A famous relation in physics relates 'moving mass' m to 'rest mass' m

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1.Units, Dimensions and Measurement

easy

A famous relation in physics relates 'moving mass' $m$ to the 'rest mass' $m_{0}$ of a particle in terms of its speed $v$ and the speed of light, $c .$ (This relation first arose as a consequence of special relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant $c$. He writes:

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

Guess where to put the missing $c$

Option A

Option B

Option C

Option D

Solution

Given the relation, $m=\frac{m_{0}}{\left(1-v^{2}\right)^{\frac{1}{2}}}$

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 L.H.S is the same as that of R.H.S.

This is only possible when the factor, $\left(1-v^{2}\right)^{1 / 2}$ is dimensionless i.e., $\left(1-v^{2}\right)$ is dimensionless. This is only possible if $v^{2}$ is divided by $c^{2} .$

Hence, the correct relation is

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

Standard 11

Physics

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Solution

Similar Questions

In the equation $\left[X+\frac{a}{Y^2}\right][Y-b]= R T, X$ is pressure, $Y$ is volume, $R$ is universal gas constant and $T$ is temperature. The physical quantity equivalent to the ratio $\frac{a}{b}$ is

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 ?

If velocity $v$, acceleration $A$ and force $F$ are chosen as fundamental quantities, then the dimensional formula of angular momentum in terms of $v,\,A$ and $F$ would be

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