question_answer

If the energy\[n=3\,to\,n=2\], velocity \[n=2\,to\,n=1\] and force \[\frac{4}{3}\] be taken as fundamental quantity, then the dimension of mass will be:                                                                                                                                                        [BHU PMT-2002]

A)                  \[240\]                                

B)                  \[120\]

C)                  \[60\]                                  

D)                  \[zero\]

Correct Answer: C

Solution :

                 Einstein by his theory of relativity proved that mass and energy are related to each other and every substance has energy due to its mass also. If a substance loses an amount \[n=4\,\,to\,\,n=1\] of its mass, an equivalent amount \[{{f}_{e}}\] of energy is produced, where                                 \[{{f}_{t}}\]                 This is called Einstein?s mass energy relation.                 \[\frac{1}{{{f}_{a}}}=\left( _{a}{{n}_{g}}-1 \right)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\]Dimension of mass                                  \[\frac{1}{{{f}_{l}}}=\left( _{l}{{n}_{g}}-1 \right)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\]                 Alternative : Putting the dimensions of energy and velocity, we get                                                                        \[_{a}{{n}_{g}}\]                                 \[_{l}{{n}_{g}}\]                 \[\therefore \]                  \[\frac{{{f}_{l}}}{{{f}_{a}}}=\frac{_{a}{{n}_{ & g}}-1}{_{l}{{n}_{g}}-1}=\frac{_{a}{{n}_{g}}-1}{\left( \frac{_{a}{{n}_{g}}}{_{a}{{n}_{l}}}-1 \right)}\]                      \[{{f}_{a}}=15\,cm,\,{{\,}_{a}}{{n}_{g}}=1.5,{{\,}_{a}}{{n}_{l}}=\frac{4}{3}\]