Units & Measurements

The formula for the time period of simple harmonic oscillator is

$\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$.

$\Rightarrow $ $ \mathrm{T}^2=(2 \pi)^2\left(\frac{\mathrm{k}}{\mathrm{~m}}\right) $

$\Rightarrow \mathrm{k}=\frac{\mathrm{T}^2 \cdot \mathrm{~m}}{4 \pi^2}$

Using propagation of error, when quantities are multiplied or divided, their relative errors (percentage errors) are added. The constants have no error.

$ \frac{\Delta \mathrm{k}}{\mathrm{k}} \times 100=\left(\frac{\Delta \mathrm{m}}{\mathrm{~m}}+2 \cdot \frac{\Delta \mathrm{~T}}{\mathrm{~T}}\right) \times 100 $

The measured value of mass is, $\mathrm{m}=10 \mathrm{~g}$

Accuracy/Error in measurement of mass is, $\Delta \mathrm{m}=10 \mathrm{mg}=0.01 \mathrm{~g}$

So, percentage error in measurement of mass $\frac{\Delta \mathrm{m}}{\mathrm{m}} \times 100 \%=\frac{0.01}{10} \times 100=0.1 \%$

The percentage error in the time period ( $T$ ) is the same as the percentage error in the total time ( $t$ ) measured for n oscillations.

Total time measured is, $\mathrm{t}=60 \mathrm{~s}$

Resolution/Error in measurement of time is, $\Delta t=2 \mathrm{~s}$

So, percentage error in measurement of time is, $\frac{\Delta t}{t} \times 100 \%=\frac{2}{60} \times 100 \%=3.33 \%$

So, using propagation formula:

Percentage error in $\mathrm{k}=(\%$ error in m$)+2 \times(\%$ error in t$)=0.1 \%+2 \times 3.33 \%=6.76 \%$

Therefore, the correct option is (D).

The formula for the speed of a longitudinal wave ( v ) in a solid metallic bar is given as

$ \mathrm{v}=\sqrt{\frac{\mathrm{Y}}{\rho}} $

Where, $\mathrm{Y}=$ Young's modulus and its $\rho$ is its density

Taking the natural $\log$ on both sides of $\mathrm{v}=\left(\frac{\mathrm{Y}}{\rho}\right)^{1 / 2}$ :

$ \ln (\mathrm{v})=\frac{1}{2} \ln (\mathrm{Y})-\frac{1}{2} \ln (\rho) $

Differentiating both sides gives the fractional change:

$ \frac{\Delta v}{v}=\frac{1}{2} \frac{\Delta Y}{Y}-\frac{1}{2} \frac{\Delta \rho}{\rho} $

Initial speed $=\mathrm{v}=400 \mathrm{~m} / \mathrm{s}$

Increase in Young's modulus $=\frac{\Delta \mathrm{Y}}{\mathrm{Y}} \times 100=+1 \%$

Increase in density $\frac{\Delta \rho}{\rho} \times 100=+0.5 \%$

So, the percentage change in speed,

$ \% \text { change in } \mathrm{v}=\frac{1}{2}(1 \%)-\frac{1}{2}(0.5 \%)=0.5 \%-0.25 \%=0.25 \% $

Since the result is positive, the speed increases by $0.25 \%$.

So, new speed $\left(\mathrm{v}^{\prime}\right)=$ Initial Speed + Change in Speed

$ \begin{aligned} \mathrm{v}^{\prime} & =\mathrm{v}+(0.25 \% \text { of } \mathrm{v}) \\ \Rightarrow \mathrm{v}^{\prime} & =400+\left(\frac{0.25}{100} \times 400\right) \\ \Rightarrow \mathrm{v}^{\prime} & =400+1=401 \mathrm{~m} / \mathrm{s} \end{aligned} $

Hence, the correct option is (C).

From Newton's law of viscosity, the force is $\mathrm{F}=\eta \mathrm{A}\left(\frac{\Delta \mathrm{v}}{\Delta \mathrm{x}}\right)$, where A is area and $\frac{\Delta \mathrm{v}}{\Delta \mathrm{x}}$ is the velocity gradient.

$ \eta=\frac{\mathrm{F}}{\mathrm{~A} \cdot\left(\frac{\Delta \mathrm{v}}{\Delta \mathrm{x}}\right)} $

$\Rightarrow $ $ [\eta]=\frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{L}^2\right] \cdot\left(\frac{\left[\mathrm{LT}^{-1}\right]}{[\mathrm{L}]}\right)}=\frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{L}^2\right] \cdot\left[\mathrm{T}^{-1}\right]}=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right] $

$ \mathrm{A} \rightarrow \mathrm{IV} $

Surface tension is defined as the force acting per unit length.

$ \mathrm{T}=\frac{\text { Force }}{\text { Length }} $

$ [\mathrm{T}]=\frac{\left[\mathrm{MLT}^{-2}\right]}{[\mathrm{L}]}=\left[\mathrm{ML}^0 \mathrm{~T}^{-2}\right] $

B $\to $ III

Pressure is defined as force acting per unit area.

$ \mathrm{P}=\frac{\text { Force }}{\text { Area }} $

$ [\mathrm{P}]=\frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{L}^2\right]}=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right] $

$ C \rightarrow I $

Surface energy $\left(\mathrm{U}_{\mathrm{s}}\right)$ is defined as the work done by surface tension in changing the surface area

$ \mathrm{U}_{\mathrm{s}}=\mathrm{T} \cdot(\Delta \mathrm{~A}) $

$\Rightarrow $ $\left[\mathrm{U}_{\mathrm{S}}\right]=[$ Surface Tension $] \cdot[$ Area $]$

$\Rightarrow $ $ \left[\mathrm{U}_{\mathrm{S}}\right]=\left[\mathrm{ML}^0 \mathrm{~T}^{-2}\right] \cdot\left[\left[\mathrm{M}^0 \mathrm{~L}^2 \mathrm{~T}^0\right]\right]=\left[\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]\right] $

$ \mathrm{D} \rightarrow \mathrm{II} $

$ \begin{aligned} &\begin{array}{|l|l|} \hline \text { List - I } & \text { List - II } \\ \hline \text { A. Coefficient of viscosity } & \text { IV. }\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right] \\ \hline \text { B. Surface tension } & \text { III. }\left[\mathrm{ML}^0 \mathrm{~T}^{-2}\right] \\ \hline \text { C. Pressure } & \text { I. }\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right] \\ \hline \text { D. Surface energy } & \text { II. }\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right] \\ \hline \end{array}\\ &\text { Hence, the correct option is (B). } \end{aligned} $

The least count of an instrument is the smallest measurement that can be taken accurately with that instrument. The least count is shown by the position of the last significant digit (the smallest decimal place) to which all readings are recorded.

The obtained readings are :

• 1. Reading 1 : 1.24 mm

• 2. Reading 2 : 1.25 mm

• 3. Reading 3 : 1.23 mm

• 4. Reading 4 : 1.21 mm

All the readings are recorded up to two decimal places.

The difference between consecutive possible readings

$ \begin{gathered} 1.25 \mathrm{~mm}-1.24 \mathrm{~mm}=0.01 \mathrm{~mm} \\ 1.24 \mathrm{~mm}-1.23 \mathrm{~mm}=0.01 \mathrm{~m} \end{gathered} $

If the least count were 0.1 mm (the smallest decimal place is 1 ), the readings would look like 1.2 mm or 1.3 mm .

If the least count were 0.001 mm (the smallest decimal place is 3 ), the readings would look like 1.240 mm .

If the least count were 0.05 mm , the readings would only end in .00 or .05 (e.g., $1.20,1.25$ ). Since we have 1.24 and 1.23, this is not possible.

Therefore, the least count of the instrument is 0.01 mm . Hence, the correct option is (A).

When jaws touch, the vernier zero is to the right of the main scale zero, which indicates a positive zero error.

Coinciding vernier division (CVD) for error $=4$.

Least Count $(\mathrm{LC})=0.1 \mathrm{~mm}$.

So, zero error $=+(\mathrm{CVD} \times \mathrm{LC})=+(4 \times 0.1)=+0.4 \mathrm{~mm}$

Main scale reading $(\mathrm{MSR})=15$ divisions.

Assuming 1 division $=1 \mathrm{~mm}$, then main scale reading is $\mathrm{MSR}=15 \mathrm{~mm}$

Vernier Scale Reading $(\mathrm{VSR})=($ Coinciding vernier division $) \times \mathrm{LC}$.

$ \mathrm{VSR}=5 \times 0.1=0.5 \mathrm{~mm} $

Total Observed Reading $=\mathrm{MSR}+\mathrm{VSR}=15+0.5=15.5 \mathrm{~mm}$

True Length $=$ Observed Reading - Zero Error

True Length $=15.5 \mathrm{~mm}-(+0.4 \mathrm{~mm})$

True Length $=15.1 \mathrm{~mm}$

Hence, the correct option is (D).

Value of one Main Scale Division, $1 \mathrm{MSD}=0.05 \mathrm{~mm}$

Number of Vernier Scale Divisions $\mathrm{m}=50$

Number of Main Scale Divisions is $\mathrm{n}=48$

$ 50 \mathrm{VSD}=48 \mathrm{MSD} $

$\Rightarrow $ $ 1 \mathrm{VSD}=\frac{48}{50} \mathrm{MSD} $

The Least Count of a Vernier instrument is the smallest measurement that can be taken accurately.

Least Count (LC) of Vernier scale $=1$ MSD -1 VSD

Using the least count formula,

$ \mathrm{LC}=1 \mathrm{MSD}-1 \mathrm{VSD} $

$\Rightarrow $ $\mathrm{LC}=1 \mathrm{MSD}-\left(\frac{48}{50}\right) \mathrm{MSD}$

$\Rightarrow $ $\mathrm{LC}=\left(1-\frac{48}{50}\right) \mathrm{MSD}$

$\Rightarrow $ $\mathrm{LC}=\left(\frac{50-48}{50}\right) \mathrm{MSD}$

$\Rightarrow $ $ \mathrm{LC}=\frac{2}{50} \mathrm{MSD}=\frac{2}{50} \times 0.05 \mathrm{~mm}=0.002 \mathrm{~mm} $

Therefore, the smallest measurement that this Vernier calliper can record is 0.002 mm .

Hence, the correct option is (A).

The circular scale lies 3 divisions above the horizontal pitch line when the studs are in contact.

When the zero mark is above the reference line, the screw has been turned past the zero point. This indicates a negative zero error. The error corresponds to 3 divisions.

Hence, the zero error is,

$\mathrm{E}_0=-(3 \times \text { Least Count }) =-(3 \times 0.01 \mathrm{~mm})=-0.03 \mathrm{~mm} $

The observed reading is the sum of the Pitch Scale Reading (PSR) and the Circular Scale Reading (CSR) multiplied by the Least Count (LC).

The given Pitch Scale Reading (PSR) is 1 mm and the Circular Scale Reading (CSR) is 51

So, the observed reading $=\mathrm{PSR}+(\mathrm{CSR} \times \mathrm{LC})$

Observed Reading $=1 \mathrm{~mm}+(51 \times 0.01 \mathrm{~mm})$

Observed Reading $=1 \mathrm{~mm}+0.51 \mathrm{~mm}$

Observed Reading $=1.51 \mathrm{~mm}$

To find the true thickness, we must subtract the zero error from the observed reading.

Correct Reading $=$ Observed Reading - Zero Error

Correct Reading $=1.51 \mathrm{~mm}-(-0.03 \mathrm{~mm})$

Correct Reading $=1.51 \mathrm{~mm}+0.03 \mathrm{~mm}$

Correct Reading $=1.54 \mathrm{~mm}$

Therefore, the correct thickness of the sheet is 1.54 mm.

Hence, the correct option is (A).

The average (mean) of the measured values is the true value for our error calculation.

$ \bar{x}=\frac{x_1+x_2+x_3+x_4}{4} $

Substituting the given values :

$ \bar{x}=\frac{20.00+19.75+17.01+18.25}{4}=\frac{75.01}{4} $

$ \overline{\mathrm{x}}=18.7525 \approx 18.75 \mathrm{~cm} $

The absolute error is the magnitude of the difference between each individual measurement and the mean value.

$ \left|\Delta \mathrm{x}_{\mathrm{i}}\right|=\left|\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right| $

$\Rightarrow $ $\left|\Delta \mathrm{x}_1\right|=|20.00-18.75|=1.25$

$\Rightarrow $ $\left|\Delta \mathrm{x}_2\right|=|19.75-18.75|=1.00$

$\Rightarrow $ $\left|\Delta x_3\right|=|17.01-18.75|=|-1.74|=1.74$

$\Rightarrow $ $ \left|\Delta \mathrm{x}_4\right|=|18.25-18.75|=|-0.50|=0.50 $

Then the average of these absolute errors.

$ \Delta \overline{\mathrm{x}}=\frac{\sum\left|\Delta \mathrm{x}_{\mathrm{i}}\right|}{4} $

$\Rightarrow $ $\Delta \bar{x}=\frac{1.25+1.00+1.74+0.50}{4}=\frac{4.494}{4}$

$\Rightarrow $ $ \Delta \overline{\mathrm{x}}=1.1225 \approx 1.12 \mathrm{~cm} $

The relative error is defined as the ratio of the mean absolute error to the mean value.

Relative Error $=\frac{\Delta \overline{\mathrm{x}}}{\overline{\mathrm{x}}}$

Relative Error $=\frac{1.1225}{18.7525}$

Relative Error $\approx 0.0598$

Rounding to two decimal places,

Relative Error $\approx 0.06$

Therefore, the relative error in the measurement is 0.06 . Hence, the correct option is (C).

The SI unit of permittivity of free space ( $\epsilon$ ) is derived from Coulomb's law $\left(F=\frac{q_1 q_2}{4 \pi \epsilon_0 r^2}\right)$. The SI unit is $C^2 / N$. $\mathrm{m}^2$.

The electric field (E) is defined as force per unit charge, so its SI unit is Newtons per Coulomb (N/C).

The standard SI unit is the second (s).

The given expression is $\frac{\epsilon \mathrm{E}}{\mathrm{t}}$.

So, the SI unit of given expression is, $\frac{\mathrm{C}^2}{\mathrm{~N} \cdot \mathrm{~m}^2} \times\left(\frac{\mathrm{N}}{\mathrm{C}}\right)=\frac{\mathrm{C}}{\mathrm{s}}=\frac{\mathrm{s}}{\mathrm{s} \cdot \mathrm{m}^2}$

Electric current is defined as the rate of flow of electric charge over time. Therefore, 1 Ampere $(\mathrm{A})=$

1 Coulomb per second $\left(\frac{\mathrm{C}}{\mathrm{s}}\right)$.

Unit of $\frac{\epsilon \mathrm{E}}{\mathrm{t}}=\left(\frac{\mathrm{C}}{\mathrm{s}}\right) \times \frac{1}{\mathrm{~m}^2}=\mathrm{A} \times \frac{1}{\mathrm{~m}^2}=\mathrm{A} / \mathrm{m}^2$

Therefore, the SI unit of $\frac{\epsilon E}{t}$ is $\mathrm{A} / \mathrm{m}^2$. Hence, the correct option is (C).

A. Spring constant (k)

According to Hooke's Law, the restoring force F of a spring is directly proportional to its displacement x :

$ \mathrm{F}=-\mathrm{kx} $

Taking the magnitude to find dimensions :

$ \mathrm{k}=\frac{\mathrm{F}}{\mathrm{x}} \Rightarrow[\mathrm{k}]=\left[\frac{\mathrm{F}}{\mathrm{x}}\right]=\frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{M}^0 \mathrm{LT}^0\right]} $

$ [\mathrm{k}]=\frac{\left[\mathrm{MLT}^{-2}\right]}{[\mathrm{L}]}=\left[\mathrm{ML}^{1-1} \mathrm{~T}^{-2}\right]=\left[\mathrm{ML}^0 \mathrm{~T}^{-2}\right] $

A → II.

B. Thermal conductivity (K)

From Law of heat conduction, the rate of heat flow $\left(\mathrm{H}=\frac{\mathrm{d} \mathrm{Q}}{\mathrm{dt}}\right)$ is given by:

$ H=\frac{K A \Delta T}{L} $

Where A is area, $\Delta \mathrm{T}$ is temperature difference, and L is length.

$ \mathrm{K}=\frac{\mathrm{H} \cdot \mathrm{~L}}{\mathrm{~A} \cdot \Delta \mathrm{~T}} $

The dimensions of rate of heat flow, $[\mathrm{H}]=\frac{[\text { Energy }]}{[\text { Time }]}=\frac{[\text { Force × distance }]}{[\text { Time }]}=\frac{\left[\mathrm{MLT}^{-2} \times \mathrm{L}\right]}{[\mathrm{T}]}=\left[\mathrm{ML}^2 \mathrm{~T}^{-3}\right]$

So the dimensions of thermal conductivity is,

$[\mathrm{K}]=\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-3}\right] \cdot[\mathrm{L}]}{\left[\mathrm{L}^2\right] \cdot[\mathrm{K}]}=\frac{\left[\mathrm{ML}^3 \mathrm{~T}^{-3}\right]}{\left[\mathrm{L}^2 \mathrm{~K}\right]}$

$\Rightarrow $ $ [\mathrm{K}]=\left[\mathrm{ML}^{3-2} \mathrm{~T}^{-3} \mathrm{~K}^{-1}\right]=\left[\mathrm{MLT}^{-3} \mathrm{~K}^{-1}\right] $

B → IV.

C. Boltzmann constant ( $\mathrm{k}_{\mathrm{B}}$ )

The average kinetic energy (E) of a gas molecule is related to its absolute temperature (T) by :

$ \mathrm{E}=\frac{3}{2} \mathrm{k}_{\mathrm{B}} \mathrm{~T} $

The dimensionless constant is $\frac{3}{2}$.

$ \left[\mathrm{k}_{\mathrm{B}}\right]=\left[\frac{\mathrm{E}}{\mathrm{~T}}\right] $

We know Energy has the same dimensions as Work ( $\mathrm{F} \times \mathrm{d}$ ):

$ [\mathrm{E}]=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right] $

So, the dimensions of Boltzmann constant is,

$ \left[\mathrm{k}_{\mathrm{B}}\right]=\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]}{[\mathrm{K}]}=\left[\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~K}^{-1}\right] $

$ \mathrm{C} \rightarrow \mathrm{I} . $

D. Inductive reactance ( $\mathrm{X}_{\mathrm{L}}$ )

Inductive reactance is the opposition to alternating current by an inductor. It acts exactly like resistance, so it must have the same dimensional formula as electrical resistance (R).

From Ohm's Law: $\mathrm{V}=\mathrm{IR}$, so $\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}$.

Potential is Work done per unit charge (W/q):

$ [\mathrm{V}]=\frac{[\text { Work }]}{[\text { Charge }]}=\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]}{[\mathrm{A} \cdot \mathrm{~T}]}=\left[\mathrm{ML}^2 \mathrm{~T}^{-3} \mathrm{~A}^{-1}\right] $

So, the dimensions of inductive reactance, i.e., resistance is,

$ \left[\mathrm{X}_{\mathrm{L}}\right]=[\mathrm{R}]=\frac{[\mathrm{V}]}{[\mathrm{I}]}=\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-3} \mathrm{~A}^{-1}\right]}{[\mathrm{A}]} $

$ \begin{array}{ll} & {\left[\mathrm{X}_{\mathrm{L}}\right]=\left[\mathrm{ML}^2 \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right]} \\ \mathrm{D} \rightarrow \mathrm{III} . & \end{array} $

Therefore, the correct match is, $\mathrm{A} \rightarrow$ II, $\mathrm{B} \rightarrow$ IV, $\mathrm{C} \rightarrow \mathrm{I}, \mathrm{D} \rightarrow$ III.

Hence, the correct option is (D).

When a spherical body falls through a viscous liquid, the forces acting on it are,

• Weight ( W ) which is acting downwards.

• Up thrust ( $\mathrm{F}_{\mathrm{b}}$ ) is acting upwards (due to buoyancy).

• Viscous Drag $\left(\mathrm{F}_{\mathrm{v}}\right)$ is acting upwards (opposing the motion).

The radius of the spherical body is $r$.

The density of the body is $\sigma$

So, the weight of the spherical body is,

$ \mathrm{W}=\mathrm{mg}=(\sigma \mathrm{V}) \mathrm{g}=\sigma\left[\frac{4}{3} \pi \mathrm{r}^3\right] \mathrm{g} $

The density of the liquid is $\rho$, and when body is completely inside the liquid, the up thrust is,

$ \mathrm{F}_{\mathrm{B}}=\rho \mathrm{Vg}=\frac{4}{3} \pi \mathrm{r}^3 \rho \mathrm{~g} $

The viscous drag acting on the body is

$ \mathrm{F}_{\mathrm{v}}=6 \pi \eta \mathrm{rv} $

Terminal velocity is reached when the net force on the body becomes zero (acceleration is zero).

$ \mathrm{W}=\mathrm{F}_{\mathrm{B}}+\mathrm{F}_{\mathrm{v}} $

$\Rightarrow $ $\frac{4}{3} \pi r^3 \sigma g=\frac{4}{3} \pi r^3 \rho g+6 \pi \eta r v_0$

$\Rightarrow $ $6 \pi \eta r v_0=\frac{4}{3} \pi r^3 g(\sigma-\rho)$

$\Rightarrow $ $\eta=\frac{\frac{4}{3} \pi r^3 g(\sigma-\rho)}{6 \pi r v_0}$

$\Rightarrow $ $\eta=\frac{2 r^2 g(\sigma-\rho)}{9 v_0}$

$\Rightarrow $ $ \eta=\frac{\left[\frac{2 g(\sigma-\rho)}{9}\right] r^2}{v_0} $

As it is given to ignore errors associated with $\sigma, \rho$, and g . Therefore, $\left[\frac{2 \mathrm{~g}(\sigma-\rho)}{9}\right]$ is a constant $(\mathrm{k})$.

$ \eta=\mathrm{k} \cdot \frac{\mathrm{r}^2}{\mathrm{v}_0} $

Taking natural logarithm on both sides:

$ \ln (\eta)=\ln (\mathrm{k})+2 \ln (\mathrm{r})-\ln \left(\mathrm{v}_0\right) $

On differentiating both sides:

$ \frac{d \eta}{\eta}=0+\frac{2 d r}{r}-\frac{d v_0}{v_0} $

For maximum error, we sum the absolute values of the individual errors because errors always propagate in a way that maximizes the uncertainty:

$ \frac{\Delta \eta}{\eta}=\frac{2 \Delta r}{r}+\frac{\Delta v_0}{v_0} \text { or } \frac{\Delta \eta}{\eta}=-\frac{2 \Delta r}{r}-\frac{\Delta v_0}{v_0} $

Therefore, the correct option is (C).

In addition, or subtraction, the final result should be reported to the same number of decimal places as the measurement with the least number of decimal places. This ensures the result is no more precise than the least precise measurement.

The given measured values are,

• $52.01 \mathrm{~m}: 2$ decimal places.

• 153.2 m : 1 decimal place.

• 0.123 m : 3 decimal places.

The measurement with the least number of decimal places is 153.2 m ( 1 decimal place). Therefore, our final answer must be rounded to one decimal place.

$ 52.01+153.2+0.123=205.333 \mathrm{~m} $

Now, we round 205.333 to one decimal place :

• The digit at the first decimal place is $3 \rightarrow 205.33$

• The digit following it is 3 (which is less than 5 ) → 205.3

Therefore, the rounded result is 205.3 m . Hence, the correct option is (A)

When two springs are connected in parallel, they both undergo the same displacement (x) when a force is applied.

Force in spring 1 is $\mathrm{F}_1=\mathrm{k}_1 \mathrm{x}$

Force in spring 2 is $\mathrm{F}_2=\mathrm{k}_2 \mathrm{x}$

The total force F is the sum of individual forces :

$ \mathrm{F}=\mathrm{F}_1+\mathrm{F}_2 $

$\Rightarrow $ $\mathrm{k}_{\mathrm{eq}} \mathrm{x}=\mathrm{k}_1 \mathrm{x}+\mathrm{k}_2 \mathrm{x}$

$\Rightarrow $ $ \mathrm{k}_{\mathrm{eq}}=\mathrm{k}_1+\mathrm{k}_2 $

Substituting the given values, $\mathrm{k}_1=10 \mathrm{~N} / \mathrm{m}, \mathrm{k}_2=20 \mathrm{~N} / \mathrm{m}$

$ \mathrm{k}_{\mathrm{eq}}=10+20=30 \mathrm{~N} / \mathrm{m} $

When two measured quantities are added, the absolute error in the result is the sum of the absolute errors of the individual quantities.

$ \Delta \mathrm{k}_{\mathrm{eq}}=\Delta \mathrm{k}_1+\Delta \mathrm{k}_2 $

Substituting the given values, $\Delta \mathrm{k}_1=0.2 \mathrm{~N} / \mathrm{m}, \Delta \mathrm{k}_2=0.3 \mathrm{~N} / \mathrm{m}$

$ \Delta \mathrm{k}_{\mathrm{eq}}=0.2+0.3=0.5 \mathrm{~N} / \mathrm{m} $

So, the percentage error in measurement of equivalent spring constant is,

Percentage Error $=\left(\frac{\Delta \mathrm{k}_{\mathrm{eq}}}{\mathrm{k}_{\mathrm{eq}}}\right) \times 100 \%=\left(\frac{0.5}{30}\right) \times 100 \% \approx 1.666 \ldots \%$

Rounding to two decimal places, we get the percentage error in measurement of spring constant as $1.67 \%$.

Therefore, the equivalent spring constant with its error is $(30 \pm 0.5) \mathrm{N} / \mathrm{m}$, which corresponds to a percentage error of $1.67 \%$.

Hence, the correct option is (C).

In the term $(h+B t), B t$ is added to $h$, where $h$ is the height or length which has the dimension: $[h]=[L]$.

Therefore, using principle of homogeneity the product Bt must also have the dimension of length: $[\mathrm{L}]$.

Given that t is time whose dimension is $[\mathrm{T}]$ :

$ [\mathrm{B}][\mathrm{t}]=[\mathrm{L}] $

$\Rightarrow $ $[\mathrm{B}][\mathrm{T}]=[\mathrm{L}]$

$\Rightarrow $ $ [\mathrm{B}]=\frac{[\mathrm{L}]}{[\mathrm{T}]}=\left[\mathrm{LT}^{-1}\right] $

Therefore, the dimensions of B is $\left[\mathrm{M}^0 \mathrm{LT}^{-1}\right]$.

In the term $\left(\mathrm{P}+\frac{\mathrm{A}}{\mathrm{Bt}^2}\right), \frac{\mathrm{A}}{\mathrm{Bt}^2}$ is added to P (Pressure).

The dimension of pressure is $\mathrm{P}=\frac{\text { Force }}{\text { Area }}$ :

$ [\mathrm{P}]=\frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{L}^2\right]}=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right] $

According to the principle of homogeneity, the dimension of $\frac{\mathrm{A}}{\mathrm{Bt}^2}$ must equal the dimension of P :

$ \left[\frac{\mathrm{A}}{\mathrm{Bt}^2}\right]=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right] $

$\Rightarrow $$\frac{[\mathrm{A}]}{\left[\mathrm{LT}^{-1}\right]\left[\mathrm{T}^2\right]}=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]$

$\Rightarrow $ $[\mathrm{A}]=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right] \times[\mathrm{LT}]$

$\Rightarrow $ $[\mathrm{A}]=[\mathrm{M}]\left[\mathrm{L}^{-1+1}\right]\left[\mathrm{T}^{-2+1}\right]$

$\Rightarrow $ $ [\mathrm{A}]=\left[\mathrm{ML}^0 \mathrm{~T}^{-1}\right] $

Therefore, dimensions of A is $\left[\mathrm{ML}^0 \mathrm{~T}^{-1}\right]$ and the dimensions of B is $\left[\mathrm{M}^0 \mathrm{LT}^{-1}\right]$. Hence, the correct option is (D).

The quantity $ Q $ is expressed as $ Q = X^{-2}Y^{+\frac{3}{2}}Z^{-\frac{2}{5}} $. The variables $ X $, $ Y $, and $ Z $ are independent parameters with fractional errors of 0.1, 0.2, and 0.5, respectively. To determine the maximum fractional error in $ Q $, we can use the following method:

$ \text{Fractional error in } Q = \left| -2 \right| \frac{\Delta X}{X} + \left| \frac{3}{2} \right| \frac{\Delta Y}{Y} + \left| -\frac{2}{5} \right| \frac{\Delta Z}{Z} $

Substituting the fractional errors into the equation:

$ = 2 \times 0.1 + \frac{3}{2} \times 0.2 + \frac{2}{5} \times 0.5 $

Calculating each term gives:

$ = 0.2 + 0.3 + 0.2 $

Summing these values results in:

$ = 0.7 $

Thus, the maximum fractional error in $ Q $ is 0.7.

The dimension of $ \sqrt{\frac{\mu_0}{\epsilon_0}} $ can be understood as follows:

Vacuum permeability ($ \mu_0 $) and vacuum permittivity ($ \epsilon_0 $) relate to the properties of inductance and capacitance, respectively, in a vacuum. We start by considering the formulas for inductance ($ L $) and capacitance ($ C $):

Inductance ($ L $) is given by:

$ L = \frac{\mu_0 \cdot \text{Number of turns} \cdot \text{Area}}{\text{Length}} $

Capacitance ($ C $) is given by:

$ C = \frac{\text{Area} \cdot \epsilon_0}{\text{Distance}} $

From these, the ratio $\frac{L}{C}$ can be expressed as $\frac{\mu_0}{\epsilon_0}$:

$ \frac{L}{C} \propto \frac{\mu_0}{\epsilon_0} $

Taking the square root of this ratio, we have:

$ \sqrt{\frac{\mu_0}{\epsilon_0}} \propto \sqrt{\frac{L}{C}} $

This simplifies further to considering the relationship between time constant ($\tau$) and resistance ($R$):

$ \frac{L}{C} = \frac{\tau R}{(\tau / R)} = R^2 $

Thus, by taking the square root:

$ \sqrt{\frac{\mu_0}{\epsilon_0}} = R $

In conclusion, the dimension of $ \sqrt{\frac{\mu_0}{\epsilon_0}} $ is equivalent to the dimension of resistance, $ R $.

(A) Mass density:

Mass density is defined as mass per unit volume.

The dimension of mass is $[M]$.

The dimension of volume is $[L^3]$.

Therefore, the dimension of mass density is $\frac{[M]}{[L^3]} = [ML^{-3}T^{0}]$.

Matches with (IV).

(B) Impulse:

Impulse is defined as the change in momentum, or force multiplied by time.

Impulse = Force x Time

The dimension of force is $[MLT^{-2}]$.

The dimension of time is $[T]$.

Therefore, the dimension of impulse is $[MLT^{-2}] \cdot [T] = [MLT^{-1}]$.

Matches with (II).

(C) Power:

Power is defined as the rate of doing work or energy per unit time.

Power = Work / Time

The dimension of work (or energy) is $[ML^2T^{-2}]$.

The dimension of time is $[T]$.

Therefore, the dimension of power is $\frac{[ML^2T^{-2}]}{[T]} = [ML^2T^{-3}]$.

Matches with (I).

(D) Moment of inertia:

Moment of inertia is defined as $I = mr^2$, where m is mass and r is the distance from the axis of rotation.

The dimension of mass is $[M]$.

The dimension of distance squared is $[L^2]$.

Therefore, the dimension of moment of inertia is $[ML^2T^{0}]$.

Matches with (III).

So, the correct matches are:

(A) - (IV)

(B) - (II)

(C) - (I)

(D) - (III)

Therefore, the correct option is D.

Electric dipole moment $(\overrightarrow{\mathrm{P}})=\mathrm{q} \times 2 \ell$

Magnetic dipole moment $(\overrightarrow{\mathrm{M}})=\mathrm{IA}$

$\left[\frac{\mathrm{P}}{\mathrm{M}}\right]=\left[\frac{\mathrm{LTA}}{\mathrm{~L}^2 \mathrm{~A}}\right]=\mathrm{L}^{-1} \mathrm{~T}=\mathrm{M}^0 \mathrm{~L}^{-1} \mathrm{~T}^1 \mathrm{~A}^0$

After compering values of $\mathrm{P} \& \mathrm{Q}$ are $0,-1$

Correct Answer : Option 4

Determine one main scale division (msd):

$ 300 \, \text{msd} = 15 \, \text{cm} $

Therefore,

$ 1 \, \text{msd} = \frac{15}{300} \, \text{cm} = 0.05 \, \text{cm} $

Determine one vernier scale division (vsd):

$ 25 \, \text{vsd} = 24 \, \text{msd} $

Therefore,

$ 1 \, \text{vsd} = \frac{24}{25} \, \text{msd} $

Calculate the least count (LC):

The least count is given by the difference between one main scale division and one vernier scale division:

$ \text{LC} = 1 \, \text{msd} - 1 \, \text{vsd} $

Substitute the expression for 1 vsd:

$ \text{LC} = 1 \, \text{msd} - \frac{24}{25} \times 1 \, \text{msd} $

This simplifies to:

$ \text{LC} = \frac{1}{25} \times 1 \, \text{msd} $

Calculate the LC in cm:

Substituting the value of 1 msd:

$ \text{LC} = \frac{1}{25} \times 0.05 \, \text{cm} = 0.002 \, \text{cm} $

Hence, the least count of the traveling microscope is 0.002 cm.

To solve this problem, we need to find the dimensions of the electric flux and the magnetic flux, and then take their ratio.

Determine the dimension of electric flux (Φₑ):

Electric flux is defined as

$\Phi_E = \int_S \mathbf{E} \cdot d\mathbf{A}.$

The electric field $\mathbf{E}$ has dimensions:

$[E] = \frac{\text{Force}}{\text{Charge}}.$

Since force has dimensions $M L T^{-2},$ and charge (in SI) has dimensions $A\,T,$ we get:

$[E] = \frac{M L T^{-2}}{A\,T} = M L T^{-3} A^{-1}.$

The area element $dA$ has dimensions $L^2.$

Thus, the electric flux has dimensions:

$[\Phi_E] = [E][A] = M L T^{-3} A^{-1} \times L^2 = M L^3 T^{-3} A^{-1}.$

Determine the dimension of magnetic flux (Φ_B):

Magnetic flux is defined as

$\Phi_B = \int_S \mathbf{B} \cdot d\mathbf{A}.$

The magnetic field $\mathbf{B}$ has dimensions determined via the Lorentz force law (using $F = q\,v\,B$), giving:

$[B] = \frac{F}{q\,v} = \frac{M L T^{-2}}{(A\,T)(L T^{-1})} = M T^{-2} A^{-1}.$

Again, the area element has dimensions $L^2.$

So the magnetic flux has dimensions:

$[\Phi_B] = [B][A] = (M T^{-2} A^{-1})(L^2) = M L^2 T^{-2} A^{-1}.$

Find the ratio of electric flux to magnetic flux:

Taking the ratio:

$\frac{\Phi_E}{\Phi_B} = \frac{M L^3 T^{-3} A^{-1}}{M L^2 T^{-2} A^{-1}}.$

Canceling like terms:

Mass $(M)$ cancels.

Current $(A^{-1})$ cancels.

For length, $L^{3-2} = L^1.$

For time, $T^{-3 -(-2)} = T^{-1}.$

Therefore, the ratio has dimensions:

$L^{1}T^{-1}.$

This means in the expression

$M^P L^Q T^R A^S,$

the exponents for length and time are $Q = 1$ and $R = -1.$

Looking at the options provided, this corresponds to Option D: $(1,-1).$

Thus, the answer is Option D.

Explanation of Dimensional Analysis

In order to match the quantities from LIST-I with their respective dimensions in LIST-II, we need to analyze their dimensional formulas:

(A) Boltzmann Constant [k]:

The equation relating pressure (P), volume (V), number of moles (N), and temperature (T) gives us:

$ [\mathrm{k}] = \frac{\mathrm{PV}}{\mathrm{NT}} = \frac{\mathrm{ML}^2 \mathrm{~T}^{-2}}{\mathrm{~K}} = \mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~K}^{-1} $

This corresponds to option III.

(B) Coefficient of Viscosity [η]:

The coefficient of viscosity can be defined using the relation:

$ [\eta] = \frac{\mathrm{F}}{6 \pi \mathrm{rv}} = \frac{\mathrm{MLT}^{-2}}{\mathrm{~L}^2 \mathrm{~T}^{-1}} = \mathrm{ML}^{-1} \mathrm{~T}^{-1} $

This corresponds to option IV.

(C) Planck's Constant [h]:

The relationship between energy (E) and frequency (f) gives the dimensional formula:

$ [\mathrm{h}] = \frac{\mathrm{E}}{\mathrm{f}} = \frac{\mathrm{ML}^2 \mathrm{~T}^{-2}}{\mathrm{~T}^{-1}} = \mathrm{ML}^2 \mathrm{~T}^{-1} $

This corresponds to option I.

(D) Thermal Conductivity [k]:

From the heat conduction equation, we get:

$ \mathrm{k} = \frac{\left(\mathrm{ML}^2 \mathrm{~T}^{-3}\right) \mathrm{L}}{\mathrm{~L}^2 \cdot \mathrm{~K}} = \mathrm{MLT}^{-3} \mathrm{~K}^{-1} $

This corresponds to option II.

To conclude, the correct matches are:

A - III

B - IV

C - I

D - II

This analysis ensures each physical quantity is aligned with its correct dimensional representation.

To calculate the total mass of the three particles, we perform the following addition:

$ \begin{aligned} & m_1 + m_2 + m_3 = 435.42 \, \text{g} + 226.3 \, \text{g} + 0.125 \, \text{g} \end{aligned} $

When adding numbers, the result should be reported with the same number of decimal places as the measurement with the fewest decimal places. Here, the second measurement (226.3 g) has only one decimal place, which is the fewest among the given values.

Thus, according to the rules for significant figures in addition, the sum should be rounded to one decimal place:

$ \begin{aligned} & \text{Total mass} = 661.8 \, \text{g} \end{aligned} $

The expression $\frac{1}{\mu_0 \epsilon_0}$ is related to the speed of light $ c $, given by the equation:

$ c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} $

Therefore, we have:

$ \frac{1}{\mu_0 \epsilon_0} = c^2 $

The speed of light $ c $ has the dimensions of $\text{L T}^{-1}$, where $\text{L}$ is the dimension of length and $\text{T}$ is the dimension of time. Thus, when squared, the dimensions become:

$ c^2 = (\text{L T}^{-1})^2 = \text{L}^2 \text{T}^{-2} $

Hence, the dimensions of $\frac{1}{\mu_0 \epsilon_0}$ are $\text{L}^2 \text{T}^{-2}$.

Let's look at the units (dimensions) of the given quantities:

Charge $ Q $ has units of ampere × time $ (A \cdot T) $

Current $ I $ has units of ampere $ (A) $

Permeability of vacuum $ \mu_0 $ has units of mass × length × time$^{-2} $ × ampere$^{-2} $ $ (MLT^{-2}A^{-2}) $

Momentum has units of mass × length × time$^{-1} $ $ (MLT^{-1}) $

We want to combine $ Q $, $ \mu_0 $, and $ I $ so that the result has the same units as momentum $ (MLT^{-1}) $.

Suppose the answer is $ Q^x\, \mu_0^y\, I^z $. Let's write out the units for this:

$ Q^x\, \mu_0^y\, I^z = (A T)^x \, (MLT^{-2}A^{-2})^y \, (A)^z $

Combine the powers of each unit:

Mass (M): exponent = $ y $

Length (L): exponent = $ y $

Time (T): exponent = $ x - 2y $

Ampere (A): exponent = $ x - 2y + z $

We want these to match the units of momentum, $ M^1L^1T^{-1} $:

So we set up:

$ y = 1 $ (so mass and length exponents match momentum)

$ x - 2y = -1 $ (time exponent)

$ x - 2y + z = 0 $ (ampere exponent cancels out)

Solve these equations:

First, $ y = 1 $.

Second, $ x - 2(1) = -1 \implies x = 1 $.

Third, $ x - 2(1) + z = 0 \implies 1 - 2 + z = 0 \implies z = 1 $.

The required combination is $ x = 1 $, $ y = 1 $, $ z = 1 $:
So, the quantity with the dimension of momentum is $ Q \mu_0 I $.

Here are the dimensional formulas:

Coefficient of viscosity, μ

$[\mu]=\frac{\text{pressure}\times\text{time}}{} =\frac{[M\,L^{-1}T^{-2}]\,[T]}{} =[M\,L^{-1}T^{-1}]\quad\longrightarrow\text{(IV)}$

Intensity of a wave, I

$[I]=\frac{\text{power}}{\text{area}} =\frac{[M\,L^2T^{-3}]}{[L^2]} =[M\,L^0T^{-3}]\quad\longrightarrow\text{(I)}$

Pressure gradient, dP/dx

$\left[\frac{dP}{dx}\right] =\frac{[M\,L^{-1}T^{-2}]}{[L]} =[M\,L^{-2}T^{-2}]\quad\longrightarrow\text{(II)}$

Compressibility, κ

$[\kappa]=\frac1{[\text{pressure}]} =[M^{-1}L\,T^2]\quad\longrightarrow\text{(III)}$

Matching gives

(A)–(IV), (B)–(I), (C)–(II), (D)–(III), i.e. Option C.

We know, Young's modulus, $E = {\sigma \over \varepsilon }$

$ \Rightarrow E = {{{F \over A}} \over {{{\Delta L} \over L}}} = {F \over A}\left( {{L \over {\Delta L}}} \right)$

$ \Rightarrow [E] = {{[F]} \over {[A]}} = {{\left[ {{M^1}{L^1}{T^{ - 2}}} \right]} \over {\left[ {{L^2}} \right]}}$

(A) $ \Rightarrow [E] = [{M^1}{L^{ - 1}}{T^{ - 2}}]$

(B) we know, Torque = $r\times F$

$[\tau ] = [r][F]$

$ = [L][{M^1}{L^1}{T^{ - 2}}]$

$[\tau ] = [{M^1}{L^2}{T^{ - 2}}]$

(C) We know, $F = \mu A{u \over y}$

where, F = force, $\mu$ = coefficient of viscosity

A = area $\frac{u}{y}$ = rate of shear deformation

$ \Rightarrow \mu = {{Fy} \over {Au}}$

$ \Rightarrow [\mu ] = {{[{M^1}{L^1}{T^{ - 2}}][L]} \over {[{L^2}][L{T^{ - 1}}]}}$

$ \Rightarrow [\mu ] = [{M^1}{L^{ - 1}}{T^{ - 1}}]$

(D) We know, $F = {{G{m_1}{m_2}} \over {{r^2}}}$ (Newton's law of gravitation)

$ \Rightarrow G = {{F{r^2}} \over {{m_1}{m_2}}}$<$ \Rightarrow G = {{F{r^2}} \over {{m_1}{m_2}}}$

$ \Rightarrow [G] = {{[{M^1}{L^1}{T^{ - 2}}][{L^2}]} \over {[{M^2}]}}$

$ \Rightarrow [G] = [{M^{ - 1}}{L^3}{T^{ - 2}}]$

Hence, option D is correct.

$\begin{aligned} & {[\tau]=[\mathrm{E}]} \\ & {[\sigma] \neq[\mathrm{I}]} \\ & {[\mathrm{L}]=[\mathrm{h}]} \\ & {[\mathrm{P}]=[\mathrm{Y}]} \end{aligned}$

$\begin{aligned} & {\left[\mathrm{LT}^{-1}\right]=[\mathrm{A}]\left[\mathrm{T}^2\right]=\frac{[\mathrm{B}][\mathrm{T}]}{[\mathrm{C}]+[\mathrm{T}]}} \\ & {[\mathrm{C}]=[\mathrm{T}]} \\ & {[\mathrm{A}]=\left[\mathrm{LT}^{-3}\right]} \\ & {[\mathrm{B}]=\left[\mathrm{LT}^{-1}\right]} \\ & {[\mathrm{ABC}]=\left[\mathrm{L}^2 \mathrm{~T}^{-3}\right]} \end{aligned}$

Matching the quantities from List - I with their correct dimensions from List - II is crucial to understanding their physical significance. Here is the correct matching:

(A) Angular Impulse: The dimensions of angular impulse are equivalent to those of angular momentum, represented as $M L^2 T^{-1}$. Hence, (A) corresponds to (III).

(B) Latent Heat: Latent heat has dimensions of energy per unit mass, which simplifies to $M^0 L^2 T^{-2}$. Hence, (B) corresponds to (I).

(C) Electrical resistivity: Electrical resistivity has the dimensions of resistance multiplied by length, which can be represented as $M L^3 T^{-3} A^{-2}$. Hence, (C) corresponds to (IV).

(D) Electromotive force: Electromotive force (emf) is similar to electric potential difference and has dimensions $M L^2 T^{-3} A^{-1}$. Hence, (D) corresponds to (II).

Therefore, the correct answer is Option C:

(A)-(III)

(B)-(I)

(C)-(IV)

(D)-(II)

$ 32.3 \quad (3\ \text{sig figs}), \quad 1125 \quad (4\ \text{sig figs}), \quad 27.4 \quad (3\ \text{sig figs}) $

In multiplication and division, the final answer must be reported with the same number of significant figures as the factor with the fewest significant figures, which here is 3.

First, calculate the unrounded result:

$ y = \frac{32.3 \times 1125}{27.4} \approx \frac{36337.5}{27.4} \approx 1326.186 $

Expressing $1326.186$ in scientific notation:

$ 1326.186 \approx 1.326186 \times 10^3 $

Rounding to 3 significant figures:

The first three significant digits are 1, 3, and 2.

Considering the fourth digit (6) causes the last significant digit to round up.

Thus,

$ 1.326186 \times 10^3 \approx 1.33 \times 10^3 $

In standard decimal form, this is written as:

$ 1330 $

Hence, the final reported value is:

$ y = 1330 $

We start with the energy function:

$ E(t)= \alpha^3 e^{-\beta t}, $

where $\beta=0.3\, \text{s}^{-1}$.

Step 1. Take the Logarithm

Taking the logarithm of both sides:

$ \ln E = 3 \ln \alpha - \beta t. $

Step 2. Differentiate to Find the Relative Error

Differentiate both sides:

$ \frac{dE}{E} = 3\,\frac{d\alpha}{\alpha} - \beta\, dt. $

For maximum error estimation, we consider the absolute values and sum the contributions:

$ \left|\frac{\Delta E}{E}\right| \approx 3\,\left|\frac{\Delta \alpha}{\alpha}\right| + \beta\, |\Delta t|. $

Step 3. Plug in the Given Errors

The percentage error in $\alpha$ is $1.2\%$ (i.e., $\Delta \alpha/\alpha = 0.012$).

The percentage error in $t$ is $1.6\%$, so for $t=5\,\text{s}$:

$ \Delta t = 0.016 \times 5 = 0.08\, \text{s}. $

Now substitute these values:

$ \left|\frac{\Delta E}{E}\right| \approx 3(0.012) + 0.3(0.08). $

Calculating each term:

$3(0.012) = 0.036$ (or $3.6\%$),

$0.3(0.08) = 0.024$ (or $2.4\%$).

Step 4. Compute the Total Maximum Percentage Error

Add the contributions:

$ \left|\frac{\Delta E}{E}\right| \approx 0.036 + 0.024 = 0.06, $

which is $6\%$.

Thus, the maximum percentage error in the energy at $t = 5\,\text{s}$ is:

$ \boxed{6\%} $

Answer: Option A (6%).

The correct matching is:

(A) Permeability of free space has dimensions

$[MLT^{-2}A^{-2}]$

which corresponds to entry (III).

(B) Magnetic field has dimensions

$[MT^{-2}A^{-1}]$

which corresponds to entry (II).

(C) Magnetic moment in SI units is expressed in Ampere-square meters, i.e.,

$[L^2A]$

which corresponds to entry (IV).

(D) Torsional constant (torque per unit angular displacement) has dimensions

$[ML^2T^{-2}]$

which corresponds to entry (I).

Thus, the correct answer is:

Option A: (A)-(III), (B)-(II), (C)-(IV), (D)-(I)

Let's analyze the position function step by step.

The particle's position is given by:

$x(t) = A \sin t + B \cos^2 t + C t^2 + D,$

where $ t $ represents time, which has the dimension $ T $.

Since the overall dimensions of position $ x(t) $ must be length $ (L) $, each term in the expression must also have dimensions of length.

Term $ A \sin t $:

The sine function, $ \sin t $, is dimensionless (its argument must be dimensionless, and by convention, if $ t $ is in an appropriate unit where the argument is dimensionless, the function itself is dimensionless).

Therefore, $ A $ must carry the dimension of length:

$ [A] = L. $

Term $ B \cos^2 t $:

Similarly, the cosine function is dimensionless, and so is its square.

Thus, $ B $ must have the dimension of length:

$ [B] = L. $

Term $ C t^2 $:

Here, $ t^2 $ has the dimension $ T^2 $.

To have the overall term with the dimension $ L $, we require:

$ [C] \times [T^2] = L. $

So, the dimension of $ C $ must be:

$ [C] = L \, T^{-2}. $

Term $ D $:

This is a constant term added to position, so it must also have the dimension of length:

$ [D] = L. $

Now, we are asked to find the dimension of:

$ \frac{ABC}{D}. $

The dimensions of $ A $, $ B $, and $ C $ are:

$ [A] = L, \quad [B] = L, \quad [C] = L \, T^{-2}. $

Multiplying them together:

$ [ABC] = L \times L \times (L \, T^{-2}) = L^3 \, T^{-2}. $

Since $ [D] = L $, dividing by $ D $ gives:

$ \frac{[ABC]}{[D]} = \frac{L^3 \, T^{-2}}{L} = L^2 \, T^{-2}. $

Thus, the dimension of $ \frac{A B C}{D} $ is:

$ \boxed{L^2 T^{-2}}. $

The correct answer is Option C.

To determine the maximum percentage error in the density of the wire, follow these steps:

The density of the wire is given by the formula for a cylinder:

$\rho = \frac{m}{\pi r^2 l}$

When calculating the maximum percentage (relative) error, you add the relative errors from each variable, taking into account the power to which each variable is raised. Thus, for density:

$\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta r}{r} + \frac{\Delta l}{l}$

Here:

$\frac{\Delta m}{m}$ is the relative error in the mass.

$2\frac{\Delta r}{r}$ is due to the radius being squared.

$\frac{\Delta l}{l}$ is the relative error in the length.

Now plug in the given values:

Mass: $m = 0.60 \, \text{g}$ with an error $\Delta m = 0.003 \, \text{g}$

$\frac{\Delta m}{m} = \frac{0.003}{0.60} = 0.005$ or 0.5%.

Radius: $r = 0.50 \, \text{cm}$ with an error $\Delta r = 0.01 \, \text{cm}$

$\frac{\Delta r}{r} = \frac{0.01}{0.50} = 0.02$ or 2%. Since the radius is squared in the formula, multiply by 2:

$2\frac{\Delta r}{r} = 2 \times 0.02 = 0.04$ or 4%.

Length: $l = 10.00 \, \text{cm}$ with an error $\Delta l = 0.05 \, \text{cm}$

$\frac{\Delta l}{l} = \frac{0.05}{10.00} = 0.005$ or 0.5%.

Sum these contributions to find the overall maximum relative error:

$\frac{\Delta \rho}{\rho} = 0.005 + 0.04 + 0.005 = 0.05$

Converting this relative error into a percentage gives:

$0.05 \times 100\% = 5\%$

Thus, the maximum percentage error in the density of the wire is 5%, which corresponds to Option C.

To determine the dimensions of $\left(\frac{B}{\mu_0}\right)$, we start with the formula for the magnetic field $B$ inside a solenoid:

$ B = \mu_0 n I $

where $n = \frac{N}{L}$ is the number of turns per unit length, $I$ is the current, and $\mu_0$ is the permeability of free space.

Rearranging the formula gives:

$ \frac{B}{\mu_0} = nI = \frac{N}{L}I $

From the above, the dimensions of $\left(\frac{B}{\mu_0}\right)$ are:

$ \left[\frac{B}{\mu_0}\right] = \left[L^{-1} A\right] $

Thus, the dimensional formula of $\left(\frac{B}{\mu_0}\right)$ corresponds to $L^{-1} \text{A}$.

In general, one vernier scale division is smaller than one main scale division but in some modified cases it may be not correct. Also least count is given by one main scale division divided by number of vernier scale division for normal vernier calliper.

Hence, option 2 is correct.

To determine the dimensional formula for Planck's constant, we will start by analyzing the given de-Broglie wavelength equation:

$\lambda = \frac{h}{\sqrt{2 m E}}$

Here, $\lambda$ is the wavelength, $h$ is the Planck's constant, $m$ is the mass of the particle, and $E$ is the energy of the particle.

First, let's derive the dimensional formula for each term involved:

1. Wavelength $\lambda$ has the dimensional formula of length $[L]$.

2. Mass $m$ has the dimensional formula $[\text{M}]$.

3. Energy $E$ has the dimensional formula of work, which is force times distance:

$[E] = [F][L] = [\text{MLT}^{-2}][L] = [\text{ML}^{2}\text{T}^{-2}]$.

Now, let's rewrite the equation in terms of the dimensions:

$[L] = \frac{[h]}{\sqrt{2 [\text{M}] [\text{ML}^{2}\text{T}^{-2}]}}$

Simplifying inside the square root:

$[L] = \frac{[h]}{\sqrt{2 [\text{M}] [\text{M}] [\text{L}^{2}\text{T}^{-2}]}}$

$[L] = \frac{[h]}{\sqrt{2 [\text{M}^{2}] [\text{L}^{2}\text{T}^{-2}]}}$

Since the constants like 2 do not affect the dimensional formula, we can simplify further:

$[L] = \frac{[h]}{[\text{M L T}^{-1}]}$

Cross multiplying to solve for the dimensional formula of $h$:

$[h] = [L] [\text{M L T}^{-1}]$

$[h] = [\text{M L}^{2} \text{T}^{-1}]$

Therefore, the dimensional formula for Planck's constant $h$ is:

$\left[\text{ML}^{2}\text{T}^{-1}\right]$

Hence, the correct option is:

Option C $\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]$

To derive the dimensional formula of latent heat, we need to understand what latent heat actually refers to. Latent heat is the amount of heat absorbed or released by a substance during a change in its physical state (phase) that occurs without changing its temperature.

The formula for latent heat ($L$) is given by:

$Q = mL$

where:

$Q$ = Heat absorbed or released (with the dimension of energy $[\mathrm{ML}^2\mathrm{T}^{-2}]$),

$m$ = Mass of the substance ($[\mathrm{M}]$),

$L$ = Latent heat.

To find the dimensions of latent heat, we rearrange the formula to solve for $L$:

$L = \frac{Q}{m}$

Knowing the dimensions of $Q$ (energy, which is equivalent to work done, with dimensions $[\mathrm{ML}^2\mathrm{T}^{-2}]$) and $m$ (mass, with dimensions $[\mathrm{M}]$), we can substitute these into the equation to find $L$'s dimensions:

$L = \frac{[\mathrm{ML}^2\mathrm{T}^{-2}]}{[\mathrm{M}]}$

This simplifies to:

$L = [\mathrm{L}^2\mathrm{T}^{-2}]$

Therefore, the dimensional formula of latent heat is:

$L = [\mathrm{M}^0 \mathrm{L}^2 \mathrm{T}^{-2}]$

So, the correct option is:

Option C $ \left[\mathrm{M}^0 \mathrm{~L}^2 \mathrm{~T}^{-2}\right] $

The least count of a vernier caliper is defined as the smallest distance that it can measure and is calculated by the difference in length between one main scale division and one vernier scale division. It can be represented as:

Given that one main scale division is equal to $\mathrm{m}$ units and the $\mathrm{n}^{\text {th }}$ division of main scale coincides with the $(n+1)^{\text {th }}$ division of the vernier scale, this means that $n$ divisions on the main scale is equal to $(n+1)$ divisions on the vernier scale.

Since one main scale division is $\mathrm{m}$ units, $n$ divisions on the main scale would be $n \times \mathrm{m}$ units. If $n$ divisions on the main scale are equal to $(n+1)$ divisions on the vernier scale, we can determine the length of one vernier scale division as

Thus, the least count, which is the difference between one main scale division and one vernier scale division, is:

This simplifies to:

Therefore, the correct option is:

Option B: $\frac{\mathrm{m}}{\mathrm{n+1}}$

To determine the dimensions of $\epsilon_{\mathrm{o}} \mathrm{E}^2$, we need to first understand the dimensional formulas of each component in the expression.

1. Permittivity of free space, $\epsilon_{\mathrm{o}}$:

The permittivity of free space has the dimensions: $\left[\epsilon_{\mathrm{o}}\right] = [\mathrm{M}^{-1} \mathrm{L}^{-3} \mathrm{T}^4 \mathrm{A}^2]$.

2. Electric field, $\mathrm{E}$:

The electric field $\mathrm{E}$ has the dimensions: $\left[\mathrm{E}\right] = [\mathrm{M} \mathrm{L} \mathrm{T}^{-3} \mathrm{A}^{-1}]$.

Now, let's calculate the dimensions of $\epsilon_{\mathrm{o}} \mathrm{E}^2$:

$\begin{equation} \left[\epsilon_{\mathrm{o}} \mathrm{E}^2 \right] = \left[\epsilon_{\mathrm{o}}\right] \left[\mathrm{E}\right]^2 = \left[\mathrm{M}^{-1} \mathrm{L}^{-3} \mathrm{T}^4 \mathrm{A}^2\right] \left([\mathrm{M} \mathrm{L} \mathrm{T}^{-3} \mathrm{A}^{-1}]^2\right] = \left[\mathrm{M}^{-1} \mathrm{L}^{-3} \mathrm{T}^4 \mathrm{A}^2\right] \left([\mathrm{M}^2 \mathrm{L}^2 \mathrm{T}^{-6} \mathrm{A}^{-2}]\right] \end{equation}$

Combining the dimensions:

$\begin{equation} \left[\epsilon_{\mathrm{o}} \mathrm{E}^2 \right] = \left[\mathrm{M}^{-1} \mathrm{L}^{-3} \mathrm{T}^4 \mathrm{A}^2\right] \left[\mathrm{M}^2 \mathrm{L}^2 \mathrm{T}^{-6} \mathrm{A}^{-2}\right] = \left[\mathrm{M}^{-1 + 2} \mathrm{L}^{-3 + 2} \mathrm{T}^{4 - 6} \mathrm{A}^{2 - 2}\right] = \left[\mathrm{M} \mathrm{L}^{-1} \mathrm{T}^{-2}\right] \end{equation}$

Therefore, the dimensions of $\epsilon_{\mathrm{o}} \mathrm{E}^2$ are $\left[\mathrm{M} \mathrm{L}^{-1} \mathrm{~T}^{-2}\right]$, which corresponds to Option A.

Hence, the correct answer is Option A: $\left[\mathrm{M} \mathrm{L}^{-1} \mathrm{~T}^{-2}\right]$.

1. Pitch of the screw gauge: 1 mm


2. Number of divisions on the circular scale: 100 divisions


3. Zero error: The zero of the circular scale lies 5 divisions below the reference line, indicating a positive zero error.


4. Measurement data:

• 4 linear scale divisions are visible.


• 60 divisions on the circular scale coincide with the reference line.

Step-by-Step Calculation

1. Least Count of the screw gauge:

$\text{Least Count} = \frac{\text{Pitch}}{\text{Number of Divisions on Circular Scale}} = \frac{1 \text{ mm}}{100} = 0.01 \text{ mm}$

1. Main Scale Reading (MSR):

The linear scale shows 4 divisions, so the main scale reading is:

$\text{MSR} = 4 \text{ mm}$

1. Circular Scale Reading (CSR):

60 divisions coincide with the reference line, so the circular scale reading is:

$\text{CSR} = 60 \times \text{Least Count} = 60 \times 0.01 \text{ mm} = 0.60 \text{ mm}$

1. Zero Error:

The zero error is 5 divisions below the reference line, indicating a positive zero error:

$\text{Zero Error} = +5 \times \text{Least Count} = +5 \times 0.01 \text{ mm} = +0.05 \text{ mm}$

1. Total Reading without considering zero error:

$\text{Total Reading (without zero error)} = \text{MSR} + \text{CSR} = 4 \text{ mm} + 0.60 \text{ mm} = 4.60 \text{ mm}$

1. Corrected Reading considering zero error:

Since the zero error is positive, we subtract it from the total reading:

$\text{Corrected Reading} = \text{Total Reading (without zero error)} - \text{Zero Error} = 4.60 \text{ mm} - 0.05 \text{ mm} = 4.55 \text{ mm}$

Conclusion

The diameter of the wire is:

Option C: 4.55 mm

In a vernier caliper, the least count is the smallest distance measurable by the instrument. It can be defined using the difference between one main scale division and one vernier scale division. Given the least count, we can relate the number of divisions on the main scale to the divisions on the vernier scale.

The given least count of the vernier caliper is:

$\frac{1}{20 \mathrm{~N}} \mathrm{~cm}$

We know that the value of one division on the main scale is:

$1 \mathrm{~mm}$

To find the number of divisions on the main scale that coincide with N divisions of the vernier scale, let’s denote:

The number of divisions on the main scale = M

The number of divisions on the vernier scale = N

The least count formula for a vernier caliper is given by:

$\text{Least Count} = \text{Value of one main scale division} - \text{Value of one vernier scale division}$

The value of one main scale division is:

$1 \mathrm{~mm}$

The value of one vernier scale division can be expressed in terms of the number of divisions M and N:

$\text{Value of one vernier scale division} = \frac{M}{N} \mathrm{~mm}$

Given the least count:

$\frac{1}{20 \mathrm{~N}} \mathrm{~cm} = \frac{1}{20 \mathrm{~N}} \cdot 10 \mathrm{~mm} = \frac{1}{2 \mathrm{~N}} \mathrm{~mm}$

Using the least count formula, we have:

$ \frac{1}{2 \mathrm{~N}} = 1 - \frac{M}{N} $

Rearranging the equation to solve for the number of main scale divisions (M), we get:

$1 - \frac{1}{2 \mathrm{~N}} = \frac{M}{N}$

Simplifying further:

$\frac{2 \mathrm{~N} - 1}{2 \mathrm{~N}} = \frac{M}{N}$

Multiplying both sides by N, we get the value of M:

$M = \frac{2 \mathrm{~N} - 1}{2}$

Therefore, the correct answer is option D:

$ \left( \frac{2 \mathrm{~N} - 1}{2} \right) $

In expression $a \times 10^b$, If $a \leq 5 ; a \approx 1$ by round off

$\Rightarrow$ Order $B$

To find the density of the sphere, we need to calculate its volume using the measured diameter. Then, using the mass of the sphere, we can calculate the density using the formula for density, which is $\text{Density} = \frac{\text{Mass}}{\text{Volume}}$. Let's start by finding the accurate measurement of the diameter using the given vernier calipers readings.

The least count (LC) of the vernier calipers can be calculated using the formula:

$\text{LC} = \frac{\text{Value of one main scale division (MSD)}}{\text{Number of vernier scale divisions (VSD) that match with the main scale}}$

Given that 9 divisions of the main scale are equal to 10 divisions of the vernier scale and the shortest division on the main scale is equal to $1\, \mathrm{mm}$, we find:

$\text{LC} = \frac{1\, \mathrm{mm}}{10} = 0.1\, \mathrm{mm} = 0.01\, \mathrm{cm}$

For the main scale reading (MSR) of $2\, \mathrm{cm}$ and the second division of the vernier scale coinciding with a division on the main scale, the vernier scale reading (VSR) can be expressed as $\text{VSR} = 2 \times \text{LC}$.

So, $\text{VSR} = 2 \times 0.01\, \mathrm{cm} = 0.02\, \mathrm{cm}$.

The total measurement of the diameter (D) can be found by adding MSR and VSR:

$D = \text{MSR} + \text{VSR}$

$D = 2\, \mathrm{cm} + 0.02\, \mathrm{cm} = 2.02\, \mathrm{cm}$

Now, we can calculate the volume (V) of the sphere using its diameter with the formula $V = \frac{4}{3}\pi r^3$, where $r$ is the radius of the sphere. Remembering that the radius is half of the diameter, $r = \frac{D}{2} = \frac{2.02\, \mathrm{cm}}{2} = 1.01\, \mathrm{cm}$.

Therefore, $V = \frac{4}{3}\pi (1.01\, \mathrm{cm})^3$,

Calculating the volume,

$V = \frac{4}{3}\pi (1.01)^3\, \mathrm{cm}^3 \approx \frac{4}{3} \times 3.1416 \times 1.0303\, \mathrm{cm}^3 = \frac{4}{3} \times 3.1416 \times 1.0303\, \mathrm{cm}^3 \approx 4.3434\, \mathrm{cm}^3$

Now to find the density ($\rho$) using the mass ($M = 8.635\, \mathrm{g}$) and volume ($V = 4.3434\, \mathrm{cm}^3$) calculated,

$\rho = \frac{M}{V} = \frac{8.635\, \mathrm{g}}{4.3434\, \mathrm{cm}^3} \approx 1.988\, \mathrm{g}/\mathrm{cm}^3$

Comparing this result to the given options, the closest value is:

Option B $2.0\, \mathrm{g}/\mathrm{cm}^3$

$\begin{aligned} & 0.04=4(\text { L. C.) } \\ & \Rightarrow \text { L.C }=0.01 \mathrm{~mm} \\ & \begin{array}{l} 1 \mathrm{MSD}-\frac{49}{50} M S D=0.01 \mathrm{~mm} \\ \Rightarrow 1 \mathrm{MSD}=50 \times 0.01 \mathrm{~mm} \\ \quad=0.5 \mathrm{~mm} \\ \Rightarrow 1 \mathrm{~cm}=20(0.5 \mathrm{~mm}) \end{array} \end{aligned}$

To find the refractive index of the glass slab, we first need to calculate the actual readings using the Main Scale Reading (MSR) and Vernier Coincidence (VC), and understand how to translate these values into a measurement of the refractive index. We will start by calculating the Least Count (LC) of the vernier calipers.

The least count (LC) of the traveling microscope is given by the formula:

$ \text{LC} = \text{MSD} - \text{VSD} $

Where MSD is the value of one main scale division and VSD is the value of one vernier scale division in terms of the main scale. We are told 50 vernier scale divisions equal 49 main scale divisions (MSD), so 1 VSD is $ \frac{49}{50} $ of an MSD.

Given that 20 divisions on the main scale represent 1 cm, each main scale division (MSD) represents:

$ 1 \, \text{MSD} = \frac{1 \, \text{cm}}{20} = 0.05 \, \text{cm} $

Therefore, the least count (LC) of the microscope is:

$ \text{LC} = 0.05 \, \text{cm} - \left( \frac{49}{50} \times 0.05 \, \text{cm} \right) = 0.05 \, \text{cm} - 0.049 \, \text{cm} = 0.001 \, \text{cm} $

Now, using the least count to find the total reading (TR) from both supplied observations:

1. For the mark on paper, the total reading (TR) is:

$ \text{TR} = \text{MSR} + (\text{VC} \times \text{LC}) $

$ \text{TR}_{\text{paper}} = 8.45 \, \text{cm} + (26 \times 0.001 \, \text{cm}) = 8.45 \, \text{cm} + 0.026 \, \text{cm} = 8.476 \, \text{cm} $

1. For the mark on the paper seen through the slab, the total reading is:

$ \text{TR}_{\text{seen through slab}} = 7.12 \, \text{cm} + (41 \times 0.001 \, \text{cm}) = 7.12 \, \text{cm} + 0.041 \, \text{cm} = 7.161 \, \text{cm} $

1. For the powder particle on the top surface of the glass slab:

$ \text{TR}_{\text{top surface}} = 4.05 \, \text{cm} + (1 \times 0.001 \, \text{cm}) = 4.05 \, \text{cm} + 0.001 \, \text{cm} = 4.051 \, \text{cm} $

The real depth (RD) observed directly is the difference between the first and third observations (mark on paper and powder particle on top surface):

$ \text{RD} = \text{TR}_{\text{paper}} - \text{TR}_{\text{top surface}} = 8.476 \, \text{cm} - 4.051 \, \text{cm} = 4.425 \, \text{cm} $

The apparent depth (AD) when viewed through the slab is the difference between the second and third observations:

$ \text{AD} = \text{TR}_{\text{seen through slab}} - \text{TR}_{\text{top surface}} = 7.161 \, \text{cm} - 4.051 \, \text{cm} = 3.11 \, \text{cm} $

The refractive index ($n$) of the glass slab can be found using the formula:

$ n = \frac{\text{Real Depth (RD)}}{\text{Apparent Depth (AD)}} = \frac{4.425 \, \text{cm}}{3.11 \, \text{cm}} $

Calculating this gives:

$ n = \frac{4.425}{3.11} \approx 1.42 $

Therefore, the refractive index of the glass slab is approximately 1.42, which corresponds to Option D.

To determine the spring constant $k$ of a spring experimentally, we can use the formula derived from Hooke's Law and the period of oscillation for a mass-spring system:

$T = 2 \pi \sqrt{\frac{m}{k}}$

Here, $T$ is the period of oscillation, $m$ is the mass, and $k$ is the spring constant. Rearranging the formula to solve for $k$, we get:

$k = \frac{4 \pi^2 m}{T^2}$

To find the error in $k$, we have to consider the errors in both the measurements of $T$ and $m$. Let's denote the percentage errors as follows:

$\Delta T / T \cdot 100\% = 2\%$ (positive error)

$\Delta m / m \cdot 100\% = -1\%$ (negative error)

According to the rules of error propagation, the relative error in $k$ can be found by adding the relative errors in the measurements, each multiplied by the respective powers to which they affect $k$. Since $T$ is squared in the denominator and $m$ is linear in the numerator, the calculation is as follows:

$\frac{\Delta k}{k} = \left| -2 \cdot \frac{\Delta T}{T} \right| + \left| 1 \cdot \frac{\Delta m}{m} \right|$

Substituting the percentage errors:

$\frac{\Delta k}{k} = \left| -2 \cdot 0.02 \right| + \left| 1 \cdot (-0.01) \right|$

$\frac{\Delta k}{k} = 0.04 + 0.01$

$\frac{\Delta k}{k} = 0.05$

Thus, the percentage error in determining the value of $k$ is:

$\frac{\Delta k}{k} \cdot 100\% = 5\%$

The correct answer is Option A: 5%

To determine the correct matches between List I and List II, we need to consider the dimensional formulas of the given physical quantities.

1. Torque (τ) is defined as the product of force (F) and distance (d):

$ \tau = F \cdot d $

The dimensional formula for torque is:

$ [\tau] = [M^1 L^2 T^{-2}] $

Thus, Torque corresponds to IV.

1. Magnetic field (B) is derived from the relation involving force (F), charge (q), and velocity (V):

$ F = qVB $

The dimensional formula for the magnetic field is:

$ [B] = [M^1 T^{-2} A^{-1}] $

Thus, Magnetic field corresponds to III.

1. Magnetic moment (M) is the product of current (I) and area (A):

$ M = I \cdot A $

The dimensional formula for the magnetic moment is:

$ [M] = [L^2 A^1] $

Thus, Magnetic moment corresponds to II.

1. Permeability of free space (μ₀) has the dimensional formula:

$ [μ₀] = [M^1 L^1 T^{-2} A^{-2}] $

Thus, Permeability of free space corresponds to I.

Based on the above analysis, the correct matches are:

• A. Torque – IV


• B. Magnetic Field – III


• C. Magnetic Moment – II


• D. Permeability of free space – I

Therefore, the correct option is Option D:

A-IV, B-III, C-II, D-I.

To determine the diameter of the wire using a screw gauge, we employ the formula:

$ \text{Total reading} = \text{MSR} + (\text{CSR} \times \text{LC}) $

where:

• MSR (Main Scale Reading) is the value read directly from the main scale in mm.


• CSR (Circular Scale Reading) is the number of divisions observed on the circular scale.


• LC (Least Count) is the value of one division on the circular scale, calculated as $\frac{\text{Pitch}}{\text{Number of divisions on the circular scale}}$.

Given:

• Main Scale Reading (MSR) = $1 \mathrm{~mm}$


• Circular Scale Reading (CSR) = 42 divisions


• Pitch of screw gauge = $1 \mathrm{~mm}$


• Number of divisions on circular scale = 100

First, we find the Least Count (LC):

$ LC = \frac{\text{Pitch}}{\text{Number of divisions on the circular scale}} = \frac{1 \mathrm{~mm}}{100} = 0.01 \mathrm{~mm} $

Then we calculate the total measurement of the diameter of the wire:

$ \text{Total reading} = \text{MSR} + (\text{CSR} \times \text{LC}) = 1 \mathrm{~mm} + (42 \times 0.01 \mathrm{~mm}) = 1 \mathrm{~mm} + 0.42 \mathrm{~mm} = 1.42 \mathrm{~mm} $

Given that the diameter of the wire is also represented as $\frac{x}{50} \mathrm{~mm}$, we can equate this to our found total reading:

$ 1.42 \mathrm{~mm} = \frac{x}{50} \mathrm{~mm} $

Solving for $x$:

$ 1.42 = \frac{x}{50} $

$ x = 1.42 \times 50 $

$ x = 71 $

Therefore, the value of $x$ is 71, which corresponds to Option B: 71.

To find the dimensional formula of $a b^{-1}$ in the equation given by $\left( P + \frac{a}{V^2}\right)(V-b) = RT$, where $P$ is the pressure, $V$ is the volume, and $T$ is the temperature, we will first understand the dimensions of each term in the equation. The variables mentioned have their usual meanings in the context of physics and chemistry, associated with the Ideal gas laws and the Van der Waals equation.

The dimensional formula for pressure ($P$) is $[M^1 L^{-1} T^{-2}]$ assuming $P = \frac{Force}{Area}$ and Force = $Mass \times Acceleration$.

The volume ($V$) has a dimensional formula of $[L^3]$.

Temperature ($T$) typically does not factor into the dimensional analysis directly in this context as we are considering the units it would be measured in (e.g., Kelvin), which doesn't directly convert into mass, length, and time. However, $RT$ suggests energy, and since the gas constant $R$ has dimensions including time, we consider energy's dimensions, $[M^1 L^2 T^{-2}]$, but this will not directly affect the dimension we are solving for.

Given these, let's analyze the term $\frac{a}{V^2}$ to deduce $a$'s dimensions:

• $\frac{a}{V^2}$ must have the same dimensions as pressure ($P$) for the equation to be dimensionally consistent, so

$ [a] [L^{-6}] = [M^1 L^{-1} T^{-2}] $

Thus, $a$ has the dimensional formula $[M^1 L^5 T^{-2}]$.

The term we're interested in is $a b^{-1}$, which requires finding the dimension of $b$ as it's being subtracted from $V$, implying it shares dimensions with $V$:

• Hence, $b$ has the dimensional formula $[L^3]$.

Putting it all together for $a b^{-1}$:

$ [a b^{-1}] = [M^1 L^5 T^{-2}] [L^{-3}] = [M^1 L^2 T^{-2}] $

Thus, the correct answer is:

Option D $[\mathrm{ML}^2 \mathrm{~T}^{-2}]$.