Introduction to Physics

Answer:

The speed of light in vacuum is 299,792,458 m/s.
Then time taken by light to cover a distance of 1 metre in vacuum = $\frac{1}{299, 792, 458}$ s
Hence, the metre is defined as the distance travelled by light in $\frac{1}{299, 792, 458}$ s.
 As 300,000,000 m/s is an approximate speed of light in vacuum, it cannot be used to define the metre.

The distance travelled by light in one second is 299,792,458 m. This is a large quantity and cannot be used as a base unit. So, the metre is not defined in terms of second.

 

Answer:

(a) Volume of a cube of edge a, $V=a\times a\times a$
i.e., [V] = L $\times$ L $\times$ L = L³

(b) Volume of a sphere of radius a, $V=\frac{4}{3}\pi (a{)}^3$
i.e., [V] = L $\times$ L $\times$ L = L³

(c) The ratio of the volume of the cube to the volume of the sphere is a dimensionless quantity.

Answer:

The validity of this statement cannot be tested by measuring sizes with a metre stick, because the size of the metre stick has also got doubled overnight.
Yes, it can be verified by using the fact that speed of light is a universal constant and has not changed.
If the linear size of everything in the universe is doubled and all the clocks in the universe starts running at half the speed, then we cannot test the validity of this statement by any method.

Answer:

Yes, if all the terms in an equation have the same units, it is necessary that they have the same dimension.

No, if all the terms in an equation have the same dimensions, it is not necessary that they have the same unit. It is because two quantities with different units can have the same dimension, but two quantities with different dimensions cannot have the same unit. For example angular frequency and frequency, both have the dimensions $\left[T^{-1}\right]$ but units of angular frequency is rad/s and frequency is Hertz.Another example is energy per unit volume and pressure.Both have the dimensions of $\left[M{L}^{-1}{T}^{-2}\right]$but units of pressure is N/m² and that of energy per unit volume is J/m³

Answer:

No, even if two quantities have the same dimensions, they may represent different physical contents.
Example: Torque and energy have the same dimension, but they represent different physical contents.

Answer:

If we use the foot of a person as a standard unit of length,features that  will not be present are  variability, destructibility and  reproducible nature and the feature that will be present is the availability of  the foot of a person to measure any length.

Answer:

(a) The thickness of a sheet of paper can roughly be determined by measuring the height of a stack of paper.
Example: Let us consider a stack of 100 sheets of paper. We will use a ruler to measure its height. In order to determine the thickness of a sheet of paper, we will divide the height of the stack with the number of sheets (i.e., 100).

(b) The distance between the Sun and the Moon can be measured by using Pythagoras theorem when the Earth makes an angle of 90^∘ with the Sun and the Moon. We already know the distances from the Sun to the Earth and from the Earth to the Moon. However, these distances keep on changing due to the revolution of the Moon around the Earth and the revolution of the Earth around the Sun.

Answer:

(b) length, time and velocity

We define length and time separately as it is not possible to define velocity without using these quantities. This means that one fundamental quantity depends on the other. So, these quantities cannot be listed as fundamental quantities in any system of units.

Answer:

(d) $n \propto \frac{1}{u}$

The larger the unit used to express the physical quantity, the lesser will be the numerical value.

Example:
1 kg of sugar can be expressed as 1000 g or 10000 mg of sugar.
Here, g (gram) is the larger quantity as compared to mg (milligram), but the numerical value used with gram is lesser than the numerical value used with milligram.

Answer:

(d) may be represented in terms of L, T and x if $a\ne 0$

If a = 0, then we cannot represent mass dimensionally in terms of L, T and x, otherwise it can be represented in terms of L, T and x.

Answer:

(a) may have a unit

Dimensionless quantities may have units.

Answer:

(a) never has a non-zero dimension

A unitless quantity never has a non-zero dimension.

Answer:

(a) 0

[ax] = [x²]
⇒ [a] = [x]    ...(1)

Dimension of LHS = Dimension of RHS
$$\begin{gathered} \Rightarrow \left[\frac{dx}{\sqrt{x^2}}\right]=\left[a^n\right] \\ \Rightarrow \left[\frac{L}{L}\right]=\left[a^n\right] ...\left(2\right) \\ \left[L^0\right]=\left[a^n\right] \\ n=0 \end{gathered}$$ 

Answer:

(c) pressure
(d) energy per unit volume

[Work done] = [ML²T⁻²]
[Linear momentum] = [MLT⁻¹]
[Pressure] = [ML⁻¹ T⁻²]
[Energy per unit volume] = [ML⁻¹ T⁻²]

From the above, we can see that pressure and energy per unit volume have the same dimension, i.e., ML⁻¹ T⁻².

Answer:

(a) A dimensionally correct equation may be correct.
(b) A dimensionally correct equation may be incorrect.
(d) A dimensionally incorrect equation may be incorrect.

It is not possible that a dimensionally incorrect equation is correct. All the other situations are possible.

Answer:

The statements which are correct are:
(a) All quantities may be represented dimensionally in terms of the base quantities.
(b) A base quantity cannot be represented dimensionally in terms of the rest of the base quantities.
(c) The dimensions of a base quantity in other base quantities is always zero.

Statement (d) is not correct because A derived quantity can exist which is dimensionless for example fine structure constant which is given by

$\alpha =\frac{2\pi e^2}{hc}=\frac{1}{137}$

where e is the electric charge and c is the speed of light and h is Planks constant.α is a derived quantity and is dimensionless.

Answer:

(a) Linear momentum = mv
Here, [m] = [M] and [v] = [LT⁻¹]
∴ Dimension of linear momentum, [mv] = [MLT⁻¹]

(b) Frequency = $\frac{1}{\mathrm{Time}}$
∴ Dimension of frequence = $\left[\frac{1}{T}\right]=\left[{\mathrm{M}}^0{\mathrm{L}}^0{\mathrm{T}}^{-1}\right]$

(c) Pressure = $\frac{\mathrm{Force}}{\mathrm{Area}}$
$$\begin{gathered} \mathrm{Dimesion} \mathrm{of} \mathrm{force} =\left[{\mathrm{MLT}}^{-2}\right] \\ \mathrm{Dimesion} \mathrm{of} \mathrm{area}=\left[{\mathrm{L}}^2\right] \\ \therefore \mathrm{Dimesion} \mathrm{of} \mathrm{pressure}=\frac{\left[{\mathrm{MLT}}^{-2}\right]}{\left[{\mathrm{L}}^2\right]}=\left[{\mathrm{ML}}^{-1}{\mathrm{T}}^{-2}\right] \end{gathered}$$ 
         

Answer:

(a) Dimensions of angular speed, $\mathrm{\omega }=\frac{\mathrm{\theta }}{t}=\left[{\mathrm{M}}^0{\mathrm{L}}^0{\mathrm{T}}^{-1}\right]$

(b) Angular acceleration, $\mathrm{\alpha }=\frac{\mathrm{\omega }}{t}$
Here, ω = [M⁰L⁰T⁻¹] and t = [T]
So, dimensions of angular acceleration = [M⁰L⁰T⁻²]
                                          
(c) Torque, τ =Frsin$\theta$   
Here, F = [MLT⁻²] and r = [L]
So, dimensions of torque = [ML²T⁻²]

(d) Moment of inertia = mr²
Here, m = [M] and r² = [L²]
So, dimensions of moment of inertia = [ML²T⁰]

Answer:

(a) Electric field is defined as electric force per unit charge.
i.e., $\mathrm{E}=\frac{F}{q}$
$$\begin{gathered} \mathrm{Also}, \left[F\right]=\left[{\mathrm{MLT}}^{-2}\right] \mathrm{and} \left[q\right]=\left[\mathrm{A}T\right] \\ \mathrm{So}, \mathrm{dimension} \mathrm{of} \mathrm{electric} \mathrm{field}, \left[E\right]=\left[{\mathrm{MLT}}^{-3}{\mathrm{A}}^{-1}\right] \end{gathered}$$

(b) Magnetic field, $B=\frac{\mathit{F}}{\mathit{q}\mathit{v}}$
$$\begin{gathered} \mathrm{Here}, \left[F\right]=\left[{\mathrm{MLT}}^{-2}\right], \left[q\right]=\left[\mathrm{AT}\right] \mathrm{and} \left[v\right]=\left[{\mathrm{LT}}^{-1}\right] \\ \mathrm{So}, \mathrm{dimension} \mathrm{of} \mathrm{magnetic} \mathrm{field}, \left[B\right] =\frac{\left[{\mathrm{MLT}}^{-2}\right]}{\left[\mathrm{A}T\right] \left[{\mathrm{LT}}^{-1}\right]}=\left[M{\mathrm{L}}^0{T}^{-2}{\mathrm{A}}^{-1}\right] \end{gathered}$$                               

(c) Magnetic permeability, ${\mathrm{\mu }}_0=\frac{B\times 2\pi r}{I}$
 $$\begin{gathered} \mathrm{Here}, \left[B\right]=\left[{\mathrm{MT}}^{-2}{\mathrm{A}}^{-1}\right] \mathrm{and} \left[r\right]=\left[\mathrm{L}\right] \\ \mathrm{So}, \mathrm{dimension} \mathrm{of} \mathrm{magnetic} \mathrm{permeability}, \left[{\mathrm{\mu }}_0\right]=\frac{\left[{\mathrm{MT}}^{-2}{\mathrm{A}}^{-1}\right] \times \left[\mathrm{L}\right]}{\left[\mathrm{A}\right]}=\left[{\mathrm{MLT}}^{-2}{\mathrm{A}}^{-2}\right] \end{gathered}$$

Answer:

(a) Electric dipole moment, P = q.(2d)
Here, [q] = [AT] and d = [L]
∴ Dimension of electric dipole moment = [LTA]

(b) Magnetic dipole moment, M = IA
Here, A = [L²]
∴ Dimension of magnetic dipole moment = [L²A]

Answer:

E = hv, where E is the energy and v is the frequency
$$\begin{gathered} \mathrm{Here}, \left[E\right]=\left[{\mathrm{ML}}^2{\mathrm{T}}^{-2}\right] \mathrm{and} \left[v\right]=\left[{\mathrm{T}}^{-1}\right] \\ \mathrm{So}, \left[h\right]=\frac{\left[E\right]}{\left[v\right]}=\frac{\left[{\mathrm{ML}}^2{\mathrm{T}}^{-2}\right]}{\left[{\mathrm{T}}^{-1}\right]}=\left[{\mathrm{ML}}^2{\mathrm{T}}^{-1}\right] \end{gathered}$$

Answer:

(a) Specific heat capacity, $C=\frac{Q}{m∆T}$
$$\begin{gathered} \left[Q\right]=\left[{\mathrm{ML}}^2{\mathrm{T}}^{-2}\right] \mathrm{and} \left[T\right]=\left[\mathrm{K}\right] \\ \mathrm{So}, \left[C\right]=\frac{\left[{\mathrm{ML}}^2{\mathrm{T}}^{-2}\right]}{\left[\mathrm{M}\right] \left[\mathrm{K}\right]}=\left[{\mathrm{L}}^2{\mathrm{T}}^{-2}{\mathrm{K}}^{-1}\right] \end{gathered}$$

(b) Coefficient of linear expansion,$\alpha =\frac{L_1-L_0}{L_0∆T}$
So, $\left[\alpha \right]=\frac{\left[\mathrm{L}\right]}{\left[\mathrm{LK}\right]}=\left[{\mathrm{K}}^{-1}\right]$

(c) Gas constant, $\mathrm{R}=\frac{\mathrm{PV}}{n\mathrm{T}}$
$$\begin{gathered} \mathrm{Here}, \left[P\right]=\left[{\mathrm{ML}}^{-1}{\mathrm{T}}^{-2}\right], \left[n\right]=\left[\mathrm{mol}\right], \left[T\right]=\left[\mathrm{K}\right] \mathrm{and} \left[V\right]=\left[{\mathrm{L}}^3\right] \\ \mathrm{So}, \left[R\right]=\frac{\left[{\mathrm{ML}}^{-1}{\mathrm{T}}^{-2}\right] \left[{\mathrm{L}}^3\right]}{\left[\mathrm{mol}\right] \left[\mathrm{K}\right]}=\left[{\mathrm{ML}}^2{\mathrm{T}}^{-2}{\mathrm{K}}^{-1}(\mathrm{mol}{)}^{-1} \right] \end{gathered}$$

Answer:

Let F be the dimension of force.

(a) Density$=\frac{m}{v}=\frac{\left(\mathrm{force}/\mathrm{acceleration}\right)}{\mathrm{volume}}$
$$\begin{gathered} \left[\mathrm{Acceleration}\right]=\left[{\mathrm{LT}}^{-2}\right] \mathrm{and} \left[\mathrm{volume}\right]=\left[{\mathrm{L}}^3\right] \\ \therefore \left[\mathrm{Density}\right]=\frac{\left[\mathrm{F}/{\mathrm{LT}}^{-2}\right]}{\left[{\mathrm{L}}^3\right]}=\frac{\mathrm{F}}{{\mathrm{L}}^4{\mathrm{T}}^{-2}}=\left[{\mathrm{FL}}^{-4}{\mathrm{T}}^2\right] \end{gathered}$$

(b) Pressure$=\frac{\mathrm{force}}{\mathrm{area}}$
$$\begin{gathered} \left[\mathrm{Area}\right]=\left[{\mathrm{L}}^2\right] \\ \therefore \left[\mathrm{Pressure}\right]=\frac{\left[\mathrm{F}\right]}{\left[{\mathrm{L}}^2\right]}=\left[{\mathrm{FL}}^{-2}\right] \end{gathered}$$

(c) Momentum = mv = (force/acceleration) × velocity
$\mathrm{Acceleration}=\left[{\mathrm{LT}}^{-2}\right] \mathrm{and} \mathrm{velocity}=\left[{\mathrm{LT}}^{-1}\right]$
$\therefore \left[\mathrm{Momentum}\right]=\frac{\left[\mathrm{F}\right]}{\left[{\mathrm{LT}}^{-2}\right]}\times \left[{\mathrm{LT}}^{-1}\right]=\left[\mathrm{FT}\right]$

(d) Energy$=\frac{1}{2}m{v}^2=\frac{\mathrm{force}}{\mathrm{acceleration}}\times {\left(\mathrm{velocity}\right)}^2$
$\therefore \left[\mathrm{Energy}\right]=\frac{\left[\mathrm{F}\right]}{\left[{\mathrm{LT}}^{-2}\right]}\times {\left[{\mathrm{LT}}^{-1}\right]}^2=\left[\mathrm{FL}\right]$  

Answer:

Acceleration due to gravity, g = 10 $m/s^2$

∴ g = 10 m/s² = 10 × 100 cm $\times \frac{1}{{\left({\displaystyle \frac{1}{60}}\min \right)}^2}$
∴ g = 1000 × 3600 cm/min² = 36 × 10⁵ cm/min²

Answer:

1 mi = 1.6 km
1 km = 1000 m

For the snail, average speed = 0.02 mi/h =$\frac{0.020\times 1.6\times 1000}{3600}=0.0089 \mathrm{m}/\mathrm{s}$

For the leopard, average speed = 70 mi/h =$\frac{70\times 1.6\times 1000}{3600}\mathrm{m}/\sec =31 \mathrm{m}/\mathrm{s}$

Answer:

Height, h = 75 cm = 0.75 m
Density of mercury = 13600 kg/m³
g = 9.8 m/s²

In SI units, pressure = hρg = 0.75 × 13600 × 9.8 = 10 × 10⁴ N/m² (approximately)

In CGS units, pressure = 10 × 10⁴ N/m²$=\frac{10\times {10}^4\times {10}^5 \mathrm{dyne}}{{10}^4 {\mathrm{cm}}^2}=10\times {10}^5 \mathrm{dyne}/{\mathrm{cm}}^2$

Answer:

In SI unit, watt = joule/s
In CGS unit, 1 joule = 10⁷ erg
So, 100 watt = 100 joule/s $=\frac{100\times {10}^7 \mathrm{erg} }{1 \mathrm{s}}={10}^9 \mathrm{erg}/\mathrm{s}$
                    

Answer:

1 microcentury = 10⁻⁶ × 100 years = 10⁻⁴ × 365 × 24 × 60 minutes
$$\begin{gathered} 1 \min =\frac{1}{{10}^{-4}\times 365\times 24\times 60} \mathrm{microcentury} \\ \Rightarrow 100 \min =\frac{1}{{10}^{-4}\times 365\times 24\times 60}\times 100=\frac{{10}^5}{365\times 144}=1.9 \mathrm{microcenturies} \end{gathered}$$

Suppose, I slept x minutes yesterday.
∴ x min = 0.019x microcenturies

Answer:

1 dyne = 10⁻⁵ N
1 cm = 10  ⁻² m

$\therefore 72 \mathrm{dyne}/\mathrm{cm}=\frac{72\times {10}^{-5} \mathrm{N}}{{10}^{-2} \mathrm{m}}=72\times {10}^{-3} \mathrm{N}/\mathrm{m}=0.072 \mathrm{N}/\mathrm{m}$

Answer:

Kinetic energy of a rotating body is K = kI ^aω^b.

Dimensions of the quantities are [K] = [ML²T⁻²], [I] = [ML²] and [ω] = [T⁻¹].

Now, dimension of the right side are [I]^a = [ML²]^a and [ω]^b = [T⁻¹]^b.

According to the principal of homogeneity of dimension, we have:
[ML²T⁻²] = [ML²]^a [T⁻¹]^b

Equating the dimensions of both sides, we get:
2 = 2a
⇒ a = 1
And,
−2 = −b
⇒ b = 2

Answer:

According to the theory of relativity, E α m^ac^b
⇒ E = km^ac^b, where k = proportionality constant

Dimension of the left side, [E] = [ML²T⁻²]

Dimension of the right side, [m^ac^b]= [M]^a [LT⁻¹]^b

Equating the dimensions of both sides, we get:
 [ML²T⁻²] = [M]^a [LT⁻¹]^b
⇒ a = 1, b = 2
∴ E = kmc²

Answer:

Dimensional formula of resistance, [R] = [ML²A⁻²T⁻³]    ...(1)
Dimensional formula of potential difference, [V] = [ML²A⁻¹T⁻³]    ...(2)
Dimensional formula of current,  I = [A]

Dividing (2) by (1), we get:
$\frac{\left[V\right]}{\left[R\right]}=\frac{\left[{\mathrm{ML}}^2{\mathrm{A}}^{-1}{\mathrm{T}}^{-3}\right]}{\left[{\mathrm{ML}}^2{\mathrm{A}}^{-2}{\mathrm{T}}^{-3}\right]}=\left[\mathrm{A}\right]$
⇒ V = IR

Answer:

Frequency, f $\propto$L^aF^bm^c
f = kL^aF^bm^c   ...(1)
Dimension of [f] = [T⁻¹]

Dimension of the right side components:
[L] = [L]
[F] = [MLT⁻²]
[m] = [ML⁻¹]

Writing equation (1) in dimensional form, we get:
[T⁻¹] = [L]^a [MLT⁻²]^b [ML⁻¹]^c
[M⁰L⁰T⁻¹] = [M^{b + c} L^{a + b − c} T^−2b]

Equating the dimensions of both sides, we get:
b + c = 0               ....(i)
a + b −c = 0      ....(ii)
−2b = −1              ....(iii)
Solving equations (i), (ii) and (iii), we get:
$a=-1, b=\frac{1}{2} \mathrm{and} c=\left(-\frac{1}{2}\right)$
∴ Frequency, f = kL⁻¹F^1/2m^−1/2$=\frac{k}{L}{F}^{1/2} m^{\left(-\frac{1}{2}\right)}=\frac{k}{L} \sqrt{\left(\frac{F}{m}\right)}$
⇒ $f=\frac{k}{L} \sqrt{\left(\frac{F}{m}\right)}$

Answer:

(a) $h=\frac{2\mathrm{S} \cos \theta }{\mathrm{\rho }rg}$
Height, [h] = [L]
Surface Tension, $\left[\mathit{S}\right]=\frac{\left[F\right]}{\left[L\right]}=\frac{\left[{\mathrm{MLT}}^{-2}\right]}{\left[\mathrm{L}\right]}=\left[{\mathrm{MT}}^{-2}\right]$
Density, $\left[\rho \right]=\frac{\left[M\right]}{\left[\mathrm{I}\right]}=\left[{\mathrm{ML}}^{-3}{\mathrm{T}}^0\right]$
Radius, [r] = [L], [g]= [LT⁻²]
Now, $\frac{2\left[S\right]\cos \theta }{\left[\rho \right]\left[r\right]\left[g\right]}=\frac{\left[{\mathrm{MT}}^{-2}\right]}{\left[{\mathrm{ML}}^{-3}{\mathrm{T}}^0\right] \left[\mathrm{L}\right] \left[{\mathrm{LT}}^{-2}\right]}=\left[{\mathrm{M}}^0{\mathrm{L}}^1{\mathrm{T}}^0\right]=\left[\mathrm{L}\right]$

Since the dimensions of both sides are the same, the equation is dimensionally correct.

(b) $\nu =\sqrt{\left(\frac{\mathit{P}}{\mathit{\rho }}\right)}$
Velcocity, [ν] = [LT⁻¹]
Pressure, $P=\frac{\left[F\right]}{\left[A\right]}=\left[{\mathrm{ML}}^{-1}{\mathrm{T}}^{-2}\right]$
Density, $\left[\rho \right]=\frac{\left[M\right]}{\left[V\right]}=\left[{\mathrm{ML}}^{-3}{\mathrm{T}}^0\right]$
Now, $\sqrt{\frac{\mathit{P}}{\mathit{\rho }}}={\left[\frac{\left[{\mathrm{ML}}^{-1}{\mathrm{T}}^{-2}\right]}{\left[{\mathrm{ML}}^{-3}\right]}\right]}^{\frac{1}{2}}={\left[{\mathrm{L}}^2{\mathrm{T}}^{-2}\right]}^{1/2}=\left[{\mathrm{LT}}^{-1}\right]$
Since the dimensions of both sides of the equation are the same, the equation is dimensionally correct.

(c) $V=\frac{\left(\pi \mathit{ }P\mathit{ }{r}^{4}t\right)}{\left(8 \eta l\right)}$
Volume, [V] = [L³]
Pressure, $P=\frac{\left[F\right]}{\left[A\right]}=\left[{\mathrm{ML}}^{-1}{\mathrm{T}}^{-2}\right]$
[r]= [L] and [t] = [T]
Coefficient of viscosity,
 $\left[\eta \right]=\frac{\left[F\right]}{6\mathrm{\pi }\left[\mathrm{r}\right]\left[\mathrm{v}\right]}=\frac{\left[{\mathrm{MLT}}^{-2}\right]}{\left[\mathrm{L}\right]\left[{\mathrm{LT}}^{-1}\right]}=\left[{\mathrm{ML}}^{-1}{\mathrm{T}}^{-1}\right]$

Now, $\frac{\pi \left[P\right] {\left[r\right]}^4\left[t\right]}{8\left[\mathrm{\eta }\right]\left[l\right]}=\frac{\left[{\mathrm{ML}}^{-1}{\mathrm{T}}^{-2}\right] \left[{\mathrm{L}}^4\right] \left[\mathrm{T}\right]}{\left[{\mathrm{ML}}^{-1}{\mathrm{T}}^{-1}\right] \left[\mathrm{L}\right]}=\left[{\mathrm{L}}^3\right]$
Since the dimensions of both sides of the equation are the same, the equation is dimensionally correct.

(d) $\mathrm{\nu }=\frac{1}{2\pi }\sqrt{\left(\frac{mgl}{I}\right)}$
Frequency, ν = [T⁻¹]

$$\begin{gathered} \sqrt{\left(\frac{mgl}{I}\right)}=\sqrt{\frac{\left[\mathrm{M}\right] \left[{\mathrm{LT}}^{-2}\right] \left[\mathrm{L}\right]}{\left[{\mathrm{ML}}^2\right]}} \\ \Rightarrow {\left[\frac{\left[{\mathrm{ML}}^2{\mathrm{T}}^{-2}\right]}{\left[{\mathrm{ML}}^2\right]}\right]}^{\frac{1}{2}}=\left[{\mathrm{T}}^{-1}\right] \end{gathered}$$

Since the dimensions of both sides of the equation are the same, the equation is dimensionally correct.

Answer:

Dimension of the left side of the equation= $\left[\int \frac{dx}{\sqrt{\left(a^2-x^2\right)}}\right]=\left[\int \frac{\mathrm{L}}{\sqrt{\left({\mathrm{L}}^2-{\mathrm{L}}^2\right)}}\right]=\left[{\mathrm{L}}^0\right]$

Dimension of the right side of the equation = $\left[\left(\frac{1}{a}\right) {\sin }^{-1} \left(\frac{a}{x}\right)\right]=\left({\mathrm{L}}^{-1}\right)$

So, $\left[\int \frac{dx}{\sqrt{\left(a^2-x^2\right)}}\right]\ne \left[\left(\frac{1}{a}\right) {\sin }^{-1} \left(\frac{a}{x}\right)\right]$

Since the dimensions on both sides are not the same, the equation is dimensionally incorrect.