Correct option is (C)

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

$\lambda =\frac{h}{\sqrt{2mE}}$

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

Correct option is (c)

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

Correct option is (A)

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

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

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

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 ${ϵ}_{\mathrm{o}}{\mathrm{E}}^2$:

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

Combining the dimensions:

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

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

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

Correct option is (D)

$\begin{array}{ll}\tau =\mathrm{F}.\mathrm{d},& [\tau ]=\left[{\mathrm{M}}^1{\mathrm{L}}^2{\mathrm{T}}^{-2}\right]\\ \mathrm{F}=\mathrm{qVB},& [B]=\left[{\mathrm{M}}^1{\mathrm{T}}^2{\mathrm{A}}^{-1}\right]\\ {\mathrm{M}}_1=\mathrm{iA},& \left[{\mathrm{M}}_1\right]=\left[{\mathrm{L}}^2{\mathrm{A}}^1\right]\end{array}$

Correct option is (C)

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 $\left[\mathrm{T}\right]$:

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

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