A metal piece of mass 160 g lies iln equilibrium inside a glass of water figure. The ppece touches the bottom of the glass at a small number of points. If the density of the metl is 8000 kg `m^-3`, find the normal fore exerted by the bottom of the glass on the metal piece.

Options

A. `2N`
B. `8N`
C. `0.16N`
D. `1.4N`

Answer

D

Explanation

As the piece touches the glass at small number of points, water would be there at the bottom of the piece also and hecne buoyancy acts on it.
From equilibrium condition `mg=F_(B)+N`
`F_(B)=Vrho_("water")g=m/(rho_("metal"))xxrho_("water")xxg`
`=0.16/8xx10=0.2N`
So, `N=mg-F_(B)=160/1000xx10-0.2=1.4N`