A new system of units is proposed in which unit of mass is $\alpha $ $kg$, unit of length $\beta $ $m$ and unit of time $\gamma $ $s$. How much will $5\,J$ measure in this new system ?
$[$ Energy $]=\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]$
Let $\mathrm{M}_{1}, \mathrm{~L}_{1}, \mathrm{~T}_{1}$ and $\mathrm{M}_{2}, \mathrm{~L}_{2}, \mathrm{~T}_{2}$ are units of mass, length and time in given two systems,
$\therefore \mathrm{M}_{1} =1 \mathrm{~kg}, \mathrm{~L}_{1}=1 \mathrm{~m}, \mathrm{~T}_{1}=1 \mathrm{~s}$
$\mathrm{M}_{2} =\alpha \mathrm{kg}, \mathrm{L}_{2}=\beta \mathrm{m}, \mathrm{T}_{2}=\gamma \mathrm{s}$
The magnitude of a physical quantity remains the same, whatever be the system of units of its measurement i.e.,
$n_{1} u_{1} =n_{2} u_{2}$
$n_{2}=n_{1} \frac{u_{1}}{u_{2}}=n_{1} \frac{\left[\mathrm{M}_{1} \mathrm{~L}_{1}^{2} \mathrm{~T}_{1}^{-2}\right]}{\left[\mathrm{M}_{2} \mathrm{~L}_{2}^{2} \mathrm{~T}_{2}^{-2}\right]}=5\left[\frac{\mathrm{M}_{1}}{\mathrm{M}_{2}}\right] \times\left[\frac{\mathrm{L}_{1}}{\mathrm{~L}_{2}}\right]^{2} \times\left[\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right]^{-2}$
$=5\left[\frac{1}{\alpha}\right] \times\left[\frac{1}{\beta}\right]^{2} \times\left[\frac{1}{\gamma}\right]^{-2}$
$=5\times \frac{1}{\alpha} \times \frac{1}{\beta^{2}} \times \frac{1}{\gamma^{-2}}$
$n_{2} =\frac{5 \gamma^{2}}{\alpha \beta^{2}}$
Thus, new unit of energy will be $\frac{\gamma^{2}}{\alpha \beta^{2}}$
Let $\mathrm{M}_{1}, \mathrm{~L}_{1}, \mathrm{~T}_{1}$ and $\mathrm{M}_{2}, \mathrm{~L}_{2}, \mathrm{~T}_{2}$ are units of mass, length and time in given two systems,
$\therefore \mathrm{M}_{1} =1 \mathrm{~kg}, \mathrm{~L}_{1}=1 \mathrm{~m}, \mathrm{~T}_{1}=1 \mathrm{~s}$
$\mathrm{M}_{2} =\alpha \mathrm{kg}, \mathrm{L}_{2}=\beta \mathrm{m}, \mathrm{T}_{2}=\gamma \mathrm{s}$
The magnitude of a physical quantity remains the same, whatever be the system of units of its measurement i.e.,
$n_{1} u_{1} =n_{2} u_{2}$
$n_{2}=n_{1} \frac{u_{1}}{u_{2}}=n_{1} \frac{\left[\mathrm{M}_{1} \mathrm{~L}_{1}^{2} \mathrm{~T}_{1}^{-2}\right]}{\left[\mathrm{M}_{2} \mathrm{~L}_{2}^{2} \mathrm{~T}_{2}^{-2}\right]}=5\left[\frac{\mathrm{M}_{1}}{\mathrm{M}_{2}}\right] \times\left[\frac{\mathrm{L}_{1}}{\mathrm{~L}_{2}}\right]^{2} \times\left[\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right]^{-2}$
$=5\left[\frac{1}{\alpha}\right] \times\left[\frac{1}{\beta}\right]^{2} \times\left[\frac{1}{\gamma}\right]^{-2}$
$=5\times \frac{1}{\alpha} \times \frac{1}{\beta^{2}} \times \frac{1}{\gamma^{-2}}$
$n_{2} =\frac{5 \gamma^{2}}{\alpha \beta^{2}}$
Thus, new unit of energy will be $\frac{\gamma^{2}}{\alpha \beta^{2}}$
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Column $-I$
Column $-II$
$(A)$ Electrical resistance
$(p)$ $M{L^3}{T^{ - 3}}{A^{ - 2}}$
$(B)$ Electrical potential
$(q)$ $M{L^2}{T^{ - 3}}{A^{ - 2}}$
$(C)$ Specific resistance
$(r)$ $M{L^2}{T^{ - 3}}{A^{ - 1}}$
$(D)$ Specific conductance
$(s)$ None of these
Match the following two coloumns
| Column $-I$ | Column $-II$ |
| $(A)$ Electrical resistance | $(p)$ $M{L^3}{T^{ - 3}}{A^{ - 2}}$ |
| $(B)$ Electrical potential | $(q)$ $M{L^2}{T^{ - 3}}{A^{ - 2}}$ |
| $(C)$ Specific resistance | $(r)$ $M{L^2}{T^{ - 3}}{A^{ - 1}}$ |
| $(D)$ Specific conductance | $(s)$ None of these |