Question

The force F is given in terms of time t and displacement x by the equation $F = A\cos Bx + C\sin Dt$. The dimensional formula of $\dfrac{D}{B}$ is
A. ${M^0}{L^0}{T^0}$
B. ${M^0}{L^0}{T^{ - 1}}$
C. ${M^0}{L^{ - 1}}{T^0}$
D. ${M^0}{L^1}{T^{ - 1}}$

Answer 1

As we all know that to find the dimensional formula of a physical quantity then its formula should be known. In the absolute units only, we express the physical quantity. The dimensional formula makes us clear that every term in the formula has suitable units or not.

Complete Step-by-step Solution
We must know that physical quantities are expressed by dimensional formulas in terms of fundamental quantities. Mass (M), Length(L), and time (T) are the three fundamental quantities expressed in the powers of M, L, T.
We know from mathematics that the elements inside cos and sin functions are the angles and are always dimensionless. Firstly, we will rewrite the given equation as:
$F = A\cos Bx + C\sin Dt$
We can see that in the above equations, the angles $Bx$ and $Dt$ are dimensionless. It means that they are having no fundamental quantities i.e. they are having the powers of $M$, $L$, and $T$ as zero.
So we will write the dimensional formula for $Bx$ and $Dt$.
Dimensional formula for $Bx$ is ${M^0}{L^0}{T^0}$ and dimensional formula for $Dt$ is ${M^0}{L^0}{T^0}$. Now here $x$ has the dimensions of the length and $t$ has the dimensions of time
So we can say that the dimensional formula for $t$ is ${M^0}{L^0}{T^1}$ since the unit of time is seconds, hours, minutes, etc. Also, we know that the dimensional formula for $x$ is ${M^0}{L^1}{T^0}$ since it has the unit of length which can be displacement, distance and it can have the units of meter, kilometer, etc.
Since we know that the dimensional formula of $Bx$ is
$\left[ B \right]\;\left[ x \right] = {M^0}{L^0}{T^0}$ ……(I)
Here $\left[ B \right]$ is the dimensional formula of $B$, and $\left[ x \right]$ is the dimensional formula for $x$ which is equal to ${M^0}{L^1}{T^0}$ as described above. Now we can substitute $\left[ x \right] = {M^0}{L^1}{T^0}$ in equation (I) to find the value of $\left[ B \right]$.
$\Rightarrow \left[ B \right]\;{M^0}{L^1}{T^0} = {M^0}{L^0}{T^0}$
Therefore, $\left[ B \right]$ can be properly modified as,
$\left[ B \right] = {M^0}{L^{ - 1}}{T^0}$ …… (II)
We will solve the above equation further and we will get,
$\Rightarrow \left[ B \right] = {\dfrac{{{M^0}{L^0}T}}{{{M^0}{L^1}{T^0}}}^0}$
$\Rightarrow \left[ B \right] = {L^{ - 1}}$
We also know the dimensional formula for $Dt$ is
$\left[ D \right]\left[ t \right] = {M^0}{L^0}{T^0}$ …… (III)
Here $\left[ D \right]$ is the dimensional formula of $B$, and $\left[ t \right]$ is the dimensional formula for $t$ which is equal to ${M^0}{L^0}{T^1}$ as described above. Now we can substitute $\left[ t \right] = {M^0}{L^0}{T^1}$ in equation (III) to find the value of $\left[ D \right]$.
$\Rightarrow \left[ D \right]\left[ {{M^0}{L^0}{T^1}} \right] = {M^0}{L^0}{T^0}$
$\Rightarrow \left[ D \right]{M^0}{L^0}{T^1} = {M^0}{L^0}{T^0}$
$\Rightarrow \left[ D \right] = \dfrac{{{M^0}{L^0}{T^0}}}{{{M^0}{L^0}{T^1}}}$
We will solve it further and we will get,
$\Rightarrow \left[ D \right] = {T^{ - 1}}$
Therefore, $\left[ D \right]$ can be properly modified as,
$\Rightarrow \left[ D \right] = {M^0}{L^0}{T^{ - 1}}$ …… (IV)
Now we can divide equation (IV) from equation (III) to find the value of $\dfrac{{\left[ D \right]}}{{\left[ B \right]}}$.
$\Rightarrow \dfrac{{\left[ D \right]}}{{\left[ B \right]}} = \dfrac{{{M^0}{L^0}{T^{ - 1}}}}{{{M^0}{L^{ - 1}}{T^0}}}$
$\therefore \dfrac{{\left[ D \right]}}{{\left[ B \right]}} = {M^0}{L^1}{T^{ - 1}}$

So therefore, we can say that the dimensional formula for $\dfrac{D}{B}$ is ${M^0}{L^1}{T^{ - 1}}$. Hence the correct option is (D).

Note:
We must know that If K is the unit of a derived quantity represented by $K = {M^a}{L^b}{T^c}$, then ${M^a}{L^b}{T^c}$ is called dimensional formula, and the powers $a$, $b$, and, $c$ care called the dimensions. We should also notice that in the above question, the dimensional formula for $Bx$ and $Dt$ is the same so as to make the whole equation dimensionally feasible.