Units and Measurements — NEET Physics PYQ MCQs with Solutions

Solution: $1$ minute $=\dfrac{1}{60}$ degree $=\dfrac{1}{60}\times\dfrac{\pi}{180}$ rad $=2.91\times10^{-4}$ rad.

Solution: The rate of heat flow through a conductor of length $L$ and area of cross-section $A$ is given by $\dfrac{dQ}{dt}=KA\dfrac{\Delta T}{L}$ J/s or watt, where $K$ = coefficient of thermal conductivity and $\Delta T$ = change in temperature. $\Rightarrow K=\dfrac{L}{A\Delta T}\dfrac{dQ}{dt}$. $\therefore$ Unit of $K=\dfrac{\text{metre}}{(\text{metre})^2\times\text{kelvin}}\times\text{watt}=\text{Wm}^{-1}\text{K}^{-1}$.

Solution: According to Coulomb's law, the electrostatic force $F=\dfrac{1}{4\pi\varepsilon_0}\times\dfrac{q_1q_2}{r^2}$, where $q_1$ and $q_2$ = charges, $r$ = distance between charges and $\varepsilon_0$ = permittivity of free space. $\Rightarrow \varepsilon_0=\dfrac{1}{4\pi}\times\dfrac{q_1q_2}{r^2F}$. Substituting the units for $q$, $r$ and $F$, we obtain unit of $\varepsilon_0=\dfrac{\text{coulomb}\times\text{coulomb}}{\text{newton-(metre)}^2}=\dfrac{(\text{coulomb})^2}{\text{newton-(metre)}^2}$.

Solution: The value of Planck's constant is $6.63\times10^{-34}$ and J-s is the unit of the Planck's constant.

Solution: Force $F=[MLT^{-2}]=(10\text{ g})(10\text{ cm})(0.1\text{ s})^{-2}$. Changing these units into the MKS system, $F=(10^{-2}\text{ kg})(10^{-1}\text{ m})(10^{-1}\text{ s})^{-2}=10^{-1}$ N $=0.1$ N.

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

Solution: Let $[E]=[F]^a[a]^b[T]^c$, so $[ML^2T^{-2}]=[MLT^{-2}]^a[LT^{-2}]^b[T]^c=[M^a L^{a+b} T^{-2a-2b+c}]$. Comparing powers: $a=1$; $a+b=2\Rightarrow b=1$; $-2a-2b+c=-2\Rightarrow c=2$. Hence energy $=[F][a][T^2]$.

Solution: $[E]=[F]\cdot[d]=[MLT^{-2}][L]=[ML^2T^{-2}]$. From $F=\dfrac{GM_1M_2}{r^2}$, $G=\dfrac{Fr^2}{M_1M_2}=\dfrac{[MLT^{-2}][L^2]}{[M^2]}=[M^{-1}L^3T^{-2}]$. Thus $\dfrac{E}{G}=\dfrac{[ML^2T^{-2}]}{[M^{-1}L^3T^{-2}]}=[M^2L^{-1}T^0]$.

Solution: Stress $=\dfrac{\text{Force}}{\text{Area}}=\dfrac{[MLT^{-2}]}{[L^2]}=[ML^{-1}T^{-2}]$.

Solution: As $F=\dfrac{e^2}{4\pi\varepsilon_0 r^2}\Rightarrow\dfrac{e^2}{4\pi\varepsilon_0}=r^2 F$, so $\left[\dfrac{e^2}{4\pi\varepsilon_0}\right]=[ML^3T^{-2}]$. Also $G=[M^{-1}L^3T^{-2}]$ and $\dfrac{1}{c^2}=[L^{-2}T^2]$. Checking option (a): $\dfrac{1}{c^2}\left[G\dfrac{e^2}{4\pi\varepsilon_0}\right]^{1/2}=[L^{-2}T^2][L^6T^{-4}]^{1/2}=[L^{-2}T^2][L^3T^{-2}]=[L]$.

Solution: Surface tension $S=\dfrac{\text{Force}}{\text{Length}}=\dfrac{[MLT^{-2}]}{[L]}=[ML^0T^{-2}]$. Let $S\propto E^a v^b T^c$ with $[E]=[ML^2T^{-2}]$, $[v]=[LT^{-1}]$. Then $[ML^0T^{-2}]=[M^a L^{2a+b} T^{-2a-b+c}]$. Equating: $a=1$; $2a+b=0\Rightarrow b=-2$; $-2a-b+c=-2\Rightarrow c=-2$. So $S=[Ev^{-2}T^{-2}]$.

Solution: With $[\eta]=[ML^{-1}T^{-1}]$, $[\rho]=[ML^{-3}]$, $[r]=[L]$ and $v_c=[M^0LT^{-1}]$: $[M^0LT^{-1}]=[ML^{-1}T^{-1}]^x[ML^{-3}]^y[L]^z$. Comparing exponents of M, L, T: $x+y=0$, $-x-3y+z=1$, $-x=-1$. Solving gives $x=1$, $y=-1$, $z=-1$.

Solution: From $F=ma=\dfrac{mv}{t}$, we get $m=\dfrac{Ft}{v}$, so $[M]=\dfrac{[F][T]}{[v]}=[Fv^{-1}T]$.

Solution: $(\mu_0\varepsilon_0)^{-1/2}$ is the expression for the velocity of light, since $c=\dfrac{1}{\sqrt{\mu_0\varepsilon_0}}$. So its dimensions are $[LT^{-1}]$.

Solution: $\dfrac{1}{2}\varepsilon_0 E^2$ is the energy density of an electric field. With $[\varepsilon_0]=[M^{-1}L^{-3}T^4A^2]$ and $[E]=[MLT^{-3}A^{-1}]$, $\left[\dfrac{1}{2}\varepsilon_0E^2\right]=[M^{-1}L^{-3}T^4A^2]\times[MLT^{-3}A^{-1}]^2=[ML^{-1}T^{-2}]$.

Solution: Dimensions of velocity $=[M^0L^1T^{-1}]$, acceleration $=[M^0L^1T^{-2}]$, force $=[M^1L^1T^{-2}]$ and pressure $=[M^1L^{-1}T^{-2}]$. Thus $a=1, b=-1, c=-2$ corresponds to pressure.

Solution: Energy density $u=\dfrac{\text{Energy}}{\text{Volume}}=\dfrac{[ML^2T^{-2}]}{[L^3]}=[ML^{-1}T^{-2}]$. Young's modulus $=\dfrac{\text{stress}}{\text{strain}}=[ML^{-1}T^{-2}]$. Refractive index and dielectric constant are dimensionless, and magnetic field $=\dfrac{F}{qv}=[MT^{-2}A^{-1}]$. Hence (i) and (iv) have the same dimensions.

Solution: By Ohm's law $R=\dfrac{V}{I}=\dfrac{W}{qI}=\dfrac{\text{work}}{\text{charge}\times\text{current}}=\dfrac{[ML^2T^{-2}]}{[IT][I]}=[ML^2T^{-3}I^{-2}]$.

Solution: By the principle of homogeneity $[a][t]=[v]$, so $[a]=\dfrac{[LT^{-1}]}{[T]}=[LT^{-2}]$. Also $[c]=[t]=[T]$, and $[b]=[v][t+c]=[LT^{-1}][T]=[L]$.