A uniform rod is 4m long and weighs 10kg. If it is supported on a knife edge at one metre from the end, what weight placed at that end keeps the rod horizontal ?

From the question,

Linear density of the rod, 

$\mathrm{\mu }=\frac{\mathrm{M}}{\mathrm{L}}=\frac{10}{4}=2.5\mathrm{kg}/\mathrm{m}$

now weight,

${\mathrm{W}}_{\mathrm{AB}}=\mathrm{\mu }\times 1=2.5\times 1=2.5\mathrm{Kg}$

${\mathrm{W}}_{\mathrm{BC}}=\mathrm{\mu }\times 3=2.5\times 3=7.5\mathrm{Kg}$

knife's centre of mass is at 0.5 metres.

1.5 metres is the centre of the BC from the knife.

Since AB's length is short, force must be applied at point A to balance the rod.

Let the force be F, 

$\mathrm{F}\times {\mathrm{L}}_{\mathrm{AB}}={\mathrm{W}}_{\mathrm{BC}}\times 1.5 - {\mathrm{W}}_{\mathrm{AB}}\times 0.5$

$\mathrm{F}=\frac{{\mathrm{W}}_{\mathrm{BC}}\times 1.5-{\mathrm{W}}_{\mathrm{AB}}\times 0.5}{{\mathrm{L}}_{\mathrm{AB}}}$

$\mathrm{F}=\frac{7.5\times 1.5+2.5\times 0.5}{1}=10\mathrm{N}$

Hence the correct answer is 10N.