens is + 5D ($\mu_g= 1.5$). When this lens is immersed in a liquid of refractive index m, it acts like a divergent lens of focal length 100 cm. Then refractive index of the liquid will be 

$\frac{5}{3}$

$\text{Focal length of lens in air is }P_a =5 =  \frac{1}{f_a} = (\mu_g - 1)(\frac{1}{R_1} - \frac{1}{R_2})$

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

$\frac{P_a}{P_l} =\frac{5}{-1}=  \frac{1.5 - 1}{\frac{1.5}{\mu_l} - 1} $

$ \frac{1.5}{\mu_l} - 1= -0.1$

$ \frac{1.5}{\mu_l} = 0.9 $
$\Rightarrow \mu_l = \frac{5}{3}$