Units And Measurements Practice Questions

The Least Count ($LC$) of a vernier callipers is defined as the difference between one main scale division ($1 MSD$) and one vernier scale division ($1 VSD$): $LC = 1 MSD - 1 VSD$.
Given that $50 VSD = 48 MSD$, we can calculate $1 VSD$ in terms of $MSD$ as $1 VSD = \frac{48}{50} MSD = 0.96 MSD$.
Since $1 MSD = 0.05 mm$, substituting this value yields $1 VSD = 0.96 \times 0.05 mm = 0.048 mm$.
Now, substituting these into the $LC$ formula gives $LC = 0.05 mm - 0.048 mm = 0.002 mm$.

The physical dimensions are derived as follows:

1. A. Coefficient of viscosity ($\eta$): Defined by $F = \eta A \frac{dv}{dx}$, so [\eta] = \frac{[Force]}{[Area][Velocity\ Gradient]} = \frac{[MLT^{-2}]\}{[L^2]$$[T^{-1}]$} = [ML^{-1}T^{-1}]$. Matches IV.
2. B. Surface tension ($S$): Defined as force per unit length, $S = \frac{F}{L}$, so [S] = \frac{[MLT^{-2}]\}{[L]} = [ML^0T^{-2}]$. Matches III.
3. C. Pressure ($P$): Defined as force per unit area, $P = \frac{F}{A}$, so [P] = \frac{[MLT^{-2}]\}{[L^2]$} = [ML^{-1}T^{-2}]$. Matches I.
4. D. Surface energy: By process of elimination, this maps to the remaining dimension II $[ML^2T^{-2}]$.

Combining these mappings gives A-IV, B-III, C-I, and D-II, which corresponds to option B.

The terminal velocity $v_0$ of a spherical body of radius $r$ in a viscous liquid is given by Stokes' Law:
$v_0 = \frac{2}{9} \frac{r^2 g (\sigma - \rho)}{\eta}$
Rearranging the equation to solve for the coefficient of viscosity $\eta$:
$\eta = \frac{2 r^2 g (\sigma - \rho)}{9 v_0}$
Given that $\sigma, \rho,$ and $g$ have no associated errors, we can write $\eta \propto \frac{r^2}{v_0}$. Using the principle of error propagation for products and quotients, the maximum relative error in $\eta$ is:
$\frac{\Delta \eta}{\eta} = 2 \left( \frac{\Delta r}{r} \right) + \frac{\Delta v_0}{v_0} = \frac{2 \Delta r}{r} + \frac{\Delta v_0}{v_0}$

The average length of the rod is $\bar{x} = \frac{20.00 + 19.75 + 17.01 + 18.25}{4} = \frac{75.01}{4} = 18.7525 \text{ cm}$.
The absolute errors for each measurement are $|20.00 - 18.7525| = 1.2475$, $|19.75 - 18.7525| = 0.9975$, $|17.01 - 18.7525| = 1.7425$, and $|18.25 - 18.7525| = 0.5025$.
The mean absolute error is $\Delta \bar{x} = \frac{1.2475 + 0.9975 + 1.7425 + 0.5025}{4} = \frac{4.49}{4} = 1.1225 \text{ cm}$.
The relative error is calculated as $\frac{\Delta \bar{x}}{\bar{x}} = \frac{1.1225}{18.7525} \approx 0.05986$.
Rounding this value gives $0.06$.

The speed of a longitudinal wave in a bar is given by $v = \sqrt{\frac{Y}{\rho}}$.
For small relative changes, the fractional change in speed is approximated by $\frac{\Delta v}{v} \approx \frac{1}{2} \frac{\Delta Y}{Y} - \frac{1}{2} \frac{\Delta \rho}{\rho}$.
Substituting the given increases $\frac{\Delta Y}{Y} = 0.01$ and $\frac{\Delta \rho}{\rho} = 0.005$, we get $\frac{\Delta v}{v} = \frac{1}{2}(0.01) - \frac{1}{2}(0.005) = 0.005 - 0.0025 = 0.0025$.
The absolute change in speed is $\Delta v = v \times 0.0025 = 400 \times 0.0025 = 1$ m/s.
Thus, the new speed is $v_{new} = 400 + 1 = 401$ m/s.

According to the principle of dimensional homogeneity, each term added together must have the same dimensions. Given the equation $(P + \frac{A}{Bt^2}) + \rho g(h+Bt) + \frac{1}{2} \rho V^2 = \text{constant}$, all terms must share the dimension of pressure $[P] = [M L^{-1} T^{-2}]$.

1. For the term $(h + Bt)$, dimensions must match length $[L]$:
   [Bt] = [L] \implies [B][T] = [L] \implies [B] = \[L T^{-1}]$ = [M^0 L T^{-1}]$.

2. For the term $\frac{A}{Bt^2}$, it must have the dimension of pressure $[M L^{-1} T^{-2}]$:
   [\frac{A}{Bt^2}] = \[M L^{-1} T^{-2}]$ \implies [A] = $[M L^{-1} T^{-2}]$ \cdot [B] \cdot [t^2]$Substituting$[B] = [L T^{-1}]$and$[t] = [T]$:$[A] = $[M L^{-1} T^{-2}]$ \cdot $[L T^{-1}]$ \cdot $[T^2]$ = $[M L^{0} T^{-2-1+2}]$ = [M L^0 T^{-1}]$.

Thus, the dimensions of $A$ and $B$ are $[M L^0 T^{-1}]$ and $[M^0 L T^{-1}]$ respectively.

The zero error is negative because the zero is above the horizontal line:
$ZE = -3 \times 0.01 \text{ mm} = -0.03 \text{ mm}$.
The observed measurement ($M_{obs}$) is given by:
$M_{obs} = \text{Pitch scale reading} + (\text{Circular scale reading} \times \text{Least count}) = 1 \text{ mm} + 51 \times 0.01 \text{ mm} = 1.51 \text{ mm}$.
The correct thickness is calculated as:
$M_{corr} = M_{obs} - ZE = 1.51 \text{ mm} - (-0.03 \text{ mm}) = 1.54 \text{ mm}$.

The arithmetic sum of the quantities is $52.01\text{ m} + 153.2\text{ m} + 0.123\text{ m} = 205.333\text{ m}$. According to the rules for significant figures in addition, the final result must be reported to the same number of decimal places as the measurement with the fewest decimal places. Among the given values, $153.2\text{ m}$ has the minimum precision with only $1$ decimal place. Therefore, we round $205.333\text{ m}$ to $1$ decimal place, which gives $205.3\text{ m}$.

The permittivity of free space $\epsilon$ has units $\frac{F}{m} = \frac{C}{V \cdot m}$, and the electric field $E$ has units $\frac{V}{m}$.
The product $\epsilon E$ represents the electric displacement field, with dimensions:
$\epsilon E = \left( \frac{C}{V \cdot m} \right) \left( \frac{V}{m} \right) = \frac{C}{m^2}$
Dividing this product by time $t$ (in seconds, $s$) gives the unit:
$\frac{\epsilon E}{t} = \frac{C/m^2}{s} = \frac{C/s}{m^2}$
Since current is defined as $I = \frac{dq}{dt}$, its unit is Amperes ($A = C/s$). Substituting this, the unit is $\frac{A}{m^2}$.
This corresponds to option C.

For springs connected in parallel, the equivalent spring constant is $k_p = k_1 + k_2 = 10 + 20 = 30 \, \text{N/m}$.
The absolute error in the equivalent constant is the sum of the individual absolute errors: $\Delta k_p = \Delta k_1 + \Delta k_2 = 0.2 + 0.3 = 0.5 \, \text{N/m}$.
The percentage error is defined as $\frac{\Delta k_p}{k_p} \times 100\%$.
Substituting the calculated values, we get $\frac{0.5}{30} \times 100\% = \frac{1}{60} \times 100\%$.
Evaluating this results in $\frac{10}{6} \% \approx 1.666... \% \approx 1.67\%$.

The Zero Error ($ZE$) is positive because the vernier zero is to the right of the main scale zero: $ZE = +(4 \times 0.1 \text{ mm}) = +0.4 \text{ mm}$.
The Observed Reading ($OR$) is calculated as: $OR = \text{MSR} + (\text{VSR} \times LC) = 15.0 \text{ mm} + (5 \times 0.1 \text{ mm}) = 15.5 \text{ mm}$.
The actual length is obtained by subtracting the zero error from the observed reading: $\text{Length} = OR - ZE$.
Substituting the values: $\text{Length} = 15.5 \text{ mm} - 0.4 \text{ mm} = 15.1 \text{ mm}$.

$\;d E=3 \alpha ^2 d \alpha e^{-\beta t}+\alpha ^3 e^{-\beta t}(-\beta ) \text{dt}$ $\;\;\frac{d E}{E}=\;\frac{3 d \alpha }{\alpha }+(-\beta ) \text{dt}$ $\;\;\frac{\Delta E}{E}=\;\frac{3 \Delta \alpha }{\alpha }+\beta \Delta t$(%) $\Delta t=1.6 \% \times 5=8 \%$ $\% (\;\frac{3 \Delta \alpha }{\alpha }) =3 \times 1.2 \%=3.6 \%$Putting values we get$\Rightarrow \;\; 3.6+8 \times 0.3=6 \%$

$\;\rho =\;{ {\text{kg}}}/{m^3} \Rightarrow [ {\text{ML}}^{-3} {\text{T}}^0]$......(iv)$\;I=F \cdot \Delta t \Rightarrow [ {\text{MLT}}^{-1}]$.......(ii)$\;P=\;\frac{w}{t} \Rightarrow [ {\text{ML}}^2 {\text{T}}^{-3}]$........(i)$\;I=\;\frac{M R^2}{n} \Rightarrow [ {\text{ML}}^2 {\text{T}}^0]$.........(iii)

$y=\;\frac{32.3 \times 1125}{27.4}=1326.18$So we need to report to three significant digit.So, $y=1330$

$\;\frac{\Delta Q}{Q}=-2 \;\frac{( ± \Delta x)}{x}+\;\frac{3}{2} \;\frac{( ± \Delta y)}{y}-\;\frac{2}{5} \;\frac{( ± \Delta z)}{z}$For maximum fractional error$\;\frac{\Delta Q}{Q}\;=\;\frac{2 \Delta x}{x}+\;\frac{3}{2} \;\frac{\Delta y}{y}+\;\frac{2}{5} \;\frac{\Delta z}{z}$ $\;=2 \times 0.1+\;\frac{3}{2} \times 0.2+\;\frac{2}{5} \times 0.5$ $\;=0.7$

[Angular momentum] $= ML ^2 T^{-1}$[Planck's Constant] $= ML ^2 T^{-1}$[Torque] $= ML ^2 T^{-2}$[Energy] $= ML ^2 T^{-2}$[Surface tension] $= MT ^{-2}$[Impulse] $= MLT ^{-1}$[Pressure] $= ML ^{-1} T^{-2}$[Young's modulus] $= ML ^{-1} T^{-2}$

The electric dipole moment $p$ is defined as $p = q \cdot d$, with dimensions $[A \cdot T \cdot L] = [M^0 L^1 T^1 A^1]$. The magnetic dipole moment $m$ is defined as $m = I \cdot A_{area}$, with dimensions $[A \cdot L^2] = [M^0 L^2 T^0 A^1]$. The ratio of the two is:
$\frac{p}{m} = \frac{[M^0 L^1 T^1 A^1]\$}{[M^0 L^2 T^0 A^1]\$} = \$[M^{0-0} L^{1-2} T^{1-0} A^{1-1}]\$ = [M^0 L^{-1} T^1 A^0]$
Comparing this with $[M^P L^Q T^R A^S]$, we identify $P = 0$ and $Q = -1$.

For a current carrying loop at centre$\;B=\;\frac{\mu _0 i}{2 R}$ $\; \therefore \;\; \;\frac{B}{\mu _0} \equiv \;\frac{i}{R} \equiv [AL^{-1}]$

$\;\;\text{ Least count }\;=0.01= (\;\frac{L}{N})$ $\;\;\text{ Now }\; \;\frac{1.75 L}{N / 2}\;=\;\text{ New least count }\;$ $\Rightarrow 3.5 \times \;\frac{L}{N}\;=3.5 \times 0.01 mm$ $\;=35 \times 10^{-3} mm$