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

Answer

A, D

Explanation

given `m=160g`
`=160xx10^-3kg`
`rho=8000kg/m^3` From the diagram
`rarr mg=U+R`
`[where U=Upward thrust]`
`rarr mg=mg-vrho_wg` ltbr. `[becaue U=vrho_wg]`
`=mg-m/pxxrho_wxxg`
`=160xx10^-3(10-(10^3xx10)/8000)`
`160xx10^-3xx10(1-1/8)`
`=1.4N`