If mass is written as mathrm{m}=mathrm{kc}^{mathrm{p}} mathrm{G}^{-1 / 2} mathrm{~h}^{1 / 2} then va

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1.Units, Dimensions and Measurement

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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

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(JEE MAIN-2018)

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)

Solution

Similar Questions

Write the dimensions of $a/b$ in the relation $P = \frac{{a – {t^2}}}{{bx}}$ , where $P$ is pressure, $x$ is the distance and $t$ is the time

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The potential energy of a particle varies with distance $x$ from a fixed origin as $V = \frac{{A\sqrt x }}{{x + B}}$,where
$A$ and $B$ are constants. The dimensions of $AB$ are