Question

A new system of unit is proposed in which unit of mass is \[{\mathbf{\alpha }}\,{\mathbf{kg}}\] ,unit of length \[{\mathbf{\beta }}\] m and unit of time \[{\mathbf{\gamma \;s}}\].How much will \[{\mathbf{5\; J}}\] measure in this new system?
A. $5\alpha {\beta ^2}{\gamma ^{ - 2}}$
B. $5{\alpha ^{ - 1}}{\beta ^2}{\gamma ^2}$
C. $5{\alpha ^{ - 2}}{\beta ^{ - 2}}{\gamma ^{ - 2}}$
D. $5{\alpha ^{ - 1}}{\beta ^{ - 2}}{\gamma ^2}$

Answer 1

Hint: Like different systems of units one can generate a new system which consist of some fundamental unit that can only be converted into these fundamental units.for example in MKS system M is fundamental unit of base mass, L is fundamental unit of base length and T is the fundamental unit of base time.

Formula used:
Dimensional formula of work done(W) is
\[W = F \times S\] , where \[F\] is force and \[S\] is displacement
\[\Rightarrow W=m \times a \times s\] , where $m$ is mass, $a$ is acceleration
\[\Rightarrow W=m \times \dfrac{v}{t} \times s\] , where v is velocity and t is time
\[\Rightarrow W=m \times \dfrac{s}{{{t^2}}} \times s\]
\[\Rightarrow W=m\dfrac{{{s^2}}}{{{t^2}}} \\
\Rightarrow W=\left[ {M{L^2}{T^{ - 2}}} \right] \]

Complete step by step solution:
For initial system of unit,
Mass \[[{M_1}] = {\text{ }}\;1kg\] ,length \[[{L_1}] = {\text{ }}1\;m\] , time \[[{T_1}] = {\text{ }}1\] second and energy \[[{n_1}] = {\text{ }}5\;J\].
For new system of unit,
Mass ${M_2} = α kg$,
$\Rightarrow$ Length ${L_2}$ = \[\beta \] meter,
$\Rightarrow$ Time \[\left[ {{T_2}} \right] = {\text{ }}\gamma \] second
We have to find \[{n_2}\]
The dimension of Energy is \[ = \left[ {M{L^2}{T^{ - 2}}} \right] \]
Therefore, comparing this with above mentioned formula we get,
\[a = 1,{\text{ }}b = {\text{ }}2,{\text{ }}c = - 2\]
Putting these values of \[{M_1},\,{L_1},\,{T_1},\,{M_2},\,{\text{ }}{L_2},{\text{ }}{T_2},\,\] \[a,{\text{ }}b,{\text{ }}c\] and \[{n_1}\] in equation
$\Rightarrow$ ${n_1}[M_1^aL_1^bT_1^c] = {n_2}[M_2^aL_2^bT_2^c] $
We get, ${n_2} = {n_1}{\left( {\dfrac{{{M_1}}}{{{M_2}}}} \right)^a}{\left( {\dfrac{{{L_1}}}{{{L_2}}}} \right)^b}{\left( {\dfrac{{{T_1}}}{{{T_2}}}} \right)^c}$
Thus, ${n_2} = 5\alpha {\beta ^2}{\gamma ^{ - 2}}$

Hence option A is correct.

Note: Dimensions of physical quantity will remain the same whatever be the system of units of its measurement.The dimension of physical quantity may be defined as the number of times the fundamental units of mass, length and time appear in the physical quantity.