We have a bullet that weights 80.0g in air. The bullet is dropped into a 100mL graduated cylinder with 50.0mL of water. Now the bullet (in water) is weighed to be 65.2g. What is the bullet density? *1.0 g/mL - water density
Archimedes' principle indicates that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to the weight of the fluid that the body displaces and it acts in the upward direction at the centre of mass of the displaced fluid. https://en.wikipedia.org/wiki/Archimedes%27_principle 80.0g in air - 65.2g in water = mass of the displaced fluid g since water density is 1g/mL we can convert the mass of water to volume of water, this is going to be equal to the volume of the bullet. Now you have the volume of the bullet and the mass of the bullet was given in the problem, you can easily calculate the density of the bullet. denstity= mass/volume
bullet's weight in water = bullet's weight in air - weight of the water displaced by bullet.
\[65.2=80-\left( d_w \times \frac{65.2}{d_b} \right)\]d_w= density of water d_b=density of bullet
so this? \[\large{\frac{Volume_{water}-Bullet_{water}}{Bullet_{air}}}\]
Wait, I got that upside down... but is it right otherwise @jiteshmeghwal9
density of bullet = 4.405 gm/ml
You have two methods to answer this. 1. You can use density formula : \[\rho = \frac{ m }{ V }\] We can spell out this formula become : |dw:1473587076563:dw| 2. You can answer this quickly using comparative equation : \[ \frac{ \rho b }{ \rho a }\ = \frac{ Wu }{ Fa }\] with, pb = object density pa = water density Wb = object wight in air Fa = the buoyant force -------------------------------------------- So what we need here is : Wb = m x g = 80 x 10 = 800 g pa = 1 g/mL Fa = 800 - (65.2 x 10) = 800 - 652 = 148 g Then plug them all in the first equation or second equation that I wrote above. You will get answer, density of bullet = 5.4054 g/mL Check it out if my answer is true or not :)
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