A metal piece of mass `160 g` lies in equilibrium inside a glass of water. The piece touches the glass at small number of points. If the density of the metal is `8000 kg//m^(3)`, then the normal force exerted by the bottom of the glass on the metal piece is

Options

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

Answer

D

Explanation

` mg = F_B + N , F _ B = V rho _ w xx g = (m ) /( rho_("metal")) xx rho _w xx g `
` rArr = (0.16 xx 10)/(8) = 0.2 N `
` therefore N = mg - F_B = 1.6 - 0.2 = 1.4 N `