Correct Answer - B
Here, `P_(a)=1.01xx10^(5)Nm^(-2)` ,
`P_(w)=10^(3)kg m^(-3) , g=10 ms^(-2)` ,
Area of base, `A_(1)=10 cm^(2)=10^(-3)m` ,
`h=10 cm=0.1 m` ,
Area of top, `A_(2)=30 cm^(2)=3xx10^(-3)m^(2)` ,
Volume `=1 litre =10^(-3)m^(3)`
Force exerted by the water on the bottom of glass is `F_(1)=(P_(a)+h P_(w)g)A_(1)`
`=(1.01xx10^(5)+0.1xx10^(3)xx10)xx10^(-3)`
`=102 N` (downward)