Units And Measurements Practice Questions

Magnetic induction Magnetic force of induction can be given as $\text{F}=\text{q} \text{v} \text{B}$ \Rightarrow $\text{B}=\frac{\text{F}}{q v}$ [\text{B}]=\frac{[\text{F}]}{[q][v]}={[\text{M} \text{L} \text{T}^{-2}]\}/{[{\text{AT}}]$[{\text{LT}}^{-1}]$}=[{\text{M}}^{1} {\text{T}}^{-2} {\text{A}}^{-1}]$Magnetic Flux Magnetic flux is the product of magnetic inductionand area. ϕ = B \cdot A$[\phi ]=[\text{B}][\text{A}]=$[\text{M}^{1} \text{T}^{-2} \text{A}^{-1}]$[\text{L}^{2}]$$=[\text{M} \text{L}^{2} \text{T}^{-2} \text{A}^{-1}]$Magnetic permeability We know that, magnetic field inside the solenoid.$\text{B}=\μ_{0} \text{n} \text{I}$\Rightarrow$\μ_{0}=\frac{\text{B}}{n l}$(where, n is number of turns per unit length)$\Rightarrow [\μ_{0}]=\frac{[\text{B}]}{[n][l]}=\frac{[\text{M} \text{T}^{-2} \text{A}^{-1}]$}{[\text{L}^{-1}]$[\text{A}]}=[\text{M} \text{L} \text{T}^{-2} \text{A}^{-2}]$Magnetisation The magnetic dipole moment acquired per unitvolume is known as magnetisation.${\text{I}}={{\text{M}}}/{{\text{V}}}$$[\text{l}]={[{\text{L}}^{2} {\text{A}}]$}/{[{\text{L}}^{3}]$}=[{\text{L}}^{-1} {\text{A}}]$ Hence, no option is correct.

Dimensional formula of $[e]=[I T]$ $[h] \;\; =[M^{1} L^{2} T^{-1}]$ $[c] \;\; =[M^{0} L^{1} T^{-1}]$ $[\;\frac{1}{4 \pi \epsilon _{0}}] \;\; =[M^{1} L^{3} T^{-4} I^{-2}]$ where, $\;\frac{1}{4 \pi \epsilon _{0}}$ is Coulomb's constant. Therefore, dimensional formula of \;\frac{1}{4 \pi \epsilon _{0}} \;\frac{| {e}|^{2}}{h c} \;\; =\[ {M}^{1} {L}^{3} {T}^{-4} {I}^{-2}]$ \cdot ;\frac{[ {IT}]^{2}}{[ {M}^{1} {L}^{2} {T}^{-1}]$$[ {M}^{0} {LT}^{-1}]$}$$;; =$[ {M}^{0} {L}^{0} {T}^{0} {I}^{0}]$ ;\text{ or };[ {M}^{0} {L}^{0} {T}^{0}]$

Given, $C$ and $V$ represent capacity and voltage, respectively. Dimensional formula of $[C]=[M^{-1} L^{-2} T^{4} I^{2}]$ Dimensional formula of $[ {V}]=[ {ML}^{2} {T}^{-3} {I}^{-1}]$ Therefore, dimensional formula of [C / V]=\;\frac{[M^{-1} L^{-2} T^{4} I^{2}]\}{[M L^{2} T^{-3}|^{-1}]$}$$=[M^{-2} L^{-4} T^{7} I^{3}]$

Young's modulus, $Y=\;\frac{\;\text{ Stress }\;}{\;\text{ Strain }\;}=\;\frac{F L}{A l}$ $\Rightarrow Y=\;\frac{m g L}{\pi R^{2} I}$.....(i) $[ \because F=m g$ and $A=\pi R^{2}]$ To determine the fractional errors, we can write Eq. (i) as follows $\;\frac{\Delta Y}{Y}=\;\frac{\Delta m}{m}+\;\frac{\Delta L}{I}+\;\frac{2 \cdot \Delta R}{R}+\;\frac{\Delta l}{I}$ $\Rightarrow \;\; \;\frac{\Delta Y}{Y} \times 100=\;\frac{\Delta m}{m} \times 100+\;\frac{\Delta L}{L} \times 100+\;\frac{2 \Delta R}{R} \times 100+\;\frac{\Delta l}{l} \times 100$ $\Rightarrow \;\; \;\frac{\Delta Y}{Y} \times 100=100[\;\frac{\Delta m}{m}+\;\frac{\Delta L}{L}+\;\frac{2 \Delta R}{R}+\;\frac{\Delta l}{I}]$ $=100[\;\frac{1}{1000}+\;\frac{1}{1000}+2(\;\frac{0.001}{0.2})+\;\frac{0.001}{0.5}]$ $=\;\frac{1}{10}+\;\frac{1}{10}+1+\;\frac{1}{5}=\;\frac{14}{10}=1.4 \%$

The dimensional formulae of given terms are Planck's constant $(h)=[ {ML}^{2} {T}^{-1}]$ Kinetic energy $(E)=[ {ML}^{2} {T}^{-2}]$ Electric potential $(V)=[M L^{2}{ }^{-1}{ }^{1} {T}^{-3}]$ Linear momentum $(p)=[ {MLT}^{-1}]$ So, the correct match is ${A} \to 2, {B} \to 3, {C} \to 4, {D} \to 1 .$

Increase in length of pendulum is 0.1%. i.e. $\frac{\Delta\text{L}}{\text{L}} \times 100=0.1$ Time period of pendulum is given by $\text{T}=2 \pi \\sqrt{\frac{\text{L}}{g}}$ Here, 2 π and g are constant. $\frac{\Delta\text{T}}{\text{T}} \times 100=\frac{1}{2} \frac{\Delta\text{L}}{\text{L}} \times 100$ $\Rightarrow \frac{\Delta\text{T}}{\text{T}} \times 100=\frac{1}{2} \times 0.1$ $\Delta\text{T}=0.05 \times \frac{\text{T}}{100}$ In one single day, the time in seconds is $\text{T}=24 \times 60 \times 60=86400 {\text{s}}$ $\therefore \Delta\text{T}=0.05 \times \frac{86400}{100}=43.2 {\text{s}}$ Thus, the error in time per day is 43.2 s.

Relative magnetic permeability $(\\mu _{r})$ It is the ratio of permeability of medium to the permeability of free space i.e., $\\mu _{r}=\frac{\\mu _{m}}{\\mu _{0}}$ As, it is a ratio of permeabilities. So, it is unitless and dimensionless quantity. Power factor $(\cos \phi )$ It is cosine of phase difference between alternating current and alternating voltage. Hence, it has only a numerical value, so it is also a unitless and dimensionless quantity. Permeability of free space $(\\mu _{0})$ It is the ratio of magnetising field induction (B) to magnetising field intensity (H) i.e. $\\mu _{0}=\frac{\text{B}}{\text{H}}$ $\Rightarrow [\\mu _{0}]=\frac{[\text{B}]}{[\text{H}]}$ ...(i) Since, $\text{H}=\text{nl}$ where, n = number of turns per unit length and I = current $\therefore [\text{H}]=[\text{n}][\text{I}]$ =\[{\text{L}}^{-1}]$[{\text{A}}]=[{\text{AL}}^{-1}]$We know that, force on a current carrying conductor in magnetic field B is given as F = IBl where, l = length of the conductor, B = magnetic field and I = current.$\therefore \text{B}=\frac{\text{F}}{\text{II}}$$\Rightarrow [\text{B}]=\frac{[\text{F}]}{[\text{I}][\text{I}]}={[{\text{MLT}}^{-2}]$}/{[{\text{A}}][{\text{L}}]}=[{\text{MT}}^{-2} {\text{A}}^{-1}]$\therefore From Eq. (i), we have$[\μ_{0}]={[{\text{MT}}^{-2}{\text{A}}^{-1}]$}/{[{\text{AL}}^{-1}]$}=[{\text{MLT}}^{-2} {\text{A}}^{-2}]$Hence, permeability of free space is not a dimensionless quantity. Quality factor It is the ratio of energy stored to the energy dissipated per cycle. Quality factor$={{ \text{Energy} \text{stored} }}/{{ \text{Energy} \text{dissipated} \text{per} c\text{ycle} }}$ As, it is ratio of energies so, it will be unitless and dimensionless quantity.

${T}=2 \pi \\sqrt{\frac{l}{{g}}} \Rightarrow{g}=\frac{4 \pi ^{2} l}{{T}^{2}}$ $\frac{\Deltag}{g}=\frac{\Deltal}{l}+\frac{2 \Delta{T}}{{T}}$ $\Delta{T}=\frac{\text{least count of time }(\Delta{T}_{0})}{\text{number of oscillations(n) }}$ $\frac{\Deltag}{g}=\frac{\Deltal}{l}+\frac{2 \Delta{T}_{0}}{{n} {T}}$ As $\Deltal$ and $\Delta{T}_{0}$ are same for all observations so${\Delta{g}}/g$ is minimum for highest value of $l, {n}$ and ${T}$ \Rightarrow Minimum percentage error in ${g}$ is for student number-1

Given, positive zero error $=0.2 {mm}=0.02 {cm}$ $[ \therefore$ Least count $L C=0.01 {cm}]$ Main scale reading $=8.5 {cm}$ Vernier scale reading $=$ Vernier scale coincidence $\times$ Least count $=6 \times 0.01=0.06 {cm}$ Final reading $=$ Main scale reading $+$ Vernier scale reading - Zero error $=8.5+0.06-0.02$ $=8.54 {cm}$

The given formula of Young's modulus of elasticity, $\text{Y}=\frac{\text{m} \text{g} \text{L}^{3}}{4 \text{b} \text{d}^{3} \delta}$ where, Y = Young's modulus of elasticity, m = mass of the bar, L = length of the bar b = breadth of the bar d = thickness of the bar and $\delta$ = depression of the bar. There is no error in the value of the g. The fractional error in the measurement of Y, $\frac{\Delta\text{Y}}{\text{Y}}$ $=\frac{\Delta\text{M}}{\text{M}}+3 \frac{\Delta\text{L}}{\text{L}}+\frac{\Delta\text{b}}{\text{b}}+3 \frac{\Delta\text{d}}{\text{d}}+\frac{\Delta\delta}{\delta}$ Substituting the values in the above expression, we get $\frac{\Delta\text{Y}}{\text{Y}}=\frac{10^{-3}}{2}+3 \frac{(1 \times 10^{-3})}{1}+\frac{0.1 \times 10^{-3}}{4 \times 10^{-2}} +3 \frac{(0.01 \times 10^{-3})}{0.4 \times 10^{-2}}+\frac{(0.01 \times 10^{-3})}{5 \times 10^{-3}}$ $\frac{\Delta\text{Y}}{\text{Y}}=0.0155$ The fractional error in the measurement of the Young's modulus is 0.0155.

For (A) Reading = MSR + CSR + Error 0.322 = 0.300 + CSR + 5 × LC 0.322 = 0.300 + CSR + 0.005 CSR = 0.017 For B Reading = MSR + CSR + Error 0.322 = 0.200 + CSR + 0.092 CSR = 0.030 Difference = 0.030 - 0.017 = 0.013 cm Division on circular scale $=\frac{0.013}{0.001}=13$

Given, radius of a sphere, $R=(7.50 ± 0.85) {cm}$ Volume of the sphere, $V=\;\frac{4 \pi r^{3}}{3}$ $\;\frac{\Delta V}{V} \times 100=3(\;\frac{\Delta r}{r}) \times 100$ $\Rightarrow \;\; \;\frac{\Delta V}{V} \times 100=3(\;\frac{0.85}{7.5}) \times 100$ $\Rightarrow \;\; \;\frac{\Delta V}{V} \times 100=34 \%$ Hence, the percentage error in its volume is $34 \%$. So, the value of $x$ near to the integer is 34 .

As we know that, Dimensional formula of volume $=[{\text{M}}^{0} {\text{L}}^{3} {\text{T}}^{0}]$ ...(i) Since, $\text{F}=6 \pi \eta r v$ $\therefore \eta \propto\frac{\text{F}}{r v}$ where, η is viscosity and F is force. \therefore [\eta ]={[{\text{ML}} {\text{T}}^{-2}]\}/{[{\text{L}} .{\text{L}} {\text{T}}^{-1}]$}=[{\text{M}} {\text{L}}^{-1} {\text{T}}^{-1}]$So,$[\frac{\pi \text{pa}^{4}}{8 \eta \text{L}}]=\frac{[\text{ML}^{-1} \text{T}^{-2}]$$[\text{L}^{4}]$}{[\text{ML}^{-1} \text{T}^{-1}]$$[\text{L}^{1}]$}$$=[\text{M}^{0} \text{L}^{3} \text{T}^{-1}]$...(ii) Since, Eq. (i) is not equal to Eq (ii), so option (a) is wrong. Now, since formula of capillary rise in tube,$h=\frac{2 s \cos \theta }{\rho g r}$Dimensional formula of LHS part, ∴ [h] = [L] Dimensional formula of RHS part$=\frac{[{s}]}{[\rho ][{g}][r]}={[{\text{MT}}^{-2}]$}/{[{\text{ML}}^{-3}]$$[{\text{LT}}^{-2}]$[{\text{L}}]}=[{\text{L}}]$Hence,$h=\frac{2 s \cos \theta }{\rho r g}$is dimensionally correct. So, option (b) will also be dimensionally correct. In option (c),$\text{J}=\epsilon \frac{\text{dE}}{\∂ t}$...(iii)$\Rightarrow \text{J}=\epsilon \frac{\text{E}}{t}$Dimension of current density J is calculated as Since,$\text{J}=\frac{\text{l}}{\text{A}}$$\therefore [\text{J}]=\frac{[\text{l}]}{[\text{A}]}={[{\text{A}}]}/{[\text{L}^{2}]$}$$\Rightarrow [\text{J}]=[{\text{AL}}^{-2}]$...(iv) Again, we know that$\text{E}=\frac{1}{4 \pi \epsilon } .\frac{q}{r^{2}}$$\Rightarrow \epsilon \text{E}=\frac{1}{4 \pi } .\frac{q}{r^{2}}$$\frac{\Rightarrow \epsilon \text{E}}{t}=\frac{1}{4 \pi } .\frac{q}{t r^{2}}$$\Rightarrow [\frac{\epsilon \text{E}}{t}]=\frac{[q]}{[t]$[r^{2}]$}={[{\text{AT}}]}/{[{\text{T}}]$[{\text{L}}^{2}]$}$$[\frac{\epsilon \text{E}}{t}]=[{\text{AL}}^{-2}]$...(v) Form Eqs. (iv) and (v), we see that Eq. (iii) is dimensionally correct. In option (d) W = \tau \theta and$\tau =r × \text{F}$So, dimensional formula of$[\tau ][\theta ]=[r][\text{F}]=[{\text{L}}]$[{\text{MLT}}^{-2}]$=[{\text{ML}}^{2} {\text{T}}^{-2}]$$=[{\text{W}}]$

Given, $R=\;\frac{V}{I}$....(i) where, &V=(50 ± 2) {V} $I=(20 ± 0.2) {A}$ From Eq. (i) $\;\frac{\Delta R}{R} \times 100=\;\frac{\Delta V}{V} \times 100+\;\frac{\Delta l}{l} \times 100$ $\%$ error in $R=[\;\frac{2}{50} \times 100+\;\frac{0.2}{20} \times 100] \%$ $=[2 \times 2+0.2 \times 5] \%$ $=5 \%$

Given, accuracy of acceleration due to gravity = 4% Accuracy of time period = 3% Energy stored in pendulum at any instant is given as $\text{E}=\frac{\text{mgL} \theta ^{2}}{2}$ ...(i) Time period of pendulum is given by expression $\text{T}=2 \pi \\sqrt{\frac{\text{L}}{g}}$ Rearrange the above expression for L, we get $\text{L}=g(\frac{\text{T}}{2 \pi })^{2}$ ...(ii) Substituting the value of L in Eq. (i), we get $\text{E}=\frac{m g^{2} \theta ^{2}}{2}({\text{T}}{2 \pi })^{2}=\frac{m g^{2} \theta ^{2} \text{T}^{2}}{8 \pi ^{2}}$ Now, the accuracy in measurement of energy can be calculated as $\frac{\Delta\text{E}}{\text{E}} \times 100=2 \frac{\Deltag}{g} \times 100+2 \frac{\Delta\text{T}}{\text{T}} \times 100$ (\because m,θ and π are constant) \Rightarrow $\frac{\Delta\text{E}}{\text{E}} \times 100=2 \times 4 \%+2 \times 3 \%=14 \%$ Thus, the accuracy to which E is known is 14%.

As we know that, Sl unit of following terms are (a) $R_{H}$ (Rydberg constant) $= {m}^{-1}$(b)h(Planck's constant)$=J-s=k g-m^{2} \cdot s^{-2} \cdot s=k g-m^{2} s^{-1}$(c) Magnetic field energy density$(\mu _{B}) \;\; =\;\frac{\;\text{ Energy }\;}{\;\text{ Volume }\;}$ $\;\; =\;\frac{k g-m^{2} s^{-2}}{m^{3}}$ $\;\; =k g m^{-1} s^{-2}$(d) $η$ (coefficient of viscosity)$\because \;\; F=η A \;\frac{d v}{d x}$ $\therefore \;\; η=\;\frac{F d x}{A d v}=\;\frac{k g- {ms}^{-2} \cdot m}{ {m}^{2} \cdot {ms}^{-1}}= {kgm}^{-1} {s}^{-1}$So, the correct match is $A-3, B-2, C-4$ and D-1.

${m} \propto{t}^{{a}} {v}^{{b}} l^{{c}}$ {m} \propto[{T}]^{{a}}\[{LT}^{-1}]^{{b}}$[{ML}^{2} {T}^{-1}]^{{c}}$${M}^{1} {L}^{0} {T}^{0}={M}^{{c}} {L}^{{b}+2 {c}} {T}^{{a}-{b}-{c}}$comparing powers${c}=1, {b}=-2, {a}=-1$${m} \propto{t}^{-1} {v}^{-2} l^{1}$

Dimension of energy, $\text{E}=[{\text{ML}}^{2} {\text{T}}^{-2}]$ Angular momentum, $\text{L}=[{\text{ML}}^{2} {\text{T}}^{-1}]$ Mass, M = [M] and gravitational constant, $\text{G}=[{\text{M}}^{-1} {\text{L}}^{3} {\text{T}}^{-2}]$ The dimension of P in formula, $\text{P}=\text{E} \text{L}^{2} \text{M}^{-5} \text{G}^{-2}$ is given as follows [\text{P}]=\frac{[\text{E}]\[\text{L}^{2}]$}{[\text{M}^{5}]$$[\text{C}^{2}]$}$$={[{\text{ML}}^{2} {\text{T}}^{-2}]$ \times $[{\text{M}}^{2} {\text{L}}^{4} {\text{T}}^{-2}]$}/{[{\text{M}}^{5}]$$[{\text{M}}^{-2} {\text{L}}^{6} {\text{T}}^{-4}]$}$$=[{\text{M}}^{0} {\text{L}}^{0} {\text{T}}^{0}]$