Units & Measurements

To determine the dimensions of

, we need to first understand the dimensional formulas of each component in the expression.

1.

Permittivity of free space,

: The permittivity of free space has the dimensions:

. 2. Electric field,

: The electric field

has the dimensions:

. Now, let's calculate the dimensions of

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

are

, which corresponds to Option A. Hence, the correct answer is Option A:

.

Pitch of the screw gauge: 1 mm Number of divisions on the circular scale: 100 divisions Zero error: The zero of the circular scale lies 5 divisions below the reference line, indicating a positive zero error.

Measurement data: 4 linear scale divisions are visible. 60 divisions on the circular scale coincide with the reference line.

Step-by-Step Calculation Least Count of the screw gauge: $\text{Least Count} = \dfrac{\text{Pitch}}{\text{Number of Divisions on Circular Scale}} = \dfrac{1 \text{ mm}}{100} = 0.01 \text{ mm}$ Main Scale Reading (MSR): The linear scale shows 4 divisions, so the main scale reading is: $\text{MSR} = 4 \text{ mm}$ Circular Scale Reading (CSR): 60 divisions coincide with the reference line, so the circular scale reading is: $\text{CSR} = 60 \times \text{Least Count} = 60 \times 0.01 \text{ mm} = 0.60 \text{ mm}$ Zero Error: The zero error is 5 divisions below the reference line, indicating a positive zero error: $\text{Zero Error} = +5 \times \text{Least Count} = +5 \times 0.01 \text{ mm} = +0.05 \text{ mm}$ Total Reading without considering zero error: $\text{Total Reading (without zero error)} = \text{MSR} + \text{CSR} = 4 \text{ mm} + 0.60 \text{ mm} = 4.60 \text{ mm}$ 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:

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

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:

The value of one main scale division is:

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

Given the least count:

Using the least count formula, we have:

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

Simplifying further:

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

Therefore, the correct answer is option D:

In expression

, If

by round off $\Rightarrow$ Order

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

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

.

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:

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

, we find:

For the main scale reading (MSR) of

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

. So,

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

Now, we can calculate the volume (V) of the sphere using its diameter with the formula

, where

is the radius of the sphere. Remembering that the radius is half of the diameter,

. Therefore,

, Calculating the volume,

Now to find the density ($\rho$) using the mass (

) and volume (

) calculated,

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

Adding these values together:

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

So, the arithmetic mean of these readings is

.

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 find the dimensional formula of

in the equation given by

, 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 $P = \dfrac{Force}{Area}$ 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 $\dfrac{a}{V^2}$ to deduce $a$'s dimensions: $\dfrac{a}{V^2}$ 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

.

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 ($\text{VSD}$): $\text{VSD} = \dfrac{(n - 1) \times \text{MSD}}{n}$ 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: $\text{LC} = \text{MSD} - \left( \dfrac{(n - 1) \times \text{MSD}}{n} \right)$ $\text{LC} = \text{MSD} \left( 1 - \dfrac{n - 1}{n} \right)$ $\text{LC} = \text{MSD} \left( \dfrac{n - (n - 1)}{n} \right)$ $\text{LC} = \text{MSD} \left( \dfrac{1}{n} \right)$ Solving for One Main Scale Division ($\text{MSD}$): Rearranging the equation: $\text{LC} = \dfrac{\text{MSD}}{n}$ $\text{MSD} = \text{LC} \times n$ 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.