A function $f(\theta )$ is defined as $f(\theta )\, = \,1\, - \theta  + \frac{{{\theta ^2}}}{{2!}} - \frac{{{\theta ^3}}}{{3!}} + \frac{{{\theta ^4}}}{{4!}} + ...$ Why is it necessary for  $f(\theta )$  to be a dimensionless quantity ?

Vedclass pdf generator app on play store
Vedclass iOS app on app store

Since, $f(\theta)$ is a sum of different power of $\theta$ and as $RHS$ is dimensionless, hence $LHS$ should also be dimensionless.

Similar Questions

If force $(F)$, length $(L)  $ and time $(T)$ are assumed to be fundamental units, then the dimensional formula of the mass will be

The potential energy of a particle varies with distance $x$ from a fixed origin as $U=\frac{A \sqrt{x}}{x^2+B}$, where $A$ and $B$ are dimensional constants then dimensional formula for $A B$ is

An expression of energy density is given by $u=\frac{\alpha}{\beta} \sin \left(\frac{\alpha x}{k t}\right)$, where $\alpha, \beta$ are constants, $x$ is displacement, $k$ is Boltzmann constant and $t$ is the temperature. The dimensions of $\beta$ will be.

  • [JEE MAIN 2022]

In electromagnetic theory, the electric and magnetic phenomena are related to each other. Therefore, the dimensions of electric and magnetic quantities must also be related to each other. In the questions below, $[E]$ and $[B]$ stand for dimensions of electric and magnetic fields respectively, while $\left[\varepsilon_0\right]$ and $\left[\mu_0\right]$ stand for dimensions of the permittivity and permeability of free space respectively. $[L]$ and $[T]$ are dimensions of length and time respectively. All the quantities are given in $SI$ units.

($1$) The relation between $[E]$ and $[B]$ is

$(A)$ $[ E ]=[ B ][ L ][ T ]$  $(B)$ $[ E ]=[ B ][ L ]^{-1}[ T ]$  $(C)$ $[ E ]=[ B ][ L ][ T ]^{-1}$  $(D)$ $[ E ]=[ B ][ L ]^{-1}[ T ]^{-1}$

($2$) The relation between $\left[\varepsilon_0\right]$ and $\left[\mu_0\right]$ is

$(A)$ $\left[\mu_0\right]=\left[\varepsilon_0\right][ L ]^2[ T ]^{-2}$  $(B)$ $\left[\mu_0\right]=\left[\varepsilon_0\right][ L ]^{-2}[ T ]^2$   $(C)$ $\left[\mu_0\right]=\left[\varepsilon_0\right]^{-1}[ L ]^2[ T ]^{-2}$  $(D)$ $\left[\mu_0\right]=\left[\varepsilon_0\right]^{-1}[ L ]^{-2}[ T ]^2$

Give the answer or quetion ($1$) and ($2$)

  • [IIT 2018]

Stokes' law states that the viscous drag force $F$ experienced by a sphere of radius $a$, moving with a speed $v$ through a fluid with coefficient of viscosity $\eta$, is given by $F=6 \pi \eta a v$. If this fluid is flowing through a cylindrical pipe of radius $r$, length $l$ and pressure difference of $p$ across its two ends, then the volume of water $V$ which flows through the pipe in time $t$ can be written as $\frac{V}{t}=k\left(\frac{p}{l}\right)^a \eta^b r^c$, where $k$ is a dimensionless constant. Correct values of $a, b$ and $c$ are

  • [KVPY 2017]