# Measurement of Bucket - Physics Essay Example

Measurement of Bucket

Objective: To carry out measurements of different attributes of a bucket

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Introduction:

Measurement is essential for any scientific study. Any scientific theory is not acceptable until and unless it is verifiable by appropriate measurement. All physical quantities are measurable. In SI system, there are seven basic physical quantities – length, mass, time, temperature, electrical current, temperature and amount of substance and rest all can be derived by suitably combining one or more of these basic quantities. To measure any physical quantity we need its unit and a number which tells us how many times the measured value is with respect to the unit. For example for length we have meter as SI unit. We also need an instrument to measure a physical quantity. There is a scale associated with any measuring instrument and this scale has a least count, which is nothing but the minimum division on the scale. Because the least count is a finite value, therefore, there is an error associated with any measurement, which is nothing but plus and minus one half of the least count. Besides, we carry out various mathematical operations like addition, subtraction, multiplication, division etc. on the measured values and accordingly the errors also get combined. In this exercise we will carry out measurement of different attributes like weight, volume, length, area etc. of a bucket. A bucket is a truncated conical container, which is used for storing a liquid – mainly water. Its volume V is given by following formula (Derivation of all these formulae is presented in Appendix 1).

V = ………………(1)

Where, r and R are bottom and top inner radius respectively, h is height of the bucket and H is given by …………….(2)

Surface area of a bucket on the inner side is given by S = ……(3)

Where, …….(4) and ……..(5)

Materials: A bucket, nonstretchable strings, measuring scale, measuring tube, a spring balance

Procedure:

The string was used for the measurement of the diameter of the top and the bottom of the bucket. For this purpose a point was marked on the string and it was kept at one point on the top circumference of the bucket and the string was pulled straight to the opposite side on the circumference. This point was marked and the distance between these two points was measured with a measuring scale and recorded as top diameter of the bucket. The experiment was repeated thrice and the average value was taken. Similarly bottom diameter of the bucket was also measured and recorded. Volume of the bucket was calculated using formula 1. Volume of the bucket was also measured by pouring water into it by the measuring tube until the water just starts flowing out of the bucket. The scale is having least count of cm.

Weight of the bucket was measured by using a spring balance having least count of 0.05 kg-wt.

Role of Approximation in Measurement:

Because, there is a finite least count associated with any measuring instrument, therefore, one has to resort to approximation, while measuring any physical quantity. Here we will discuss measurement of the top diameter of the bucket. The measurement is done using a measuring scale. The measuring scale one meter long, it has 100 main divisions at 1 cm distance and each main division has 10 subdivisions, each subdivision is 1 mm or 0.1 cm long. While measuring the diameter of the bucket at the top, one end was kept at ‘0’ mark on the measuring scale and the other end was lying between the marks 35.9 and 36.0 on the scale. Now the question is what value should be assigned to the top diameter of the bucket. Here approximation comes to our help. The gap between 25.9 and 36.0 is mentally divided in 10 equal parts and then it is decided as to which of these marks (in our mind) coincides the other end. As it appeared to coincide to the 5th mark so the measured value of the top diameter was taken as 35.95 mm. Because, we are not sure about the accuracy of the last digit so the last digit in this measurement is termed as insignificant digit. This example illustrates use of approximation in the measurement. The same procedure was followed for all the measurements.

Data:

Top diameter = 35.95 cm, therefore, top radius R= 17.48 cm

Bottom Diameter = 12.07 cm therefore, Bottom radius r = 6.04 cm

Height of the bucket h = 30.52 cm

Volume of the bucket as measured by poring water into it = 14.255 liters

Weight of the bucket = 5.450 kg wt

Calculations:

Using formula 2, we get H = 46.67 cm; Using formula 1, we get V = 14.308 liters

Using formula 4, we get L = 49.84 cm; Using formula 5, we get l = 32.62 cm

Using formula 4, we get S = 2523.54 cm2

Results:

We get following results for the different attributes of the bucket as a result of the measurements.

Weight of the bucket W = 5.450 kg-wt

Volume of the bucket V = 14.308 liters (By length measurements)

= 14.255 liters (Using water and measuring tube)

Surface area of the bucket S = 2.2523.54 cm2

D

L R X M

l h H

L d

N r Y P

O

Figure 1: Cross-section (schematic) of a bucket

Appendix 1: Volume and surface area of a bucket:

Figure 1 shows a cross-section LNPM of a bucket through axis of the bucket. The sides of the cross-section LN and MP has been extended and they meet at point O to form two cones LOM and NOP. It is very clear from the picture that the bucket LNPM is formed by truncating a small cone (here NOP) from a large cone (here LOM).

Now to get volume of the bucket, one needs to subtract volume of cone NOP from that of cone LOM and the same is true for the surface area of the bucket.

Volume of cone LOM = , where R is radius of the bucket at the top and H is height of the cone LOM (figure 1).

Volume of cone NOP = , where r is the bottom radius of the bucket and h is height of the bucket, there H-h is height of the small cone NOP.

Therefore, volume of the bucket LNPM = vol(LOM) – vol(NOP) =

Here, we can measure r, R and h from the bucket, but we need to know H and this is found by using similar triangles LOX and NOY,

From these similar triangles we get,

Similarly,

Area of the curved surface of a cone is given by A = prl , where r is radius of the cone and l is slant length of the cone = sqrt (r2 + h2)

Therefore, area of the curved surface of cone LOM = pRL

And area of the curved surface of cone NOP = pr(L-l)

Besides, the bucket has a circular bottom, area of which is pr2

Surface area of the bucket LNPM

= area of bottom of the bucket + surface area(LOM) – surface area(NOP)

=

Again to calculate L, the slant length of the cone LOM, we need to compare similar triangles LOX and NOY, which gives,

, here l is slant length of the bucket, r is bottom radius of the bucket and R is top radius of the bucket. All these are measurable from the given bucket.