Question

A cube 4.8 cm on each side is made of a metal alloy. After you drill a cylindrical hole 2.2 cm in diameter all the way through and perpendicular to one face, you find that the cube weighs 7.35 N .

What is the density of this metal?

What did the cube weigh before you drilled the hole in it?

Answer 1

let L = 4.8 cm = 0.048 m
d = 2.2 cm
r = d/2 = 2.2/2 = 1.1 cm = 0.011 m

A) Volume of the cube = L^3

= 0.048^3

= 1.10592*10^-4 m^3

volume of the cyllinder = pi*r^2*L

= pi*0.011^2*0.048

= 1.8246*10^-5 m^3

effective volume of the metal = 1.10592*10^-4 - 1.8246*10^-5

= 9.2346*10^-5 m^3

mass of remaining metal alloy, m = W/g

= 7.35/9.8

= 0.75 kg

density of metal alloy, rho = mass/volume

= 0.75/(9.2346*10^-5)

= 8.12*10^3 kg/m^3 <<<<<<<<<<---------------Answer

B) weight of the cube before drilled, W = M*g

= density*Volume*g

= rho*L^3*g

= 8.12*10^3*1.10592*10^-4*9.8

= 8.80 N <<<<<<<<<<---------------Answer