JEE MAIN - Physics (2024 - 30th January Evening Shift - No. 19)
If mass is written as $$m=k \mathrm{c}^{\mathrm{P}} G^{-1 / 2} h^{1 / 2}$$ then the value of $$P$$ will be : (Constants have their usual meaning with $k a$ dimensionless constant)
1/3
$$-$$1/3
1/2
2
Explanation
$$\begin{aligned} & \mathrm{m}=\mathrm{kc}^{\mathrm{P}} \mathrm{G}^{-1 / 2} \mathrm{~h}^{1 / 2} \\ & \mathrm{M}^1 \mathrm{~L}^0 \mathrm{~T}^0=\left[\mathrm{LT}^{-1}\right]^{\mathrm{P}}\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]^{-1 / 2}\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]^{1 / 2} \end{aligned}$$
By comparing $$\mathrm{P}=1 / 2$$
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