Question

Measurement and Error Analysis Data Sheet Table 1. Measure the dimensions of the cylinder using the vernier caliper Zero correction:Indication error O-, Unit: Trial Outer diameter Inner diameter Height 2 ?1008-26.oel n? toon :n.07| so t o.og: O.? 4 5, , 35.471 Di Measure the diameter of the ball using the micrometer caliper 0.004 Table 2. Zero correction:Indication error:Unit: m Trial 0-e2 Diameter D-3.2 4-24 4.24 14.26 A-75 4. 26 -1 X16 o-ol 4 0.01

Answer 1

Given

$\bar D_1=35.472mm$ , $\bar D_2=22.038mm$ , $\bar H=50.09mm$

Assuming diameter and height, both are measured with vernier calliper.

So we get

$\Delta D_1= \Delta D_2$ = indication error of the vernier calliper= 0.02mm

$\Delta H$ = indication error of the vernier calliper= 0.02mm

Volume of the cylinder given by

$V=\frac{\pi}{4}(\bar D_1^2-\bar D_2^2)\bar H$

   $=\frac{\pi}{4}(35.472^2-22.038^2)\times 50.09= 30394 mm^3$

$\Delta V=\sqrt{(\frac{\pi}{2}\bar D_1\bar H \Delta D_1)^2+(\frac{\pi}{2}\bar D_2\bar H \Delta D_2)^2+[\frac{\pi}{4}(\bar D_1^2-\bar D_2^2) \Delta H ]^2}$

$=\sqrt{55.82^2+34.68^2+12.14^2}=66.827 (mm^3)$

Given diameter of the ball $\bar D = 14.25 mm$

$\Delta D$ = indication error of the vernier calliper= 0.004mm

So, Volume of the ball is

$\bar V= \frac{\pi \bar D^3}{6}=1515 mm^3$

$\Delta V= \frac{3 \Delta D}{D} \bar V=\frac{3 \times 0.004}{14.25} \times 1515=1.275 mm^3$

3. Given Mass of the cylinder M= 86.8 gm

Density of the cylinder, $\bar \rho=\frac{M}{\bar V}=\frac{86.8}{30394}(gm/mm^3)= 2855 (gm/cm^3)$

Taking logarithm of the above equation

$ln \bar \rho=ln M- ln \bar V$

and differentaiting both side we get

$\frac{\Delta \bar \rho}{ \bar \rho}=\frac{\Delta M}{M}-\frac{\Delta \bar V}{\bar V}$

since errors are always add up we can write

$\frac{\Delta \bar \rho}{ \bar \rho}=\frac{\Delta M}{M}+\frac{\Delta \bar V}{\bar V}$

Or

$\Delta \bar \rho=[\frac{\Delta M}{M}+\frac{\Delta \bar V}{\bar V}]\bar \rho$

$\Delta M$ = Let indication error of the vernier calliper= 0.01mg( As it is not given)

So

$\Delta \bar \rho=[\frac{0.01}{86.8}+\frac{1.275}{30394}] \times 2855= 0.448 (gm/cm^3)$

So

$\rho= \bar \rho \pm \Delta \bar \rho$

$\rho=2855 \pm 0.448 (gm/cm^3)$