A tank of base area `4 m^2` is initially filled with water up to height 2m. An object of uniform cross-section `2m^2` and height 1m is now suspended by wire into the tank, keeping distance between base of tank and that of object 1m. Density of the object is 2000kg/`m^3`. Take atmospheric pressure `1xx10^5 N//m^2 , g = 10m//s^2`.

The buyont force on the object is

Options

A. `0.1xx10^(5) N`
B. `0.2xx10^(5) N`
C. `0.3xx10^(5)N`
D. `0.4xx10^(5)N`

Explanation

(i) By conservation of volume
4xh=4x2+2x1=10
h=2.5 m
Pressure at top of the object
`=P_(0)+0.5xx1000xx10`
=1.05x `10^(5) N//m^(2)`
`F=P_(1)A`
`=1.05xx10^(5)xx2=2.1xx10^(5) N`
By F.B.D. `T+P_(2)A=mg=P_(1)A`
`T=mg+(P_(2)+P_(1))A`
`=mg-(P_(2)-P_(1))A`
`=2xx2000xx10-(.2xx1^(5)) N`
`=.4xx10^(5)-0.2xx10^(5)=0.2xx10^(5) N`
`F_(b)=V.rho_(w)g`
`=2xx1000xx10=0.2xx10^(5)N`
It is also equal to net contact force by the liquid `=P_(2)A-P_(1)A`
`=0.2xx10^(5) N`
Note: Net contact force and buyont force are same.