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).

To find the distance between two points when an object moves at a constant speed, we use formula,

Distance $=$ Speed × Time

$\alpha$ is a new unit of length such that 1 unit $(\alpha)=$ speed of light in vacuum (c).

The speed of light c is the distance light travels per unit of time (usually meters per second).

$ 1 \alpha=c \times(1 \text { second }) $

The time taken for light to travel from Venus to Earth is 6 minutes and 40 seconds.

1 minute $=60$ seconds

6 minutes $=6 \times 60=360$ seconds

Total time $\mathrm{t}=360 \mathrm{~s}+40 \mathrm{~s}=400$ seconds

Using distance formula:

Distance $=$ Speed of light × Time

Distance $=\mathrm{c} \times 400$ seconds

Since $1 \alpha$ is the distance light travels in 1 second ( $\mathrm{c} \times 1 \mathrm{~s}$ ), we can substitute c with our new unit,

Distance $=400 \times(\mathrm{c} \times 1 \mathrm{~s})$

Distance $=400 \alpha$

Therefore, the distance between Venus and Earth is $400 \alpha$. Hence, the correct option is (B).

To find the values of the exponents $p, q, r$, and $s$, we will use dimensional analysis. Both sides of the equation must have the same fundamental dimensions.

The dimension of distance $(\mathrm{x})$ is the dimension of length

$ [\mathrm{x}]=\mathrm{L} $

The dimension is time $[\mathrm{t}]=\mathrm{T}$

The dimension of electric field ( E ) can be derived from the formula for force on a charge, $\mathrm{F}=\mathrm{qE}$. Therefore, $\mathrm{E}=\mathrm{F} / \mathrm{q}$.

The dimensions of force is $[\mathrm{F}]=\left[\mathrm{MLT}^{-2}\right]$

Charge $(\mathrm{q})=$ Current $\times$ Time

So, the dimensions of charge is, $[\mathrm{q}]=[\mathrm{AT}]$

So, the dimensions of electric field is, $[\mathrm{E}]=\frac{\left[\mathrm{MLT}^{-2}\right]}{[\mathrm{AT}]}=\left[\mathrm{MLT}^{-3} \mathrm{~A}^{-1}\right]$

Using relationship between electric field (E) and and magnetic field (H) in electromagnetic waves. In a vacuum, the ratio of the magnitudes of the electric field to the magnetic field (B) is the speed of light (c),

$ \mathrm{E} / \mathrm{B}=\mathrm{c} $

Using the formula of magnetic force on a current-carrying conductor:

$ \mathrm{F}=\mathrm{BIL} $

Where, $\mathrm{F}=$ force $\mathrm{B}=$ magnetic field (flux density), $\mathrm{I}=$ current and $\mathrm{L}=$ length

$ \mathrm{B}=\frac{\mathrm{F}}{\mathrm{IL}} $

Substituting the dimensions,

$ [\mathrm{B}]=\frac{\left[\mathrm{MLT}^{-2}\right]}{[\mathrm{LA}]} $

$\Rightarrow $ $ [\mathrm{B}]=\left[\mathrm{M} \mathrm{~T}^{-2} \mathrm{I}^{-1}\right] $

The relation between B and H is $\mathrm{B}=\mu \mathrm{H} \Rightarrow \frac{\mathrm{B}}{\mu}=\mathrm{H}$. Where $\mu=$ permeability.

Using the formula of magnetic field, $B=\frac{\mu i}{2 \pi r} \Rightarrow\left[\frac{B}{\mu}\right]=\left[\frac{i}{r}\right]$, here $r$ is distance from straight conductor carrying current i.

$ \left[\frac{\mathrm{B}}{\mu}\right]=[\mathrm{H}]=\frac{[\mathrm{A}]}{[\mathrm{L}]}=\left[\mathrm{L}^{-1} \mathrm{~A}\right] $

Putting the dimensions of electric field and speed of light, the dimensions of magnetic field is,

$ [\mathrm{H}]=\frac{\left[\mathrm{MLT}^{-3} \mathrm{~A}^{-1}\right]}{\left[\mathrm{LT}^{-1}\right]}=\left[\mathrm{MT}^{-2} \mathrm{~A}^{-1}\right] $

We can derive the dimensions of permittivity ( $\epsilon$ ) from Coulomb's Law, $\mathrm{F}=\frac{1}{4 \pi \epsilon} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}^2}$.

$[\epsilon]=\left[\frac{q^2}{\mathrm{Fr}^2}\right]$

$ [\epsilon]=\frac{[\mathrm{AT}]^2}{\left[\mathrm{MLT}^{-2}\right]\left[\mathrm{L}^2\right]}=\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{~A}^2\right] $

Substituting the dimensions into the given equation,

$ [\mathrm{H}]=\frac{[\mathrm{x}]^{\mathrm{p}}[\epsilon]^{\mathrm{q}}[\mathrm{E}]^{\mathrm{r}}}{[\mathrm{t}]^{\mathrm{s}}} $

$\Rightarrow $ $\left[L^{-1} A\right]=\frac{[L]^p\left[M^{-1} L^{-3} T^4 A^2\right]^q\left[M L T^{-3} A^{-1}\right]^r}{[T]^s}$

$ \left[L^{-1} A\right]=\left[M^{-q+r} L^{p-3 q+r} T^{4 q-3 r-s} A^{2 q-r}\right] $

On comparing the powers of $\mathrm{M}, \mathrm{L}, \mathrm{T}$, and A on both sides,

For $\mathrm{M},-\mathrm{q}+\mathrm{r}=0 \ldots$ (i)

For $\mathrm{A}, 2 \mathrm{q}-\mathrm{r}=1$. Since $\mathrm{r}=\mathrm{q}$, then $2 \mathrm{q}-\mathrm{q}=1$, so $\mathrm{q}=1$ and $\mathrm{r}=1$.

For $\mathrm{L}, \mathrm{p}-3 \mathrm{q}+\mathrm{r}=-1 \ldots$ (ii).

$ \begin{gathered} p-3(1)+1=-1 \\ p-2=-1 \Rightarrow p=1 \end{gathered} $

For T, $4 q-3 r-s=0 \ldots$ (iii)

$ \begin{gathered} 4(1)-3(1)-s=0 \\ 1-s=0 \Rightarrow s=1 \end{gathered} $

Therefore, the values are $\mathrm{p}=1, \mathrm{q}=1, \mathrm{r}=1, \mathrm{~s}=1$.

Hence, the correct option is (A).

The volume $V$ of a sphere is given by the formula $V=\frac{4}{3} \pi r^3$, where $r$ is the radius of the sphere.

Since the error is given in terms of diameter (D), we express the radius as $r=\frac{D}{2}$.

Substituting this into the volume formula:

$ \mathrm{V}=\frac{4}{3} \pi\left(\frac{\mathrm{D}}{2}\right)^3 $

$\Rightarrow $ $\mathrm{V}=\frac{4}{3} \pi\left(\frac{\mathrm{D}^3}{8}\right)$

$\Rightarrow $ $ V=\frac{1}{6} \pi D^3 $

To find the relative error, we take the natural logarithm $(\ln )$ on both sides of the equation :

$ \ln (V)=\ln \left(\frac{\pi}{6}\right)+\ln \left(D^3\right) $

$\Rightarrow $ $ \ln (V)=\ln \left(\frac{\pi}{6}\right)+3 \ln (D) $

Now, on differentiating both sides.

$ \frac{\mathrm{dV}}{\mathrm{~V}}=0+3\left(\frac{\mathrm{dD}}{\mathrm{D}}\right) $

$\Rightarrow $ $ \frac{\mathrm{dV}}{\mathrm{~V}}=3 \frac{\mathrm{dD}}{\mathrm{D}} $

To express this as a percentage error, we multiply both sides by 100 :

$ \left(\frac{\mathrm{dV}}{\mathrm{~V}} \times 100\right)=3 \times\left(\frac{\mathrm{dD}}{\mathrm{D}} \times 100\right) $

It is given that the percentage error in the diameter is $2 \%$.

$ \frac{d D}{D} \times 100=2 \% $

So, percentage error in Volume $=3 \times 2 \%$

Percentage error in Volume $=6 \%$

Therefore, the correct option is (C).

For the dimensional equation of Boltzmann Constant $\left(k_B\right)$

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

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

To find the dimensions of $\mathrm{k}_{\mathrm{B}}$ :

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

Using the formula Work $=$ Force × Distance, dimensions of energy is $[\mathrm{E}]=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]$

The dimensions of temperature is $[\mathrm{T}]=[\mathrm{K}]$

So,

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

Thus, (A) Boltzmann constant → (III) [ $\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~K}^{-1}$ ]

For the dimensions of Stefan's Constant ( $\sigma$ )

According to Stefan-Boltzmann law, the energy radiated per unit area per unit time $(\mathrm{P})$ is:

$ \mathrm{P}=\sigma \mathrm{T}^4 $

Where $\mathrm{P}=\frac{\text { Energy }}{\text { Area× Time }}$

Dimensions of

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

Dimensions of $\mathrm{T}^4$ is $\left[\mathrm{K}^4\right]$

So, the dimensions of Stefan's constant is

$ [\sigma]=\frac{[\mathrm{P}]}{\left[\mathrm{T}^4\right]}=\frac{\left[\mathrm{MT}^{-3}\right]}{\left[\mathrm{K}^4\right]}=\left[\mathrm{MT}^{-3} \mathrm{~K}^{-4}\right]=\left[\mathrm{ML}^0 \mathrm{~T}^{-3} \mathrm{~K}^{-4}\right] $

Thus, (B) Stefan's constant → (IV)[ $\mathrm{ML}^0 \mathrm{~T}^{-3} \mathrm{~K}^{-4}$ ]

For the dimensions of Planck's Constant (h):

The energy (E) of a photon is given by $E=h \nu$, where $\nu$ is frequency (Time ${ }^{-1}$ ).

So, the dimension of Planck's constant is

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

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

Thus, (C) Planck's constant → (II)[ $\mathrm{ML}^2 \mathrm{~T}^{-1}$ ]

For the dimensions of Gravitational Constant (G):

From Newton's law of gravitation $F=\frac{G m_1 m_2}{r^2} \Rightarrow G=\frac{F r^2}{m_1 m_2}$

The dimensions of Force is $[\mathrm{F}]=\left[\mathrm{MLT}^{-2}\right]$

The dimensions of Distance squared is $\left[\mathrm{r}^2\right]=\left[\mathrm{L}^2\right]$

The dimensions of mass squared is $\left[\mathrm{m}_1 \mathrm{~m}_2\right]=\left[\mathrm{M}^2\right]$

So, the dimensions of Gravitational constant is

$ [\mathrm{G}]=\frac{\left[\mathrm{MLT}^{-2}\right]\left[\mathrm{L}^2\right]}{\left[\mathrm{M}^2\right]}=\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right] $

Thus (D) Gravitational constant $\rightarrow$ (I) $\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]$

$ \mathrm{A} \rightarrow \mathrm{III}, \mathrm{~B} \rightarrow \mathrm{IV}, \mathrm{C} \rightarrow \mathrm{II}, \mathrm{D} \rightarrow \mathrm{I} $

Therefore, the correct option is (C).

The density $(\rho)$ of a body is defined as the ratio of its mass (m) to its volume (V).

$ \rho=\frac{m}{V} $

For a uniform cylinder of length l and diameter D , the volume is,

$ V=\pi r^2 l=\pi\left(\frac{D}{2}\right)^2 l=\frac{\pi D^2 l}{4} $

Substituting the volume into the density formula:

$ \rho=\frac{\mathrm{m}}{\mathrm{~V}}=\frac{4 \mathrm{~m}}{\pi \mathrm{D}^2 \mathrm{l}} $

To find the fractional error, we take the natural logarithm $(\ln )$ on both sides:

$ \ln (\rho)=\ln (4)+\ln (m)-\ln (\pi)-2 \ln (D)-\ln (l) $

Differentiating both sides to find the relative change

$ \frac{\mathrm{d} \rho}{\rho}=\frac{\mathrm{dm}}{\mathrm{~m}}-\frac{2 \mathrm{dD}}{\mathrm{D}}-\frac{\mathrm{dl}}{\mathrm{l}} $

The maximum possible fractional error is the sum of the absolute values of the individual relative errors:

$ \frac{\Delta \rho}{\rho}=\frac{\Delta \mathrm{m}}{\mathrm{~m}}+\frac{2 \Delta \mathrm{D}}{\mathrm{D}}+\frac{\Delta \mathrm{l}}{\mathrm{l}} $

For the percentage fractional error,

$ \% \text { Error in } \rho=\left(\frac{\Delta \mathrm{m}}{\mathrm{~m}}+\frac{2 \Delta \mathrm{D}}{\mathrm{D}}+\frac{\Delta \mathrm{l}}{\mathrm{l}}\right) \times 100 $

The measured value of mass is $97.42 \pm 0.02 \mathrm{~g} \Rightarrow \mathrm{~m}=97.42 \mathrm{~g}$ and the absolute error is $\Delta \mathrm{m}=0.02 \mathrm{~g}$ So, relative error in mass is,

$ \frac{\Delta \mathrm{m}}{\mathrm{~m}}=\frac{0.02}{97.42} \approx 0.000205 $

The measured value of length is $8.35 \pm 0.05 \mathrm{~mm} \Rightarrow \mathrm{l}=8.35 \mathrm{~mm}$ and the absolute error is $\Delta \mathrm{l}=0.05 \mathrm{~mm}$

So, relative error in length is,

$ \frac{\Delta \mathrm{l}}{\mathrm{l}}=\frac{0.05}{8.35} \approx 0.005988 $

The measured value of diameter is $20.20 \pm 0.02 \mathrm{~mm} \Rightarrow \mathrm{D}=20.20 \mathrm{~mm}$ and the absolute error is $\Delta \mathrm{D}=$ 0.02 mm

So, relative error in diameter is,

$ \frac{2 \Delta \mathrm{D}}{\mathrm{D}}=\frac{2 \times 0.02}{20.20}=\frac{0.04}{20.20} \approx 0.001980 $

Putting the values in the expression for the percentage error in density,

$ \frac{\Delta \rho}{\rho} \times 100 \%=(0.000205+0.005988+0.001980) \times 100 \%=0.008173 \times 100 \% $

$\Rightarrow $ $ \frac{\Delta \rho}{\rho} \times 100 \%=0.8173 \% \approx 0.82 \% $

Therefore, the calculated percentage fraction error in density is approximately $0.82 \%$.

Hence, the correct option is (B).

According to the principle of homogeneity, dimensions of each term on both sides of a equation must be the same. Also, only physical quantities of the same dimensions can be added or subtracted.

The potential energy of the particle is given as $V=\frac{A \sqrt{x}}{x+B}$, where $A$ and $B$ are constant.

In the denominator of the expression $\mathrm{x}+\mathrm{B}$, the constant B is added to distance x . Therefore, the dimensions of $B$ must be the same as the dimensions of $x$.

$ [\mathrm{B}]=[\mathrm{x}]=\left[\mathrm{L}^1\right] $

The potential energy $V$ has the same dimensions as work or energy.

$ [\mathrm{V}]=[\mathrm{F} \times \mathrm{x}]=\left[\mathrm{M}^1 \mathrm{~L}^1 \mathrm{~T}^{-2} \times \mathrm{L}^1\right]=\left[\mathrm{M}^1 \mathrm{~L}^2 \mathrm{~T}^{-2}\right] $

On rearranging the formula of potential energy for A :

$ A=\frac{V(x+B)}{\sqrt{x}} $

Substituting the dimensions:

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

$\Rightarrow $ $[\mathrm{A}]=\frac{\left[\mathrm{M}^1 \mathrm{~L}^3 \mathrm{~T}^{-2}\right]}{\left[\mathrm{L}^{1 / 2}\right]}$

$\Rightarrow $$ [\mathrm{A}]=\left[\mathrm{M}^1 \mathrm{~L}^{3-\frac{1}{2}} \mathrm{~T}^{-2}\right]=\left[\mathrm{M}^1 \mathrm{~L}^{5 / 2} \mathrm{~T}^{-2}\right] $

To find the dimensions of the product AB :

$ [\mathrm{AB}]=[\mathrm{A}] \times[\mathrm{B}] $

$\Rightarrow $ $[\mathrm{AB}]=\left[\mathrm{M}^1 \mathrm{~L}^{5 / 2} \mathrm{~T}^{-2}\right] \times\left[\mathrm{L}^1\right]$

$\Rightarrow $ $[\mathrm{AB}]=\left[\mathrm{M}^1 \mathrm{~L}^{5 / 2+1} \mathrm{~T}^{-2}\right]$

$\Rightarrow $ $ [\mathrm{AB}]=\left[\mathrm{M}^1 \mathrm{~L}^{7 / 2} \mathrm{~T}^{-2}\right] $

Therefore, the dimensions of $A B$ is $\left[M^1 L^{7 / 2} T^{-2}\right]$.

Thus, the correct option is (D).

We compare the dimensions of each expression in List-II with the required fundamental quantities in List-I.

Given:

• $h$ = Planck's constant

• $G$ = gravitational constant

• $c$ = speed of light

Their dimensions are:

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

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

$ [c] = L T^{-1} $

Now let us check each expression.

I. $\sqrt{\dfrac{hc}{G}}$

First find dimensions inside the root:

$ \frac{hc}{G} = \frac{(M L^2 T^{-1})(L T^{-1})}{M^{-1} L^3 T^{-2}} $

$ = \frac{M L^3 T^{-2}}{M^{-1} L^3 T^{-2}} = M^2 $

So,

$ \sqrt{\frac{hc}{G}} = M $

Hence, expression I represents kilogram.

So,

$ C \to I $

II. $\sqrt{\dfrac{Gh}{c^5}}$

Dimensions inside root:

$ \frac{Gh}{c^5} = \frac{(M^{-1}L^3T^{-2})(M L^2 T^{-1})}{(L T^{-1})^5} $

$ = \frac{L^5 T^{-3}}{L^5 T^{-5}} = T^2 $

Therefore,

$ \sqrt{\frac{Gh}{c^5}} = T $

So expression II represents second.

Thus,

$ B \to II $

III. $\sqrt{\dfrac{K^2 L^2 c^3}{Gh}}$

Here $K$ stands for temperature dimension and $L$ for length dimension.

Now,

$ \frac{K^2 L^2 c^3}{Gh} $

We already know:

$ [Gh] = (M^{-1}L^3T^{-2})(M L^2 T^{-1}) = L^5 T^{-3} $

Also,

$ [c^3] = L^3 T^{-3} $

So numerator dimensions:

$ K^2 L^2 \cdot L^3 T^{-3} = K^2 L^5 T^{-3} $

Thus,

$ \frac{K^2 L^5 T^{-3}}{L^5 T^{-3}} = K^2 $

Therefore,

$ \sqrt{\frac{K^2 L^2 c^3}{Gh}} = K $

So expression III represents kelvin.

Thus,

$ D \to III $

IV. $\sqrt{\dfrac{Gh}{c^3}}$

Dimensions inside root:

$ \frac{Gh}{c^3} = \frac{L^5 T^{-3}}{L^3 T^{-3}} = L^2 $

Hence,

$ \sqrt{\frac{Gh}{c^3}} = L $

So expression IV represents meter.

Thus,

$ A \to IV $

Final matching

$ A \to IV,\quad B \to II,\quad C \to I,\quad D \to III $

So the correct option is:

$ \boxed{\text{Option B}} $

Using Ohm’s law, the resistance is

$ R = \frac{V}{I} = \frac{10}{5} = 2 \ \Omega $

Now, for error estimation in a quotient, NCERT method gives:

$ \frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I} $

Given least count of voltmeter:

$ \Delta V = 500 \text{ mV} = 0.5 \text{ V} $

Given least count of ammeter:

$ \Delta I = 200 \text{ mA} = 0.2 \text{ A} $

So,

$ \frac{\Delta R}{R} = \frac{0.5}{10} + \frac{0.2}{5} $

$ \frac{\Delta R}{R} = 0.05 + 0.04 = 0.09 $

Hence absolute error in resistance:

$ \Delta R = R \times 0.09 = 2 \times 0.09 = 0.18 \ \Omega $

Therefore, the estimated error in resistance measurement is

$ \boxed{0.18 \ \Omega} $

So, the correct option is D.

When the jaws are fully closed, the zero of the Vernier scale is shifted to the right of the zero of the main scale.

This means the Vernier caliper has a positive zero error.

Now let us find its magnitude.

Given:

• $1$ main scale division $= 1 \, \text{mm}$

• Total Vernier divisions $= 10$

For a standard Vernier calipers:

$ 10 \, \text{VSD} = 9 \, \text{MSD} $

So,

$ 1 \, \text{VSD} = \frac{9}{10} \, \text{MSD} $

Since

$ 1 \, \text{MSD} = 1 \, \text{mm} $

therefore,

$ 1 \, \text{VSD} = 0.9 \, \text{mm} $

Hence the least count is

$ \text{L.C.} = 1 \, \text{MSD} - 1 \, \text{VSD} = 1 - 0.9 = 0.1 \, \text{mm} $

Now, it is given that the $7^\text{th}$ Vernier division coincides with a main scale division.

So zero error

$ = 7 \times \text{L.C.} = 7 \times 0.1 = 0.7 \, \text{mm} $

Convert into cm:

$ 0.7 \, \text{mm} = 0.07 \, \text{cm} $

Since the zero of Vernier scale is to the right, the zero error is positive.

Therefore, the Vernier caliper has

$ \boxed{0.07 \, \text{cm positive zero error}} $

So, the correct option is Option C.

Write the dimensions of each quantity first.

We know:

• Resistance, $R$:

$ [R] = M L^2 T^{-3} A^{-2} $

• Inductance, $L$:

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

• Capacitance, $C$:

$ [C] = M^{-1} L^{-2} T^4 A^2 $

We need:

$ M L^2 T^{-4} A^{-2} $

Now check each option.

Option A: $\dfrac{R}{\sqrt{LC}}$

First find $[LC]$:

$ [LC] = \left(M L^2 T^{-2} A^{-2}\right)\left(M^{-1} L^{-2} T^4 A^2\right)=T^2 $

So,

$ [\sqrt{LC}] = T $

Hence,

$ \left[\frac{R}{\sqrt{LC}}\right] = \frac{M L^2 T^{-3} A^{-2}}{T} = M L^2 T^{-4} A^{-2} $

This matches the required dimension.

Option B: $\dfrac{R}{LC}$

Since

$ [LC]=T^2 $

we get

$ \left[\frac{R}{LC}\right] = \frac{M L^2 T^{-3} A^{-2}}{T^2} = M L^2 T^{-5} A^{-2} $

Not matching.

Option C: $\dfrac{C}{\sqrt{LR}}$

First,

$ [LR] = \left(M L^2 T^{-2} A^{-2}\right)\left(M L^2 T^{-3} A^{-2}\right) = M^2 L^4 T^{-5} A^{-4} $

So,

$ [\sqrt{LR}] = M L^2 T^{-5/2} A^{-2} $

Then,

$ \left[\frac{C}{\sqrt{LR}}\right] = \frac{M^{-1} L^{-2} T^4 A^2}{M L^2 T^{-5/2} A^{-2}} = M^{-2} L^{-4} T^{13/2} A^4 $

Not matching.

Option D: $\dfrac{1}{R}\sqrt{\dfrac{L}{C}}$

First,

$ \left[\frac{L}{C}\right] = \frac{M L^2 T^{-2} A^{-2}}{M^{-1} L^{-2} T^4 A^2} = M^2 L^4 T^{-6} A^{-4} $

So,

$ \left[\sqrt{\frac{L}{C}}\right] = M L^2 T^{-3} A^{-2} $

Also,

$ \left[\frac{1}{R}\right] = M^{-1} L^{-2} T^3 A^2 $

Therefore,

$ \left[\frac{1}{R}\sqrt{\frac{L}{C}}\right] = 1 $

This is dimensionless, so not matching.

Hence, the correct answer is

$ \boxed{\text{Option A } \frac{R}{\sqrt{LC}}} $

For matching, we find the dimensional formula of each quantity one by one.

A. Planck's constant

Using the relation

$ E = h\nu $

So,

$ h = \frac{E}{\nu} $

Dimension of energy is

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

Dimension of frequency is

$ [\nu] = \mathrm{T}^{-1} $

Therefore,

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

So, $A \to \text{III}$.

B. Stopping potential

Stopping potential is a potential difference.

We know

$ V = \frac{\text{work}}{\text{charge}} $

Dimension of work is

$ \mathrm{ML}^2\mathrm{T}^{-2} $

Dimension of charge is

$ \mathrm{AT} $

Hence,

$ [V] = \frac{\mathrm{ML}^2\mathrm{T}^{-2}}{\mathrm{AT}} = \mathrm{ML}^2\mathrm{T}^{-3}\mathrm{A}^{-1} $

So, $B \to \text{IV}$.

C. Work function

Work function is the minimum energy needed to remove an electron. So its dimension is same as energy:

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

So, $C \to \text{I}$.

D. Threshold frequency

Frequency has dimension

$ [\nu_0] = \mathrm{T}^{-1} $

So, $D \to \text{II}$.

Final matching

$ A \to \text{III}, \quad B \to \text{IV}, \quad C \to \text{I}, \quad D \to \text{II} $

So the correct option is:

$ \boxed{\text{Option A}} $

Given:

• In $5$ complete rotations, the screw moves linearly by $2.5 \, \text{mm}$

• Circular scale has $100$ divisions

First, find the pitch of the screw gauge.

Pitch means the linear distance moved in one complete rotation.

$ \text{Pitch} = \frac{2.5}{5} = 0.5 \, \text{mm} $

Now, least count is given by:

$ \text{Least Count} = \frac{\text{Pitch}}{\text{Number of circular scale divisions}} $

So,

$ \text{Least Count} = \frac{0.5}{100} = 0.005 \, \text{mm} $

$ 0.005 \, \text{mm} = 5 \times 10^{-3} \, \text{mm} $

So, the correct answer is:

$ \boxed{5 \times 10^{-3} \, \text{mm}} $

Option D

Using the dimensional formula of gravitational constant:

$ [G] = \frac{[F]\,[r^2]}{[m_1][m_2]} $

Since,

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

So,

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

Now, we need to express this in terms of $h$, $L$, $M$, and $T$.

We know Planck’s constant has dimensions:

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

Assume

$ [G] = [h^a L^b M^c T^d] $

Then,

$ [h^a L^b M^c T^d] = [M^a L^{2a} T^{-a}] [L^b] [M^c] [T^d] = [M^{a+c} L^{2a+b} T^{-a+d}] $

Compare with

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

So,

$ a+c = -1 $

$ 2a+b = 3 $

$ -a+d = -2 $

Now check options.

Option A:

$ [h T L M^{-2}] $

Dimensions:

$ [h][T][L][M^{-2}] = [M L^2 T^{-1}] [T] [L] [M^{-2}] = [M^{-1} L^3 T^0] $

Not correct.

Option B:

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

Dimensions:

$ [h][T^{-1}][L][M^{-2}] = [M L^2 T^{-1}] [T^{-1}] [L] [M^{-2}] = [M^{-1} L^3 T^{-2}] $

This matches $[G]$.

So, the correct answer is:

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

Hence, Option B is correct.

For a screw gauge, first we find the least count.

$ \text{Least count}=\frac{\text{Pitch}}{\text{Number of circular scale divisions}} =\frac{0.1\ \text{mm}}{100}=0.001\ \text{mm} $

Now let us find the zero error.

When the two studs are in contact, the zero of main scale reference line coincides with the $5^\text{th}$ division of circular scale.

This means the screw gauge shows a reading of

$ 5 \times 0.001 = 0.005\ \text{mm} $

So the instrument has a positive zero error of

$ +0.005\ \text{mm} $

Hence, zero correction is

$ -0.005\ \text{mm} $

Now for the sphere:

• Main scale reading $= 5\ \text{mm}$

• Circular scale reading $= 50$

So observed reading is

$ 5 + 50 \times 0.001 = 5 + 0.050 = 5.050\ \text{mm} $

Applying zero correction:

$ \text{Correct reading} = 5.050 - 0.005 = 5.045\ \text{mm} $

Therefore, the diameter of the sphere is

$ \boxed{5.045\ \text{mm}} $

So, the correct option is A.

For Young’s modulus of a wire,

$ Y=\frac{FL}{A\,\Delta L} $

For a circular wire, area

$ A=\frac{\pi d^2}{4} $

So,

$ Y=\frac{4FL}{\pi d^2 \Delta L} $

Since we have to find the error in $Y$, and error due to load is ignored, the fractional error is

$ \frac{\Delta Y}{Y}=\frac{\Delta L}{L}+2\frac{\Delta d}{d}+\frac{\Delta (\Delta L)}{\Delta L} $

Now substitute the given measurements.

Diameter:

$ d=0.08\ \text{cm}, \qquad \Delta d=0.001\ \text{cm} $

So,

$ 2\frac{\Delta d}{d}=2\times \frac{0.001}{0.08}=0.025 $

Length:

$ L=150\ \text{cm}, \qquad \Delta L=0.1\ \text{cm} $

Hence,

$ \frac{\Delta L}{L}=\frac{0.1}{150}=0.00067 $

Extension:

$ \Delta l=0.5\ \text{cm}, \qquad \Delta (\Delta l)=0.001\ \text{cm} $

Thus,

$ \frac{\Delta (\Delta l)}{\Delta l}=\frac{0.001}{0.5}=0.002 $

Therefore total fractional error is

$ \frac{\Delta Y}{Y}=0.00067+0.025+0.002=0.02767 $

Now calculate $Y$.

Convert to SI units:

$ L=150\ \text{cm}=1.5\ \text{m} $

$ d=0.08\ \text{cm}=8\times 10^{-4}\ \text{m} $

$ \Delta l=0.5\ \text{cm}=5\times 10^{-3}\ \text{m} $

Given force,

$ F=100\ \text{N} $

Area:

$ A=\frac{\pi d^2}{4} =\frac{\pi (8\times 10^{-4})^2}{4} \approx 5.03\times 10^{-7}\ \text{m}^2 $

So,

$ Y=\frac{FL}{A\Delta l} =\frac{100\times 1.5}{(5.03\times 10^{-7})(5\times 10^{-3})} \approx 5.97\times 10^{10}\ \text{N/m}^2 $

Hence absolute error:

$ \Delta Y=Y\left(\frac{\Delta Y}{Y}\right) $

$ \Delta Y=(5.97\times 10^{10})(0.02767) \approx 1.65\times 10^9\ \text{N/m}^2 $

Therefore,

$ \alpha=1.65 $

So the correct answer is:

$ \boxed{1.65} $

Option B.

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.