Gujarati
Hindi
1.Units, Dimensions and Measurement
hard

The density of a solid ball is to be determined in an experiment. The diameter of the ball is measured with 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 ball has a relative error of $2 \%$, the relative percentage error in the density is

A

$0.9 \%$

B

$2.4 \%$

C

$3.1 \%$

D

$4.2 \%$

(IIT-2011)

Solution

Given

Screw gauge readings

Pitch $=0.5 \ mm$

Circular scale division $=50$

Main scale reading $=2.5 \ mm$

Least count $=\frac{0.5}{50}=0.01 \ mm$

Circular scale division reading $=2$ odivisions

Relative error $=2 \%$

So,

Then dimensions reading from screw gauge $=$ Diameter of a solid ball

S.G. Reading $=$ main scale reading $+\left(\frac{\text { pitch }}{\text { circular scale division }}\right) \times$ circular scale division reading

$\text { Diameter of solid ball }=2.5+\frac{0.5}{50} \times 20=2.7 \ mm$

As density $\rho=\frac{\text { mass }}{\text { Volume }}=\frac{M}{\frac{4 \pi}{3}\left(\frac{D}{2}\right)^3}$

The relative percentage error in the density is

$\frac{\Delta \rho}{\rho} \times 100=\left(\frac{\Delta M }{ M }+\frac{3 \Delta D }{ D }\right) \times 100$

$\% \text { error }=\left(\frac{\Delta M }{ M } \times 100+\frac{3 \Delta D }{ D }\right) \times 100$

$=\left(2+3 \times \frac{0.01}{2.7}\right) \times 100$

$=3.1 \%$

Standard 11
Physics

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