A famous relation in physics relates 'moving | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given the relation,$m = \frac{m_{0}}{(1 - v^{2})^{1/2}}$.Dimension of $m = [M^{1} L^{0} T^{0}]$.Dimension of $m_{0} = [M^{1} L^{0} T^{0}]$.Dimensi

Vedclass · Accepted Answer

$m = \frac{m_{0}}{(1 - v^{2}/c^{2})^{1/2}}$

Vedclass · Answer

(B) Given the relation,$m = \frac{m_{0}}{(1 - v^{2})^{1/2}}$.Dimension of $m = [M^{1} L^{0} T^{0}]$.Dimension of $m_{0} = [M^{1} L^{0} T^{0}]$.Dimension of $v = [M^{0} L^{1} T^{-1}]$.Dimension of $v^{2} = [M^{0} L^{2} T^{-2}]$.Dimension of $c = [M^{0} L^{1} T^{-1}]$.The given formula will be dimensionally correct only when the dimension of the $L$.$H$.$S$. is the same as that of the $R$.$H$.$S$.This is only possible when the factor $(1 - v^{2})^{1/2}$ is dimensionless,i.e.,$(1 - v^{2})$ must be dimensionless.Since $1$ is dimensionless,$v^{2}$ must also be dimensionless. This is only possible if $v^{2}$ is divided by $c^{2}$,as the ratio $v^{2}/c^{2}$ is dimensionless.Hence,the correct relation is $m = \frac{m_{0}}{(1 - v^{2}/c^{2})^{1/2}}$.

Similar Questions

The frequency of vibration $f$ of a mass $m$ suspended from a spring of spring constant $K$ is given by a relation of the type $f = C\,{m^x}{K^y}$,where $C$ is a dimensionless quantity. The values of $x$ and $y$ are:

$A$ force $F$ acting on an object is given by $F = a \sqrt{x} - bt$,where $x$ is the distance in meters and $t$ is the time in seconds. The units of $a$ and $b$ are respectively:

The velocity of water waves $v$ may depend upon their wavelength $\lambda$,the density of water $\rho$,and the acceleration due to gravity $g$. The method of dimensions gives the relation between these quantities as:

If $A, B, C$ and $D$ represent velocity,acceleration,inductance and capacitance respectively,then $A^{-1} BCD$ has the dimensions of

If $E$,$L$,$m$ and $G$ denote the quantities as energy,angular momentum,mass and constant of gravitation respectively,then the dimensions of $P$ in the formula $P = EL^2 m^{-5} G^{-2}$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module