1.Units, Dimensions and Measurement
hard

If mass is written as $\mathrm{m}=\mathrm{kc}^{\mathrm{p}} \mathrm{G}^{-1 / 2} \mathrm{~h}^{1 / 2}$ then the value of $P$ will be : (Constants have their usual meaning with $\mathrm{k}$ a dimensionless constant)

A$1 / 2$
B$1 / 3$
C$2$
D$-1 / 3$
(JEE MAIN-2024)

Solution

$\mathrm{m}=\mathrm{kc}^{\mathrm{P}} \mathrm{G}^{-1 / 2} \mathrm{~h}^{1 / 2}$
$\mathrm{M}^1 \mathrm{~L}^0 \mathrm{~T}^0=\left[\mathrm{LT}^{-1}\right]^{\mathrm{P}}\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]^{-1 / 2}\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]^{1 / 2}$
$\text { By comparing } P=1 / 2$
Standard 11
Physics

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