
From 1st lens
1/v + 1/6 = 1/24
1/v = 1/24 - 1/6 = -1/8
v = -8 cm
From 2nd lens
1/v + 1/18 = 1/9
1/v = 1/9 - 1/18 = 1/18
v = 18 cm
So distance between object and its image:
6 + 10 + 18 = 34 cm
In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 
Method used for separation of mixture of products (B and C) obtained in the following reaction is: 