Correct option is (B)

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:

$\text{Least Count}=\text{Main scale division}-\text{Vernier scale division}$

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

$\text{Length of one vernier scale division}=\frac{n\times \mathrm{m}}{n+1}$

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

$\text{Least Count}=\mathrm{m}-\frac{n\times \mathrm{m}}{n+1}=\mathrm{m}(1-\frac{n}{n+1})=\mathrm{m}\left(\frac{n+1-n}{n+1}\right)=\mathrm{m}\left(\frac{1}{n+1}\right)$

This simplifies to:

$\text{Least Count}=\frac{\mathrm{m}}{\mathrm{n}+1}$

Therefore, the correct option is:

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

Correct option is (D)

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:

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

Correct option is (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 $2.0\mathrm{g}/{\mathrm{cm}}^3$

Correct option is (D)

$\begin{array}{rl}& 0.04=4(\text{ L. C.) }\\ & \Rightarrow \text{ L.C }=0.01\mathrm{mm}\\ & \begin{array}{c}1\mathrm{MSD}-\frac{49}{50}MSD=0.01\mathrm{mm}\\ \Rightarrow 1\mathrm{MSD}=50\times 0.01\mathrm{mm}\\ =0.5\mathrm{mm}\\ \Rightarrow 1\mathrm{cm}=20(0.5\mathrm{mm})\end{array}\end{array}$

Correct option is (D)

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:

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

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.