First slide

Accuracy; precision of instruments and errors in measurements

Question

In an experimental set up, the density of a small sphere is to be determined. The diameter of the small sphere is measured with the help of a screw gauge, whose pitch is 0.5 mm and there are 50 divisions on the circular scale. The reading on the main scale is 2.5 mm and that on the circular scale is 20 divisions. If the measured mass of the sphere has a relative error of 2%, the relative percentage error in the density is

Difficult

A

0.03%
B

3.11%
C

0.08%
D

8.2%

Solution

$= \frac{P i t c h}{T o t a l d i v i s i o n s o n c i r c u l a r s c a l e}$
$∴ L e a s t c o u n t = \frac{0.5}{50} = 0.01 m m = Δ r$
Diameter = Main scale + Circular scale $\times$ Least count
$= 2.5 + 20 \times \frac{0.5}{50} = 2.70 m m$
$∵ \frac{Δ r}{r} = \frac{0.01}{2.70}$
$\frac{Δ r}{r} \times 100 = \frac{1}{2.7}$
Density, $D = \frac{m}{V} = \frac{m}{\frac{4}{3} π {(\frac{r}{2})}^{3}}$
Here, r is diameter
$∴ \frac{Δ D}{D} \times 100 = {\frac{Δ m}{m} + 3 (\frac{Δ r}{r})} \times 100$
$= \frac{Δ m}{m} \times 100 + 3 \times \frac{Δ r}{r} \times 100$
$= 2 % + 3 \times \frac{1}{2.7} = 3.11 %$

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