JEE MAIN - Physics (2023 - 1st February Evening Shift - No. 8)
If the velocity of light $$\mathrm{c}$$, universal gravitational constant $$\mathrm{G}$$ and Planck's constant $$\mathrm{h}$$ are chosen as fundamental quantities. The dimensions of mass in the new system is :
$$\left[\mathrm{h}^{1} \mathrm{c}^{1} \mathrm{G}^{-1}\right]$$
$$\left[\mathrm{h}^{-1 / 2} \mathrm{c}^{1 / 2} \mathrm{G}^{1 / 2}\right]$$
$$\left[\mathrm{h}^{1 / 2} \mathrm{c}^{1 / 2} \mathrm{G}^{-1 / 2}\right]$$
$$\left[\mathrm{h}^{1 / 2} \mathrm{c}^{-1 / 2} \mathrm{G}^{1}\right]$$
Explanation
$$
\begin{aligned}
& {[\mathrm{M}]=[\mathrm{G}]^{\mathrm{x}}[\mathrm{h}]^{\mathrm{y}}[\mathrm{c}]^{\mathrm{z}}} \\\\
& {[\mathrm{M}]=\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]^x\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]^y\left[\mathrm{LT}^{-1}\right]^z} \\\\
& {\left[\mathrm{M}^1 \mathrm{~L}^0 \mathrm{~T}^0\right]=\left[\mathrm{M}^{-x+y}\right]\left[\mathrm{L}^{3 x+2 y+z}\right]\left[\mathrm{T}^{-2 x-y-z}\right]} \\\\
& \mathrm{y}-\mathrm{x}=1 ......(1) \\\\
& 3 \mathrm{x}+2 \mathrm{y}+\mathrm{z}=0 .......(2) \\\\
& -2 \mathrm{x}- \mathrm{y}-\mathrm{z}=0 ........(3)
\end{aligned}
$$
On solving, $\mathrm{x}=-\frac{1}{2}, \mathrm{y}=\frac{1}{2}, \mathrm{z}=\frac{1}{2}$
So $\mathrm{m}=\sqrt{\frac{\mathrm{hc}}{\mathrm{G}}}$
On solving, $\mathrm{x}=-\frac{1}{2}, \mathrm{y}=\frac{1}{2}, \mathrm{z}=\frac{1}{2}$
So $\mathrm{m}=\sqrt{\frac{\mathrm{hc}}{\mathrm{G}}}$
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