Units & Measurements

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.

S is emf per unit temperature ⇒

$[S]=\frac{[V]}{[K]}=\frac{M\,L^2\,T^{-3}\,I^{-1}}{K} =M\,L^2\,T^{-3}\,I^{-1}\,K^{-1}.$

σ is electrical conductivity ⇒

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

κ is thermal conductivity ⇒

$[κ]=\frac{\dot Q/(A)}{∇T} =\frac{M\,L^2\,T^{-3}/L^2}{K\,L^{-1}} =M\,L\,T^{-3}\,K^{-1}.$

Now

$[Z] =[S]^2\,[σ]\,[κ]^{-1} =(M\,L^2\,T^{-3}\,I^{-1}\,K^{-1})^2 \times M^{-1}\,L^{-3}\,T^3\,I^2 \times (M\,L\,T^{-3}\,K^{-1})^{-1}$

Simplifying exponents gives

$[Z]=M^0\,L^0\,T^0\,I^0\,K^{-1}.$

Answer: Option B.

For vernier scale,

$ \begin{aligned} & 10 \mathrm{div}=7 \mathrm{~mm} \\ & 1 \mathrm{div}=0.7 \mathrm{~mm} \end{aligned} $

Reading: main scale $=1 \mathrm{~mm}$

VS 1 marking matches with MS div

So VS reads $(1-0.7 \mathrm{~mm})=0.3 \mathrm{~mm}$

$ \begin{aligned} \text { So total } & =(1+0.3) \mathrm{mm} \\ & =1.3 \mathrm{~mm} \\ & =0.13 \mathrm{~cm} \text { Ans. } \end{aligned} $

$ \begin{aligned} & l=10.5 \mathrm{~cm} \rightarrow 3 \mathrm{SF} \\ & b=0.05 \mathrm{~mm}=5 \times 10^{-3} \mathrm{~cm} \rightarrow 1 \mathrm{SF} \\ & h=6.0 \mu \mathrm{~m}=6.0 \times 10^{-4} \mathrm{~cm} \rightarrow 2 \mathrm{SF} \end{aligned} $

we know, $V=$ lbh

$ \begin{aligned}So\,\,\,\, v & =10.5 \times 5 \times 10^{-3} \times 6.0 \times 10^{-4} \\ & =315 \times 10^{-7} \mathrm{~cm}^3 \\ & =3.15 \times 10^{-5} \mathrm{~cm}^3 \end{aligned} $

The final answer should have the same number of significant figures as the factor with the fewest significant figures.

$ \text { So } V=3 \times 10^{-5} \mathrm{~cm}^3 \quad \rightarrow 1 \mathrm{SF} $

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

To determine the value of one main scale division (MSD) for the given vernier calipers, we'll analyze the relationship between the main scale divisions, vernier scale divisions, and the least count of the instrument.

Given:

Number of vernier scale divisions (n): 20

Vernier scale coincides with the 19th division on the main scale.

Least count (LC): $ 0.1 \, \text{mm} $

Understanding the Vernier Calipers Configuration:

Relation between Vernier and Main Scale Divisions:

The total length of the vernier scale (comprising 20 divisions) coincides with the length of 19 main scale divisions.

Mathematically, this is expressed as:

$ n \times \text{VSD} = (n - 1) \times \text{MSD} $

where:

$ \text{VSD} $ is the value of one vernier scale division.

$ \text{MSD} $ is the value of one main scale division.

Calculating the Value of One Vernier Scale Division ($ \text{VSD} $):

$ \text{VSD} = \frac{(n - 1) \times \text{MSD}}{n} $

Least Count (LC):

The least count is the smallest measurement that can be accurately read using the instrument.

It is calculated as:

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

Substituting $ \text{VSD} $ from step 2:

$ \text{LC} = \text{MSD} - \left( \frac{(n - 1) \times \text{MSD}}{n} \right) $

$ \text{LC} = \text{MSD} \left( 1 - \frac{n - 1}{n} \right) $

$ \text{LC} = \text{MSD} \left( \frac{n - (n - 1)}{n} \right) $

$ \text{LC} = \text{MSD} \left( \frac{1}{n} \right) $

Solving for One Main Scale Division ($ \text{MSD} $):

Rearranging the equation:

$ \text{LC} = \frac{\text{MSD}}{n} $

$ \text{MSD} = \text{LC} \times n $

Substituting the given values ($ \text{LC} = 0.1 \, \text{mm}, n = 20 $):

$ \text{MSD} = 0.1 \, \text{mm} \times 20 $

$ \text{MSD} = 2 \, \text{mm} $

Conclusion:

The value of one main scale division is 2 mm, which corresponds to Option B.

To find the arithmetic mean of the time periods recorded, we need to sum up the values and then divide by the number of readings. Let's calculate the sum first:

$4.62 \mathrm{~s} + 4.632 \mathrm{~s} + 4.6 \mathrm{~s} + 4.64 \mathrm{~s}$

Adding these values together:

$4.62 + 4.632 + 4.6 + 4.64 = 18.492 \mathrm{~s}$

Now, we divide this sum by the number of readings, which is 4:

$ \frac{18.492 \mathrm{~s}}{4} = 4.623 \mathrm{~s} $

So, the arithmetic mean of these readings is $4.623 \mathrm{~s}$. However, we need to consider the significant figures. The least number of significant figures among the readings is 2 (from 4.6 s). Hence, the mean should also be represented with 2 significant figures.

In this case, the correct answer with proper significant figures is:

Option B

4.6 s

To determine which option is correct based on the principle of homogeneity of dimensions, we need to ensure that both sides of the equation have the same dimensions. The time period ($T$) is measured in units of time ($T$), the gravitational constant ($G$) has units of $\text{m}^3\text{kg}^{-1}\text{s}^{-2}$, mass ($M$) has units of mass ($M$), and the radius of orbit ($r$) has units of length ($L$).

Let's analyze each option:

Option A: $T^2=\frac{4 \pi^2 r}{G M^2}$

The dimensions of the right-hand side of the equation are $\frac{L}{\left(\frac{L^3}{MT^2}\right)M^2}=\frac{L}{L^3M^{-1}T^{-2}M^2}=\frac{L}{L^3T^{-2}}=L^{-2}T^2$, which do not match with $T^2$ (time squared). Thus, Option A is incorrect.

Option B: $T^2=4 \pi^2 r^3$

The dimensions of the right-hand side are $L^3$, which clearly do not match the dimensions $T^2$ of the squared time period. So, Option B is incorrect.

Option C: $T^2=\frac{4 \pi^2 r^3}{G M}$

The dimensions of the right-hand side of the equation are $\frac{L^3}{\left(\frac{L^3}{MT^2}\right)M}=\frac{L^3}{L^3T^{-2}}=T^2$, which match the dimensions of the squared time period $T^2$. Therefore, Option C is correct based on the principle of homogeneity of dimensions.

Option D: $T^2=\frac{4 \pi^2 r^2}{G M}$

The dimensions of the right-hand side are $\frac{L^2}{\left(\frac{L^3}{MT^2}\right)M}=\frac{L^2}{L^3M^{-1}T^{-2}M}=L^{-1}T^2$, which do not match with $T^2$ (time squared). Hence, Option D is incorrect.

Thus, based on the principle of homogeneity of dimensions, Option C is the correct one: $T^2=\frac{4 \pi^2 r^3}{G M}$. This equation also corresponds to Kepler's third law of planetary motion, which relates the orbital period of a planet to its orbital radius, considering the mass of the central body.

To determine which of the options is NOT correct, we need to analyze the dimensional consistency of each term in the given equation of the stationary wave:

$y = 2a \sin \left(\frac{2\pi nt}{\lambda}\right) \cos \left(\frac{2\pi x}{\lambda}\right).$

Let's break down the dimensions for each relevant term:

1. Analyzing $\mathrm{nt}$:

The argument of the sine function $\frac{2\pi nt}{\lambda}$ must be dimensionless. Therefore, the dimensions of $\mathrm{nt}$ should be the same as the dimensions of $\lambda$ (wavelength), which is [L].

Thus, the dimensions of $nt$ should be [L].
Hence, Option A is correct.

2. Analyzing $n$:

From the above analysis, since $\mathrm{nt}$ has the dimension [L] and $t$ (time) has the dimension [T], it follows that:

$n = \frac{\text{Dimension of } nt}{\text{Dimension of } t} = \frac{[L]}{[T]} = [\mathrm{LT}^{-1}].$

Hence, Option B is also correct.

3. Analyzing $x$:

In the argument of the cosine function $\frac{2\pi x}{\lambda}$, since it must be dimensionless, the dimensions of $x$ should be the same as the dimensions of $\lambda$ (wavelength), which is [L].

Hence, the dimensions of $x$ should be [L].
Therefore, Option C is correct.

4. Analyzing $\frac{n}{\lambda}$:

From the dimensions we have determined for $n$ and $\lambda$:

$\frac{n}{\lambda} = \frac{[\mathrm{LT}^{-1}]}{[L]} = [\mathrm{T}^{-1}].$

Thus, the dimensions of $\frac{n}{\lambda}$ should be [T^-1], not [T].
This indicates that Option D is NOT correct.

Conclusion:

Option D is the correct answer since it is NOT dimensionally correct.

Significant figures in a number represent the digits that carry meaning contributing to its precision. This includes all digits except:

• All leading zeros.


• Trailing zeros only when they are merely placeholders to indicate the scale of the number (they must not be significant if there is no decimal point).

Here is the explanation for the significant figures of each number in List I:

(A) 1001 - All the digits in this number are significant because there are no leading or trailing zeros acting as placeholders. Hence, this number has 4 significant figures.

(B) 010.1 - The leading zero is not significant, but the zero between 1 and the decimal point, the 1 itself, and the 1 after the decimal point are all significant. So, this number has 3 significant figures.

(C) 100.100 - Here, the number has zeros which are significant since they are between significant digits or after the decimal point. This makes all the zeros and the 1s significant. Therefore, the number has 6 significant figures.

(D) 0.0010010 - The leading zeros are not significant, but the three zeros within and at the end of the number are significant because they are between significant digits or at the end of the number after the decimal. Therefore, this number has 5 significant figures.

Given List I and List II, the correct matching based on the explanation of significant figures is as follows:

Option A

(A) - II (1001 has 4 significant figures)

(B) - I (010.1 has 3 significant figures)

(C) - IV (100.100 has 6 significant figures)

(D) - III (0.0010010 has 5 significant figures)

Thus, the correct answer is Option A.

The least count of a Vernier caliper is the smallest measurement that can be obtained with it and is determined by the difference in the measurement of one main scale division and one Vernier scale division.

In this case, 10 divisions on the main scale coincide with 11 divisions on the Vernier scale. This means that 11 divisions on the Vernier scale are equal in length to 10 divisions on the main scale. Since each division on the main scale is 5 units, this is equal to:

$ 10 \text{ divisions on main scale} \times 5 \text{ units per division} = 50 \text{ units} $

The length of 10 divisions on the main scale (or 50 units) is therefore equal to the length of 11 divisions on the Vernier scale. This means that:

$ 1 \text{ division on the Vernier scale} = \frac{50 \text{ units}}{11} $

Therefore, the least count of the Vernier caliper, which is the difference between one division on the main scale and one division on the Vernier scale, can be calculated as follows:

$ \text{Least count} = \text{value of one main scale division} - \text{value of one Vernier scale division} $

$ \text{Least count} = 5 \text{ units} - \frac{50 \text{ units}}{11} $

$ \text{Least count} = \frac{55 \text{ units}}{11} - \frac{50 \text{ units}}{11} $

$ \text{Least count} = \frac{5 \text{ units}}{11} $

Thus, the least count of the Vernier caliper is

$ \frac{5 \text{ units}}{11} $

Therefore, the correct answer is:

Option A: $ \frac{5}{11} $

Angular impulse is given when a torque is applied for a certain amount of time. The angular impulse changes the angular momentum of an object and has the same dimensions as angular momentum.

The dimensional formula for torque $\tau$ is the same as that for work, since torque is a kind of rotational work, which is given by force times distance (or in rotational terms, it can be considered as force times lever arm). The dimensional formula for force is $\left[\mathrm{M} \mathrm{L} \mathrm{T}^{-2}\right]$, and when multiplied by distance $\left[\mathrm{L}\right]$, we get:

$\left[\mathrm{Torque}\right] = \left[\mathrm{M} \mathrm{L} \mathrm{T}^{-2}\right] \times \left[\mathrm{L}\right] = \left[\mathrm{M} \mathrm{L}^2 \mathrm{T}^{-2}\right]$

Now, angular impulse is torque times time, so we multiply the dimension of torque by time $\left[\mathrm{T}\right]$:

$\left[\mathrm{Angular\ Impulse}\right] = \left[\mathrm {M} \mathrm{L}^2 \mathrm{T}^{-2}\right] \times \left[\mathrm{T}\right] = \left[\mathrm{M} \mathrm{L}^2 \mathrm{T}^{-1}\right]$

So the correct dimensional formula for angular impulse is given by Option C, which is $\left[\mathrm{M} \mathrm{L}^2 \mathrm{T}^{-1}\right]$.

To calculate the percentage error in the resistivity of the material of the wire, we need to understand the formula for resistivity. The resistivity $ \rho $ of a wire is given by:

$\rho = \frac{RA}{l}$

where:

• $ R $ is the resistance


• $ A $ is the cross-sectional area of the wire


• $ l $ is the length of the wire

The cross-sectional area $ A $ of the wire with radius $ r $ is:

$A = \pi r^2$

We can plug this into the equation for resistivity to get:

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

Now, to find the percentage error in resistivity, we need to find the percentage errors in $ R $, $ r $, and $ l $ and then use the following rule for combining errors:

For a given function, $ f = f(x,y,z,...) $, where $ x, y, z,... $ are the measured quantities with possible errors, the percentage error in $ f $, denoted as $ (\delta f)_{\%} $, can be approximated by adding the relative percentage errors of the input quantities. If $ f $ has the form of a product and quotient of the measured quantities as in our case ($ \rho = \frac{R \pi r^2}{l} $), the percentage error in $ f $ is given by:

$(\delta f)_{\%} = (\delta x)_{\%} + (\delta y)_{\%} + (\delta z)_{\%} + ...$

Where $ (\delta x)_{\%} $, $ (\delta y)_{\%} $, and $ (\delta z)_{\%} $ are the percentage errors in each measured quantity $ x, y, z, ... $ respectively.

For our case:

• The percentage error in radius $ (\delta r)_{\%} $ is given by the error in $ r $ divided by the average radius and then multiplied by 100:

$(\delta r)_{\%} = \left(\frac{0.05}{0.35}\right) \times 100$

• The percentage error in resistance $ (\delta R)_{\%} $ is:

$(\delta R)_{\%} = \left(\frac{10}{100}\right) \times 100$

• The percentage error in length $ (\delta l)_{\%} $ is:

$(\delta l)_{\%} = \left(\frac{0.2}{15}\right) \times 100$

Now let's calculate each:

$(\delta r)_{\%} = \left(\frac{0.05}{0.35}\right) \times 100 \approx 14.29\%$

$(\delta R)_{\%} = \left(\frac{10}{100}\right) \times 100 = 10\%$

$(\delta l)_{\%} = \left(\frac{0.2}{15}\right) \times 100 \approx 1.33\%$

However, since the area $ A $ is proportional to $ r^2 $, the percentage error in $ A $ will be twice the percentage error in $ r $. Thus:

$(\delta A)_{\%} = 2 \times (\delta r)_{\%} = 2 \times 14.29\% \approx 28.58\%$

Finally, we add the percentage errors to find the percentage error in resistivity:

$(\delta \rho)_{\%} = (\delta R)_{\%} + (\delta A)_{\%} + (\delta l)_{\%}$

$(\delta \rho)_{\%} = 10\% + 28.58\% + 1.33\% \approx 39.91\%$

This calculation gives us a value close to 39.91%, which means the correct option is closest to this value. Thus, the best answer is:

Option D $39.9\%$

$\begin{aligned} & {[\mathrm{B}]=\mathrm{L}^2} \\ & \mathrm{~A}=\frac{\mathrm{x}^2}{\mathrm{tE}}=\frac{\mathrm{L}^2}{\mathrm{TML}^2 \mathrm{~T}^{-2}}=\frac{1}{\mathrm{MT}^{-1}} \\ & {[\mathrm{~A}]=\mathrm{M}^{-1 \mathrm{~T}}} \\ & {[\mathrm{AB}]=\left[\mathrm{L}^2 \mathrm{M}^{-1} \mathrm{~T}^1\right]} \end{aligned}$

$\begin{aligned} & \mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}}} \\ & \mathrm{g}=\frac{4 \pi^2 \ell}{\mathrm{T}^2} \\ & \frac{\Delta \mathrm{g}}{\mathrm{g}}=\frac{\Delta \ell}{\ell}+\frac{2 \Delta \mathrm{T}}{\mathrm{T}} \\ & =\frac{0.2}{20}+2\left(\frac{1}{40}\right) \\ & =\frac{1.2}{20} \end{aligned}$

Percentage change $=\frac{1.2}{20} \times 100=6 \%$

$\begin{aligned} & \mathrm{R}=\frac{\rho \mathrm{L}}{\pi \frac{\mathrm{d}^2}{4}} \\ & \frac{\Delta \mathrm{R}}{\mathrm{R}}=\frac{\Delta \mathrm{L}}{\mathrm{L}}+\frac{2 \Delta \mathrm{d}}{\mathrm{d}} \\ & \frac{\Delta \mathrm{L}}{\mathrm{L}}=0.1 \% \text { and } \frac{\Delta \mathrm{d}}{\mathrm{d}}=0.1 \% \\ & \frac{\Delta \mathrm{R}}{\mathrm{R}}=0.3 \% \end{aligned}$

To determine the dimensions of $\frac{b^2}{a}$, let's start by identifying the dimensions of each term in the equation $F=a x^2+b t^{\frac{1}{2}}$, where $F$ represents force, $x$ represents distance, and $t$ represents time.

The dimension of force ($F$) is given by [MLT^-2], where $M$ is mass, $L$ is length, and $T$ is time.

The term $ax^2$ has the same dimension as force, so:

$[a] = \frac{[F]}{[x]^2} = \frac{MLT^{-2}}{L^2} = M L^{-1} T^{-2}$

The term $bt^{\frac{1}{2}}$ also has the same dimension as force, which gives:

$[b] = \frac{[F]}{[t]^{\frac{1}{2}}} = \frac{MLT^{-2}}{T^{\frac{1}{2}}} = M L T^{-\frac{5}{2}}$

Now, to find $\frac{b^2}{a}$, we substitute the dimensions of $b$ and $a$:

$\left[\frac{b^2}{a}\right] = \frac{\left(M^2 L^2 T^{-5}\right)}{\left(M L^{-1} T^{-2}\right)} = [M^1 L^3 T^{-3}]$

Therefore, the dimensions of $\frac{b^2}{a}$ are $\left[\mathrm{ML}^3 \mathrm{~T}^{-3}\right]$, which corresponds to mass times length cubed per time cubed.

$\begin{aligned} & 50 \mathrm{~V}+\mathrm{S}=49 \mathrm{~S}+\mathrm{S} \\ & \mathrm{S}=50(\mathrm{~S}-\mathrm{V}) \\ & 5=50(\mathrm{~S}-\mathrm{V}) \\ & \mathrm{S}-\mathrm{V}=\frac{0.5}{50}=\frac{1}{100}=0.01 \mathrm{~mm} \end{aligned}$

$\begin{aligned} & \mathrm{m}=\mathrm{kc}^{\mathrm{P}} \mathrm{G}^{-1 / 2} \mathrm{~h}^{1 / 2} \\ & \mathrm{M}^1 \mathrm{~L}^0 \mathrm{~T}^0=\left[\mathrm{LT}^{-1}\right]^{\mathrm{P}}\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]^{-1 / 2}\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]^{1 / 2} \end{aligned}$

By comparing $\mathrm{P}=1 / 2$

$\begin{aligned} & F=\eta A \frac{d v}{d y} \\ & {\left[M L T^{-2}\right]=\eta\left[L^2\right]\left[T^{-1}\right]} \\ & \eta=\left[M L^{-1} T^{-1}\right] \\ & S . T=\frac{F}{\ell}=\frac{\left[M L T^{-2}\right]}{[L]}=\left[M L^0 T^{-2}\right] \\ & L=m v r=\left[M L^2 T^{-1}\right] \\ & K . E=\frac{1}{2} I \omega^2=\left[M L^2 T^{-2}\right] \end{aligned}$

$\begin{aligned} & \mathrm{Q}=\frac{\mathrm{a}^4 \mathrm{~b}^3}{\mathrm{c}^2} \\ & \frac{\Delta \mathrm{Q}}{\mathrm{Q}}=4 \frac{\Delta \mathrm{a}}{\mathrm{a}}+3 \frac{\Delta \mathrm{b}}{\mathrm{b}}+2 \frac{\Delta \mathrm{c}}{\mathrm{c}} \end{aligned}$

$\frac{\Delta \mathrm{Q}}{\mathrm{Q}} \times 100=4\left(\frac{\Delta \mathrm{a}}{\mathrm{a}} \times 100\right)+3\left(\frac{\Delta \mathrm{b}}{\mathrm{b}} \times 100\right)+2\left(\frac{\Delta \mathrm{c}}{\mathrm{c}} \times 100\right)$

$\begin{aligned} \% \text { error in } \mathrm{Q} & =4 \times 3 \%+3 \times 4 \%+2 \times 5 \% \\ & =12 \%+12 \%+10 \% \\ & =34 \% \end{aligned}$

$\mathrm{R}=\frac{\mathrm{V}}{1}$

According to error analysis

$\begin{aligned} & \frac{\mathrm{dR}}{\mathrm{R}}=\frac{\mathrm{dV}}{\mathrm{V}}+\frac{\mathrm{dI}}{\mathrm{I}} \\ & \frac{\mathrm{dR}}{\mathrm{R}}=\frac{5}{200}+\frac{0.2}{20} \\ & \frac{\mathrm{dR}}{\mathrm{R}}=\frac{7}{200} \\ & \% \text { error } \frac{\mathrm{dR}}{\mathrm{R}} \times 100=\frac{7}{200} \times 100=3.5 \% \end{aligned}$

The provided statements (A) Assertion and (R) Reason have to be analyzed to determine their correctness and also whether Reason (R) correctly explains Assertion (A).

Assertion (A) is addressing a scenario where a positive zero error occurs in a Vernier caliper. Positive zero error refers to the condition when the zero mark on the Vernier scale is to the right of the zero mark on the main scale when the jaws of the caliper are completely closed. In this case, even when measuring a zero length, the Vernier caliper will show a positive reading, indicating an error. This error is constant and will get added to actual measurements, causing the instrument to give readings that are more than the actual measurement. So, Assertion (A) is true.

Reason (R) explains the possible causes of zero error in Vernier calipers. Zero errors can indeed occur due to imperfect manufacturing processes, where the scales are not perfectly aligned. They can also happen due to rough handling, for instance, if the instrument is dropped, which might cause a permanent deformation leading to a continuous zero error. Hence, Reason (R) is true as well.

Finally, we need to analyze whether Reason (R) is the correct explanation of Assertion (A). Although both statements are correct, the Reason (R) does not explain why a positive zero error would lead to the reading being more than the actual reading, it merely states the possible causes of zero errors. Therefore, the correct relationship between the statements is that they are both true, but (R) does not provide the correct explanation for (A).

The correct option is :

$\begin{aligned} & {[\mathrm{h}]=\mathrm{ML}^2 \mathrm{~T}^{-1}} \\ & {[\mathrm{~L}]=\mathrm{ML}^2 \mathrm{~T}^{-1}} \\ & {[\mathrm{P}]=\mathrm{MLT}^{-1}} \\ & {[\tau]=\mathrm{ML}^2 \mathrm{~T}^{-2}} \end{aligned}$

(Here $\mathrm{h}$ is Planck's constant, $\mathrm{L}$ is angular momentum, $\mathrm{P}$ is linear momentum and $\tau$ is moment of force)

To find the values of $(\alpha, \beta, \gamma, \delta)$ for which the expression $e^\alpha \varepsilon_0{ }^\beta h^\gamma c^\delta$ is dimensionless, we need to ensure that the product of the quantities has no overall physical dimensions. This means the dimensions of charge ($e$), permittivity ($\varepsilon_0$), Planck's constant ($h$), and speed of light ($c$) need to cancel each other out.

Let's look at the dimensions of each of the physical quantities:

Electronic charge ($e$): The dimension of charge in terms of fundamental dimensions is:

$[e] = [A \cdot T]$

where $A$ is the dimension of current and $T$ is the dimension of time.

Permittivity of free space ($\varepsilon_0$): The dimension is given by:

$[\varepsilon_0] = [\text{M}^{-1} \text{L}^{-3} \text{T}^4 \text{A}^2]$

Planck's constant ($h$): The dimension is:

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

Speed of light ($c$): The dimension is:

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

Now, we write the expression for the dimensionless quantity and set its dimensions to zero:

$[e^\alpha \varepsilon_0{ }^\beta h^\gamma c^\delta] = [A^\alpha T^\alpha] [\text{M}^{-\beta} \text{L}^{-3\beta} \text{T}^{4\beta} \text{A}^{2\beta}] [\text{M}^\gamma \text{L}^{2\gamma} \text{T}^{-\gamma}] [\text{L}^\delta \text{T}^{-\delta}]$

Combining the dimensions, we get:

$[e^\alpha \varepsilon_0{ }^\beta h^\gamma c^\delta] = [\text{M}^{-\beta + \gamma} \text{L}^{-3\beta + 2\gamma + \delta} \text{T}^{\alpha + 4\beta - \gamma - \delta} \text{A}^{\alpha + 2\beta}]$

For the quantity to be dimensionless, the exponents of $\text{M}$, $\text{L}$, $\text{T}$, and $\text{A}$ must all be zero:

• $-\beta + \gamma = 0$
• $-3\beta + 2\gamma + \delta = 0$
• $\alpha + 4\beta - \gamma - \delta = 0$
• $\alpha + 2\beta = 0$

Solving the above set of equations, we start with the fourth equation:

$\alpha + 2\beta = 0 \Rightarrow \alpha = -2\beta$

Substituting $\alpha = -2\beta$ into the third equation, we get:

$-2\beta + 4\beta - \gamma - \delta = 0 \Rightarrow 2\beta - \gamma - \delta = 0 \Rightarrow \gamma + \delta = 2\beta$

From the first equation:

$-\beta + \gamma = 0 \Rightarrow \gamma = \beta$

Then, substituting $\gamma = \beta$ into $\gamma + \delta = 2\beta$:

$\beta + \delta = 2\beta \Rightarrow \delta = \beta$

Thus, we have:

$\alpha = -2\beta$

$\gamma = \beta$

$\delta = \beta$

Let $\beta = -n$ where $n$ is a non-zero integer. Then, we can write the values as:

$\alpha = -2(-n) = 2n$

$\beta = -n$

$\gamma = -n$

$\delta = -n$

Hence, the tuple $(\alpha, \beta, \gamma, \delta)$ is given by:

$ (2n, -n, -n, -n) $

So, the correct option is Option A: $(2n,-n,-n,-n)$