Question

Part A A cube 5.5 cm on each side is made of a metal alloy. After you drill a cylindrical hole 2.5 cm in diameter all the way through and perpendicular to one face, you find that the cube weighs 6.70 N What is the density of this metal? Express your answer with the appropriate units. ρ.1 Value ! Units Submit My Answers Give Up Part B What did the cube weigh before you drilled the hole in it? Express your answer with the appropriate units. wValue Units Submit My Answers Give Up

Answer 1

Data Given

Volume of the cube $V = L^{3}= (5.5 \times 10^{-2}m)^{3}= 1.66375\times 10^{-4}m^{3}$

Volume of the hole as hole will be cylinder of length L $V' =\pi r^{2} L= \pi \times \left ( \frac{2.5 \times 10^{-2}m}{2} \right )\times (5.5 \times 10^{-2}m)= 0.26998\times 10^{-4}m^{3}$

Now Effective Volume

$V_{eff}= V-V' =1.66375\times 10^{-4}m^{3}- 0.26998\times 10^{-4}m^{3}= 1.39377\times 10^{-4}m^{3}$

Mass of the object

$m = \frac{W}{g}= \frac{6.70 N}{9.8 m/s^{2}}= 0.68674 kg$

Part A) density of the object

$\rho = \frac{m}{V_{eff}}= \frac{0.68674 kg}{1.39377\times 10^{-4}m^{3}}=4927.2 kg/m^{3}$

Part B) mass of the cube before drilling the hole

$m = \rho V = 4927.2 kg/m^{3}\times1.66375\times 10^{-4}m^{3} = 0.8198 kg$

So weight will be

$w = mg = 0.8198 kg\times 9.8 m/s^{2}= 8.034 N$