Errors in Measurement and Significant Figures — NEET Physics PYQ MCQs with Solutions

Solution: Given, main scale reading, MSR $=0$ and circular scale reading, CSR $=52$ divisions. Least count $\text{LC}=\dfrac{p}{n}$, where $p$ is the pitch and $n$ is the number of circular divisions in one complete revolution. $\text{LC}=\dfrac{1}{100}$ mm $=0.01$ mm $=0.001$ cm. $\therefore$ Diameter $D=\text{MSR}+(\text{CSR}\times\text{LC})=0+(52\times0.001)=0.052$ cm.

Solution: Mean time interval $\bar{T}=\dfrac{1.25+1.24+1.27+1.21+1.28}{5}=\dfrac{6.25}{5}=1.25$ s. Mean absolute error $\Delta\bar{T}=\dfrac{|\Delta T_1|+|\Delta T_2|+|\Delta T_3|+|\Delta T_4|+|\Delta T_5|}{5}=\dfrac{0+0.01+0.02+0.04+0.03}{5}=\dfrac{0.1}{5}=0.02$ s. $\therefore$ Percentage relative error $=\dfrac{\Delta\bar{T}}{\bar{T}}\times100=\dfrac{0.02}{1.25}\times100=1.6\%$.

Solution: Given, least count $=0.01$ mm and number of divisions on circular scale $=50$. Pitch of the screw gauge $=$ least count $\times$ number of divisions on circular scale $=0.01\times50=0.5$ mm.

Solution: The difference between $9.99$ m and $0.0099$ m is $9.99-0.0099=9.9801$ m. Taking significant figures into account, as both values have two significant figures after the decimal, their difference will also have two significant figures after the decimal, i.e. $9.98$ m.

Solution: Given, $n$ divisions of vernier scale coincide with $(n-1)$ divisions of main scale, i.e. $n(\text{VSD})=(n-1)\text{MSD}\Rightarrow 1\text{VSD}=\dfrac{n-1}{n}\text{MSD}$. Least count $=1\text{MSD}-1\text{VSD}=1\text{MSD}-\dfrac{n-1}{n}\text{MSD}=\left(1-\dfrac{n-1}{n}\right)\text{MSD}=\dfrac{1}{n}\text{MSD}$. Here $1\text{MSD}=\dfrac{1}{n}$ cm $\Rightarrow \text{LC}=\dfrac{1}{n}\times\dfrac{1}{n}=\dfrac{1}{n^2}$ cm.

Solution: Given, least count LC $=0.001$ cm, main scale reading MSR $=5$ mm $=0.5$ cm, vernier (circular) scale reading VSR $=25$ and zero error $=-0.004$ cm. Final reading $=\text{MSR}+\text{VSR}\times\text{LC}-\text{zero error}=0.5+25\times0.001-(-0.004)=0.5+0.025+0.004=0.529$ cm.

Solution: As given, $P=\dfrac{a^3b^2}{cd}$. $\therefore \dfrac{\Delta P}{P}\times100=\left(\dfrac{3\Delta a}{a}+\dfrac{2\Delta b}{b}+\dfrac{\Delta c}{c}+\dfrac{\Delta d}{d}\right)\times100=3\times1+2\times2+3+4=14\%$.

Solution: Volume of a sphere, $V=\dfrac{4}{3}\pi r^3$. $\therefore \dfrac{\Delta V}{V}\times100=\dfrac{3\times\Delta r}{r}\times100$. Here $\dfrac{\Delta r}{r}\times100=2\%$. $\therefore \dfrac{\Delta V}{V}\times100=3\times2\%=6\%$.

Solution: As density $\rho=\dfrac{m}{V}=\dfrac{m}{l^3}$, so $\dfrac{\Delta\rho}{\rho}\times100=\pm\left(\dfrac{\Delta m}{m}+3\dfrac{\Delta l}{l}\right)\times100\%=\pm(4+3\times3)=\pm13\%$.

Solution: Kinetic energy $K=\dfrac{1}{2}mv^2$, so $\dfrac{\Delta K}{K}\times100=\dfrac{\Delta m}{m}\times100+2\dfrac{\Delta v}{v}\times100=2\%+2\times3\%=8\%$.

Solution: $N\,VSD=(N-1)MSD\Rightarrow 1\,VSD=\dfrac{(N-1)}{N}MSD$. $LC=1\,MSD-1\,VSD=\left(1-\dfrac{N-1}{N}\right)MSD=\dfrac{1}{N}MSD=\dfrac{1}{N}\times0.1\text{ cm}=\dfrac{1}{10N}$ cm.

Solution: Density $\rho=\dfrac{m}{V}$, so $\dfrac{\Delta\rho}{\rho}=\dfrac{\Delta m}{m}+\dfrac{\Delta V}{V}$. With $\Delta m=0.01$, $m=22.42$, $\Delta V=0.1$, $V=4.7$: $\dfrac{\Delta\rho}{\rho}\times100=\left(\dfrac{0.01}{22.42}+\dfrac{0.1}{4.7}\right)\times100=2\%$.