In the beginning of this experiment I calculated the volume of both the given objects by two different methods each. I decided to first take the spherical object. The amount of water displaced by the spherical object is equal to the volume of said object. The rise in the level of water came out to be 9cm3. The error in this case can be calculated as + 1 (because the LC of the cylinder is 0.5 mm). Next we calculate the volume using the formula to calculate the volume of a sphere i.e.Volume of a sphere = 4/3 πr3. The radius of the spherical object is half the diameter, which I found using the Vernier Calipers. The final volume is 9.525 cm3. The error of the volume is thrice the percentage error of the radius/diameter. This is because the radius is cubed; hence the error has to be multiplied by 3. The final value of the error of the spherical object is +0.108 cm. Thus V1 = 10 cm3 + 1 cm and V2= 9.525cm3 + 0.108cm.
We use similar methods for calculating the volume of the cylindrical object. First, by calculating the amount of water held in the object (with the help of a measuring cylinder). The volume = 310 cm3 + 5cm3(as the least count of the cylinder is 2.5 cm). Now we calculate the volume by using the formula for volume of a cylinderVolume = πr2h. The final value I got was305.614 cm3. The error in this case is twice the percentage error of the radius (as the radius was squared). This gives us a resultant error of + 0.096 cm3. Thus we get the final volume of the cylinder to be =305.61 cm3 + 0.096 cm3.
My data and calculations are as follows-
FOR THE SPHERICAL OBJECT:-
Therefore, the final volume is (12+11+12+10)/ 4
= 11.25 cm3.
The error in this case is + 2 cm3.
So volume for the first case is 11.25cm3 + 2 cm3
Now with the diameter we can calculate the volume of the sphere using the formula for volume of a sphere:
V=4/3 π r3
V = 4/3 x πx (2.46/2)3
Volume = 12.9cm3
The error is equal to thrice the percentage error of the radius as it is cubed. The percentage error is hence calculated as the percentage of the error of the true value. %Error = (0.01/2.43)(100)
Thus, Error = 3(%error in diameter)
= 3(0.41)
= 1.21% of volume
So volume with the second method is 12.9ml +_1.21 %.
FOR THE CYLINDRICAL OBJECT:
Thus, the final volume (average) is (311+310+310+309)/ 4 = 310 cm3.
The error is = + 5 cm.
Hence the Volume = 310cm3 +5cm.
For volume of the cylinder
We calculate the volume by the formula V = πr2h
Thus, Volume = πx (9.5/2)2 x 6.4
= 305.614 cm3
The error is twice the percent error of the radius. (Since the radius is squared)
Error = 2 (% error of radius)
Error = 2 x (0.01/6.4) x 100
= 0.03125 % of the volume
Thus, error = (0.03125/100) x 305.614
= 0.0955 cm3
Volume of Cylinder = 305.60 +0.096 cm3
I noticed that there is a big difference between both the methods (with the sphere and the cylinder). E.g.: - volume of sphere with water displacement is 10 +1 cm2, but when we use the formula it comes out to be 9.525 + 0.108 cm2. This shows us that with the correct precautions and methods, we can get an answer that has the least possible error and is therefore the most accurate. My expectation regarding the difference was hence right. I found this to be an engaging and eye-opening experiment and I enjoyed doing it. I look forward to more experiments of this type in the future